.gitignore

 *~
 .ipynb_checkpoints
 site
-docs
 *.pyc
 __pycache__

.gitlab-ci.yml

@@ -7,9 +7,8 @@ stages:
 build lectures:
   stage: build
   before_script:
-    - pip install -U mkdocs mkdocs-material python-markdown-math notedown
+    - pip install -U mkdocs mkdocs-material python-markdown-math
   script:
-    - python execute.py
     - mkdocs build
   artifacts:
     paths:
@@ -37,10 +36,10 @@ build lectures:
     ## Create the SSH directory and give it the right permissions
     - mkdir -p ~/.ssh
     - chmod 700 ~/.ssh
-    - ssh-keyscan tnw-tn1.tudelft.net >> ~/.ssh/known_hosts
+    - ssh-keyscan qt4.tudelft.net >> ~/.ssh/known_hosts
     - chmod 644 ~/.ssh/known_hosts
   script:
-    - "rsync -rv site/* mathforquantum@tnw-tn1.tudelft.net:$DEPLOY_PATH"
+    - "rsync -rv site/* uploader@qt4.tudelft.net:$DEPLOY_PATH"
 
 deploy master version:
   <<: *prepare_deploy
@@ -72,7 +71,7 @@ undeploy test version:
     DEPLOY_PATH: "test_builds/$CI_COMMIT_REF_NAME"
   script:
     - mkdir empty/
-    - "rsync -rlv --delete empty/ mathforquantum@tnw-tn1.tudelft.net:$DEPLOY_PATH"
+    - "rsync -rlv --delete empty/ uploader@qt4.tudelft.net:$DEPLOY_PATH"
   environment:
     name: $CI_COMMIT_REF_NAME
     action: stop

README.md

-# mathforquantum
-
-Lecture notes and teaching material used for the Delft University of Technology course awesome course.
+# Mathematics for Quantum Physics
+Lecture notes and teaching material used for the Delft University of Technology course TN3105.
 
 The compiled materials are available at https://mathforquantum.quantumtinkerer.tudelft.nl
 
@@ -8,3 +7,41 @@ The compiled materials are available at https://mathforquantum.quantumtinkerer.t
 
 This repository is based on a template for publishing lecture notes, developed
 by Anton Akhmerov, who also hosts such repositories for other courses.
+
+# Version
+
+This a minimal stable version of the website from the branch "enabling search" based on mkdocs-material with a funcitoning website-wide search (without the support for inline jupyter notebook conversion by thebe)
+
+# HOWTOs
+
+## How to add new material to the lecture notes
+
+1. First, create a new merge request. In this way, your edits
+   will be pushed to a separate folder, and not directly appear on the website.
+   
+   Detailed information on how to create a merge request can be found
+   [here](https://docs.gitlab.com/ee/user/project/merge_requests/creating_merge_requests.html), but in most cases these two simple steps are sufficient:
+    - create a new branch in the repository on gitlab (either using the gitlab UI, or on the command line and then push to gitlab)
+    - on top of the gitlab page you will see a blue "Create merge request" button associated with your new branch. Fill out the information, and don't forget to
+start the name of the merge request with "WIP:"
+
+2. Write the new material using [markdown](https://en.wikipedia.org/wiki/Markdown#Example). The markdown files are stored in the `src` folder and have the
+ending `.md`. In particular, in markdown you can
+    - write math using latex syntax. `$...$` is used for math in the text,
+      `$$...$$` for separate equations.
+    - highlight certain blocks using the `!!!` syntax. For examples, use
+      ```
+      !!! check "Example: optional title"
+          The text of the example (could have math in it
+	        $f(x)$), which must be indented by 4 spaces
+      ```
+      Other useful blocks are `!!! warning` and `!!! info`
+
+3. Place figures in `docs/figures`
+
+4. If you added a new markdown file that should be linked in the index, you need
+   to add it to `mkdocs.yml` under the `nav:` entry.
+
+5. Whenever you push a commit to the branch/merge request, it will automatically be deployed on a preview webpage. This process may take a few minutes. You can find the preview website by going to your merge request. There will be on top a box with the label "Pipeline", and in the box a button "View app". Clicking on "View app" will bring you to the preview webpage.
+
+6. When you are done with the merge request, remove "Draft:" from the title, and notify an instructor.

docs/1_complex_numbers.md

+---
+title: Complex Numbers
+---
+
+# 1. Complex Numbers
+
+The lecture on complex numbers consists of three parts, each with their own video:
+
+- [1.1. Definition and basic operations](#11-definition-and-basic-operations)
+- [1.2. Complex functions](#12-complex-functions)
+- [1.3. Differentiation and integration](#13-differentiation-and-integration)
+
+**Total video length: 38 minutes and 53 seconds**
+
+## 1.1 Definition and basic operations
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/fLMdaMuEp8s?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+### Complex numbers 
+!!! info "Definition I"
+    Complex numbers are numbers of the form $$z = a + b {\rm i}.$$ 
+    Here $\rm i$ is the square root of -1: $${\rm i} = \sqrt{-1},$$ 
+    or equivalently: $${\rm i}^2 = -1.$$
+
+Usual operations on numbers have their natural extension for complex
+numbers, as we shall see below.
+
+Some useful definitions:
+
+!!! info "Definition II" 
+    For a complex number $z = a + b {{\rm i}}$, $a$ is called the *real part*, and $b$ the *imaginary part*.
+
+!!! info "Complex conjugate"
+    The *complex conjugate* $z^*$ of $z = a + b {{\rm i}}$ is defined as
+    $$z^* = a - b{{\rm i}},$$ 
+
+i.e., taking the complex conjugate means flipping the sign of the imaginary part.
+
+### Addition
+
+!!! info "Addition"
+    For two complex numbers, $z_1 = a_1 + b_1 {{\rm i}}$ and $z_2 = a_2 + b_2 {{\rm i}}$, 
+    the sum $w = z_1 + z_2$ is given as 
+    $$w = w_1 + w_2 {{\rm i}}= (a_1 + a_2) + (b_1 + b_2) {{\rm i}}$$ 
+
+where the parentheses in the rightmost expression have been added to group the real and the imaginary part. A consequence of this definition is that the sum of a complex number and its complex conjugate is real:
+$$z + z^* = a + b {{\rm i}}+ a - b {{\rm i}}= 2a,$$ i.e., this results in twice the real part of $z$. 
+
+Similarly, subtracting $z^*$ from $z$ yields $$z - z^* = a + b {{\rm i}} - a + b {{\rm i}}= 2b{\rm i},$$ i.e., twice the imaginary part of $z$ (times $\rm i$).
+
+### Multiplication
+
+!!! info "Multiplication"
+    For the same two complex numbers $z_1$ and $z_2$ as above, their product is calculated as
+    $$w = z_1 z_2 = (a_1 + b_1 {{\rm i}}) (a_2 + b_2 {{\rm i}}) = (a_1 a_2 - b_1 b_2) + (a_1 b_2 + a_2 b_1) {{\rm i}},$$
+    where the parentheses have again beèn used to indicate the real and imaginary parts.
+
+A consequence of this definition is that the product of a complex number
+$z = a + b {{\rm i}}$ with its conjugate is real:
+$$z z^* = (a+b{{\rm i}})(a-b{{\rm i}}) = a^2 + b^2.$$ 
+The square root of this number is called the *norm* $|z|$ of $z$:
+$$|z| = \sqrt{z z^*} = \sqrt{a^2 + b^2}.$$
+
+### Division
+
+The quotient $z_1/z_2$ of two complex numbers $z_1$ and $z_2$ defined above can be evaluated by multiplying the numerator and denominator by the complex conjugate of $z_2$:
+!!! info "Division"
+    $$\frac{z_1}{z_2} = \frac{z_1 z_2^*}{z_2 z_2^*} = \frac{(a_1 a_2 + b_1 b_2) + (-a_1 b_2 + a_2 b_1) {{\rm i}}}{a_2^2 + b_2^2}.$$
+
+Try this yourself!
+
+!!! check "Example:"
+    $$\begin{align} 
+    \frac{1 + 2{\rm i}}{1 - 2{\rm i}} &= \frac{(1 + 2{\rm i})(1 + 2{\rm i})}{1^2 + 2^2} = \frac{1+4{\rm i} -4}{5}\\
+    & = -\frac{3}{5} + {\rm i} \frac{4}{5}
+    \end{align}$$
+
+### Visualization: the complex plane
+
+Complex numbers can be rendered on a two-dimensional (2D) plane, the
+*complex plane*. This plane is spanned by two unit vectors, one
+horizontal representing the real number 1 and the vertical
+unit vector representing ${\rm i}$.
+
+<figure markdown>
+  ![image](figures/complex_numbers_5_0.svg)
+  <figcaption>The norm of $z$ is the length of its vector spanned in the complex plane.</figcaption>
+</figure>
+
+#### Addition in the complex plane
+
+Adding two numbers in the complex plane corresponds to adding their
+respective horizontal and vertical components:
+
+<figure markdown>
+  ![image](figures/complex_numbers_8_0.svg)
+  <figcaption>The sum of two complex numbers is found as the diagonal of a parallelogram spanned by the vectors of those two numbers.</figcaption>
+</figure>
+
+## 1.2. Complex functions
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/7XtR_wDSqRc?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+Real functions can (most of the times) be written in terms of a Taylor series expanded at a point $x_{0}$:
+$$f(x) = \sum \limits_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}$$
+We can write something similar for complex functions by replacing the *real* variable $x$ with its *complex* counterpart $z$:
+$$f(z) = \sum \limits_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (z-x_{0})^{n}$$
+
+For this course, the most important function is the *complex exponential function*, at which we will have a closer look below.
+
+### The complex exponential function
+The complex exponential is used *extremely often*. 
+It occurs in Fourier transforms and it is very convenient for doing calculations involving cosines and sines. 
+It also makes many common operations on complex number a lot easier to perform.
+
+!!! info "The exponential function and Euler identity"
+    The exponential function $f(z) = \exp(z) = e^z$ is defined as:
+    $$\exp(z) = e^{x + {\rm i}y} = e^{x} e^{{\rm i} y} = e^{x} \left( \cos y + {\rm i} \sin y\right).$$
+The last expression is called the *Euler identity*.
+
+!!! note "**Exercise**" 
+    Check that this function obeys
+    $$\exp(z_1) \exp(z_2) = \exp(z_1 + z_2).$$ 
+    *You will need sum and difference formulas of cosine and sine.*
+
+### The polar form
+
+A complex number $z$ can be represented by two real numbers, $a$ and $b$, which correspond to the real and imaginary part of the complex number.
+Another representation of $z$ is a *vector* in the complex plane with a horizontal component that corresponds to the real part of $z$ and a vertical component that corresponds to the imaginary part of $z$. 
+It is also possible to characterize that vector by its *length* and *direction*, where the latter can be represented by the
+angle that the vector makes with the horizontal axis:
+
+<figure markdown>
+  ![image](figures/complex_numbers_10_0.svg)
+  <figcaption>The angle with the horizontal axis is denoted by $\varphi$ 
+  like in the case of conventional polar coordinates, 
+  but in the context of complex numbers, this angle is called as the <b>argument</b>.</figcaption>
+</figure>
+
+!!! info "Polar form of complex numbers"
+    A complex number can be represented either by its real and imaginary part 
+    corresponding to the Cartesian coordinates in the complex plane,
+    or by its *norm* and its *argument* corresponding to polar coordinates. 
+    The norm is the length of the vector, and the argument is the angle it makes with the horizontal axis.
+
+We can conclude that for a complex number $z = a + b {\rm i}$, its real and imaginary parts
+can be expressed  in polar coordinates as $$a = |z| \cos\varphi$$ $$b = |z| \sin\varphi$$
+
+!!! info "Inverse equations"
+    The inverse equations are $$|z| = \sqrt{a^2 + b^2}$$
+    $$\varphi = \arctan(b/a)$$ for $a>0$. 
+    In general:
+    $$\varphi = \begin{cases} \arctan(b/a) &{\rm for ~} a>0; \\
+    \pi + \arctan(b/a) & {\rm for ~} a<0 {\rm ~ and ~} b>0;\\
+    -\pi + \arctan(b/a) &{\rm for ~} a<0 {\rm ~ and ~} b<0. \end{cases}$$ 
+ 
+It turns out that by using the magnitude $|z|$ and phase $\varphi$, we can write any complex number as
+$$z = |z| e^{{\rm i} \varphi}$$
+By increasing $\varphi$ by $2 \pi$, we make a full circle around the origin and reach the same point on the complex plane. In other words, by adding $2 \pi$ to the argument of $z$, we get the same complex number $z$!
+As a result, the argument $\varphi$ is defined up to $2 \pi$, and we are free to make any choice we like, such as in the examples in the figure below:
+
+<figure markdown>
+  ![image](figures/complex_numbers_11_0.svg)
+  <figcaption> $-\pi < \varphi < \pi$ (left) and (right) $-\frac{\pi}{2} < \varphi < \frac{3 \pi}{2}$ </figcaption>
+</figure>
+
+Some useful values of the complex exponential to know by heart are:
+
+!!! tip "Useful identities:"
+    $$e^{2{\rm i } \pi} = 1$$
+    $$e^{{\rm i} \pi} = -1 $$ 
+    $$e^{{\rm i} \pi/2} = {\rm i}$$
+    From the first expression, it also follows that 
+    $$e^{{\rm i} (y + 2\pi n)} = e^{{\rm i}y} {\rm ~ for ~} n \in \mathbb{Z}$$
+    As a result, $y$ is only defined up to $2\pi$.  
+
+Furthermore, we can define the sine and cosine in terms of complex exponentials:
+
+!!! info "Complex sine and cosine"
+    $$\cos(x) = \frac{e^{{\rm i} x} + e^{-{\rm i} x}}{2}$$
+    $$\sin(x) = \frac{e^{{\rm i} x} - e^{-{\rm i} x}}{2i}$$
+
+Most operations on complex numbers become easier when complex numbers are converted to their *polar form* using the complex exponential.
+Some functions and operations, which are common in real analysis, can be easily derived for their complex counterparts by substituting the real variable $x$ with the complex variable $z$ in its polar form:
+!!! info "Examples of some complex functions stated using polar form"
+    $$z^{n} = \left(r e^{{\rm i} \varphi}\right)^{n} = r^{n} e^{{\rm i} n \varphi}$$
+    $$\sqrt[n]{z} = \sqrt[n]{r e^{{\rm i} \varphi} } = \sqrt[n]{r} e^{{\rm i}\varphi/n} $$
+    $$\log(z) = log \left(r e^{{\rm i} \varphi}\right) = log(r) + {\rm i} \varphi$$
+    $$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2})}$$
+
+Use of polar form lets us notice immediately that for example, as a result of multiplication, the norm of the new number is the *product* of the norms of the multiplied numbers and its argument is the *sum* of the arguments of the multiplied numbers. 
+
+In the complex plane, this looks as follows:
+
+<figure markdown>
+  ![image](figures/complex_numbers_12_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+!!! check "Example: Find all solutions solving $z^4 = 1$."
+    Of course, we know that $z = \pm 1$ are two solutions, but which other solutions are possible? We take a systematic approach:
+    $$\begin{align} z = e^{{\rm i} \varphi} & \Rightarrow z^4 = e^{4{\rm i} \varphi} = 1 \\
+    & \Leftrightarrow 4 \varphi = n 2 \pi \\
+    & \Leftrightarrow \varphi = 0, \varphi = \frac{\pi}{2}, \varphi = -\frac{\pi}{2}, \varphi = \pi \\
+    & \Leftrightarrow z = 1, z = i, z = -i, z = -1 \end{align}$$
+
+## 1.3. Differentiation and integration
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/JyftSqmmVdU?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+**We only consider differentiation and integration over *real* variables.** 
+
+We can then regard the complex ${\rm i}$ as another constant, and use our usual differentiation and integration rules:
+!!! info "Differentiation and Integration rules"
+    $$\frac{d}{d\varphi} e^{{\rm i} \varphi} = e^{{\rm i} \varphi} \frac{d}{d\varphi} ({\rm i} \varphi) ={\rm i} e^{{\rm i} \varphi} .$$
+    $$\int_{0}^{\pi} e^{{\rm i} \varphi} = \frac{1}{{\rm i}} \left[ e^{{\rm i} \varphi} \right]_{0}^{\pi} = -{\rm i}(-1 -1) = 2 {\rm i}$$
+
+## 1.4. Bonus: the complex exponential function and trigonometry
+
+Let us show some tricks in the following examples where the simple properties of the exponential
+function help in re-deriving trigonometric identities.
+
+!!! example "Properties of the complex exponential function I"
+    Take $|z_1| = |z_2| = 1$, and $\arg{(z_1)} = \varphi_1$ and
+    $\arg{(z_2)} = \varphi_2$. 
+    It is easy to see then that $z_i = \exp({\rm i} \varphi_i)$, $i=1, 2$. Then:
+    $$z_1 z_2 = \exp[{\rm i} (\varphi_1 + \varphi_2)].$$ 
+    The left hand side can be written as
+    $$\begin{align}
+    z_1 z_2 & = \left[ \cos(\varphi_1) + {\rm i} \sin(\varphi_1) \right] \left[ \cos(\varphi_2) + {\rm i} \sin(\varphi_2) \right] \\
+    & = \cos\varphi_1 \cos\varphi_2 - \sin\varphi_1 \sin\varphi_2 + {\rm i} \left( \cos\varphi_1 \sin\varphi_2 + 
+    \sin\varphi_1 \cos\varphi_2 \right).
+    \end{align}$$
+
+    Also, the right hand side can be written as
+    $$\exp[{\rm i} (\varphi_1 + \varphi_2)] = \cos(\varphi_1 + \varphi_2) + {\rm i} \sin(\varphi_1 + \varphi_2).$$
+    Comparing the two expressions, equating their real and imaginary parts, we find
+    $$\cos(\varphi_1 + \varphi_2) = \cos\varphi_1 \cos\varphi_2 - \sin\varphi_1 \sin\varphi_2;$$
+    $$\sin(\varphi_1 + \varphi_2) = \cos\varphi_1 \sin\varphi_2 + 
+    \sin\varphi_1 \cos\varphi_2.$$ 
+    Note that we used the Euler formula in order to derive the identities of trigonometric function.
+    The point is to show you that you can use the properties of the complex exponential to quickly find the form of trigonometric formulas, which are often easily forgotten.
+
+!!! example "Properties of the complex exponential function II" 
+    In this example, let's see what we can learn from the derivative of the exponential function:
+    $$\frac{d}{d\varphi} \exp({\rm i} \varphi) = {\rm i} \exp({\rm i} \varphi) .$$
+    Writing out the exponential in terms of cosine and sine, we see that
+    $$\cos'\varphi + {\rm i} \sin'\varphi = {\rm i} \cos\varphi - \sin\varphi.$$
+    where the prime $'$ denotes the derivative as usual. Equating real and imaginary parts leads to 
+    $$\cos'\varphi = - \sin\varphi;$$
+    $$\sin'\varphi = \cos\varphi.$$
+
+## 1.5. Summary
+
+1.  A complex number $z$ has the form $$z = a + b \rm i$$ where $a$ and
+    $b$ are both real, and $\rm i^2 = 1$. The real number $a$ is called
+    the *real part* of $z$ and $b$ is the *imaginary part*. Two complex
+    numbers can be added, subtracted and multiplied straightforwardly.
+    The quotient of two complex numbers $z_1=a_1 + \rm i b_1$ and
+    $z_2=a_2 + \rm i b_2$ is
+    $$\frac{z_1}{z_2} = \frac{z_1 z_2^*}{z_2 z_2^*} = \frac{(a_1 a_2 + b_1 b_2) + (-a_1 b_2 + a_2 b_1) {{\rm i}}}{a_2^2 + b_2^2}.$$
+
+2.  Complex numbers can also be characterised by their *norm*
+    $|z|=\sqrt{a^2+b^2}$ and *argument* $\varphi$. These parameters
+    correspond to polar coordinates in the complex plane. For a complex
+    number $z = a + b {\rm i}$, its real and imaginary parts can be
+    expressed as $$a = |z| \cos\varphi$$ $$b = |z| \sin\varphi$$ The
+    inverse equations are $$|z| = \sqrt{a^2 + b^2}$$
+    $$\varphi = \begin{cases} \arctan(b/a) &{\rm for ~} a>0; \\
+     \pi + \arctan(b/a) & {\rm for ~} a<0 {\rm ~ and ~} b>0;\\
+     -\pi + \arctan(b/a) &{\rm ~ for ~} a<0 {\rm ~ and ~} b<0.
+     \end{cases}$$
+    The complex number itself then becomes
+    $$z = |z| e^{{\rm i} \varphi}$$
+
+3.  The most important complex function for us is the complex exponential function, which simplifies many operations on complex numbers
+    $$\exp(z) = e^{x + {\rm i}y} = e^{x} \left( \cos y + {\rm i} \sin y\right).$$
+    where $y$ is defined up to $2 \pi$.\\
+    The $\sin$ and $\cos$ can be rewritten in terms of this complex exponential as
+    $$\cos(x) = \frac{e^{{\rm i} x} + e^{-{\rm i} x}}{2}$$
+    $$\sin(x) = \frac{e^{{\rm i} x} - e^{-{\rm i} x}}{2i}$$
+    Because we only consider *differentiation* and *integration* over *real variables*, the usual rules apply:
+    $$\frac{d}{d\varphi} e^{{\rm i} \varphi} = e^{{\rm i} \varphi} \frac{d}{d\varphi} ({\rm i} \varphi) ={\rm i} e^{{\rm i} \varphi} .$$
+    $$\int_{0}^{\pi} e^{{\rm i} \varphi} = \frac{1}{{\rm i}} \left[ e^{{\rm i} \varphi} \right]_{0}^{\pi} = -{\rm i}(-1 -1) = 2 {\rm i}$$
+
+## 1.6. Problems
+
+1.  [:grinning:] Given $a=1+2\rm i$ and $b=-3+4\rm i$, calculate and draw in the complex plane the numbers:
+    1.  $a+b$,
+    2.  $ab$, 
+    3.  $b/a$.
+
+2.  [:grinning:] Evaluate:  
+    1. $\rm i^{1/4}$, 
+    2. $\left(1+\rm i \sqrt{3}\right)^{1/2}$,
+    3. $\exp(2\rm i^3)$.
+
+3.  [:grinning:] Find the three 3rd roots of $1$ and ${\rm i}$. </br>
+    (i.e. all possible solutions to the equations $x^3 = 1$ and $x^3 = {\rm i}$, respectively).
+
+4.  [:grinning:] *Quotients*</br>
+    1. Find the real and imaginary part of $$ \frac{1+ {\rm i}}{2+3{\rm i}} \, .$$
+    2. Evaluate for real $a$ and $b$:$$\left| \frac{a+b\rm i}{a-b\rm i} \right| \, .$$ 
+
+5.  [:sweat:] For any given complex number $z$, we can take the inverse $\frac{1}{z}$. 
+    1.  Visualize taking the inverse in the complex plane. 
+    2.  What geometric operation does taking the inverse correspond to? </br>
+    (Hint: first consider what geometric operation $\frac{1}{z^*}$ corresponds to.)
+
+6.  [:grinning:]  *Differentation and integration* </br>
+    1. Compute $$\frac{d}{dt} e^{{\rm i} (kx-\omega t)},$$
+    2. Calculate the real part of $$\int_0^\infty e^{-\gamma t  +\rm i \omega t} dt$$ 
+    ($k$, $x$, $\omega$, $t$ and $\gamma$ are real; $\gamma$ is positive).
+
+7.  [:smirk:] Compute by making use of the Euler identity.
+    $$\int_{0}^{\pi}\cos(x)\sin(2x)dx$$
+    

docs/2_coordinates.md

+---
+title: Coordinate systems
+---
+# 2. Coordinate systems
+
+The lecture on coordinate systems consists of 3 parts, each with their own video:
+
+- [2.1. Introduction to coordinate systems: Cartesian and polar](#21-introduction-to-coordinate-systems-cartesian-and-polar)
+- [2.2. Converting derivatives between coordinate systems](#22-converting-derivatives-between-coordinate-systems)
+- [2.3. Coordinate systems in 3D](#23-coordinate-systems-in-3d)
+
+**Total video length: 35 minutes and 13 seconds**
+
+## 2.1. Introduction to coordinate systems: Cartesian and polar
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/CPMrsQlNxS8?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+### Cartesian coordinates
+
+The most common coordinates are *Cartesian coordinates*, where we use a
+number $n$ of perpendicular axes. The coordinates corresponding to these
+axes are $x_j$ where $j=1, \ldots, n$.
+
+Cartesian coordinates are simple to describe and operate on. The coordinate axes are 
+straight lines perpendicular to each other. It is therefore
+very easy to do calculations in Cartesian coordinates. For example,
+the distance $\Delta s$ between two points $(x_1, x_2, \ldots, x_n)$
+and $(x'_1, x'_2, \ldots, x'_n)$ can be quickly computed using a general formula for n-dimensions:
+
+$$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
+
+(A space with such a distance definition is called an *Euclidean
+space*.)
+
+In mathematics, we are often dealing with so-called *infinitesimally* small
+distances, for example in the definition of derivatives and integrals.
+In Cartesian coordinates, the expressions for infinitesimal distances $ds$ and
+infinitesimal volumes $dV$ are given as:
+
+!!! info "Infinitesimal segment and volume elements in n-dimensional Cartesian coordinates"
+    $$ds = \sqrt{dx_1^2 + dx_2^2 + \ldots + dx_n^2}$$
+    $$dV = dx_1 dx_2 \ldots dx_N.$$
+
+The formula for $dV$ also indicates that in Cartesian coordinates, the integral
+over a volume can be expressed as individual integrals over all coordinate directions:
+$$\int dV = \idotsint dx_1 dx_2 \ldots dx_N.$$
+
+Cartesian coordinates are used a lot and they are particularly suitable for
+infinite spaces or for rectangular volumes.
+
+<figure markdown>
+  ![image](figures/Coordinates_5_1.svg)
+  <figcaption>An example of a vector drawn in a 2D Cartesian plane</figcaption>
+</figure>
+
+### Polar coordinates
+
+#### Definition
+
+It often turns out that a change to a different type of coordinate
+system makes mathematics easier. For example, if you want to describe vibrations of a
+circular drum, polar coordinates become very convenient. These are
+defined for a two-dimensional space (a plane). The position on this plane is characterised by two
+coordinates: the *distance* $r$ between the point and the origin, and by the
+angle ($\varphi$) between the line connecting the point to the origin and the $x$-axis. The radius is therefore always a non-negative number $r \geq 0$, and the range for the polar angle is $\varphi \in \left< 0,2\pi \right)$
+
+Note that each Cartesian coordinate has a *dimension* of length. 
+In polar coordinates, the radius $r$ has a dimension of *length*, but
+the angular coordinate $\varphi$ is dimensionless.
+
+<figure markdown>
+  ![polarplot](figures/Coordinates_7_0.svg)
+  <figcaption>In this example of a polar plot, you can distinguish the radial coordinate $( 0.2, 0.4, \text{ etc.. })$ from the angular one expressed in degrees $( 0^\circ, 45^\circ, \text{ etc..} )$</figcaption>
+</figure>
+
+
+The plot below shows a point on a curve with the polar coordinates
+$(r,\varphi)$ indicated. From this, we can see that the *Cartesian*
+coordinates $(x,y)$ of the point are related to the polar ones as
+follows:
+
+$$\begin{equation} x = r \cos\varphi; \end{equation}$$
+$$\begin{equation} y = r \sin \varphi.\end{equation}$$
+
+<figure markdown>
+  ![image](figures/Coordinates_9_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+#### The inverse relation
+
+!!! info "Inverse relation between polar and Cartesian coordinate systems"
+    \begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}
+    \begin{equation} \varphi=\begin{cases}
+    \arctan(y/x) & \text{$x>0$,}\\
+    \pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\
+    -\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.}
+    \end{cases} \label{phixy}\end{equation}
+
+The last formula for $\varphi$ warrants a closer explanation: It is easy
+to see that $\tan(\varphi)=y/x$, but this is not a unique relation, due to
+the fact that the $\tan$ has different branches. Convince yourself that
+the expression above is correct for all the four sectors!
+
+#### Distances and areas
+
+Now suppose we want to calculate the distance between two points, one
+with polar coordinates $(r_1, \varphi_1)$, and the other with
+$(r_2, \varphi_2)$. This looks like a difficult exercise. One possible
+way to perform this is by translating the polar coordinates into
+Cartesian coordinates and using the expression given above for this
+distance: $$\Delta s^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2,$$ so
+$$\Delta s^2 = (r_1\cos\varphi_1 - r_2 \cos \varphi_2)^2 + (r_1\sin\varphi_1 - r_2 \sin \varphi_2)^2$$
+which is not a very convenient expression.
+
+If we consider two points which are *very close*, the analysis
+simplifies however. We can use the geometry of the problem to find the
+distance (see the figure below).
+
+<figure markdown>
+  ![image](figures/Coordinates_11_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+When going from point 1 to point 2, we first traverse a small circular
+arc of radius $r_1$ and then we move a small distance radially outward
+from $r_1$ to $r_2$. Provided the difference between the angles
+$\varphi_1$ and $\varphi_2$ is (very) small, these paths are
+approximately perpendicular and we can use Pythagoras’ theorem to find
+the distance $d s$. Note that the arc is approximately straight –
+it has a length $r_1 d \varphi$, where
+$d \varphi = \varphi_2-\varphi_1$. So we have:
+$$d s^2 = (d r)^2 + (arc~length)^2   = (d r)^2 + r_1^2 (d \varphi)^2 .$$
+
+We can use the same arguments also for the area: since the different
+segments are approximately perpendicular, we find the area by simply
+multiplying them:
+
+!!! info "Infinitesimal surface element in polar coordinates"
+    	$$dA = r dr d\varphi.$$
+
+This is an important formula to remember for integrating in polar
+coordinates!  The extra $r$ that appears here can be intuitively
+understood: the area swept by an angle difference $d\varphi$
+*increases* as we move further away from the origin.
+
+!!! check "Example: Integrating over a circular area"
+    To check the area element we just derived, let us compute a simple
+    integral. We compute the integral over a circle with radius $r_0$
+    with a very simple function that equals to $1$. In this case,
+    we expect to get as a result the are of the region we integrate over.
+
+    We find:
+
+    $$
+    \int_0^{2\pi} d\varphi \int_0^{r_0} r dr =\\
+    2\pi \int_0^{r_0} r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2,
+    $$
+
+    which is indeed the area of a circle with radius $r_0$.
+
+## 2.2. Converting derivatives between coordinate systems
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/NGQWGx71w98?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+Important equations in physics often involve derivatives given in terms
+of Cartesian coordinates. One prominent example are equations of the form
+$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)
+f(x, y) = \ldots.$$
+The derivative operator $\left(\frac{\partial^2}{\partial x^2} +
+\frac{\partial^2}{\partial y^2}\right)$ is so common it has its own name:
+the Laplacian (here for two-dimensional space).
+
+This equation is universal, but for particular situations it might be
+advantageous to use a different coordinate system, such as polar coordinates
+for a system with rotational symmetry. The question then is: How does the
+corresponding equation look like in a different coordinate system?
+
+There are different ways to find the answer. Here, we will focus on
+directly deriving the transformed equation through an explicit calculation
+involving the chain rule for a function of several variables.
+
+!!! info "Chain rule for a multi-variable function"
+    Let $f$ be a function of $n$ variables: $f(y_1, y_2, \ldots, y_n)$,
+    as well as $g_i(x_1, x_2, \ldots, x_n)$ for $i=1,2,\ldots, n$. Then
+
+    $$\frac{\partial}{\partial x_i} = \sum_{j=1}^n
+    \frac{\partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$
+
+We start by replacing the function $f(x, y)$ by a function in polar coordinates
+$f(r, \varphi)$, and ask what is $\frac{\partial}{\partial x} f(r, \varphi)$. When
+we look at this expression, we need to understand what it *means* to take the derivative
+of a function of $r, \varphi$ in terms of $x$?
+
+For this, we need to realize that there are relations between the coordinate systems.
+In particular, $r = r(x,y)$ and $\varphi = \varphi(x, y)$ as defined in equations
+of [the inverse relations](#the-inverse-relation). In fact, we have been rather sloppy in our notation above,
+as the functions $f(x,y)$ and $f(r, \varphi)$ do not mean that I substitute $x=r$
+and $y=\varphi$! It is more precise to state that there are two diferent
+functions $f_\text{cart}(x,y)$ and $f_\text{polar}(r, \varphi)$ that are equivalent,
+in the sense that
+
+$$f_\text{cart}(x, y) = f_\text{polar}(r(x,y), \varphi(x,y))$$
+
+In physics, we usually never write this down explicitly, but we are aware that these
+are two different functions from the fact that they use different coordinates.
+
+With this information, we can now apply the chain rule:
+
+$$ \frac{\partial}{\partial x} f(r, \varphi) =
+\frac{\partial f}{\partial r} \frac{\partial r(x, y)}{\partial x} +
+\frac{\partial f}{\partial \varphi} \frac{\partial \varphi(x,y)}{\partial x}
+$$
+
+and it is now a matter of (tedious) calculus to arrive at the right result.
+This is the task of exercises 3 and 4, which lead you to compute the Laplacian
+in polar coordinates.
+
+!!! warning "Inverse function theorem"
+    In this calculation, one might be tempted to use the inverse
+    function theorem to compute derivatives like
+    $\frac{\partial \varphi}{\partial x}$ from the much simpler
+    $\frac{\partial x}{\partial \varphi}$. However, note that here we
+    are dealing with functions depending on several variables, so an appropriate
+    *Jacobian* has to be used (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_function_theorem)). A direct calculation is in this particular case considerably easier.
+
+Note that this procedure also applies to transformations to other coordinate systems,
+although the calculations can become quite tedious. In conventional cases,
+it is usually advised to look up the correct form.
+
+## 2.3. Coordinate systems in 3D
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/VjUbnZN1BvA?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+### Cylindrical coordinates
+
+Three dimensional systems may have axial symmetry. An example is an
+electrically charged wire of which we would like to calculate the
+electric field, or a current-carrying wire for which we would like to
+calculate the magnetic field. For such problems, the most convenient
+coordinates are *cylindrical coordinates*. For a further convenience, we choose
+the symmetry-axis as the $z$-axis. Note that this allowed, because we may
+choose the coordinate system ourselves - it is not imposed by the
+problem.
+
+Cylindrical coordinates are defined straightforwardly: we use polar
+coordinates $r$ and $\varphi$ in the $xy$ plane, and the distance $z$
+along the symmetry-axis as the third coordinate. The radius is therefore again always defined as a non-negative number $r \in \left<0, \infty \right)$, and the range for the azimuthal angle is analogically $\varphi \in \left< 0,2\pi \right)$. The *height* $z$ along the cylinder axis can take any real value, hence $z \in \mathbb{R}$ . If the axis system is
+chosen in physical space, we have two coordinates which have the
+dimension of a distance: $r$ and $z$. The other coordinate,
+$\varphi$ is of course dimensionless.
+
+What is the distance traveled along a path when we express this in
+cylindrical coordinates? Let’s consider an example shown in the figure below.
+
+<figure markdown>
+  ![image](figures/Coordinates_13_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+We want to find the length of the (small) red segment $d s$. By
+inspecting the figure, we see that the horizontal (i.e. parallel to the
+$xy$-plane) segment $d l$ is perpendicular to the vertical segment
+$dz$. Using for $d l$ the length we obtained before for a line
+segment in the $xy$ plane expressed in polar coordinates, we
+immediately find:
+$$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$
+The volume element is consequently given as:
+
+!!! info "Infinitesimal volume element in cylindrical coordinates"
+    $$dV = r dr d\varphi dz.$$
+
+### Spherical coordinates
+
+For problems with spherical symmetry, we use *spherical coordinates*.
+These work as follows. For a point $\bf r$ in 3D space, we can specify
+the position of that point by specifying its (1) distance to the origin
+and (2) the direction of the line connecting the origin to our point.
+The specification of this direction can be identified with a point on a
+sphere which is centered at the origin:
+
+<figure markdown>
+  ![image](figures/Coordinates_15_0.svg)
+  <figcaption>The position of a point on the sphere is specified using the radius $r$ and two angles (azimuthal)
+$\varphi$ and (polar) $\theta$ in the given order </figcaption>
+</figure>
+
+!!! warning "Parameter ranges in spherical coordinates"
+    - The radius ($r$) is defined for $r \in \left<0, \infty \right)$ </br>
+    - The azimuthal angle ($\varphi$) has the range of $\varphi \in \left< 0, 2\pi \right)$  </br>
+    - The polar angle ($\theta$) has the range $\theta \in \left<0, \pi \right>$ 
+
+!!! warning
+    In mathematics, the angles are often labeled the other way
+    around: there, $\varphi$ is used for the angle between a line running from
+    the origin to the point of interest and the $z$-axis, and $\theta$ for
+    the angle of the projection of that line with the $x$-axis. The
+    convention used here is customary in physics. 
+
+The relation between Cartesian and spherical coordinates is defined by:
+!!! info "The relation between Cartesian and spherical coordinates"
+    $$x = r \cos \varphi \sin \theta$$
+    $$y = r \sin\varphi \sin \theta$$ $$z = r \cos\theta$$ 
+
+The inverse transformation is easy to find: 
+!!! info "The inverse relation between Cartesian and spherical coordinates"
+    $$r = \sqrt{x^2+y^2+z^2}, \qquad r \in \left<0, \infty \right)$$
+    $$\varphi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
+    \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\
+    -\pi + \arctan(y/x) &{\rm for ~} x<0 {\rm ~ and ~} y<0.
+    \end{cases}, \qquad \varphi \in \left< 0,2\pi \right)$$ 
+    $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2}), \qquad \theta \in \left< 0,\pi \right> $$
+    
+These relations can be derived from the following figure:
+
+<figure markdown>
+  ![image](figures/Coordinates_17_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+The distance related to a change in the spherical coordinates is
+calculated using Pythagoras’ theorem. The length $ds$ of a short segment
+on the sphere with radius $r$ corresponding to the changes in the polar
+angles of $d\theta$ and $d\varphi$ is given as
+$$dl^2 = r^2 \left(\sin^2 \theta d\varphi^2 + d\theta^2\right).$$
+In order to verify this, it is important to realize that all points with
+*the same* coordinate $\theta$ span a circle in a horizontal plane
+with a radius $r\sin\theta$ as shown in the figure below.
+
+From this, we can also infer that for a segment with a radial component
+$dr$ in addition to the displacement on the surface of the sphere, the combined displacement is:
+$$ds^2 = r^2 \left(\sin^2 \theta d\varphi^2 + d\theta^2\right) + dr^2.$$
+
+The picture below shows the geometry behind the calculation of this
+displacement.
+
+<figure markdown>
+  ![image](figures/Coordinates_19_0.svg)
+  <figcaption></figcaption>
+</figure>
+
+
+From these arguments we can again also find the volume element, it is
+here given as
+
+!!! info "Infinitesimal volume element in spherical coordinates"
+    $$dV = r^2 \sin\theta dr d\theta d\varphi.$$
+
+## 2.4. Summary
+
+We have discussed four different coordinate systems:
+
+1.  !!! tip "Cartesian coordinates" 
+    $${\bf r} = (x_1, \ldots, x_n)$$
+    $$ x_{n} \in \mathbb{R}$$
+    This systems can be used for any dimension $n$. It is particularly convenient for: infinite spaces, systems
+    with rectangular symmetry.
+    Distance between two points ${\bf r} = (x_1, \ldots, x_n)$ and
+    ${\bf r}' = (x'_1, \ldots, x'_n)$:
+    $$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
+
+2.  !!! tip "Polar coordinates" 
+    $${\bf r} = (r, \varphi)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right) $$
+    This system can be used in two dimensions. It is particularly suitable for systems with circular symmetry or functions
+    given in terms of these coordinates. <br/>
+    Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\varphi^2.$$
+    Infinitesimal area: $$dA = r dr d\varphi.$$
+
+3.  !!! tip "Cylindrical coordinates" 
+    $${\bf r} = (r, \varphi, z)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right), \quad z \in \mathbb{R} $$ 
+    This system can be used in three dimensions. It is particularly suitable for systems with axial symmetry
+    or functions given in terms of these coordinates. <br/>
+    Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\varphi^2 + dz^2.$$
+    Infinitesimal volume: $$dV = r dr d\varphi dz.$$
+
+4.  !!! tip "Spherical coordinates" 
+    $${\bf r} = (r, \varphi, \theta)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right), \quad \theta \in \left< 0,\pi \right> $$
+    This system can be used in three dimensions. It is particularly suitable for systems with spherical
+    symmetry or functions given in terms of these coordinates. <br/>
+    Infinitesimal distance: 
+    $$ds^2 =r^2 (\sin^2 \theta d\varphi^2 + d\theta^2) +  dr^2 .$$
+    Infinitesimal volume:
+    $$dV = r^2 \sin(\theta) dr d\varphi d\theta.$$ 
+
+## 2.5. Problems
+
+1.  [:grinning:] *Warm-up*
+
+    1.  Find the polar coordinates of the point with Cartesian
+        coordinates $${\bf r} = \sqrt{2} (1,1).$$
+
+    2.  Find the cylindrical coordinates of the point with Cartesian
+        coordinates $${\bf r} = \frac{3}{2} (\sqrt{3}, 1, 1).$$
+
+    3.  Find the spherical coordinates of the points
+        $${\bf r} = (3/2, \sqrt{3}/2, 1).$$
+
+2.  [:grinning:] *Geometry and different coordinate systems*
+
+    What geometric objects do the following boundary conditions describe?
+   
+    1. $r<1$ in cylindrical coordinates,
+    2. $\varphi=0$ in cylindrical coordinates,
+    3. $r=1$ in spherical coordinates,
+    4. $\theta = \pi/4$ in spherical coordinates,
+    5. $r=1$ and $\theta=\pi/4$ in spherical coordinates,
+    6. $\varphi=\pi/2$ and $\theta=\pi/2$ in spherical coordinates.
+
+3.  [:smirk:] *Partial derivatives*
+
+    (a) Consider the function $f(r,\varphi,\theta)=\frac{1}{r^2}$ defined
+        using spherical coordinates.
+        Compute $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
+
+    (b) Now let us consider a function defined using cylindrical coordinates
+        as $f(r, \varphi, z) = \frac{1}{r^2}$ (i.e.~very similar to the previous
+        question).
+        Compute again $\frac{\partial}{\partial z} f(r, \varphi, z)$.
+ 
+4.  [:smirk:] *Chain rule practice* 
+
+    From the transformation from polar to Cartesian
+    coordinates, show that
+    $$\frac{\partial}{\partial x} = \cos\varphi \frac{\partial}{\partial r} - \frac{\sin\varphi}{r} \frac{\partial}{\partial \varphi}$$
+    and
+    $$\frac{\partial}{\partial y} = \sin\varphi \frac{\partial}{\partial r} + \frac{\cos\varphi}{r} \frac{\partial}{\partial \varphi}.$$
+    (Use the chain rule for differentiation).
+
+5.  [:sweat:] *Laplace operator in spherical coordinates*
+
+    Using the result of problem 4, show that the Laplace
+    operator acting on a function $\psi({\bf r})$ in polar coordinates
+    takes the form
+    $$\nabla^2 \psi({\bf r}) =\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \psi({\bf r}) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi(r,\varphi)}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi(r,\varphi)}{\partial \varphi^2}.$$
+
+    In a similar fashion it can be shown that for spherical coordinates,
+    the Laplace operator acting on a function $\psi({\bf r})$ becomes:
+    $$\begin{align} \nabla^2 \psi (r,\varphi, \theta) &= 
+    \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\varphi,\theta)}{\partial r} \right) \\ &+ \frac{1}{r^2\sin^2\theta} \frac{\partial^2 \psi(r,\varphi, \theta)}{\partial \varphi^2} \\ &+ \frac{1}{r^2\sin\theta} 
+    \frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial\psi(r,\varphi, \theta)}{\partial \theta}\right).
+    \end{align}$$
+    This is however even more tedious (you do not have to show this).
+
+6.  [:grinning:] *Integration and coordinates I*
+
+    We define $f(r, \varphi) = \frac{1}{r}$ in polar coordinates. Explain how
+    a circular region, centered at the origin and with radius $r_0$, can be described
+    using polar coordinates. Then compute the integral of $f(r,\varphi)$ over
+    this region.
+
+7.  [:grinning:] *Integration and coordinates II*
+
+    Compute the area of the spherical cap defined by $r=r_0$ and $\theta <\theta_0$. 
+
+8.  [:smirk:] *Integration and coordinates III*
+
+    In 2D, we can define a shape by specifying a function $r(\varphi)$:
+
+    ![image](figures/shape_polar.svg)
+    
+    (Of course, here we need to have $r(0) = r(2\pi)$.)
+
+    Show that the area of this shape is given by
+    $$
+    \int_0^{2\pi} \frac{1}{2}\left[r(\varphi)\right]^2 d\varphi
+    $$
+

docs/3_vector_spaces.md

+---
+title: Vector Spaces
+---
+
+# 3. Vector spaces
+
+The lecture on vector spaces consists of **three parts**:
+
+- [3.1. Definition and basis dependence](#31-definition-and-basis-dependence)
+
+- [3.2. Properties of a vector space](#32-properties-vector-space)
+
+- [3.3. Matrix representation of vectors](#33-matrix-representation-vectors)
+
+and at the end of this lecture note, there is a set of corresponding exercises
+
+- [3.4 Problems](#34-problems)
+
+---
+
+The contents of this lecture are summarised in the following **videos**:
+
+1. [Vector spaces: Introduction](https://www.dropbox.com/s/evytrbb55fgrcze/linear_algebra_01.mov?dl=0)
+
+2. [Operations in vector spaces](https://www.dropbox.com/s/1530xb7zbuhwu6u/linear_algebra_02.mov?dl=0)
+
+3. [Properties of vector spaces](https://www.dropbox.com/s/5lwkxd8lw5uwri9/linear_algebra_03.mov?dl=0)
+
+**Total video lentgh: ~16 minutes**
+
+## 3.1. Definition and basis dependence
+
+A vector $\vec{v}$ is a mathematical object characterised by both a **magnitude** and a **direction**, that is, an orientation in a given space.
+  
+We can express a vector in terms of its individual **components**. Let's assume we have an $n$-dimensional space, meaning that the vector $\vec{v}$ can be oriented in different ways along each of $n$ dimensions. The expression of $\vec{v}$ in terms of its components is
+
+$$\vec{v} = (v_1, v_2,\ldots, v_n) \, ,$$
+ 
+We will denote by ${\mathcal V}^n$ the **vector space** composed by all possible vectors of the above form.
+
+The components of a vector, $\{ v_i\}$ can be **real numbers** or **complex numbers**,
+depending on whether we have a real or a complex vector space. 
+
+!!! info "Vector basis" 
+    Note that the above expression of $\vec{v}$ in terms of its components assume that we are using a specific **basis**. 
+    It is important   to  recall that the same vector can be expressed in terms of different bases. 
+    A **vector basis** is a set of $n$ vectors that can be used to generate all the elements of a vector space.
+
+For example, a possible basis of  ${\mathcal V}^n$ could be denoted by $\vec{a}_1,\vec{a}_2,\ldots,\vec{a_n}$,
+and we can write a generic vector  $\vec{v}$  as
+
+$$\vec{v} = (v_1, v_2, \ldots, v_n) = v_1 \vec{a}_1 + v_2 \vec{a}_2 + \ldots v_n \vec{a}_n \, .$$
+
+However, one could choose a different basis, denoted by $\vec{b}_1,\vec{b}_2,\ldots,\vec{b_n}$, where the same vector would be expressed in terms of a different set of components
+
+$$ \vec{v} = (v'_1, v'_2, \ldots, v'_n) = v'_1 \vec{b}_1 + v'_2 \vec{b}_2 + \ldots v'_n \vec{b}_n \, .$$
+
+Thus, while the vector remains the same, the values of its components depend on the specific choice of basis.
+
+The most common basis is the **Cartesian basis**, where for example for $n=3$:
+
+$$\vec{a}_1 = (1, 0, 0) \, ,\qquad \vec{a}_2 = (0, 1, 0)\, ,\qquad \vec{a}_3 = (0, 0, 1) \, .$$
+  
+!!! warning ""
+    The elements of a vector basis must be **linearly independent** from one another, meaning
+    that none of them can be expressed as a linear combination of the other basis vectors.
+
+We can consider one example in the two-dimensional real vector space $\mathbb{R}$, namely the $(x,y)$ coordinate plane, shown below.
+
+<figure markdown>
+  ![image](figures/3_vector_spaces_1.jpg)
+  <figcaption></figcaption>
+</figure>
+  
+In this figure, you can see how the same vector $\vec{v}$ can be expressed in two different bases. In the first one (left panel), the Cartesian basis is used and its components are $\vec{v}=(2,2)$. In the second basis (right panel), the components are different, namely $\vec{v}=(2.4 ,0.8)$, while the magnitude and direction of the vector remain unchanged.
+
+For many problems, both in mathematics and in physics, the appropriate choice of the vector space basis may significantly simplify the
+solution process.
+    
+## 3.2. Properties of a vector space
+
+You might be already familiar with the concept of performing a number of various **operations** between vectors, so in this course, let us review some essential operations that are relevant to start working with quantum mechanics:
+
+!!! info "Addition" 
+    I can add two vectors to produce a third vector, $$\vec{a} + \vec{b}= \vec{c}.$$
+    As with scalar addition, also vectors satisfy the commutative property, $$\vec{a} + \vec{b} = \vec{b} + \vec{a}.$$
+    Vector addition can be carried out in terms of their components,
+    $$ \vec{c} = \vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n) =  (c_1, c_2, \ldots, c_n).$$
+
+!!! info "Scalar multiplication" 
+    I can multiply a vector by a scalar number (either real or complex) to produce another vector, $$\vec{c} = \lambda \vec{a}.$$ 
+    Addition and scalar multiplication of vectors are both *associative* and *distributive*, so the following relations hold
+    $$\begin{align} &1. \qquad (\lambda \mu) \vec{a} = \lambda (\mu \vec{a}) = \mu (\lambda \vec{a})\\
+    &2. \qquad \lambda (\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}\\
+    &3. \qquad (\lambda + \mu)\vec{a} = \lambda \vec{a} +\mu \vec{a} \end{align}$$
+ 
+### Vector products
+
+In addition to multiplying a vector by a scalar, as mentioned above, one can also multiply two vectors among them. 
+There are two types of vector products; where the end result is a scalar (so just a number) and where the end result is another vector. 
+
+!!! info "Scalar product of vectors" 
+    The scalar product of vectors is given by $$ \vec{a}\cdot \vec{b} = a_1b_1 + a_2b_2 + \ldots + a_nb_n \, .$$
+    Note that since the scalar product is just a number, its value will not depend on the specific
+    basis in which we express the vectors: the scalar product is said to be *basis-independent*. The scalar product is also found via 
+    $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$$ with $\theta$ the angle between the vectors.
+
+!!! info "Cross product"
+    The vector product (or cross product) between two vectors $\vec{a}$ and $\vec{b}$ is given by 
+    $$ \vec{a}\times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \hat{n}$$
+    where $|\vec{a}|=\sqrt{ \vec{a}\cdot\vec{a} }$ (and likewise for $|\vec{b}|$) is the norm of the vector $\vec{a}$, $\theta$ is the angle between the two vectors, and $\hat{n}$ is a unit vector which is *perpendicular* to the plane that contains $\vec{a}$ and $\vec{b}$. 
+    Note that this cross-product can only be defined in *three-dimensional vector spaces*. The resulting vector 
+    $\vec{c}=\vec{a}\times \vec{b} $ will have as components $c_1 = a_2b_3-a_3b_2$, $c_2= a_3b_1 - a_1b_3$, and $c_3= a_1b_2 - a_2b_1$.
+
+### Unit vector and orthonormality
+
+!!! info "Unit vector"
+    A special vector is the **unit vector**, which has a norm of 1 *by definition*. A unit vector is often denoted with a hat, rather than an arrow ($\hat{i}$ instead of $\vec{i}$). To find the unit vector in the direction of an arbitrary vector $\vec{v}$, we divide by the norm: $$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
+
+!!! info "Orthonormality"    
+    Two vectors are said to be **orthonormal** of they are perpendicular (orthogonal) *and* both are unit vectors.
+
+Now we are ready to define in a more formal way what vector spaces are,
+an essential concept for the description of quantum mechanics.
+
+### The main properties
+
+The main properties of **vector spaces** are the following:
+
+!!! info ""
+    A vector space is **complete upon vector addition**.
+    This property means that if two arbitrary vectors  $\vec{a}$ and $\vec{b}$
+    are elements of a given vector space ${\mathcal V}^n$,
+    then their addition should also be an element of the same vector space 
+    $$\vec{a}, \vec{b} \in {\mathcal V}^n, \qquad \vec{c} = (\vec{a} + \vec{b}) \in {\mathcal V}^n  \, ,\qquad \forall\,\, \vec{a}, \vec{b} \,.$$
+
+!!! info "" 
+    A vector space is **complete upon scalar multiplication**.
+    This property means that when I multiply one arbitrary vector  $\vec{a}$,
+    element of the vector space ${\mathcal V}^n$, by a general scalar $\lambda$, the result is another vector which also belongs to the same vector space $$\vec{a} \in {\mathcal V}^n, \qquad \vec{c} = \lambda \vec{a}
+    \in {\mathcal V}^n \qquad \forall\,\, \vec{a},\lambda \, .$$
+    
+The property that a vector space is complete upon scalar multiplication and vector addition is also known as the **closure condition**.
+
+!!! info ""
+    There exists a **null element** $\vec{0}$ such that $\vec{a}+\vec{0} =\vec{0}+\vec{a}=\vec{a} $.
+
+!!! info ""
+    **Inverse element**: for each vector $\vec{a} \in \mathcal{V}^n$ there exists another
+    element of the same vector space, $-\vec{a}$, such that their addition results
+    in the null element, $\vec{a} + ( -\vec{a}) = \vec{0}$. This element it called the **inverse element**.
+
+A vector space comes often equipped with various multiplication operations between vectors, such as the scalar product mentioned above
+(also known as *inner product*), but also many other operations such as *vector product* or *tensor product*. There are also many other properties, but for what we are interested in right now, these are sufficient.
+
+
+## 3.3. Matrix representation of vectors
+
+It is advantageous to represent vectors with a notation suitable for matrix manipulation and operations. As we will show in the next lectures, the operations involving states in quantum systems can be expressed in the language of linear algebra.
+
+First of all, let us remind ourselves how we express vectors in the standard Euclidean space. In two dimensions, the position of a point $\vec{r}$ when making explicit the Cartesian basis vectors reads
+$$ \vec{r}=x \hat{i}+y\hat{j} \, .$$
+As mentioned above, the unit vectors $\hat{i}$ and $\hat{j}$ form an *orthonormal basis* of this vector space, and we call $x$ and $y$ the *components* of $\vec{r}$ with respect to the directions spanned by the basis vectors.
+
+Recall also that the choice of basis vectors is not unique, we can use any other pair of orthonormal unit vectors $\hat{i}$ and $\hat{j}$, and express the vector $\vec{r}$ in terms of these new basis vectors as 
+$$ \vec{r}=x'\hat{i}'+y'\hat{j}'=x\hat{i}+y\hat{i} \, ,$$
+with $x'\neq x$ and $y'\neq y$. So, while the vector itself does not depend on the basis, the values of its components are basis dependent.
+
+We can also express the vector $\vec{r}$ in the following form
+$$ \vec{r} =  \begin{pmatrix}x\\y\end{pmatrix} \, ,$$
+which is known as a *column vector*. Note that this notation assumes a specific choice of basis vectors, which is left
+implicit and displays only the information on its components along this specific basis.
+
+For instance, if we had chosen another set of basis vectors $\hat{i}'$ and $\hat{j}'$, the components would be $x'$ and $y'$, and the corresponding column vector representing the same vector $\vec{r}$ in such case would be given by
+$$ \vec{r}= \begin{pmatrix}x'\\y'\end{pmatrix}.$$
+
+We also know that Euclidean space is equipped with a scalar vector product. 
+The scalar product $\vec{r_1}\cdot\vec{r_2}$ of two vectors in 2D Euclidean space is given by
+$$ \vec{r_1}\cdot\vec{r_2}=r_1\,r_2\,\cos\theta \, ,$$
+where $r_1$ and $r_2$ indicate the *magnitude* (length) of the vectors and $\theta$ indicates its relative angle. Note that the scalar product of two vectors is just a number, and thus it must be *independent of the choice of basis*.
+
+The same scalar product can also be expressed in terms of components of $\vec{r_1}$ and $\vec{r_2}$. When using the $\{ \hat{i}, \hat{j} \}$ basis, the scalar product will be given by
+$$ \vec{r_1}\cdot\vec{r_2}=x_1\,x_2\,+\,y_1\,y_2 \, .$$
+Note that the same result would be obtained if the basis $\{ \hat{i}', \hat{j}' \}$ 
+had been chosen instead 
+$$ \vec{r_1}\cdot\vec{r_2}=x_1'\,x_2'\,+\,y_1'\,y_2' \, .$$
+
+The scalar product of two vectors can also be expressed, taking into
+account the properties of matrix multiplication, in the following form
+$$ \vec{r_1}\cdot\vec{r_2} = \begin{pmatrix}x_1, y_1\end{pmatrix}\begin{pmatrix}x_2\\y_2\end{pmatrix} = x_1x_2+y_1y_2 \, ,$$
+where here we say that the vector $\vec{r_1}$ is represented by a *row vector*.
+
+Therefore, we see that the scalar product of vectors in Euclidean space can be expressed as the matrix multiplication of row and column vectors. The same formalism, as we will see in the next class, can be applied for the case of Hilbert spaces in quantum mechanics.
+
+***
+
+## 3.4. Problems
+
+**1)** [:grinning:] Find a unit vector parallel to the sum of $\vec{r}_1$ and $\vec{r}_2$, where we have defined
+$$\vec{r}_1=2\vec{i}+4\vec{j}-5\vec{k} \, , $$ and $$\vec{r}_2=\vec{i}+2\vec{j}+3\vec{k} \, .$$.
+
+***
+
+**2)** [:grinning:] If the vectors $\vec{a}$ and $\vec{b}$ may be written in the parametric form
+as a function of the parameter $t$ as follows
+$$\vec{a}=3t^3\,\vec{i}-2t\,\vec{j}+t^2\,\vec{k}$$ and $$\vec{b}=3\sin{t}\,\vec{i}+2\cos{t}\,\vec{k}$$
+Evaluate the following derivatives with respect to the parameter $t$:
+
+**(a)** $ d(\vec{a}\cdot\vec{b}) / dt$.
+
+**(b)** $d \left( \vec{a} \times \vec{b}\right)/dt$.
+
+***
+
+**3)** [:sweat:] Three non-zero vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are such that $(\vec{a}+\vec{b})$ is perpendicular to $(\vec{a}+\vec{c})$ and $(\vec{a}-\vec{b})$ is perpendicular to $(\vec{a}-\vec{c})$. Show that $\vec{a}$ is perpendicular to $\vec{b}+\vec{c}$. If the magnitude of the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are in the ratio 1:2:4, find the angle between $\vec{b}$ and $\vec{c}$.
+
+***
+
+**4)** [:grinning:] Find the vector product $\vec{b} \times \vec{c}$ and the triple product $\vec{a}\cdot(\vec{b} \times \vec{c})$, where these three vectors are defined as
+$$\vec{a}=\vec{i}+4\vec{j}+\vec{k}\,,$$ and $$\vec{b}=-\vec{i}+2\vec{j}+2\vec{k}\,,$$ and $$\vec{c}=2\vec{i}-\vec{k}\,.$$
+
+***
+

docs/4_vector_spaces_QM.md

+---
+title: Vector spaces in quantum mechanics
+---
+
+# 4. Vector spaces in quantum mechanics
+
+The lecture on vector spaces in quantum mechanics consists of the following parts:
+
+- [4.1. Dirac notation and Hilbert spaces](#41-dirac-notation-and-hilbert-spaces)
+
+- [4.2. Inner product of state vectors](#42-inner-product-of-state-vectors)
+
+- [4.3. Matrix representation of ket and bra vectors](#43-matrix-representation-ket-and-bra-vectors)
+
+- [4.4. A two-dimensional Hilbert space](#44-a-two-dimensional-hilbert-space)
+
+and at the end of the lecture there is a set of exercises 
+
+- [4.5. Problems](#45-problems)
+
+---
+
+The contents of this lecture are summarised in the following **videos**:
+
+- [1. Dirac notation and properties of Hilbert spaces](https://www.dropbox.com/s/mnccmpff33pre9r/linear_algebra_04.mov?dl=0)
+
+- [2. Algebra with Dirac notation - bras and kets](https://www.dropbox.com/s/709bh9j083y7d0s/linear_algebra-06.mov?dl=0)
+
+- [3. Finding expansion coefficients for Dirac notation](https://www.dropbox.com/s/k9plspkonnk3nc0/linear_algebra-07.mov?dl=0)
+
+**Total length of the videos: ~14 minutes**
+
+---
+
+## 4.1. Dirac notation and Hilbert spaces
+
+In the previous lecture, we reviewed the basic properties of linear vector spaces. Next, we will discuss how the same formalism
+can be applied to describe physical states in quantum mechanics.
+
+The state of a physical system in quantum mechanics is represented by a vector belonging to a *complex vector space*.
+This vector space is known as the *state space* of the system.
+
+### Ket
+
+!!! info "Ket"
+     A physical state of a quantum system is represented by a symbol $$|~~\rangle$$ known as a **ket**. 
+     This notation is known as the *Dirac notation*, and it is very prominent in the description of quantum mechanics. 
+     Note that a *ket* is also referred to as a state vector, *ket* vector, or just a state. 
+
+### Hilbert space
+
+The set of all possible state vectors describing a given physical system forms a complex vector space $\mathcal{H}$, which is known as the *Hilbert space* of the system. You can think of the Hilbert space as the space populated by all possible states that a quantum system can be found on. Hilbert spaces inherit a number of the important properties of general vector spaces:
+    
+!!! info "Superposition" 
+     A linear combination (or superposition) of two or more state vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$, is also a state of the quantum system. Therefore, a linear combination $|{\Psi}\rangle$ of the form $$|{\Psi}\rangle=c_1|{\psi_1}\rangle+c_2|{\psi_1}\rangle+c_3|{\psi_3}\rangle+...+c_n|{\psi_n}\rangle = \sum_{i=1}^n c_i|{\psi_i}\rangle$$ 
+     where $c_1, c_2, c_3, ...$ are general complex numbers will also be a physically allowed state vector of the quantum system.
+     
+!!! info "Normalisation"
+     If a physical state of the system is given by a vector $|{\Psi}\rangle$, then the same physical state can also be represented by the vector $c|{\Psi}\rangle$ where $c$ is a non-zero complex number. The reason for this is that the overall normalisation of the state vector *does not change the physics* of the system (or in other words, does not modify the *information content* of the state vector). As we will discuss below, in quantum mechanics it is advantageous to work with  *normalised vectors*, that is, whose *length* is one. 
+     We will define in a while what do we mean by length.
+
+!!! info "Completeness"
+     A set of vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$ is said to be  *complete* if every state 
+     of the quantum system can be represented as a linear combination of them.
+     In such a case, it becomes possible to express  *any* state vector $|{\Psi}\rangle$ of the system's Hilbert space as a superposition of these $n$ vectors,
+     $$ |{\Psi}\rangle=\sum_{i=1}^n c_i|{\psi_i}\rangle$$
+     for some specific choice of coefficients $c_i$. The set of vector \{$|{\psi_i}\rangle$\} are then said to *span* the Hilbert space of the quantum system.
+    
+!!! info "Basis"
+     A set of vectors \{$|{\psi_i}\rangle$\} is said to form a basis for the state space if the set of vectors is *complete* and if in addition they are  *linearly independent*. The latter condition means essentially that one cannot express a given basis vector as a linear combination of the rest of basis vectors.
+     Linear independence can also be expressed as the requirement that if one has that
+     $$\sum_{i=1}^n c_i |{\psi_i}\rangle=0\;\text{then}\; c_i=0\;\text{for all}\; i$$
+
+!!! info "Dimensionality"
+     The minimum number of vectors needed to form a complete set of basis states is known as the *dimensionality* of the state space. In quantum mechanics you will encounter systems whose Hilbert spaces have very different dimensionality, from the spin-1/2 particle (a $n=2$ vector space) to the free particle (whose state vectors live in an infinite vector space).
+
+### Bra vectors
+
+We need now to extend the Dirac notation to describe other elements of this vector space. We need to introduce a quantity $\langle{\Psi}|$, known as a  *bra vector*, which represents the  *complex conjugates* of the corresponding ket vector. Bra vectors are elements of the vector space $\mathcal{H}^{*}$, called the *dual space* of the original Hilbert space $\mathcal{H}$.
+
+!!! info "Bra vector"
+     If a ket vector is given by $$| \Psi\rangle= c_1 |\psi_1\rangle+c_2|\psi_2\rangle \, ,$$ 
+     then the corresponding bra vector will be given by 
+     $$\langle{\Psi}|= c_1^*\langle{\psi_1}|+c_2^*\langle{\psi_2}| \, .$$
+
+As mentioned above, the vector space spanned by all bra vectors $\langle{\Psi}|$ is referred to as the dual space and is represented by $\mathcal{H}^*$. For each ket vector belonging to $\mathcal{H}$, there will exist an associated bra vector belonging to the dual space $\mathcal{H}^*$.
+
+Below, we will further discuss the concept of bra vectors when presenting the matrix representation of elements of the Hilbert space.
+      
+## 4.2. Inner product of state vectors
+
+Assume that $|{\psi}\rangle$ and $|{\phi}\rangle$ are any two state vectors belonging to the
+state (Hilbert) space $\mathcal{H}$, then we can define the  *inner product*
+between them, $\langle{\psi}|{\phi}\rangle$, as follows. 
+
+The inner product in quantum mechanics is the analog of the usual scalar product that one encounters in vector spaces, and which we reviewed in the previous lecture. As in usual vector spaces, the inner product of two state vectors is a  *scalar* and in this case a complex number in general. 
+
+!!! tip "Interpretation of the inner product in quantum mechanics"
+     1.   The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the **probability amplitude** (not the probability) of measuring a system, which characterised by the state $|{\phi}\rangle$, to be in the state $|{\psi}\rangle$. 
+     2.   This inner product can also be understood as measuring the **overlap** between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. 
+     3.   Then the  **probability of observing the system to be in the state $|\psi\rangle$** given that it is in the state $|\phi\rangle$ will be given by $$|\langle \psi | \phi \rangle|^2 \, .$$ Since the latter quantity is a probability, we know that it should satisfy the condition that 
+     $$0 \le |\langle \psi | \phi \rangle|^2 \le 1 \, .$$
+
+### Properties of the inner product
+
+The inner product (probability amplitude) $\langle \psi | \phi \rangle$ exhibits the following properties:
+      
+!!! info "Properties of the inner product"
+     1. **Complex conjugate:** $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
+     2. **Distributivity and associativity:** $\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$
+     3. **Positivity:** $\langle \psi | \psi \rangle\geq0 \, .$
+          If $\langle \psi | \psi \rangle = 0$ then, this implies that the state vector $|\psi\rangle=0$ is the null element of the Hilbert space.
+     4. **Orthogonality:** Two states $|\psi \rangle$ and $|\phi \rangle$ are said to be *orthogonal* if $\langle \psi | \phi\rangle=0 \, .$
+          By analogy with regular vector spaces, we can think of these two state vectors $|\psi \rangle$ and $|\phi \rangle$ as being *perpendicular* to each other. Note that for a quantum system occupying a certain state, there is a vanishing probability of it being observed in a state orthogonal to it.
+     5. **Norm:** The quantity $\sqrt{\langle \psi | \psi \rangle}$ is known as the  *length* or the *norm* of the state vector $|\psi\rangle$. 
+          You can see from the properties of complex algebra that this length must be a real number. A physically valid state $|\psi \rangle$ must be normalized to unity, that is $\langle \psi | \psi \rangle=1$. Note that a state that cannot be normalized to unity does not represent a physically acceptable state.
+     6. **Orthonormality:** A set of orthonormal basis vectors $\{|\psi_i\rangle\text{;}\; i=1,2,3,...,n\}$ will have the property $\langle \psi_i |\psi_j \rangle=\delta_{ij}$ where $\delta_{ij}$ is a mathematical symbol known as the *Kronecker delta*, which equals unity if $i=j$ and zero if $i\neq j$.
+   
+From all the above conditions, we see that a Hilbert space is a so-called *complex inner product space*, which is nothing else but a complex vector space equipped with a inner product. All the vectors belonging to a Hilbert space $\mathcal{H}$ have a finite norm, which means that they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic interpretation of the state vectors described above.
+
+## 4.3. Matrix representation of ket and bra vectors
+ 
+As we have discussed, in quantum mechanics a general state vector $|\psi\rangle$ can be represented in terms of the basis vectors, $\{|\phi_i\rangle;i=1,2,...,n\}$, as 
+$$ |\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle $$
+for some values of the complex coefficients $\{ c_i\}$. To determine the values of these coefficients, we can take the inner product between the bra basis vector $\langle \phi_j|$ and the ket state vector $|\psi\rangle$ and use the orthogonality properties of the basis vectors:
+$$ \langle \phi_j|\psi\rangle = \langle \phi_j|\sum_{i=1}^n c_i |\phi_i\rangle = \sum_{i=1}^n c_i\langle \phi_j|\phi_i\rangle = \sum_{i=1}^n c_i\delta_{ij} = c_j \, .$$
+Therefore, if we now denote the coefficients $\{ c_i\}$ of the state vector $|\psi\rangle$ by $\{ \psi_i\}$, we have the expansion
+$$ |\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle= \sum_{i=1}^n \left( \langle \phi_i|\psi\rangle \right) |\phi_i\rangle \, .$$
+By analogy with the Euclidean case, we can understand the coefficients $\psi_i$ as the *components* of the state vector $ |\psi\rangle$ along the $n$ directions spanned by the basis vectors. Here, note also that in this notation $\psi_i$ is an *scalar* (just a number) and not a vector. Furthermore, note that, as opposed to the Euclidean space, the coefficients $\psi_i$ will generally be complex numbers.
+
+This analogy with the case of ordinary vectors allows us to write the state $|\psi\rangle$ as a *column vector* with respect to the set of basis vectors $\{|\phi_i\rangle;i=1,2,...,n\}$, which are kept implicit: 
+$$ |\psi\rangle= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\vdots\\\psi_n\end{pmatrix} \, . $$
+
+We can also express the basis vectors in this manner. Given that the basis vectors are *orthonormal* among themselves,
+the basis state $|\phi_i\rangle$ will have as component in the $j$ direction
+$$ (\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,$$ 
+and thus the vector column expression of the basis vectors will be very simple
+$$ |\phi_1\rangle= \begin{pmatrix} 1\\0\\0 \\\vdots\end{pmatrix} \;, \quad |\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots $$
+
+!!! note "Evaluating the inner product"
+     Let us show how we can use the matrix representation to evaluate the inner product (bracket) between two state vectors when expanded in terms of their components in the same basis:
+     $$ |\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle \,, \qquad |\chi\rangle=\sum_{i=1}^n \chi_i |\phi_i\rangle\, .$$
+     First of all, we note that we can write the above expansions in the following way
+     $$
+     |\psi\rangle=\sum_{i=1}^n |\phi_i \rangle \langle \phi_i | \psi \rangle \, ,
+     $$
+     and thus we see that the basis vectors provide a very useful representation of the *identity operator*:
+     $$
+     \hat{I} = \sum_i |\phi_i\rangle \langle\phi_i| \, ,
+     $$
+     We can insert this  identify operator within the bracket to evaluate the inner
+     product $\langle \chi|\psi\rangle$ between the two state vectors to evaluate the inner product $\langle \chi|\psi\rangle$:
+     $$
+     \langle \chi|\psi\rangle=
+     \langle\chi|\hat{I} |\psi\rangle=\sum_{i=1}^n \langle\chi| \phi_i \rangle \langle\phi_i|\psi\rangle \, .
+     $$
+     Next, using that $\chi_i = \langle \phi_i|\chi \rangle$ are the components of the
+     state vector $|\chi\rangle$ and that  $\langle \chi| \phi_n \rangle=(\langle\phi_i|\chi\rangle)^*$,
+     we have that $\langle \chi |\phi_i\rangle =\chi_i^*$
+     and therefore the inner product of the two state vectors $|\psi\rangle$
+     and  $|\chi\rangle$ can be expressed in terms of their components
+     as follows
+     $$\langle\chi|\psi\rangle=\sum_{i=1}^n\chi_i^*\psi_i.$$
+     which in the matrix representation of state vectors can also be written as
+     $$\langle \chi|\psi\rangle=\begin{pmatrix} \chi^*_1 , \chi^*_2 &,\ldots \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix} \, .$$ Therefore, we can present  bra vector $\langle \chi|$ as row vectors and ket vectors as column vector. 
+     The row vector can thus be treated as the *complex conjugate* of the corresponding column vector.
+
+## 4.4. A two-dimensional Hilbert space
+
+As a practical example to illustrate the basic ideas of vector spaces applied to quantum physics presented above, we will consider a quantum system which is fundamental for quantum mechanics and its applications. This system corresponds to the possible states that the intrinsic angular momentum of an electron, known as *spin*, can occupy. As you will see in following courses, the Hilbert space for the electron spin has dimension $n=2$, meaning that we can find an electron spin *pointing* either in the up direction, denoted by $|+\rangle$, or the down direction, denoted by $|-\rangle$.
+
+The general state vector of this system will be expressed as a linear superposition of the *up* and *down* states,
+$$
+|\Psi\rangle = c_+ | + \rangle + c_- | - \rangle \, , \quad c_+ = \langle +|\Psi \rangle
+\, , \quad c_- = \langle -|\Psi \rangle \, .
+$$  
+In terms of the matrix representation, if we take $| + \rangle$
+and  $| - \rangle$ as the *basis* for this vector space, we can express $|\Psi\rangle$
+as a column vector
+$$
+|\Psi\rangle = \left( \begin{array}{c} c_+ \\ c_- \end{array}\right) \, ,
+$$
+and similarly for the basis vectors
+$$
+| + \rangle = \left( \begin{array}{c} 1 \\ 0 \end{array}\right) \, ,\quad
+| - \rangle = \left( \begin{array}{c} 0 \\ 1 \end{array}\right) \, .
+$$
+
+We can likewise express the inner product between  $ |\Psi\rangle$ and some other state vector
+$$
+|\Psi'\rangle = c_+' | + \rangle + c_-' | - \rangle \, , \quad c_+ '= \langle +|\Psi' \rangle
+\, , \quad c_-' = \langle -|\Psi' \rangle \, . $$
+as a multiplication of a row vector and a column vector,
+$$
+\langle \Psi'|\Psi\rangle = \left(  (c'_+)^{*} , (c'_-)^{*} \right)\left( \begin{array}{c} c_+ \\ c_- \end{array}\right) =  (c'_+)^{*}c_+  + (c'_-)^{*} c_- \, .
+$$ 
+Note that one needs to take the complex conjugate of the components when expressing a state vector as a bra vector.
+
+Examples of elements of this Hilbert space are the following:
+$$
+\left( \begin{array}{c}3 \\ -2i \end{array} \right) \, ,\quad
+\left( \begin{array}{c}i \\ -4 \end{array} \right) \, ,\quad
+\left( \begin{array}{c}2 \\ 5 \end{array} \right) \, .
+$$
+The values of the coefficients $c_+$ and $c_-$ for these examples above are, respectively,
+$$
+(c_+,c_-) = (3,-2i) \, ,\qquad
+(c_+,c_-) = (i,-4) \, ,\qquad
+(c_+,c_-) = (2,5) \, .\qquad 
+$$
+!!! warning ""
+     Note however that many other bases are possible, and that the physics of a quantum system do not depend on the basis that we choose.
+
+The bra vectors associated to these ket vectors will be given by
+$$ |{\Psi}\rangle=   \left( \begin{array}{c}3 \\ -2i \end{array} \right) \, , \qquad
+\langle{\Psi}|=\left( 3, 2i \right) \, ,$$
+$$ |{\Psi}\rangle = \left( \begin{array}{c}i \\ -4 \end{array} \right) \, , \qquad
+\langle{\Psi}|=\left(  -i , -4  \right)\, , $$
+$$|{\Psi}\rangle=  \left( \begin{array}{c}2 \\ 5 \end{array} \right) \, , \qquad
+\langle{\Psi}|=  \left( 2 , 5  \right) \, .$$
+Note however that the above vectors are not normalised (the inner product with themselves is different from unity), and thus
+cannot represent physical states. We show below an explicit example of a normalised state vector belonging to this Hilbert space.
+
+!!! done "Example: Evaluating the inner product"
+     We also know how we can evaluate the inner product between any two state vectors belonging to this Hilbert space. If we have two state vectors given by
+     $$
+     |\psi\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
+     |\phi \rangle = \left( \begin{array}{c}0 \\1 \end{array} \right)
+     $$
+     then their inner product is
+     $$
+     \langle \psi | \phi \rangle =
+     \frac{1}{\sqrt{2}} \left( 1 , i  \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
+     \frac{i}{\sqrt{2}} 
+     $$
+     and the associated probability will be given by
+     $$
+     |\langle \psi|\phi\rangle|^2 = \frac{1}{2}
+     $$
+     meaning that if I measure the state $| \phi \rangle $, I will have a 50% probability
+     of finding it in the state $| \psi \rangle$. Recall that probabilities must always be smaller than 1 to make physical sense. Note also that I am using normalised vectors, you can check yourselves that
+     $$
+     \langle \psi |\psi \rangle =  \langle \phi | \phi\rangle = 1 \, ,
+     $$
+     as required to ensure the probabilistic description of the state vector.
+
+## 4.5. Problems
+
+**1)** [:grinning:] *The inner product*
+
+Two vectors in a three-dimensional complex vector space are defined by:
+$$
+|A\rangle =\begin{pmatrix}2\\-7i\\1\end{pmatrix},~~|B\rangle=\begin{pmatrix}1+3i\\4\\8\end{pmatrix}
+$$
+Let $a=6+5i$ and answer the following questions:
+
+*(a)*. Calculate $a|A\rangle$, $a |B\rangle$, and $a( |A\rangle+|B\rangle)$. Show that $a(|A\rangle+|B\rangle)=a|A\rangle+a|B\rangle$.
+
+*(b)*. Find the inner products $\langle A | B\rangle$ and $\langle B | A\rangle $. What is the relationship between them?
+
+*(c)* Find the norm of each vector.
+
+***
+  
+**2)**  [:grinning:] *A normalised vector*
+
+*(a)* Show that the two state vectors 
+$$
+|\Psi\rangle=\begin{pmatrix} 1/\sqrt{2}\\ 1/\sqrt{2}\end{pmatrix},~~|\Phi\rangle=\begin{pmatrix} 1/\sqrt{2}\\ -1/\sqrt{2}\end{pmatrix}
+$$
+are orthogonal to each other. Is the state vector $ |\Psi \rangle$ normalised?
+
+*(b)* Assume a vector
+$$
+|u \rangle =\begin{pmatrix} x\\ 3x\\-2x\end{pmatrix}
+$$
+where $x$ represents an unknown real number. Find the value of $x$ such that the state vector $|u\rangle$ is normalised.
+
+***
+    
+**3)** [:sweat:]  The *Cauchy-Schwartz inequality* states that
+$$
+\mid{\langle \Phi|\Psi\rangle}\mid^2 \leq \langle\Psi|\Psi\rangle \langle \Phi | \Phi\rangle
+$$
+Demonstrate this result using the properties of the inner product that we have covered in this lecture.
+
+***
+
+**4)** [:grinning:] *Basis vectors*
+
+*(a)* Show that the following vectors are linearly dependent:
+
+$$
+|a\rangle=\begin{pmatrix} 1\\ 2\\1\end{pmatrix},~~|b\rangle=\begin{pmatrix} 0\\ 1\\0\end{pmatrix},~~|c\rangle=\begin{pmatrix} -1\\ 0\\-1\end{pmatrix}
+$$
+
+*(b)* Is the following set of vectors linearly independent?
+
+$$
+|a\rangle=\begin{pmatrix} 2\\ 0\\0\end{pmatrix},~~|b\rangle=\begin{pmatrix} 0\\ -1\\0\end{pmatrix},~~|c\rangle=\begin{pmatrix} 0\\ 0\\-1\end{pmatrix}
+$$
+
+***
+
+**5)** [:grinning:] *Dirac algebra with bras and kets*
+
+Suppose that $|a\rangle $, $|b\rangle$, $|c\rangle$ is an orthonormal basis. In this basis let us define the following two state vectors:
+$$
+|\Psi\rangle=2i |a\rangle-3|b\rangle+i|c\rangle
+$$
+$$
+|\Phi\rangle=3|a\rangle-2|b\rangle+4|c\rangle
+$$
+
+*(a)* Find $\langle \Psi|$ and $\langle \Phi|$.
+
+*(b)* Compute the inner product $\langle \Phi | \Psi \rangle$.
+
+*(c)* Show that $\langle \Phi | \Psi\rangle=\langle \Psi|\Phi\rangle^*$.
+
+*(d)* Write the column vector representing the vector $|\Psi\rangle$ in the given basis. Then write down the row vector that represents $\langle \Psi|$ in the given basis as well.
+
+***
+
+**6)** [:sweat:] *The state vector for a spin half particle*
+
+The state vector for a spin half particle that passes through a magnetic field oriented in the direction $\hat{n}$ and exists with its spin component in the direction of the magnetic field, i.e. $S=\vec{S}\cdot\hat{n}=\frac{1}{2}\hbar$ is given by
+$$
+|S\rangle =\cos(\theta/2) |+\rangle +\sin(\theta/2) \, e^{i\phi} |-\rangle
+$$
+where $\hat{n}=\sin\theta \,\cos\phi \, \hat{i} +\sin\theta \, \sin\phi \,\hat{j}+\cos\theta \, \hat{k}$. 
+
+*(a)* What is the corresponding bra vector?
+
+*(b)* Show that this state is normalized to unity.
+
+*(c)* Identify the state $|S\rangle $ if $\hat{n}=\hat{i}, \hat{j},$ and $\hat{k}$.
+
+*(d)* Express $|S\rangle $ in terms of the basis states $|-\rangle ,\,|+\rangle $ in each case.
+
+***
+
+**7)** [:sweat:] *A particle in an infinite well*
+
+A particle of mass $m$ confined to move in an infinite well of width $L$ can have the energies $E_n=\pi^2\hbar^2n^2/2mL^2$ where $n=1,2,...$ We can specify the states of the particle in the well by the kets $| 1\rangle $, $|2\rangle $, $|3\rangle $, ... where $|n\rangle $ is the ket corresponding to the particle having the energy $E_n$. These states form a complete orthonormal set of basis states for the particle in the well.
+
+*(a)* What is the dimension of the state space for the particle?
+
+*(b)* State the orthonormality conditions for the kets $\{| 1\rangle,|2\rangle,|3\rangle,...\}$
+
+*(c)* A particle is prepared in the state
+$$
+|\psi\rangle=\frac{1}{3}|1\rangle+\frac{1}{3}(2+i)|2\rangle+\alpha|3\rangle.
+$$
+This state is normalized to unity. If the experiment is repeated 500 times under identical conditions, and the energy of the particle in the well is measured, roughly how many times will the particle be observed to have the energy $E_3$?
+  
+

docs/5_operators_QM.md

+---
+title: Operators in quantum mechanics
+---
+
+# 5. Operators in quantum mechanics
+
+The lecture on operators in quantum mechanics consists of the following parts:
+
+- [5.1. Definition and properties of operators](#51-definition-properties-operators)
+
+- [5.2. Manipulating operators](#52-manipulating-operators)
+
+- [5.3. Projection operators](#53-projection-operators)
+
+- [5.4. The Hermitian adjoint](#54-the-hermitian-adjoint)
+
+- [5.5. Matrix representation of operators](#55-matrix-representation-of-operators)
+
+and at the end of the lecture notes, there is a set the corresponding exercises:
+
+- [5.6. Problems](#56-problems)
+
+*** 
+
+The contents of this lecture are supplemented with the following **videos**:
+
+- [1. Representation of an operator](https://www.dropbox.com/s/cg9s5hb8dtlkydh/linear_algebra-05.mov?dl=0)
+
+- [2. The action of an operator on kets in matrix representation](https://www.dropbox.com/s/74w7bvwq3o8xiam/linear_algebra-08.mov?dl=0)
+
+**The total length of the videos: ~5 minutes**
+
+***
+
+In the previous lecture, we presented the mathematical language to describe the *quantum states* of a physical system. We saw that the state of a quantum system is described by its *vector state* $|\Psi\rangle$, an element of a special complex vector space called the *Hilbert space*. We presented the Dirac notation and discussed that we can assign a *probabilistic interpretation* to vector states and their inner products. We also discussed their matrix representation and how we can express a state vector in terms of its components in a specific basis.
+
+Now, we need to introduce a concept and a mathematical language required to *extract information* about the physical properties of a system from its state vector, which we will denote by *observables*. We emphasize that this distinction between the *state of a quantum system* (given by the wave function) and the *observables*, which we can extract from it, is the novelty of quantum mechanics with respect to classical physics where this notion is absent. With this motivation, in order to represent fundamental physical quantities of a quantum system that we can measure; such as position, momentum, or energy, we need to introduce a special mathematical entity known as an *operator*.
+
+## 5.1. Definition and properties of operators
+
+Operators in quantum mechanics are mathematical entities  used to represent physical processes that result in the *change* of the state vector of the system, such as the evolution of these states with time. These operators can also represent physical properties of a system that can be experimentally measured (for example position, momentum, or energy), the **observables** associated to this quantum system. Note that each quantum system will have in general a different set of physical observables associated to it.
+
+!!! info "Operators in quantum mechanics"
+    An operator is a mathematical object that *acts* on the state vector of the system and produces another state vector. To be precise, if we denote an operator by $\hat{A}$ and $|\psi\rangle$ is an element of the Hilbert space of the system, then
+    $$\hat{A} |\psi\rangle = |\phi\rangle \, ,$$
+    where the state vector $|\phi\rangle$ *also* belongs to the same Hilbert space.
+
+There are many types of important operators in quantum mechanics. In this lecture, we will present some of these, such as the
+**unitary operators** that determine the time evolution of a quantum system and the **Hermitian operators** which can be assigned to
+physically observable properties of a system, such as momentum or energy. We will also discuss how we can manipulate operators and combine them in various ways.
+
+!!! info "Linear operators"
+    In this course, we are interested in the so-called *linear operators*, which are those operators $\hat{A}$ such that for any arbitrary pair of state vectors $|\psi_1\rangle$ and $|\psi_2\rangle$ and for any complex numbers $c_1$ and $c_2$ they satisfy *associative* and *distributive* properties, for instance
+    $$\hat{A}[c_1|\psi_1\rangle+c_2|\psi_2\rangle]=c_1\hat{A}|\psi_1\rangle+c_2\hat{A}|\psi_2\rangle \, .$$
+
+Linearity of operators has several important consequences. Recall that in the previous lecture we discussed that any state vector $|\psi\rangle$ can be expressed as a linear combination of a complete set of basis states $\{|\phi_i\rangle,i=1,2,3,...,n\}$ associated to this Hilbert space:
+$$|\psi\rangle=\sum_{i=1}^nc_i|\phi_i\rangle \, , \quad  c_i = \langle \phi_i | \psi\rangle \, ,$$
+where the values of the coefficients $c_i$ can be fixed thanks to the orthogonality properties of the basis, $\langle \phi_i | \phi_j\rangle=\delta_{ij} $.
+Then one can see that for linear operators the following applies 
+$$ \hat{A}|\psi\rangle=  \hat{A}\sum_{i=1}^nc_i|\phi_i\rangle =  \sum_{i=1}^nc_i ( \hat{A}|\phi_i\rangle ) \, .$$
+This result tells us that if we know the effects of the operator $\hat{A}$ for each of the elements of the basis $|\phi_i\rangle$,
+we can easily determine its effect on a *general state vector* $|\psi\rangle$ belonging to the same Hilbert space. 
+In other words, the action of the operator $\hat{A}$ on the basis vectors $\{\phi_i\rangle \}$ correlates with its action on any other state vector $|\psi\rangle$ to which the operator was applied. 
+
+Some other important properties of the operators can be stated as follows.
+!!! info "Properties of the operators"
+    1. If two operators $\hat{A}$ and $\hat{B}$ are such that
+        $$\hat{A}|\psi\rangle=\hat{B}|\psi\rangle \, , $$
+        for all state vectors $|\psi\rangle$ belonging to the Hilbert
+        space of the system, then these two operators must be *identical*:
+        $$ \hat{A}=\hat{B} \, .$$
+        Note that this is true only if the action of two operators
+        is identical for all elements of the Hilbert space.
+    2. Like in general vector spaces, in Hilbert spaces, we also have the identity (or unit) and zero (or null) operators defined as
+        - The **unit (or identity) operator** $\hat{I}$ is the operator that satisfies
+            $$\hat{I}|\psi\rangle=|\psi\rangle $$
+        - The **zero (or null) operator** $\hat{0}$ is the operator that satisfies
+            $$ \hat{0}|\psi\rangle=0$$
+        In both cases, these relations hold for all state vectors $|\psi\rangle$ of the Hilbert space $\mathcal{H}$.
+    
+## 5.2. Manipulating operators 
+
+We can combine and manipulate operators in various ways. In doing so, we should be careful because manipulations
+of operators can be quite different compared with manipulations of scalar complex numbers. 
+For example, if you have two complex numbers, the result of their multiplication does not depend on the order in which you multiply them, but for operators it does! As we will show, **in general, operators are non-commutative**, meaning that the order in which they are applied will vary the result of the operation. 
+
+As mentioned above, in these lectures, we will be focusing only on linear operators. For this class of operators, following operations are possible:
+
+!!! info "Addition of operators:" 
+    The sum of two operators $\hat{A}$ and $\hat{B}$ is defined by
+    $$(\hat{A}+\hat{B})|\psi\rangle=\hat{A}|\psi\rangle+\hat{B}|\psi\rangle \, ,$$
+    for all state vectors $|\psi\rangle$. The sum of two operators defines another operator, $\hat{C}$:
+    $$ 
+    \hat{C}|\psi\rangle=(\hat{A}+\hat{B})|\psi\rangle=\hat{A}|\psi\rangle+\hat{B}|\psi\rangle
+    $$
+    for all states $|\psi\rangle$, and so we can write
+    $$
+    \hat{C} = \hat{A} + \hat{B} \, .
+    $$
+    Note that when we express operator relations, such as this one, we are implicitly stating that they hold when these operators are applied to any of the state vectors $| \Psi \rangle$ of the Hilbert space $\mathcal{H}$.
+    
+!!! info "Multiplication of an operator by a complex number:"
+    If we have an operator that acts on a state vector as
+    $$\hat{A}|\psi\rangle=|\phi\rangle \, ,$$
+    then we can define the operator $\hat{C}=\lambda \hat{A}$, where $\lambda$ is a complex number as follows
+    $$\hat{C} |\psi\rangle= (\lambda \hat{A})|\psi\rangle=\lambda(\hat{A}|\psi\rangle)=\lambda |\phi\rangle \, .$$
+
+!!! info "Multiplication of operators"
+    Assume that an operator $\hat{A}$ acting on a ket vector $|\psi\rangle$ maps it into another ket vector $|\phi\rangle$ and that the operator $\hat{B}$  acting on $|\phi\rangle$ results in a third ket vector $|\rho\rangle$:
+    $$
+    \hat{B} \left( \hat{A}|\psi\rangle \right)=\hat{B}|\phi\rangle=|\rho\rangle \, .
+    $$
+    One can then define the product of the two operators as a new operator, $\hat{C}=\hat{B}\hat{A}$, such that its action on the initial ket vector $|\psi\rangle$ is defined as
+    $$ 
+    \hat{C}|\psi\rangle=\left( \hat{B}\hat{A} \right) |\psi\rangle = |\rho\rangle \, .
+    $$
+    Note that in general, multiplication of two operators is *non-commutative*, so the order in which we multiply $\hat{A}$ and $\hat{B}$ is important, and the operator $\hat{C}=\hat{B}\hat{A}$ will be different from $\hat{D}=\hat{A}\hat{B}$.
+
+!!! info "Commutator of two operators" 
+    The difference between a product of operators $\hat{B}\hat{A}$ and the product in the opposite order, namely $\hat{B}\hat{A}$, is defined as the *commutator* of these two operators: 
+    $$ [\hat{A},\hat{B}]\equiv \hat{A}\hat{B}-\hat{B}\hat{A} \, . $$
+    The commutator plays a fundamental role in the physical interpretation of quantum mechanics. 
+    In a nutshell, it tells us whether or not two observable properties of a system can be determined simultaneously with arbitrary precision according to the Heisenberg uncertainty relations. 
+    
+!!! note "Exercise"
+    Using the properties of the addition and multiplication of operators that we just discussed, you can check that the commutator satisfies the following properties:
+
+    1. $[\hat{A},\hat{B}]= -[\hat{B},\hat{A}] $
+
+    2. $[\hat{A},\alpha\hat{B}+\beta\hat{C}]=\alpha[\hat{A},\hat{B}]+\beta[\hat{A},\hat{C}] $
+
+    3. $[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}] +[\hat{A},\hat{C}] \hat{B}$ 
+
+    4. $[\hat{A},[\hat{B},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=0$
+
+## 5.3. Projection operators
+
+!!! info "Projection operator:"
+    An operator $\hat{A}$ with the property
+    $$ \hat{A}^2= \hat{A}\hat{A}= \hat{A} \, ,$$
+    which means that acting twice on a given state vector produces the same result as acting just once, is described as a *projection operator*.
+    Let us give an explicit example of such operator. Assume that we have an $n$-dimensional Hilbert space with a basis given by $\{|\phi_i\rangle\}$. We can then define the operator $\hat{B}_i$ as 
+    $$\hat{B}_i|\phi_j \rangle \equiv \delta_{ij}|\phi_j \rangle \, .$$ 
+
+Recall that as demonstrated above, once we indicate the behaviour of an operator
+for the basis vectors, we automatically know how it will act for any general
+state vector of the Hilbert space.
+
+!!! check "Example"
+    Let's demonstrate that this operator is a projection operator: 
+    $$ \left( \hat{B}_i\right)^2|\phi_j \rangle=\hat{B}_i\hat{B}_i|\phi_j \rangle=\delta_{ij}\hat{B}_i|\phi_j \rangle=\delta_{ij}^2|\phi_j \rangle = \delta_{ij}|\phi_j \rangle = \hat{B}_i|\phi_j \rangle  \, .$$
+    From this we can conclude that $\hat{B}_i^2=\hat{B}_i$, as expected for a projection operator, and where we have used that the square of the Kronecker delta is the same Kronecker delta itself.
+
+This projection operator has one important property. Let's act with $\hat{B}_i$ on an arbitrary vector $|\psi\rangle$ expanded in terms of the basis vectors $\{|\phi_j \rangle\}$:
+$$
+\hat{B}_i|\psi\rangle=\hat{B}_i\sum_{j=1}^n|\phi_j \rangle\langle \phi_j | \psi\rangle=\sum_{j=1}^n\left(\hat{B_i}|\phi_j \rangle\right) \langle \phi_j | \psi \rangle
+$$
+which implies that
+$$
+\hat{B}_i |\psi\rangle=\sum_{j=1}^n \delta_{ij}|\phi_j \rangle \langle \phi_ | \psi\rangle=|\phi_i\rangle\langle \phi_i | \psi\rangle \, .
+$$
+In other words, the operator $ \hat{B}_i$ *projects* the state vector $|\psi\rangle$ onto the direction given by the basis vector $|\phi_i\rangle$. We can clearly see in this case why these operators are called *projection operators*; They allow us to single out specific directions in the Hilbert space of our quantum system.
+
+## 5.4. The Hermitian adjoint
+
+When discussing vector spaces in quantum mechanics, we learned that for each vector state ket $|\Psi\rangle$, there exists the corresponding bra vector $\langle \Psi|$ which can be understood as its complex conjugate. When expressing $|\Psi\rangle$ as a column vector in terms of its components, $\langle \Psi|$ was the associated row
+ vector expressing the complex conjugate of its components. A similar discussion is required now in the case of operators.
+
+!!! warning "Warning" 
+    For an operator $\hat{A}$ such that $$\hat{A}|\psi\rangle=|\phi\rangle$$ holds true, the already familiar complex conjugation is in general not valid
+    $$\langle\psi|\hat{A}\neq \langle\phi |$$ 
+
+!!! info "Hermitian adjoint operator"
+    We can introduce another operator related to $\hat{A}$ and written as $\hat{A}^\dagger$ which has the following defining property
+    $$\hat{A}|\psi\rangle=|\phi\rangle\; \text{then}\; \langle \psi|\hat{A}^\dagger= \langle \phi|$$
+    The operator $\hat{A^\dagger}$ is known as the *Hermitian adjoint* of $\hat{A}$. 
+    
+    What is then the action of this Hermitian adjoint operator on a ket vector? We can consider the following product
+    $$
+    \langle\rho|\hat{A}|\psi\rangle=\langle \rho|(\hat{A}|\psi\rangle)=\langle\rho|\phi\rangle
+    $$
+    using that $\hat{A}|\psi\rangle=|\phi\rangle$. The complex conjugate of the previous expression yields:
+    $$
+    \langle \rho |\hat{A}|\psi\rangle^*=\langle \rho |\phi\rangle^*=\langle \phi |\rho\rangle
+    $$
+    and if $\langle\psi|\hat{A^\dagger}= \langle \phi|$, then
+    $$
+    \langle \rho|\hat{A}|\psi\rangle^*=\langle \phi|\rho\rangle=(\langle \psi |\hat{A^\dagger})|\rho\rangle=\langle \psi |\hat{A^\dagger}|\rho\rangle \, .
+    $$
+
+!!! warning ""
+    As we will see next, in quantum mechanics, we are interested in operators for which $\hat{A}=\hat{A}^\dagger$, that is where the operator coincides with its Hermitian adjoint.
+
+## 5.5. Matrix representation of operators
+ 
+In the previous lecture, we discussed the matrix representation of state vectors, a notation describing the elements of the Hilbert space assigned to a given quantum system. We will now show how one can also construct a matrix representation of *operators*.
+
+The starting point is an operator equation of the form
+$$
+\hat{A}|\psi\rangle=|\varphi\rangle \, ,
+$$
+where $|\psi\rangle$ and $|\varphi\rangle $ are state vectors. This equation can be rewritten using the identity operator given by the basis elements 
+$$
+\hat{I} = \sum_{i=1}^n |\phi_i\rangle\langle \phi_i| \, ,
+$$
+and rewritten as
+$$
+|\varphi\rangle=\hat{A}|\psi\rangle=\hat{A}\sum_{i=1}^n |\phi_i\rangle\langle \phi_i|\psi\rangle=
+\sum_{i=1}^n \left( \hat{A}|\phi_i\rangle\right) \langle \phi_i|\psi\rangle
+$$
+
+We can now evaluate the inner product between the basis vector $| \phi_j\rangle$
+and the state vector $ |\varphi \rangle$ to obtain
+$$
+\langle {\phi_j}|\varphi\rangle=\sum_{i=1}^n \left(\langle \phi_j| \hat{A}|\phi_i\rangle
+\right) \langle \phi_i|\psi\rangle \, ,
+$$
+which can also be expressed as a matrix multiplication equation in terms of its components using
+$$
+\varphi_j=\sum_{i=1}^n {A}_{ji}\, \psi_i \, ,
+$$
+where we define
+$$
+{A}_{ji} \equiv \langle \phi_j| \hat{A}|\phi_i\rangle \, .
+$$
+
+!!! info "Matrix representation of operators"
+    The derivation above demonstrates that an operation equation of the form $|\varphi\rangle = \hat{A}|\psi\rangle$ can be expressed in terms of a matrix representation
+    $$
+    \begin{pmatrix} \varphi_1\\\varphi_2\\\varphi_3 \\\vdots\end{pmatrix} = \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \begin{pmatrix} \psi_1\\\psi_2\\\psi_3 \\\vdots\end{pmatrix}
+    $$
+    where the operator $\hat{A}$ is represented by a (square) matrix
+    $$
+    \hat{A} = \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} 
+    $$
+
+Like in the case of regular vector spaces, when we represent state vectors as operators in terms of components (matrix representation), we are implicitly or explicitly assuming a choice of basis. If we change the basis, the values of the components will change too. 
+In other words, while the operator equation $|\varphi\rangle = \hat{A}|\psi\rangle$ is identical in any chosen basis, once we express it as
+$$
+\begin{pmatrix} \varphi_1\\\varphi_2\\\varphi_3 \\\vdots\end{pmatrix} = \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \begin{pmatrix} \psi_1\\\psi_2\\\psi_3 \\\vdots\end{pmatrix}
+$$
+then the specific values of  *e.g.* $A_{11}, A_{12}, ...$ will be basis-dependent.
+
+Having demonstrated that we can represent state vectors and operators in term of vectors and matrices, we can now highlight some of the most important *properties of the matrix representation*.
+
+!!! info "Properties of matrix representation of operators"
+    1. **Equality**: two operators are said to be equal if their corresponding operator matrix elements are equal, for instance $\hat A = \hat B$ if $A_{ij}=B_{ij}$ for all possible values of $i$ and $j$.
+
+    2. **Identity  operator**: the unit (or identity) operator is represented by $\hat1$ and satisfies $\hat1|\psi\rangle=|\psi\rangle$ for all possible state vectors $|\psi\rangle$. The diagonal elements of the  matrix representation of the identity operator $\hat1$ are all unity while the off-diagonal elements vanish:  $1_{ij}=\delta_{ij}$. This means that for a $n$-dimensional Hilbert space the unit operator is the $n$-dimensional identity matrix. Note that the unit operator has the same form in all representations irrespective of the specific choice of basis states.
+
+    3. **Null operator**: the zero operator $\hat0$ is such as $\hat{0}|\psi\rangle=0$ for all $\psi$. Its matrix elements are all zero, $0_{ij}=0$.
+
+    4. **Addition of operators**: given two operators $\hat{A}$ and $\hat{B}$ with matrix elements $A_{ij}$ and $B_{ij}$, then the matrix elements of their sum $\hat{A}+\hat{B}$ are given by
+    $$
+    C_{ij}=A_{ij}+B_{ij} \, .
+    $$
+    
+    5. **Multiplication by a complex number**: if $\lambda$ is a complex number, then the matrix elements of the operator $\hat{C}=\lambda \hat{A}$ are given by
+    $$
+    C_{ij}=\lambda A_{ij} \, .
+    $$
+
+    6. **Product of operators**: given two operators $\hat{A}$ and $\hat{B}$ with matrix elements $A_{ij}$ and $B_{ij}$, then the matrix elements of their operator product $\hat{P}=\hat{A}\hat{B}$ are given by
+    $$
+    P_{ij}=\sum_{k=1}^nA_{ik}B_{kj} \, ,
+    $$
+    which as you might recall is nothing but the standard rule for matrix multiplication. So once we express operators in their matrix representation, we can multiply them by following standard matrix multiplication. 
+
+    7. **Commutator**: in the same way as matrix multiplication is not commutative, also  operator multiplication is *not commutative*:
+    $$
+    \hat{A}\hat{B}\;\neq\hat{B}\hat{A}
+    $$
+    As mentioned above, the difference, $\hat{A}\hat{B}-\hat{B}\hat{A}$ is known as the *commutator* of $\hat{A}$ and $\hat{B}$ and is represented by $[\hat{A},\hat{B}]$. In terms of the matrix representation, the  components of the commutator of $\hat{A}$ and $\hat{B}$
+    (which is also a matrix itself) will be given by
+    $$
+    [\hat{A},\hat{B}]_{ij}= \sum_{k=1}^nA_{ik}B_{kj} - \sum_{k=1}^nB_{ik}A_{kj} =
+    \sum_{k=1}^n\left( A_{ik}B_{kj} - B_{ik}A_{kj} \right) \, .
+    $$  
+
+### Matrix representation of Hermitian operators
+
+As discussed above, Hermitian operators play a central role in the physical interpretation of quantum systems.
+We can now provide the explicit expression of the Hermitian operators in the matrix representation. 
+
+!!! info "Matrix representation of Hermitian operators"
+    Let us assume that we have an operator $\hat{A}$ in an $n$-dimensional Hilbert space
+    with matrix elements $A_{ij}$ defined with respect to a set of orthonormal basis states, $\{ |\phi_i\rangle; i=1,2,\ldots\,n\}$. 
+    From its matrix representation one can construct a new operator by taking the transpose and complex conjugate of the original matrix, that is:
+    $$
+    \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}\rightarrow \begin{pmatrix} A_{11}^* & A_{21}^* & A_{31}^* & \ldots \\ A_{12}^* & A_{22}^* & A_{32}^* & \ldots\\A_{13}^* & A_{23}^* & A_{33}^* & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}
+    $$
+    This new matrix corresponds to the matrix representation of a new operator that will be denoted by $\hat{A}^\dagger$.
+     
+    This new operator is called the **Hermitian adjoint** of the operator $\hat{A}$.
+
+!!! warning "Hermitian operators are equal to their adjoint version"
+    At the operator level, we can write that Hermitian operators satisfy $\hat{A}=\hat{A}^\dagger$.
+    From the matrix representation giving the expression in terms of its components, the condition for a Hermitian operator will therefore read
+    $$
+    \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} = \begin{pmatrix} A_{11}^* & A_{21}^* & A_{31}^* & \ldots \\ A_{12}^* & A_{22}^* & A_{32}^* & \ldots\\A_{13}^* & A_{23}^* & A_{33}^* & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \, ,
+    $$
+    which can be expressed in a more compact way as
+    $$
+    A_{ij} = A_{ji}^* \, , \quad i,j=1,\ldots,n \, .
+    $$
+    Any operator which satisfies this condition will be a Hermitian operator. Note that the Hermiticity of an operator is a property which is independent of the specific choice of basis.
+
+***
+## 5.6. Problems
+
+1. [:grinning:] *Commutator algebra*
+
+    Prove that $[\hat{A},\hat{B} \hat{C}]=[\hat{A}, \hat{B}] \hat{C}+\hat{B} [\hat{A}, \hat{C}]$.
+
+2. [:grinning:] *The commutator*
+    
+    Let 
+    $$
+    \hat{A}=\begin{pmatrix} -1&2i&0\\0&4&0\\1&0&1\end{pmatrix},~~~~\hat{B}=\begin{pmatrix} 0&2&i\\-i&2i&0\\0&1&4\end{pmatrix}
+    $$
+
+    Do $\hat{A}$ and $\hat{B}$ commute?
+
+3. [:grinning:] *Linear operators*
+
+    Suppose that in an orthonormal basis given by $\{|a\rangle, |b\rangle, |c \rangle\}$ an operator $\hat{A}$ acts as follows:
+    $$\hat{A}|{a}\rangle=2|{a}\rangle\,,\quad \hat{A}|{b}\rangle=3|{a}\rangle-i|{c}\rangle\, , \quad \hat{A}|{c}\rangle=-|{b}\rangle$$
+    Determine the matrix representation of the operator $\hat{A}$.
+
+4. [:grinning:] *Calculating expectation values of operators*
+
+    The expectation value of an operator $\hat{A}$ with respect to a state $|{\Psi}\rangle$ is given by 
+    $$<\hat{A}>=\langle{\Psi}|\hat{A}|{\Psi}\rangle \, .$$
+    A particle is in a state $|{\Psi}\rangle=2i|{a}\rangle-|{b}\rangle+4i|{c}\rangle$
+    and an operator $\hat{A}=|{a}\rangle\langle{a}|-2i|{b}\rangle\langle{b}|+|{c}\rangle\langle{c}|$. 
+    
+    (The basis is orthonormal).   
+    
+    Find the expectation value $<\hat{A}>$ in this state.
+
+5. [:smirk:] *The Hermitian conjugate of an operator*
+
+    Consider that the matrix representation of the operator $\hat{A}$ is given by:
+    $$\hat{A}=\begin{pmatrix}1&0&-3i\\3&5&0\\3i&0&-2 \end{pmatrix}$$
+    and the following two state vectors from the same Hilbert space are given by:
+    $$|{\Psi}\rangle=\begin{pmatrix}2\\3i\\-1 \end{pmatrix}\, ,\quad |{\Phi}\rangle=\begin{pmatrix}0\\-1\\1 \end{pmatrix}\, .$$
+
+    **(a)** Find the result of $\hat{A}|{\Psi}\rangle$ and $\hat{A}|{\Phi}\rangle$.
+
+    **(b)** Find the Hermitian conjugates $|{\Psi}\rangle^\dagger$ and $|{\Phi}\rangle^\dagger$, and use these to calculate the inner products between the two state vectors $\langle{\Psi}|{\Phi}\rangle$ and $\langle{\Phi}|{\Psi}\rangle$.
+
+6. [:grinning:] *The Hadamard gate*
+
+    An important operator used in quantum computation is the *Hadamard gate*, which is represented by the matrix:
+    $$\hat{H}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$. 
+
+    **(a)** Discuss whether or not $\hat{H}$ is Hermitian or unitary.
+
+    **(b)** Evaluate how the Hadamard gate acts upon the *up* and *down* basis states of the system
+    $$|+\rangle=\begin{pmatrix}1 \\ 0  \end{pmatrix}\, ,\quad |-\rangle=\begin{pmatrix}0\\1 \end{pmatrix}\, ,$$
+    and discuss the interpretation of these results.
+
+7. [:smirk:] *Spin-$1/2$ system*
+
+    Consider two vectors
+    $$|a\rangle=\frac{1}{\sqrt 2} \left( |-\rangle-i|+\rangle \right) $$
+    $$|b\rangle=\frac{1}{\sqrt 2} \left(|-\rangle +i|+\rangle \right)$$
+    where $|\pm \rangle$ are the basis vectors for a spin-half system, and the operator $\hat{A}$ defined by
+    $$\hat{A}|\pm\rangle=\pm \frac{1}{2} i \hbar|\mp\rangle$$
+
+    Note: the basis states are $$|+\rangle=\begin{pmatrix}1 \\ 0  \end{pmatrix}\, ,\quad |-\rangle=\begin{pmatrix}0\\1 \end{pmatrix}\, ,$$
+
+    **(a)** Express the state vectors $|a\rangle$ and $|b\rangle$ as column vectors.
+
+    **(b)** Write down the corresponding bra vectors as row vectors.
+
+    **(c)** Calculate the inner products $\langle a|b\rangle$ and $\langle b|a\rangle$.
+
+    **(d)** Write this operator as a matrix in the $\{|+\rangle,|-\rangle\}$ representation.
+
+    **(e)** Calculate $\hat{A}|a\rangle$ using the matrix representation of $\hat{A}$ and the column representation of $|a\rangle$.
+
+    **(f)** Calculate $\langle b|\hat{A}$ using the matrix representation of $\hat{A}$ and the row representation of $\langle b|$.
+
+  
+  
+  

docs/6_eigenvectors_QM.md

+---
+title: Eigenvalues and eigenvectors
+---
+
+# 6. Eigenvalues and eigenvectors
+
+The lecture on eigenvalues and eigenvectors consists of the following parts:
+
+- [6.1. Eigenvalue equations in linear algebra](#61-eigenvalue-equations-in-linear-algebra)
+
+- [6.2. Eigenvalue equations in quantum mechanics](#62-eigenvalue-equations-in-quantum-mechanics)
+
+and at the end of the lecture notes, there is a set of corresponding exercises:
+
+- [6.3. Problems](#63-problems)
+
+***
+
+The contents of this lecture are summarised in the following **video**:
+
+- [Eigenvalues and eigenvectors](https://www.dropbox.com/s/n6hb5cu2iy8i8x4/linear_algebra_09.mov?dl=0)
+
+*The total length of the videos: ~3 minutes 30 seconds*
+
+***
+
+In the previous lecture, we discussed a number of *operator equations*, which have the form
+$$
+\hat{A}|\psi\rangle=|\varphi\rangle \, ,
+$$
+where $|\psi\rangle$ and $|\varphi\rangle$ are state vectors
+belonging to the Hilbert space of the system $\mathcal{H}$.
+
+!!! info "Eigenvalue equation:"
+    A specific class of operator equations, which appear frequently in quantum mechanics, consists of equations in the form
+    $$
+    \hat{A}|\psi\rangle= \lambda_{\psi}|\psi\rangle \, ,
+    $$
+    where $\lambda_{\psi}$ is a scalar (in general complex). These are equations where the action of the operator $\hat{A}$
+    on the state vector $|\psi\rangle$ returns *the same state vector* multiplied by the scalar $\lambda_{\psi}$. 
+    This type of operator equations are known as *eigenvalue equations* and are of great importance for the description of quantum systems.
+
+In this lecture, we present the main ingredients of these equations and how we can apply them to quantum systems.
+
+##6.1. Eigenvalue equations in linear algebra
+
+First of all, let us review eigenvalue equations in linear algebra. Assume that we have a (square) matrix $A$ with dimensions $n\times n$ and $\vec{v}$ is a column vector in $n$ dimensions. The corresponding eigenvalue equation will be of form
+$$
+A \vec{v} =\lambda \vec{v} .
+$$
+with $\lambda$ being a scalar number (real or complex, depending on the type
+of vector space). We can express the previous equation in terms of its components,
+assuming as usual some specific choice of basis, by using
+the rules of matrix multiplication:
+
+!!! info "Eigenvalue equation: Eigenvalue and Eigenvector"
+    $$
+    \sum_{j=1}^n A_{ij} v_j = \lambda v_i \, .
+    $$
+    The scalar $\lambda$ is known as the *eigenvalue* of the equation, while the vector $\vec{v}$ is known as the associated *eigenvector*.
+    The key feature of such equations is that applying a matrix $A$ to the vector $\vec{v}$ returns *the original vector* up to an overall rescaling, $\lambda \vec{v}$. 
+
+!!! warning "Number of solutions"
+    In general, there will be multiple solutions to the eigenvalue equation $A \vec{v} =\lambda \vec{v}$, each one characterised by an specific eigenvalue and eigenvectors. Note that in some cases one has  *degenerate solutions*, whereby a given matrix has two or more eigenvectors that are equal.
+
+!!! tip "Characteristic equation:"
+    In order to determine the eigenvalues of the matrix $A$, we need to evaluate the solutions of the so-called *characteristic equation*
+    of the matrix $A$, defined as
+    $$
+    {\rm det}\left( A-\lambda \mathbb{I} \right)=0 \, ,
+    $$
+    where $\mathbb{I}$ is the identity matrix of dimensions $n\times n$, and ${\rm det}$ is the determinant.
+
+This relation follows from the eigenvalue equation in terms of components
+$$
+\begin{align}
+\sum_{j=1}^n A_{ij} v_j &= \lambda v_i \, , \\
+\to \quad \sum_{j=1}^n A_{ij} v_j - \sum_{j=1}^n\lambda \delta_{ij} v_j &=0 \, ,\\
+\to \quad \sum_{j=1}^n\left( A_{ij} - \lambda \delta_{ij}\right) v_j &=0 \, .
+\end{align}
+$$
+Therefore, the eigenvalue condition can be written as a set of coupled linear equations
+$$
+\sum_{j=1}^n\left( A_{ij} - \lambda \delta_{ij}\right) v_j =0 \, , \qquad i=1,2,\ldots,n\, ,
+$$
+which only admit non-trivial solutions if the determinant of the matrix $A-\lambda\mathbb{I}$ vanishes
+(the so-called Cramer's condition), thus leading to the characteristic equation.
+
+Once we have solved the characteristic equation, we end up with $n$ eigenvalues $\lambda_k$, $k=1,\ldots,n$.
+  
+We can then determine the corresponding eigenvector
+$$
+\vec{v}_k = \left( \begin{array}{c} v_{k,1}  \\ v_{k,2} \\ \vdots \\ v_{k,n} \end{array} \right) \, ,
+$$
+by solving the corresponding system of linear equations
+$$
+\sum_{j=1}^n\left( A_{ij} - \lambda_k \delta_{ij}\right) v_{k,j} =0 \, , \qquad i=1,2,\ldots,n\, ,
+$$
+
+Let us remind ourselves that in $n=2$ dimensions the determinant of  a matrix
+is evaluated as
+$$
+{\rm det}\left( A \right) = \left|  \begin{array}{cc} A_{11}  & A_{12} \\ A_{21}  &  A_{22} \end{array} \right|
+= A_{11}A_{22} - A_{12}A_{21} \, ,
+$$
+while the corresponding expression for a matrix belonging to a vector
+space in $n=3$ dimensions in terms of the previous expression will be given as
+$$
+{\rm det}\left( A \right) = \left|  \begin{array}{ccc} A_{11}  & A_{12}  & A_{13} \\ A_{21}  &  A_{22}
+&  A_{23} \\ A_{31}  &  A_{32}
+&  A_{33}  \end{array} \right| = 
+\begin{array}{c} 
++ A_{11} \left|  \begin{array}{cc} A_{22}  & A_{23} \\ A_{32}  &  A_{33} \end{array} \right| \\
+- A_{12} \left|  \begin{array}{cc} A_{21}  & A_{23} \\ A_{31}  &  A_{33} \end{array} \right| \\
++ A_{13} \left|  \begin{array}{cc} A_{21}  & A_{22} \\ A_{31}  &  A_{32} \end{array} \right|
+\end{array}
+$$
+
+!!! check "Example"
+    Let us illustrate how to compute eigenvalues and eigenvectors by considering a $n=2$ vector space. 
+    
+    Consider the following matrix
+    $$
+    A = \left( \begin{array}{cc} 1  &  2 \\ -1  &  4 \end{array} \right) \, ,
+    $$
+    which has associated the following characteristic equation
+    $$
+    {\rm det}\left( A-\lambda\cdot I \right)  = \left| \begin{array}{cc} 1-\lambda  &  2 \\ -1  &  4-\lambda \end{array} \right| = (1-\lambda)(4-\lambda)+2 = \lambda^2 -5\lambda + 6=0 \, .
+    $$
+    This is a quadratic equation which we know how to solve exactly; the two eigenvalues are $\lambda_1=3$ and $\lambda_2=2$.
+
+    Next, we can determine the associated eigenvectors $\vec{v}_1$ and $\vec{v}_2$. For the first one, the equation to solve is
+    $$
+    \left( \begin{array}{cc} 1  &  2 \\ -1  &  4 \end{array} \right)
+    \left( \begin{array}{c} v_{1,1}  \\ v_{1,2}  \end{array} \right)=\lambda_1
+    \left( \begin{array}{c} v_{1,1}  \\ v_{1,2}  \end{array} \right) = 3 \left( \begin{array}{c} v_{1,1}  \\ v_{1,2}  \end{array} \right) 
+    $$
+    from where we find the condition that $v_{1,1}=v_{1,2}$. 
+    
+    An important property of eigenvalue equations is that the eigenvectors are only fixed up to an  *overall normalisation condition*. 
+    
+    This should be clear from its definition: if a vector $\vec{v}$ satisfies $A\vec{v}=\lambda\vec{v} $,
+    then the vector $\vec{v}'=c \vec{v}$ with $c$ some constant will also satisfy the same equation. So then we find that the eigenvalue $\lambda_1$ has an associated eigenvector
+    $$
+    \vec{v}_1 = \left( \begin{array}{c} 1   \\ 1 \end{array} \right) \, ,
+    $$
+    and indeed one can check that
+    $$
+    A\vec{v}_1 = \left( \begin{array}{cc} 1  &  2 \\ -1  &  4 \end{array} \right)
+    \left( \begin{array}{c} 1   \\ 1 \end{array} \right) = \left( \begin{array}{c} 3  \\ 3 \end{array} \right)=
+    3 \vec{v}_1 \, ,
+    $$
+    as we intended to demonstrate.
+
+!!! note "Exercise"
+    As an exercise, try to obtain the expression of the eigenvector
+    corresponding to the second eigenvalue $\lambda_2=2$.
+
+
+##6.2. Eigenvalue equations in quantum mechanics
+
+We can now extend the ideas of eigenvalue equations from linear algebra to the case of quantum mechanics.
+The starting point is the eigenvalue equation for the operator $\hat{A}$,
+$$
+\hat{A}|\psi\rangle= \lambda_{\psi}|\psi\rangle \, ,
+$$
+where the vector state $|\psi\rangle$ is the eigenvector of the equation
+and $ \lambda_{\psi}$ is the corresponding eigenvalue, in general a complex scalar.
+    
+In general this equation will have multiple solutions, which for a Hilbert space $\mathcal{H}$ with $n$ dimensions can be labelled as
+$$
+\hat{A}|\psi_k\rangle= \lambda_{\psi_k}|\psi_k\rangle \, , \quad k =1,\ldots, n \, .
+$$
+  
+In order to determine the eigenvalues and eigenvectors of a given operator $\hat{A}$, we will have to solve the
+corresponding eigenvalue problem for this operator, what we called above as the *characteristic equation*.
+This is most efficiently done in the matrix representation of this operation, where we have
+that the above operator equation can be expressed in terms of its components as
+$$
+\begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}=  \lambda_{\psi_k}\begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix} \, .
+$$
+
+As discussed above, this condition is identical to solving a set of linear equations
+for the form
+$$
+\begin{pmatrix} A_{11}- \lambda_{\psi_k} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi_k} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi_k} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}
+\begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}=0 \, .
+$$
+
+!!! info "Cramer's rule"
+    This set of linear equations only has a non-trivial set of solutions provided that
+    the determinant of the matrix vanishes, as follows from the Cramer's condition:
+    $$
+    {\rm det} \begin{pmatrix} A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}=
+    \left|  \begin{array}{cccc}A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{array} \right| = 0
+    $$
+    which in general will have $n$ independent solutions, which we label as $\lambda_{\psi,k}$.
+
+Once we have solved the $n$ eigenvalues $\{ \lambda_{\psi,k} \} $, we can insert each
+of them in the original evolution equation and determine the components of each of the eigenvectors,
+which we can express as columns vectors
+$$
+|\psi_1\rangle = \begin{pmatrix} \psi_{1,1} \\  \psi_{1,2} \\  \psi_{1,3} \\ \vdots \end{pmatrix} \,, \quad
+|\psi_2\rangle = \begin{pmatrix} \psi_{2,1} \\  \psi_{2,2} \\  \psi_{2,3} \\ \vdots \end{pmatrix} \,, \quad \ldots \, , |\psi_n\rangle = \begin{pmatrix} \psi_{n,1} \\  \psi_{n,2} \\  \psi_{n,3} \\ \vdots \end{pmatrix} \, .
+$$
+
+!!! tip "Orthogonality of eigenvectors"
+    An important property of eigenvalue equations is that if you have two eigenvectors
+    $ |\psi_i\rangle$ and $ |\psi_j\rangle$ that have associated *different* eigenvalues,
+    $\lambda_{\psi_i} \ne \lambda_{\psi_j}  $, then these two eigenvectors are orthogonal to each
+    other, that is
+    $$
+    \langle \psi_j | \psi_i\rangle =0 \, \quad {\rm for} \quad {i \ne j} \, .
+    $$
+    This property is extremely important, since it suggest that we could use the eigenvectors
+    of an eigenvalue equation as a *set of basis elements* for this Hilbert space.
+
+Recall from the discussions of eigenvalue equations in linear algebra that
+the eigenvectors $|\psi_i\rangle$ are defined *up to an overall normalisation constant*. Clearly, if $|\psi_i\rangle$ is a solution of $\hat{A}|\psi_i\rangle = \lambda_{\psi_i}|\psi_i\rangle$
+then $c|\psi_i\rangle$ will also be a solution, with $c$ being a constant. In the context of quantum mechanics, we need to choose this overall rescaling constant to ensure that the eigenvectors are normalised, thus they satisfy
+$$
+\langle \psi_i | \psi_i\rangle = 1 \, \quad {\rm for~all}~i \, .
+$$
+With such a choice of normalisation, one says that the eigenvectors in a set
+are *orthogonal* among them.
+
+!!! tip "Eigenvalue spectrum and degeneracy"
+    The set of all eigenvalues of an operator is called the *eigenvalue spectrum* of an operator. Note that different eigenvectors can also have the same eigenvalue. If this is the case the eigenvalue is said to be *degenerate*.
+
+***
+
+##6.3. Problems
+
+1. *Eigenvalues and eigenvectors I* 
+
+    Find the characteristic polynomial and eigenvalues for each of the following matrices,
+    $$A=\begin{pmatrix} 5&3\\2&10 \end{pmatrix}\,  \quad
+    B=\begin{pmatrix} 7i&-1\\2&6i \end{pmatrix} \, \quad C=\begin{pmatrix} 2&0&-1\\0&3&1\\1&0&4 \end{pmatrix}$$
+
+2. *Hamiltonian*
+
+    The Hamiltonian for a two-state system is given by 
+    $$H=\begin{pmatrix} \omega_1&\omega_2\\  \omega_2&\omega_1\end{pmatrix}$$
+    A basis for this system is 
+    $$|{0}\rangle=\begin{pmatrix}1\\0  \end{pmatrix}\, ,\quad|{1}\rangle=\begin{pmatrix}0\\1  \end{pmatrix}$$
+    Find the eigenvalues and eigenvectors of the Hamiltonian $H$, and express the eigenvectors in terms of $\{|0 \rangle,|1\rangle \}$
+
+3. *Eigenvalues and eigenvectors II*
+
+    Find the eigenvalues and eigenvectors of the matrices
+    $$A=\begin{pmatrix} -2&-1&-1\\6&3&2\\0&0&1 \end{pmatrix}\, \quad B=\begin{pmatrix} 1&1&2\\2&2&2\\-1&-1&-1 \end{pmatrix} $$.
+
+4. *The Hadamard gate*
+
+    In one of the problems of the previous section we discussed that an important operator used in quantum computation is the *Hadamard gate*, which is represented by the matrix:
+    $$\hat{H}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix} \, .$$
+    Determine the eigenvalues and eigenvectors of this operator.
+
+5. *Hermitian matrix*
+
+    Show that the Hermitian matrix
+    $$\begin{pmatrix} 0&0&i\\0&1&0\\-i&0&0 \end{pmatrix}$$
+    has only two real eigenvalues and find and orthonormal set of three eigenvectors.
+
+6. *Orthogonality of eigenvectors*
+
+    Confirm, by explicit calculation, that the eigenvalues of the real, symmetric matrix
+    $$\begin{pmatrix} 2&1&2\\1&2&2\\2&2&1 \end{pmatrix}$$
+    are real, and its eigenvectors are orthogonal.

docs/7_differential_equations_1.md

+---
+title: Differential Equations: Part 1
+---
+
+#7. Differential equations: Part 1
+
+The first lecture on differential equations consists of three parts, each with a video embedded in the paragraph:
+
+- [7.1. First examples of differential equations](#71-first-examples-of-differential-equations-definitions-and-strategies)
+- [7.2. Theory of systems of first-order differential equations](#72-theory-of-systems-of-differential-equations)
+- [7.3. Solving homogeneous first-order differential equations with constant coefficients](#73-solving-homogeneous-linear-system-with-constant-coefficients)
+
+**Total video length: 1 hour 15 minutes 4 seconds**
+
+and at the end of the lecture notes, there is a set of corresponding exercises:
+
+- [7.4. Problems](#74-problems)
+
+***
+
+##7.1. First examples of differential equations: Definitions and strategies
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/IUr38H4dcWI?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+###7.1.1. Definitions
+
+A differential equation or DE is any equation which involves both a function and a
+derivative of that function. In this lecture, we will be focusing on 
+*Ordinary Differential Equations* (ODEs), meaning that our equations will involve 
+functions of one independent variable and hence any derivatives will be full 
+derivatives. Equations which involve a function of several independent variables
+and their partial derivatives are called *Partial Differential Equations* (PDEs); they will 
+be introduced in the follow-up lecture. 
+
+We consider functions $x(t)$ and define $\dot{x}(t)=\frac{dx}{dt}$, 
+$x^{(n)}(t)=\frac{d^{n}x}{dt^{n}}$. An $n$*-th* order differential equation is 
+an equation of the form:
+
+$$x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t).$$ 
+
+Typically, $n \leq 2$. Such an equation will usually be presented with a set of 
+initial conditions,
+
+$$x^{(n-1)}(t_{0}) = x^{(n-1)}_{0}, \cdots, x(t_0)=x_0. $$
+
+This is because to fully specify the solution of an $n$*-th* order differential 
+equation, $n$ initial conditions are necessary (we need to specify the value of $n-1$ derivatives
+of $x(t)$ and as well the value of the function $x(t)$ for some  $t_0$). To understand why we need 
+initial conditions, look at the following example.
+
+!!! check "Example: Initial conditions"
+
+    Consider the following calculus problem,
+    
+    $$\dot{x}(t)=t. $$ 
+
+    By integrating, one finds that the solution to this equation is 
+
+    $$\frac{1}{2}t^2 + c,$$
+
+    where $c$ is an integration constant. In order to specify the integration 
+    constant, an initial condition is needed. For instance, if we know that when 
+    $t=2$ then $x(2)=4$, we can plug this into the equation to get 
+
+    $$\frac{1}{2}\cdot 4 + c = 4, $$
+
+    which implies that $c=2$. 
+    
+Essentially, initial conditions are needed when solving differential equations so
+that the unknowns resulting from integration may be determined.
+
+!!! info "Terminology for Differential Equations"
+
+    1. If a differential equation does not explicitly contain the 
+        independent variable $t$, it is called an *autonomous equation*.
+    2. If the largest derivative in a differential equation is of the first order, 
+        i.e. $n=1$, then the equation is called a first order differential 
+        equation.
+    3. Often you will see differential equations presented using $y(x)$ 
+        instead of $x(t)$. This is just a different nomenclature. 
+            
+In this course, we will be focusing on *Linear Differential Equations*, meaning 
+that we consider differential equations $x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t)$
+where the function $f$ is a linear polynomial function of the unknown function
+$x(t)$. A simple way to spot a non-linear differential equation is to look for 
+non-linear terms, such as $x(t) \cdot \dot{x}(t)$ or $x^{(n)}(t) \cdot x^{(2)}(t)$. 
+
+Often, we will be dealing with several coupled differential equations. In this 
+situation, we can write the entire system of differential equations as a vector 
+equation, involving a linear operator. For a system of $m$ equations, denote 
+
+$$\vec{x}(t) = \begin{bmatrix}
+x_1(t) \\
+\vdots \\
+x_{m}(t) \\
+\end{bmatrix}.$$
+
+A system of first order linear equations is then written as 
+
+$$\dot{\vec{x}}(t) = \vec{f}(\vec{x}(t),t) $$
+
+with the initial condition $\vec{x}(t_0) = \vec{x}_0$.
+
+###7.1.2. Basic examples and strategies for a (single) first-order differential equation
+
+Before focusing on systems of first order equations, we will first consider 
+examplary cases of single first-order equations with only one unknown function $x(t)$.
+In this case, we can distinguish important cases.
+
+#### Type 1: $\dot{x}(t) = f(t)$
+
+The simplest type of differential equation is the type usually learned about in the 
+integration portion of a calculus course. Such equations have the form,
+
+$$\dot{x}(t) = f(t). $$
+
+When $F(t)$ is an anti-derivative of $f(t)$ i.e. $\dot{F}=f$, then the solutions
+to this type of equation are 
+
+$$x(t) = F(t) + c. $$
+
+
+!!! info "What is the antiderivative?"
+    
+    You may know the antiderivative $F(t)$ of a function $f(t)$ under a different name -
+    it is the same as the indefinite integral: $F(t) = \int f(t) dt$. Remember that taking
+    an integral is essentially the opposite of differentiation, and indeed taking an integral
+    means finding a function $F(t)$ such that $\dot{F}(t) = \frac{dF}{dt} = \frac{d}{dt}
+    \int f(t) dt = f(t)$. In the context of differential equations we prefer to call this the
+    antiderivative as solving the differential equation means essentially undoing the derivative.
+
+    Note that the antiderivative is only defined up to a constant (as is the indefinite integral).
+    In practice, you will thus find some particular expression for $F(t)$ through integration. To
+    capture all possible solutions, don't forget the integration constant $c$ in the expression
+    above!
+
+!!! check "Example"
+
+    Given the equation
+    
+    $$\dot{x}(t)=t, $$
+    
+    one finds by integrating that the solution is $\frac{1}{2}t^2 + c$. 
+    
+
+#### Type 2: $\dot{x}(t) = f(x(t))$
+
+The previous example was easy, as the function $x(t)$ did not enter in the right-hand side.
+A second important case that we can solve explicitly is when the right-hand side is some
+function of $x(t)$:
+    
+$$\dot{x}(t)=f(x(t)).$$ 
+    
+This implies that $\frac{\dot{x}(t)}{f(x)} = 1$. Let $F(x)$ be the 
+anti-derivative of $\frac{1}{f(x)}$. Then, by making use of the chain rule: 
+    
+$$\frac{d}{dt} F(x(t)) = \frac{dx}{dt}\,\frac{dF}{dx} = \frac{\dot{x}(t)}{f(x(t))} = 1$$
+    
+$$\Leftrightarrow F(x(t)) = t + c.$$
+    
+From this, we notice that if we can solve for $x(t)$, then we have the 
+solution! Having a specific form for the function $f(x)$ can often make it 
+possible to solve either implicitly or explicitly for the function $x(t)$.
+
+!!! check "Example"
+
+    Given the equation
+    
+    $$\dot{x} = \lambda x, $$
+    
+    re-write the equation to be in the form 
+    
+    $$\frac{\dot{x}}{\lambda x} = 1.$$
+    
+    Now, applying the same process which was shown through just above, let $f(x)=\lambda x$ 
+    and $F(x)$ be the anti-derivative of the $\frac{1}{f(x)}$. Integrating 
+    allows us to find the form of this anti-derivative. 
+    
+    $$F(x):= \int \frac{dx}{\lambda x} = \frac{1}{\lambda}\log{\lambda x} $$
+    
+    Now, making use of the general solution we also have that $F(x(t)) =t+c$. 
+    These two equations can be combined to form an equation for $x(t)$,
+    
+    $$\log(\lambda x)  = \lambda t + c$$
+    $$x(t) = \frac{1}{\lambda} e^c e^{\lambda t} $$
+    $$x(t) = c_0 e^{\lambda t}$$
+    
+    where in the last line we defined a new constant $c_0 =\frac{1}{\lambda}e^c$.
+    Given an initial condition, we could immediately determine this constant $c_0$.
+    
+
+#### Type 3: $\dot{x}(t) = g(t) f(x(t))$
+
+So far we have considered onle DE's where the right-hand side is either a function of $t$
+*or* of $x(t)$. We can still solve a more generic case, if we can separate the two dependencies
+as:
+
+$$\dot{x}(t)=g(t)f(x(t)).$$
+
+This type of differential equation is called a first order differential equation 
+with non-constant coefficients. If $f(x(t))$ is linear in $x$ then it is also 
+said to be a linear equation.  
+
+This equation can be re-written to isolate the coefficient function, g(t)
+
+$$\frac{\dot{x}(t)}{f(x(t))} = g(t). $$
+
+Now, define $F(x)$ to be the anti-derivative of $\frac{1}{f(x)}$, and $G(t)$ to
+be the anti-derivative of $g(t)$. Without showing again the use of chain rule on
+the left side of the equation, we have
+
+$$\frac{d}{dt} F(x(t)) = g(t) $$
+$$\Rightarrow F(x(t)) = G(t) + c $$
+
+Given this form of a general solution, the knowledge of specific functions $f, g$ would
+make it possible to solve for $x(t)$. 
+
+!!! check "Example"
+
+    Let us apply the above strategy to the following equation,
+    
+    $$\dot{x}= t x^2 .$$
+    
+    The strategy indicates that we should define $f(x)=x^2$ and $g(t)=t$. 
+    As before, we can re-arrange the equation into the form:
+    
+    $$\frac{\dot{x}}{x^2} = t. $$
+    
+    It is then necessary to find $F(x)$, the anti-derivative of $\frac{1}{f(x)}$,
+    or the left hand side of the above equation, as well as $G(t)$, the 
+    anti-derivative of $g(t)$, or the right hand side of the previous equation.
+    
+    By integrating, one finds
+    
+    $$F(x) = - \frac{1}{x} $$
+    $$G(t)=\frac{1}{2}t^2 + c. $$
+    
+    Accordingly then, the intermediate equation we have is
+    
+    $$- \frac{1}{x} = \frac{1}{2} t^2 + c. $$
+    
+    At this point, it is possible to solve for $x(t)$ by re-arrangement
+    
+    $$x(t)= \frac{-2}{t^2 + c_0}, $$
+    
+    where in the last line we have defined $c_0 = 2c$. Once again, specification
+    of an initial condition would enable determination of $c_0$ directly. To see 
+    this, suppose $x(0) = 2$. By inserting this into the equation for $x(t)$, we get
+    
+    $$2 = \frac{-2}{c_0} $$
+    $$ \Rightarrow c_0 = -1.$$
+    
+    When solved for $c_0$, with the choice of initial condition $x(0)=2$, the 
+    full equation for $x(t)$ becomes 
+    
+    $$x(t)=\frac{-2}{t^2 -1}. $$
+
+!!! check "Example: First order linear differential equation with general non-constant coefficient function"
+
+    Let us apply the above strategy of dealing with non-constant coefficient functions
+    to the more general equation
+
+    $$\dot{x}= g(t) \cdot x. $$
+
+    This equation suggests that we first define $f(x)=x$ and then find $F(x)$ and 
+    $G(t)$, the anti-derivatives of $\frac{1}{f(x)}$ and $g(t)$, respectively. By doing
+    so, we determine that 
+
+    $$F(x) = log(x) \, .$$
+
+    Follow the protocol subsequently, we arrive at the equation
+
+    $$log(x) = G(t) + c.$$
+
+    Exponentiating and defining $c_0:=e^c$ delivers the equation for $x(t)$,
+
+    $$x(t)= c_0 e^{G(t)} .$$
+    
+So far, we have only considered first order differential equations. If we consider
+extending the strategies which we have developed to higher order equations such as
+
+$$x^{(2)}(t)=f(x), $$
+
+with f(x) being a linear function, then our work will swiftly become more tedious. Later on,
+we will develop a general theory for linear equations which will enable us to 
+tackle such higher order equations. For now, we move on to considering systems 
+of coupled first order linear DE's. 
+
+##7.2. Theory of systems of differential equations
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/4VoSMc08nQA?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+
+An intuitive presentation of a system of coupled first order differential 
+equations can be given by a phase portrait. Before demonstrating such a portrait,
+let us introduce a useful notation for working with systems of DE's. Several
+coupled DE's can be written down concisely as a single vector equation:
+
+$$\dot{\vec{x}}=\vec{f}(\vec{x}). $$
+
+In such an equation, the vector $\dot{\vec{x}}$ is the rate of change of a vector 
+quantity, for example; the velocity which is the rate of change of the position 
+vector. The term $\vec{f}(\vec{x})$ describes a vector field, which has one vector 
+per point $\vec{x}$. This type of equation can also be extended to include a time 
+varying vector field, $\vec{f}(\vec{x},t)$. 
+
+In the phase portrait below, the velocities of the cars are determined by 
+the vector field $\vec{f}(\vec{x})$, where their velocity corresponds to the slope of 
+each arrow. The position of each of the little cars is determined by an initial 
+condition. Since the field lines do not cross and the cars begin on different 
+field lines, they will remain on different field lines. 
+
+![image](figures/Phase_portrait_with_cars.png)
+
+!!!info "Properties of a system of 1st order linear DEs"
+    If $\vec{f}(\vec{x})$ is not *crazy*, for example - if it is continuous and 
+    differentiable, then it is possible to prove the following two properties for 
+    a system of first order linear DE's
+
+    1. **Existence of solution**: For any specified initial condition, there is a solution.
+    2. **Uniqueness of solution**: Any point $\vec{x}(t)$ is uniquely determined by the
+        initial condition and the equation i.e. we know where each point "came from"
+        $\vec{x}(t'<t)$. 
+
+###7.2.1. Systems of linear first order differential equations
+
+####7.2.1.1. Homogeneous systems
+
+Any homogeneous system of first order linear DE's can be written in the form 
+
+$$\dot{\vec{x}} = A(t) \vec{x} \, ,$$
+
+where $A$ is a linear operator. The system is called homogeneous because it 
+does not contain any additional term which is not dependent on $\vec{x}$ (for 
+example an additive constant or an additional function depending only on t). 
+
+!!! info "Linearity of a system of DEs"
+    An important property of such a system is *linearity*, which has the following 
+    implications
+
+    1. If $\vec{x}(t)$ is a solution ,then $c \vec{x}(t)$ is a solution too, for any constant c
+    2. If $\vec{x}(t)$ and $\vec{y}(t)$ are both solutions, then so is $a \vec{x}(t)+ b \vec{y}(t)$,
+    where $a$ and $b$ are both constants. 
+
+These properties have special importance for modelling physical systems, due to
+the principle of superposition which is especially important in quantum physics,
+as well as electromagnetism and fluid dynamics. For example, in electromagnetism, 
+when there are four charges arranged in a square acting on a test charge 
+located within the square, it is sufficient to sum the individual forces in 
+order to find the total force. Physically, this is the principle of superposition, 
+and mathematically, superposition is linearity and applies to linear models.
+
+!!! info "General Solution"
+
+    For a system of $n$ linear first order DE's with $n \times n$ linear operator 
+    $A(t)$, the general solution can be written as 
+
+    $$\vec{x}(t) = c_1 \vec{\phi}_1 (t) + c_2 \vec{\phi}_2 (t) + \cdots + c_n \vec{\phi}_n (t),$$
+
+    where $\{\vec{\phi}_1 (t), \vec{\phi}_2(t), \cdots, \vec{\phi}_n (t) \}$ are $n$ independent solutions which form a basis for the solution space, and $c_1, c_2, \cdots c_n$ are constants. 
+
+    $\{\vec{\phi}_1 (t), \vec{\phi}_2(t), \cdots, \vec{\phi}_n (t) \}$ are a basis if and 
+    only if they are linearly independent for fixed $t$:
+
+    $$\det \big{(}\vec{\phi}_1 (t) | \vec{\phi}_2 (t) | \cdots | \vec{\phi}_n (t) \big{)} \neq 0.$$
+
+    If this condition holds for one $t$, it holds for all $t$.
+
+####7.2.2.2 Inhomogeneous systems
+
+Compared to the homogeneous equation, an inhomogeneous equation has an 
+additional term, which may be a function of the independent variable. 
+
+$$ \dot{\vec{x}}(t) = A(t) \vec{x}(t) + \vec{b}(t).$$
+
+!!! info "Relation between a solutions of a homogeneous and inhomogeneous equations" 
+    There is a simple connection between the general solution of an inhomogeneous 
+    equation and the corresponding homogeneous equation. If $\vec{\psi}_1$ and $\vec{\psi}_2$
+    are two solutions of the inhomogeneous equation, then their difference is a 
+    solution of the homogeneous equation 
+
+    $$(\dot{\vec{\psi}_1}-\dot{\vec{\psi}_2}) = A(t) (\vec{\psi}_1 - \vec{\psi}_2). $$
+
+    The general solution of the inhomogeneous equation can be written in terms of 
+    the basis of solutions for the homogeneous equation, plus one particular solution
+    to the inhomogeneous equation,
+
+    $$\vec{x}(t) = \vec{\psi}(t) + c_1 \vec{\phi}_1 (t) + c_2 \vec{\phi}_2 (t) + \cdots + c_n \vec{\phi}_n (t). $$
+
+    In the above equation, $\{\vec{\phi}_1 (t), \vec{\phi}_2(t), \cdots, \vec{\phi}_n (t) \}$
+    form a basis for the solution space of the homogeneous equation and $\vec{\psi}(t)$
+    is a particular solution of the inhomogeneous system. 
+
+!!! tip "Strategy of finding the solution of the inhomogeneous equation"
+    Now we need a strategy for finding the solution of the inhomogeneous equation. 
+    Begin by making an ansatz that $\vec{x}(t)$ can be written as a linear combination 
+    of the basis functions for the homogeneous system, with coefficients that are 
+    functions of the independent variable.
+    
+    1. Ansatz:
+        $$\vec{x}(t) = c_1(t) \vec{\phi}_1 (t)+ c_2(t) \vec{\phi}_2(t) + \cdots + c_n(t) \vec{\phi}_n (t) $$
+
+    2. Define the vector $\vec{c}(t)$ and matrix $\vec{\Phi}(t)$ as
+
+        $$\vec{c}(t) = \begin{bmatrix}
+        c_1(t) \\
+        \vdots \\
+        c_n(t) \\
+        \end{bmatrix} $$
+        $$\vec{\Phi}(t) = \big{(} \vec{\phi}_1 (t) | \cdots | \vec{\phi}_n (t) \big{)} $$
+
+    3. With these definitions, it is possible to re-write the ansatz for $\vec{x}(t)$,
+
+        $$ \vec{x}(t) = \vec{\Phi}(t) \vec{c}(t).$$
+
+    4. Using the Leibniz rule, we then have the following expanded equation,
+
+        $$\dot{\vec{x}}(t) = \dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t).$$
+
+    5. Substituting the new expression into the differential equation gives,
+
+        $$\dot{\vec{\Phi}}(t) \vec{c}(t) + \vec{\Phi}(t) \dot{\vec{c}}(t) = A(t) \vec{\Phi}(t) \vec{c}(t) + \vec{b}(t) $$
+        $$\vec{\Phi}(t) \dot{\vec{c}}(t) = \vec{b}(t). $$
+
+        In order to cancel terms in the previous line, we made use of the fact that $\vec{\Phi}(t)$ solves the homogeneous equation $\dot{\vec{\Phi}} = A \vec{\Phi}$.
+        
+    6. By way of inverting and integrating, we can write the equation for the coefficient vector $\vec{c}(t)$
+
+        $$\vec{c}(t) = \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt.$$
+
+    7. With access to a concrete form of the coefficient vector, we can then write down the particular solution,
+
+        $$\vec{\psi}(t)= \vec{\Phi}(t) \cdot \int \vec{\Phi}^{-1}(t) \vec{b}(t) dt .$$
+
+!!! check "Example: Inhomogeneous first order linear differential equation"
+
+    The technique for solving a system of inhomogeneous equations also works for a 
+    single inhomogeneous equation. Let us apply the technique to the equation
+
+    $$ \dot{x} = \lambda x + a. $$
+
+    In this particular inhomogenous equation, the function $g(t)=a$. As discussed in
+    an earlier example, the solution to the homogenous equation is
+    $c e^{\lambda t}$. Hence, we define $\phi(t)=e^{\lambda t}$ and make the ansatz
+
+    $$\psi(t) = c(t) e^{\lambda t}. $$
+
+    Solving for $c(t)$ results in 
+
+    $$c(t) = \int e^{- \lambda t} a  dt$$
+    $$c(t) = - \frac{ a }{\lambda} e^{- \lambda t} $$
+
+    Overall then, the solution (which can be easily verified by substitution) is 
+
+    $$\psi(t) = - \frac{a}{\lambda}.  $$ 
+    
+##7.3. Solving homogeneous linear system with constant coefficients
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/GGIDjgUpsH8?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+The type of equation under consideration in this section looks like 
+
+$$ \dot{\vec{x}}(t) = A \vec{x}(t),$$
+
+where, throughout this section, $A$ will be a constant matrix. It is possible 
+to define a formal solution using the *matrix exponential*, 
+$\vec{x}(t) = e^{A t}$.
+
+!!! info "Definition: Matrix Exponential"
+
+    Before defining the matrix exponential, recall the definition of the regular 
+    exponential function in terms of Taylor series,
+    
+    $$e^{x} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{x^n}{n!},$$
+    
+    in which it is agreed that $0!=1$. The matrix exponential is defined in 
+    exactly the same way, only now instead of taking powers of a number or 
+    function, powers of a matrix are calculated with
+    
+    $$e^{A} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{{A}^n}{n!}.$$
+     
+    It is important to use caution when translating the properties of the normal exponential function over to the matrix exponential, because not all of the regular properties hold generally. In particular,      
+    $$e^{X + Y} \neq e^{X} e^{Y},$$
+    
+    unless it happens that 
+     
+    $$[X, Y] = 0.$$
+     
+    The necessary condition for this property to hold, stated on the previous line, is called *commutativity*. Recall that in general, matrices are not commutative so such a condition is only met for particular choices of matrices. The property of *non-commutativity* (what happens when the condition is not met) is of central importance in the mathematical structure of quantum mechanics. *For example, mathematically, non-commutativity is responsible for the Heisenberg uncertainty relations.*
+     
+    On the other hand, one property that does hold, is that $e^{- A t}$ is the inverse of the matrix exponential of $A$. 
+    
+    Furthermore, it is possible to derive the derivative of the matrix exponential by making use of the Taylor series formulation,
+    $$\begin{align}
+    \frac{d}{dt} e^{A t} &= \frac{d}{dt} \overset{\infty}{\underset{n=0}{\Sigma}} \frac{(A t)^n}{n!} \\
+    ... &= \overset{\infty}{\underset{n=0}{\Sigma}} \frac{1}{n!} \frac{d}{dt} (A t)^n \\
+    ... &= \overset{\infty}{\underset{n=0}{\Sigma}} \frac{n A}{n!}(A t)^{n-1} \\
+    ... &= \overset{\infty}{\underset{n=1}{\Sigma}} \frac{A}{(n-1)!}(A t)^{n-1} \\
+    ... &= \overset{\infty}{\underset{n=0}{\Sigma}} \frac{A}{n!}(A t)^n \\
+    \frac{d}{dt} e^{A t} &= A e^{A t}.
+    \end{align}$$
+
+Armed with the matrix exponential and it's derivative, 
+$\frac{d}{dt} e^{A t} = A e^{A t}$, it is simple to verify that 
+the matrix exponential solves the differential equation. 
+
+!!! info "Properties of the solution using the matrix exponential:"
+    1. The columns of $e^{A t}$ form a basis for the solution space.
+    2. Accounting for initial conditions, the full solution of the equation is $\dot{\vec{x}}(t) = e^{A t} {\vec{x}}_{0}$, with initial condition $\vec{x}(0) = e^{A 0}{\vec{x}}_0 = \mathbb{I} {\vec{x}}_{0} = {\vec{x}}_{0}$. (here $\mathbb{I}$ is the $n\times n$ identity matrix)
+
+Next, we will discuss how to determine a solution in practice, beyond the 
+formal solution just presented. 
+
+### Case 1: $A$ is diagonalizable
+
+For an $n \times n$ matrix $A$, denote the $n$ distinct eigenvectors as $\{\vec{v}_1, \cdots, \vec{v}_n \}$. By definition, the eigenvectors satisfy the equation
+
+$$A \vec{v}_i = \lambda_i \vec{v}_i, \qquad \forall i \epsilon \{1, \cdots, n \}. $$
+
+Here, we give consideration to the case of distinct eigenvectors, in which case 
+the $n$ eigenvectors form a basis for $\mathbb{R}^{n}$. 
+
+!!! info "Strategy for finding solution when $A$ is diagonizable"
+
+    1.  To solve the equation $\dot{\vec{x}}(t) = A \vec{x}(t)$, define a set of scalar functions $\{u_{1}(t), \cdots u_{n}(t) \}$ and make the following ansatz:
+        $$\vec{\phi}_{i}(t) = u_{i}(t) \vec{v}_{i}.$$
+    2.  Then, by differentiating,
+        $$\dot{\vec{\phi}_i}(t) = \dot{u_i}(t) \vec{v}_{i}.$$
+
+    3.  The above equation can be combined with the differential equation for 
+        $\vec{\phi}_{i}(t)$, 
+        $$\dot{\vec{\phi}_{i}}(t)=A \vec{\phi}_{i}(t) \, ,$$
+        to derive the following equations,
+
+        $$\dot{u_i}(t) \vec{v}_{i} = A u_{i}(t) \vec{v}_{i}$$
+        $$\dot{u_i}(t) \vec{v}_{i} = u_{i}(t) \lambda_{i} \vec{v}_{i} $$
+        $$\vec{v}_{i} (\dot{u_i}(t) - \lambda_i u_{i}(t)) = 0, $$
+
+        where in the second last line, we make use of the fact that $\vec{v}_i$ is an eigenvector of $A$. 
+    4.  The obtained relation implies that 
+
+        $$\dot{u_i}(t) = \lambda_i u_{i}(t).$$
+
+        This is a simple differential equation, of the type dealt with in the third example. 
+    5.  The solution is found to be
+
+        $$u_{i}(t) = c_i e^{\lambda_i t},$$
+
+        with $c_i$ being a constant. 
+    6.  The general solution is found by adding all $n$ of the
+        solutions $\vec{\phi}_{i}(t)$,
+
+        $$\vec{x}(t) = c_{1} e^{\lambda_1 t} \vec{v}_{1} + c_{2} e^{\lambda_2 t} \vec{v}_{2} + \cdots + c_{n} e^{\lambda_n t} \vec{v}_{n}.$$
+
+        and the vectors $\{e^{\lambda_1 t} \vec{v}_{1}, \cdots, e^{\lambda_n t} \vec{v}_{n} \}$
+        form a basis for the solution space since $\det(\vec{v}_1 | \cdots | \vec{v}_n) \neq 0$
+        (the $n$ eigenvectors are linearly independent). 
+
+!!! check "Example: Homogeneous first order linear system with diagonalizable constant coefficient matrix"
+
+    Define the matrix
+    $$A = \begin{bmatrix}
+    0 & -1 \\
+    1 & 0 
+    \end{bmatrix},$$
+    and consider the DE 
+
+    $$\dot{\vec{x}}(t) = A \vec{x}(t), \quad \vec{x}_0 = \begin{bmatrix}
+    1 \\
+    0 
+    \end{bmatrix}. $$
+ 
+    To proceed by following the solution technique, we determine the eigenvalues of 
+    $A$, 
+ 
+    $$\det {\begin{bmatrix} 
+    -\lambda & -1 \\
+    1 & - \lambda \\
+    \end{bmatrix}} = \lambda^2 + 1 = 0. $$
+    
+    By solving the characteristic polynomial, one finds the two eigenvalues 
+    $\lambda_{\pm} = \pm i$. 
+    
+    Focusing first on the positive eigenvalue, we can determine the first 
+    eigenvector,
+    
+    $$\begin{bmatrix} 
+    0 & -1 \\
+    1 & 0 \\
+    \end{bmatrix} \begin{bmatrix}
+    a \\
+    b\\
+    \end{bmatrix} = i \begin{bmatrix}
+    a \\
+    b \\
+    \end{bmatrix}.$$
+
+    A solution to this eigenvector equation is given by $a=1$, $b=-i$, altogether
+    implying that
+    $$\lambda_1=i, \vec{v}_{1} = \begin{bmatrix} 
+    1 \\
+    -i \\
+    \end{bmatrix}.$$
+    
+    As for the second eigenvalue, $\lambda_{2} = -i$, we can solve the analogous
+    eigenvector equation to determine
+    $$\vec{v}_{2} = \begin{bmatrix}
+    1 \\
+    i \\
+    \end{bmatrix}.$$
+    
+    Hence, two independent solutions of the differential equation are:
+    
+    $$\vec{\phi}_{1} = e^{i t}\begin{bmatrix}
+    1 \\
+    -i \\
+    \end{bmatrix}, \vec{\phi}_{2}  = e^{-i t} \begin{bmatrix}
+    1 \\
+    i \\
+    \end{bmatrix}.$$
+ 
+    Before we can obtain the general solution of the equation, we must find coefficients for the linear combination of the two solutions which would satisfy the initial condition. To this end, we must solve:
+    
+    $$c_1 \vec{\phi}_{1}(t) + c_2 \vec{\phi}_{2}(t) = 
+    \begin{bmatrix}
+    1 \\
+    0 \\
+    \end{bmatrix}$$
+    
+    $$\begin{bmatrix}
+    c_1 + c_2 \\
+    -i c_1 + i c_2 \\
+    \end{bmatrix} = \begin{bmatrix} 
+    1 \\
+    0 \\
+    \end{bmatrix}.$$
+    
+    The second row of the vector equation for $c_1, c_2$ implies that $c_1=c_2$. 
+    The first row then implies that $c_1=c_2=\frac{1}{2}$. 
+    
+    Overall then, the general solution of the DE can be summarized
+    
+    $$\dot{\vec{x}}(t) = \begin{bmatrix}
+    \frac{1}{2}(e^{i t} + e^{-i t}) \\
+    \frac{1}{2 i}(e^{i t} - e^{-i t}) \\
+    \end{bmatrix} = \begin{bmatrix}
+    \cos(t) \\
+    \sin(t) \\
+    \end{bmatrix}. $$
+
+### Case 2: $A (2 $ by $ 2)$ is defective
+
+In this case, we consider the situation where $\det(A- \lambda I)$ 
+has a root $\lambda$ with multiplicity 2, but only one eigenvector $\vec{v}_1$. 
+
+!!! check "Example: Matrix with eigenvalue of multiplicity 2 and only a single eigenvector. (Part 1)"
+    
+    Consider the matrix
+    
+    $$A = \begin{bmatrix}
+    1 & 1 \\
+    0 & 1 \\
+    \end{bmatrix}$$
+    
+    The characteristic polynomial can be found by evaluating
+    
+    $$\det \big{(} \begin{bmatrix}
+    1-\lambda & 1 \\
+    0 & 1-\lambda \\
+    \end{bmatrix} \big{)} = 0$$
+    $$(1-\lambda)^2 = 0$$
+    
+    Hence, the matrix $A$ has the single eigenvalue $\lambda=1$ with multiplicity 2. As for finding an eigenvector, we solve
+    
+    $$\begin{bmatrix} 
+    1 & 1 \\
+    0 & 1 \\
+    \end{bmatrix} \begin{bmatrix} 
+    a \\
+    b \\
+    \end{bmatrix} = \begin{bmatrix} 
+    a \\
+    b \\
+    \end{bmatrix}$$
+    $$\begin{bmatrix} 
+    a+b \\
+    b \\
+    \end{bmatrix} = \begin{bmatrix} 
+    a \\
+    b \\
+    \end{bmatrix}.$$
+    
+    These equations, $a+b=a$ and $b=b$ imply that $b=0$ and $a$ can be chosen arbitrarily, for example as $a=1$. Then, the only eigenvector is
+    
+    $$\vec{v}_1 = \begin{bmatrix} 
+    1 \\
+    0 \\
+    \end{bmatrix}.$$
+    
+What is the problem in this case? Since there are $n$ equations to be solved and an $n \times n$ linear operator $A$, the solution space for the equation requires a basis of $n$ solutions. In this case however, there are $n-1$ eigenvectors, so we cannot use only these eigenvectors in forming a basis for 
+the solution space. 
+
+!!! info "Strategy for finding a solution when $A (2 $ by $ 2)$ is defective"
+
+    1. Suppose that we have a system of $2$ coupled equations, so that $A$ is a $2 \times 2$ matrix, which has eigenvalue $\lambda_1$ with multiplicity $2$. As in the previous section, we can form one solution using the single eigenvector $\vec{v}_1$,
+
+        $$\vec{\phi}_1(t) = e^{\lambda_1 t} \vec{v}_1.$$
+
+    2. To determine the second linearly independent solution, make the following ansatz:
+
+        $$\vec{\phi}_2(t) = t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2.$$
+
+    3. With this ansatz, it is then necessary to determine an appropriate vector $\vec{v}_2$ such that $\vec{\phi}_2(t)$ is really a solution of this problem. To achieve that, take the derivative of $\vec{\phi}_2(t)$,
+
+        $$\dot{\vec{\phi}_2}(t) = e^{\lambda_1 t} \vec{v}_1 + \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + \lambda_1 e^{\lambda_1 t} \vec{v}_2 $$
+
+    4. Also, write the matrix equation for $\vec{\phi}_2(t)$,
+
+        $$A \vec{\phi}_2(t) = A t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t} \vec{v}_2 $$
+        $$A \vec{\phi}_2(t) = \lambda_1 t e^{\lambda_1 t} \vec{v}_1 + A e^{\lambda_1 t}\vec{v}_2$$
+
+    5. Since $\vec{\phi}_2(t)$ must solve the equation $\dot{\vec{\phi}_2(t)} = A \vec{\phi}_2(t)$, we can combine and simplify the previous equations to write 
+
+        $$A \vec{v}_2 - \lambda_1  \vec{v}_2 = \vec{v}_1$$
+        $$(A- \lambda_1 I) \vec{v}_2 = \vec{v}_1 $$
+
+    6. With this condition, it is possible to write the general solution as
+
+        $$\vec{x}(t) = c_1  e^{\lambda_1 t} \vec{v}_1 + c_2(t e^{\lambda_1 t} \vec{v}_1 + e^{\lambda_1 t} \vec{v}_2).$$
+
+!!! check "Example: Continuation of the example with $A$ defective (Part 2)"
+    
+    Now, our task is to apply the condition derived just above in order to solve for $\vec{v}_2$,
+    
+    $$\begin{bmatrix}
+    1-1 & 1 \\
+    0 & 1-1 \\
+    \end{bmatrix} \begin{bmatrix}
+    a \\
+    b \\
+    \end{bmatrix} = \begin{bmatrix}
+    1 \\
+    0 \\
+    \end{bmatrix}$$
+    $$\begin{bmatrix} 
+    b \\
+    0 \\
+    \end{bmatrix} = \begin{bmatrix} 
+    1 \\
+    0 \\
+    \end{bmatrix}$$
+    
+    Hence, $b=1$ and $a$ is undetermined, so may be taken as $a=0$. Then, 
+    $$\vec{v}_{2} = \begin{bmatrix} 0 \\1 \end{bmatrix}.$$ 
+    
+    Overall then, the general solution is 
+    
+    $$\vec{x}(t) = c_1 e^t \begin{bmatrix} 
+    1 \\
+    0 \\
+    \end{bmatrix} + c_2 e^t \big{(} t \begin{bmatrix} 
+    1 \\
+    0 \\ 
+    \end{bmatrix}  + \begin{bmatrix} 
+    0 \\
+    1 \\
+    \end{bmatrix}\big{)}.$$
+
+### Bonus case 3: Higher multiplicity eigenvalues
+
+In this case, we consider the situation where the matrix $A$ has an 
+eigenvalue $\lambda$ with multiplicity $m>2$, and only one eigenvector $\vec{v}$ 
+corresponding to $\lambda$, $(A - \lambda I)\vec{v}=0$. Notice here 
+that $A$ must be at least an $m \times m$ matrix. 
+
+To solve such a situation, we will expand upon the result of the previous 
+section and define the vectors $\vec{v}_2$ through $\vec{v}_{m}$ by
+
+$$(A- \lambda I) \vec{v}_2 = \vec{v}_1$$
+$$\vdots$$
+$$(A- \lambda I) \vec{v}_m = \vec{v}_{m-1}.$$
+
+Then, the subset of the basis of solutions corresponding to eigenvalue $\lambda$
+is formed by the vectors
+
+$$\vec{\phi}_{k}(t) = e^{\lambda t} \big{(} \frac{t^{k-1}}{(k-1)!}\vec{v}_1 + \cdots + t \vec{v}_{k-1} + \vec{v}_{k} \big{)} \quad \forall k \epsilon \{1, \cdots, m \}.$$
+
+To prove this, first take the derivative of $\vec{\phi}_{k}(t)$,
+
+$$\dot{\vec{\phi}_{k}(t)} = \lambda \vec{\phi}_{k}(t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!}\vec{v}_1 + \cdots + \vec{v}_{k-1} \big{)}.$$
+
+Then, for comparison, multiply $\vec{\phi}_k(t)$ by $A$
+
+$$\begin{align} 
+A \vec{\phi}_k (t) &= e^{\lambda t} \big{(} \frac{t^{k-1}}{(k-1)!}\lambda \vec{v}_1 + \frac{t^{k-2}}{(k-2)!} A \vec{v}_2 + \cdots + A \vec{v}_{k-1} + A \vec{v}_k \big{)}\\
+&= \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!}(A- \lambda I)\vec{v}_2 + \cdots + t (A- \lambda I)\vec{v}_{k-1} + (A- \lambda I)\vec{v}_{k}  \big{)}\\
+&=  \lambda \vec{\phi}_k (t) + e^{\lambda t} \big{(} \frac{t^{k-2}}{(k-2)!} \vec{v}_1 + \cdots + t \vec{v}_{k-2} + \vec{v}_{k-1} \big{)}\\
+&= \dot{\vec{\phi}}_{k}(t).
+\end{align}$$
+
+Notice that in the second last line we made use of the relations 
+$(A- \lambda I)\vec{v}_{i} = \vec{v}_{i-1}$. 
+
+This completes the proof since we have demonstrated that $\vec{\phi}_{k}(t)$ is a solution of the DE.
+
+*** 
+##7.4. Problems
+
+1. [:grinning:] Solve:
+
+    (a)  $\dot{x}(t) = t^4$
+
+    (b)  $\dot{x}(t) = \sin(t)$
+
+2. [:grinning:] Solve, subject to the initial condition $x(0)=\frac{1}{2}$:
+
+    (a) $\dot{x}(t) = x^2$
+
+    (b) $\dot{x}(t) = t x$
+
+    (c) $\dot{x}(t) = t x^{4}$
+
+3. [:smirk:] Solve, subject to the given initial condition:
+
+    (a) $\dot{x}(t)=-\tan(x)\sin(x)$, subject to $x(0)=1$. 
+
+    (b) $\dot{x(t)}=\frac{1}{3} x^2+3$, subject to $x(0)=3$.
+
+    Hint: it is fine if you use a computer algebra program to solve the integrals for these problems.
+
+4. [:smirk:] Solve the following equation and list all possible solutions:
+
+    $$\dot{x}=\cos^2(x)$$
+
+    Hint: $\int \frac{1}{\cos^2(x)} dx = \tan(x) $
+      
+5. [:grinning:] Identify which of the following systems of equations is linear.
+
+    *Note that you do not need to solve them!*  
+
+    (a) $$\dot{x_1}= t x_1 -t x_2$$
+        $$\dot{x}_2 = x_1 x_2 - x_2$$
+
+    (b) $$\dot{x}_1 = e^{-t}x_1$$
+        $$\dot{x}_2 = \sqrt{t + \cos(t)-1}x_1 + \frac{\sin(t)}{t^2+t-1}x_2$$
+
+    (c) $$x^{(2)}_1 x_1 + \dot{x}_1 = 8 x_2$$
+        $$\dot{x}_2=5tx_2 + x_1$$
+
+6. [:grinning:] Take the system of equations:
+
+    $$\dot{x}_1 = \frac{1}{2} (t-1)x_1 + \frac{1}{2} (t+1)x_2$$
+
+    $$\dot{x}_2 = \frac{1}{2}(t+1)x_1 + \frac{1}{2}(t-1)x_2.$$
+
+    Show that 
+
+    $$\vec{\Phi}_1(t) = \begin{bmatrix} 
+    e^{- t} \\
+    -e^{- t} \\
+    \end{bmatrix}$$
+    and
+    $$\vec{\Phi}_2(t)=\begin{bmatrix}
+    e^{\frac{1}{2}(t^2)} \\
+    e^{\frac{1}{2}(t^2)} \\
+    \end{bmatrix}$$
+
+    constitute a basis for the solution space of this system of equations. 
+    To this end, first verify that they are indeed solutions and then that 
+    they form a basis. 
+
+7. [:grinning:] Take the system of equations: 
+
+    $$\dot{x}_1=x_1$$
+
+    $$\dot{x}_2=x_1.$$
+
+    Re-write this system of equations into the general form
+
+    $$\dot{\vec{x}} = A \vec{x}$$
+
+    and then find the general solution. Specify the general solution for the 
+    following initial conditions
+
+    (a) $$\vec{x}(0) = \begin{bmatrix} 
+        1 \\
+        0 \\
+        \end{bmatrix}$$
+
+    (b) $$\vec{x}(0) = \begin{bmatrix}
+        0 \\
+        1 \\ 
+        \end{bmatrix}$$
+
+8. [:smirk:] Find the general solution of 
+
+    $$\begin{bmatrix}
+    \dot{x}_1 \\
+    \dot{x}_2 \\
+    \dot{x}_3 \\
+    \end{bmatrix} = \begin{bmatrix} 
+    1 & 1 & 0 \\
+    1 & 1 & 0 \\
+    0 & 0 & 3 \\
+    \end{bmatrix} \begin{bmatrix} 
+    x_1 \\
+    x_2 \\
+    x_3 \\
+    \end{bmatrix}.$$
+
+    Then, specify the solution for the initial conditions 
+
+    (a) $$\begin{bmatrix} 
+        0 \\
+        0 \\
+        1 \\
+        \end{bmatrix}$$
+
+    (b) $$\begin{bmatrix}
+        1 \\
+        0 \\
+        0 \\
+        \end{bmatrix}$$
+
+9. [:sweat:] Find the general solution of the system of equations:
+
+    $$\dot{x}_1 = 3 x_1 + x_2$$
+    $$\dot{x}_2 = - x_1 + x_2$$  
+
+ 
+

docs/8_differential_equations_2.md

+---
+title: Differential equations: Part 2
+---
+
+#8. Differential equations: Part 2
+
+The second lecture on differential equations consists of three parts, each with their own video:
+
+- [8.1. Higher order linear differential equations](#81-higher-order-linear-differential-equations)
+- [8.2. Partial differential equations: Separation of variables](#82-partial-differential-equations-separation-of-variables)
+- [8.3. Self-adjoint differential operators](#83-self-adjoint-differential-operators)
+
+**Total video length: 1 hour 9 minutes**
+
+and at the end of the lecture notes, there is a set of corresponding exercises:
+
+- [8.4. Problems](#84-problems)
+
+***
+
+##8.1. Higher order linear differential equations
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/ucvIiLgJ2i0?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+###8.1.1 Definitions
+
+In the previous lecture, we focused on first order linear differential equations
+and systems of such equations. In this lecture, we switch focus to DE's 
+which involve higher derivatives of the function that we would like to solve for. To
+facilitate this shift, we are going to change notation. 
+
+!!! warning "Change of notation"
+    In the previous lecture, we wrote differential equations for $x(t)$. In this lecture we will write DE's 
+    of $y(x)$, where $y$ is the unknown function and $x$ is the independent variable. 
+    For this purpose, we make the following definitions,
+
+    $$y' = \frac{dy}{dx}, \ y'' = \frac{d^2 y}{dx^2}, \ \cdots, \ y^{(n)} = \frac{d^n y}{dx^n}.$$
+
+    In the new notation, a linear $n$-th order differential equation with constant
+    coefficients reads 
+
+    $$y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0. $$
+
+!!! info "Linear combination of solutions are still solutions"
+
+    Note that, like it was the case for first order linear DE's, the property of 
+    linearity once again means that if $y_{1}(x)$ and $y_{2}(x)$ are both 
+    solutions, and $a$ and $b$ are constants, 
+    
+    $$a y_{1}(x) + b y_{2}(x)$$
+    
+    then a linear combination of the solutions is also a solution.
+
+###8.1.2 Mapping to a linear system of first-order DEs
+
+In order to solve a higher order linear DE, we will present a trick that makes it
+possible to map the problem of solving a single $n$-th order linear DE into a
+related problem of solving a system of $n$ first order linear DE's. 
+
+To begin, define:
+
+$$y_{1} = y, \ y_{2} = y', \ \cdots, \ y_{n} = y^{(n-1)}.$$
+
+Then, the differential equation can be re-written as
+
+$$\begin{split}
+y_1 ' & = y_2 \\
+y_2 ' & = y_3 \\
+& \vdots \\
+y_{n-1} '& = y_{n} \\
+y_{n} ' & = - a_{0} y_{1} - a_{1} y_{2} - \cdots - a_{n-1} y_{n}.
+\end{split}$$
+
+Notice that these $n$ equations together form a linear first order system, of which the 
+first $n-1$ equations are trivial. Note that this trick can be used to 
+reduce any system of $n$-th order linear DE's to a larger system of first order 
+linear DE's. 
+
+Since we already discussed the method of solution for first order linear 
+systems, we will outline the general solution to this system. As before, the 
+general solution will be the linear combination of $n$ linearly independent 
+solutions $f_{i}(x)$, $i \epsilon \{1, \cdots, n \}$, which make up a basis for 
+the solution space. Thus, the general solution has the form
+
+$$y(x) = c_1 f_1 (x) + c_2 f_2 (x) + \cdots + c_n f_{n}(x). $$
+
+!!! info "Wronskian"
+    To check that the $n$ solutions form a basis, it is sufficient to verify
+
+    $$ \det \begin{bmatrix} 
+    f_1(x) & \cdots & f_{n}(x) \\
+    f_1 ' (x) & \cdots & f_{n}'(x) \\
+    \vdots & \vdots & \vdots \\
+    f^{(n-1)}_{1} (x) & \cdots & f^{(n-1)}_{n} (x) \\
+    \end{bmatrix}  \neq 0.$$
+
+    The determinant in the preceding line is called the *Wronskian* or *Wronski determinant*.
+
+###8.1.3. General solution
+
+To determine particular solutions, we need to find the eigenvalues of 
+
+$$A = \begin{bmatrix} 
+0 & 1 & 0 & \cdots & 0 \\
+0 & 0 & 1 & \cdots & 0 \\
+\vdots & \vdots & \vdots & \cdots & \vdots \\
+0 & 0 & 0 & \cdots & 1 \\
+-a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \\
+\end{bmatrix}.$$
+
+It is possible to show that 
+
+$$\det(A - \lambda I) = -P(\lambda),$$
+
+in which $P(\lambda)$ is the characteristic polynomial of the system matrix $A$,
+
+$$P(\lambda) = \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0.$$
+
+
+??? info "Proof of $\det(A - \lambda I) = -P(\lambda)$"
+
+    As we demonstrate below, the proof relies on the co-factor expansion
+    technique for calculating a determinant. 
+
+    $$\begin{align} -\det(A - \lambda I) &=  \det \begin{bmatrix} 
+    \lambda & -1 & 0 & \cdots & 0 \\
+    0 & \lambda & -1 & \cdots & 0 \\
+    \vdots & \vdots & \vdots & \cdots & \vdots \\
+    a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
+    \end{bmatrix} \\
+    &= \lambda \det \begin{bmatrix}
+    \lambda & -1 & 0 & \cdots & 0 \\
+    0 & \lambda & -1 & \cdots & 0 \\
+    \vdots & \vdots & \vdots & \cdots & \vdots \\
+    a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+    \end{bmatrix} + (-1)^{n+1}a_0 \det \begin{bmatrix} 
+    -1 & 0 & 0 & \cdots & 0 \\
+    \lambda & -1 & 0 & \cdots & 0 \\
+    \vdots & \vdots & \vdots & \cdots & \vdots \\
+    0 & 0 & \cdots & \lambda & -1 \\
+    \end{bmatrix} \\
+    &= \lambda \det \begin{bmatrix}
+    \lambda & -1 & 0 & \cdots & 0 \\
+    0 & \lambda & -1 & \cdots & 0 \\
+    \vdots & \vdots & \vdots & \cdots & \vdots \\
+    a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+    \end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1} \\
+    &= \lambda \det \begin{bmatrix}
+    \lambda & -1 & 0 & \cdots & 0 \\
+    0 & \lambda & -1 & \cdots & 0 \\
+    \vdots & \vdots & \vdots & \cdots & \vdots \\
+    a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
+    \end{bmatrix} + a_0 \\
+    &= \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0 \\
+    &= P(\lambda).
+    \end{align}$$
+    
+    In the second last line of the proof, we indicated that the method of
+    co-factor expansion demonstrated above is repeated an additional $n-2$ times.
+    This completes the proof. 
+
+With the characteristic polynomial, it is possible to write the differential 
+equation as 
+
+$$P(\frac{d}{dx})y(x) = 0.$$
+
+To determine solutions, we need to find $\lambda_i$ such that $P(\lambda_i) = 0$. 
+By the fundamental theorem of algebra, we know that $P(\lambda)$ can be written 
+as
+
+$$P(\lambda) = \overset{l}{\underset{k=1}{\prod}} (\lambda - \lambda_k)^{m_k}.$$
+
+In the previous equation $\lambda_k$ are the k roots of the equations, and $m_k$
+is the multiplicity of each root. Note that the multiplicities satisfy 
+$\overset{l}{\underset{k=1}{\Sigma}} m_k = n$. 
+
+If the multiplicity of each eigenvalue is one, then solutions which form the 
+basis are then given as:
+
+$$f_{n}(x) = e^{\lambda_1 x}, \ e^{\lambda_2 x}, \ \cdots, \ e^{\lambda_n x}.$$
+
+If there are eigenvalues with multiplicity greater than one, the the solutions
+which form the basis are given as 
+
+$$f_{n}(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lambda_1 x}, \ etc.$$
+
+??? info "Proof that basis solutions to $P(\frac{d}{dx})y(x) = 0$ are given by $f_{k}(x) = x^{m_{k}-1} e^{\lambda_k x}$"
+
+    In order to prove that basis solutions to the differential equation rewritten using the characteristic polynomial into the form 
+    $$P(\frac{d}{dx})y(x) = 0$$
+    are given by a general formula, taking into account the multiplicity of each eigenvalue:
+    $$f_{k}(x) = x^{m_{k}-1} e^{\lambda_k x}$$ 
+    let us first recollect some definitions:
+
+    1. A linear $n$-th order differential equation with constant coefficients reads 
+        $$y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0. $$
+
+    2. The general solution will be a linear combination of $n$ linearly independent solutions $f_{i}(x)$, $i \epsilon \{1, \cdots, n \}$, which make up a basis for the solution space. Thus, the general solution has the form:
+        $$y(x) = c_1 f_1 (x) + c_2 f_2 (x) + \cdots + c_n f_{n}(x). $$
+            
+    3.  The key to finding the suitable basis is to rewrite the DEG in terms of its basis solutions using the properties of the characteristic polynomial and the differential operator as its variable:
+        $$P(\frac{d}{dx})f_{k}(x) = 0$$
+        and thus, in the general form using the fundamental theorem of algebra: 
+        $$ P(\frac{d}{dx}) f_{k}(x) = \Biggl( \overset{l}{\underset{k=1}{\prod}} \left(\frac{d}{dx} - \lambda_k \right)^{m_k} \Biggr) f_{k}(x) = 0 \, .$$
+
+    4. The solutions to this equation are given as:
+        $$f_{k}(x) = e^{\lambda_1 x}, \ e^{\lambda_2 x}, \ \cdots, \ e^{\lambda_n x} \qquad (1 \leq k \leq l \leq n) $$
+        and for each eigenvalue $\lambda_{k}$ with multiplicity greater than one, $m>1$, there is a subset of size $m$ with solutions corresponding to that eigenvalue;
+        $$f_{k,m_{k}}(x) = e^{\lambda_k x}, \ x e^{\lambda_k x} , \ \cdots, \ x^{m_{k}-1} e^{\lambda_k x}.$$
+        These solve the differential equation above in the general form: 
+        $$ P(\frac{d}{dx}) x^{m_{k}-1} e^{\lambda_k x} = \Biggl( \overset{l}{\underset{k=1}{\prod}} \left(\frac{d}{dx} - \lambda_k \right)^{m_k} \Biggr) x^{m_{k}-1} e^{\lambda_k x} = 0 \, .$$
+    
+    5.  The solutions given above can form the basis if their Wronskian is non-zero on an interval (it may vanish at isolated points);
+        $$ \det \begin{bmatrix} 
+        f_1(x) & \cdots & f_{n}(x) \\
+        f_1 ' (x) & \cdots & f_{n}'(x) \\
+        \vdots & \vdots & \vdots \\
+        f^{(n-1)}_{1} (x) & \cdots & f^{(n-1)}_{n} (x) \\
+        \end{bmatrix}  \neq 0 \, ,$$
+        and correspondingly, if any eigenvalue has a multiplicity higher than one: 
+        $$ \det \begin{bmatrix} 
+        f_1(x) & \cdots & f_{k}(x) &x f_{k}(x) & \cdots & x^{m_{k}-1} f_{k}(x)& \cdots & f_{l}(x) \\
+        f_1 ' (x) & \cdots & f_{k}(x)' &[x f_{k}(x)]' & \cdots & [x^{m_{k}-1} f_{k}(x)]' & \cdots & f_{l}'(x) \\
+        \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
+        f^{(n-1)}_{1} (x) & \cdots & f^{(n-1)}_{k}(x) &[x f_{k}(x)]^{(n-1)} & \cdots & [x^{m_{k}-1} f_{k}(x)]^{(n-1)}& \cdots & f^{(n-1)}_{l} (x) \\
+        \end{bmatrix}  \neq 0 \, .$$
+        Computation of the Wronskian can quickly become a tedious task in general. In this case, we can easily observe that the basis functions are linearly independent, because is not possible to obtain any of the solutions from a linear combination of the others!
+        
+        For example, $x e^{\lambda_{1}x}$ cannot be obtained from $x^2 e^{\lambda_{1}x}, \, e^{\lambda_{1}x}, \, x e^{\lambda_{2}x}, \cdots \, .$
+
+
+!!! check "Example: Second order homogeneous linear DE with constant coefficients"
+
+    Consider the equation 
+    
+    $$y'' + Ey = 0.$$ 
+    
+    The characteristic polynomial of this equation is 
+    
+    $$P(\lambda) = \lambda^2 + E.$$
+    
+    There are three cases for the possible solutions, depending upon the value 
+    of E.
+    
+    **Case 1: $E>0$**
+    For ease of notation, define $E=k^2$ for some constant $k$. The 
+    characteristic polynomial can then be factored as
+    
+    $$P(\lambda) = (\lambda+ i k)(\lambda - i k). $$
+    
+    Following our formulation for the solution, the two basis functions for the 
+    solution space are 
+    
+    $$f_1(x) = e^{i k x}, \ f_2=e^{- i k x}.$$
+    
+    Alternatively, the trigonometric functions can serve as basis functions, 
+    since they are linear combinations of $f_1$ and $f_2$ which remain linearly
+    independent,
+    
+    $$\tilde{f_1}(x)=\cos(kx), \tilde{f_2}(x)=\sin(kx).$$
+    
+    **Case 2: $E<0$**
+    This time, define $E=-k^2$, for constant $k$. The characteristic polynomial 
+    can then be factored as 
+    
+    $$P(\lambda) = (\lambda+ k)(\lambda -  k).$$
+
+    The two basis functions for this solution are then 
+    
+    $$f_1(x)=e^{k x}, \ f_2(x) = e^{-k x}.$$
+    
+    **Case 3: $E=0$**
+    In this case, there is a repeated eigenvalue (equal to $0$), since the 
+    characteristic polynomial reads
+    
+    $$P(\lambda) = (\lambda-0)^2.$$
+    
+    Hence, the basis functions for the solution space read 
+    
+    $$f_1(x)=e^{0 x} = 1, \ f_{2}(x) = x e^{0 x} = x. $$
+
+
+##8.2. Partial differential equations: Separation of variables
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/I4ghpYsFLFY?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+###8.2.1. Definitions and examples
+
+A partial differential equation (PDE) is an equation involving a function of two or 
+more independent variables and derivatives of said function. These equations
+are classified similarly to ordinary differential equations (the subject of
+our earlier study). For example, they are called linear if no terms such as
+
+$$\frac{\partial y(x,t)}{\partial x} \cdot \frac{d y(x,t)}{\partial t} \ or $$
+$$\frac{\partial^2 y(x,t)}{\partial x^2} y(x,t)$$ 
+
+occur. A PDE can be classified as $n$-th order according to the highest 
+derivative order of either variable occurring in the equation. For example, the 
+equation
+
+$$\frac{\partial^3 f(x,y)}{\partial x^3} + \frac{\partial f(x,t)}{\partial t} = 5$$
+
+is a $3^{rd}$ order equation because of the third derivative with respect to x
+in the equation.
+
+To begin with a context, we demonstrate that PDEs are of fundamental importance in physics, 
+especially in quantum physics. In particular, the Schrödinger equation, 
+which is of central importance in quantum physics, is a partial differential 
+equation with respect to time and space. This equation is essential 
+because it describes the evolution in time and space of the entire description
+of a quantum system $\psi(x,t)$, which is known as the wave function. 
+
+For a free particle in one dimension, the Schrödinger equation is 
+
+$$i \hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}. $$
+
+When we studied ODEs, an initial condition was necessary in order to fully 
+specify a solution. Similarly, in the study of PDEs an initial condition is 
+required but now also boundary conditions are required. Going back to the 
+intuitive discussion from the lecture on ODEs, each of these conditions is 
+necessary in order to specify an integration constant that occurs in solving 
+the equation. In partial differential equations at least one such constant will
+arise from the time derivative and likewise at least one from the spatial 
+derivative. 
+
+For the Schrödinger equation, we could supply the initial conditions
+$$\psi(x,0)=\psi_0(x)$$
+together with the boundary conditions
+$$\psi(0,t) = \psi(L, t) = 0$$
+
+This particular set of boundary conditions corresponds to a particle in a box,
+a situation which is used as the base model for many derivations in quantum 
+physics. 
+
+Another example of a partial differential equation common in physics is the 
+Laplace equation
+
+$$\frac{\partial^2 \phi(x,y)}{\partial x^2}+\frac{\partial^2 \phi(x,y)}{\partial y^2}=0.$$
+
+In quantum physics, Laplace's equation is important for the study of the hydrogen
+atom. In three dimensions and using spherical coordinates, the solutions to 
+Laplace's equation are special functions called spherical harmonics. In the 
+context of the hydrogen atom, these functions describe the wave function of the 
+system and a unique spherical harmonic function corresponds to each distinct set
+of quantum numbers.
+
+In the study of PDEs, there are no comprehensive overall treatment methods to the same 
+extent as there is for ODEs. There are several techniques which can be applied 
+to solving these equations and the choice of technique must be tailored to the
+equation at hand. Hence, we focus on some specific examples that are common in
+physics.
+
+###8.2.2. Separation of variables
+
+Let us focus on the one-dimensional Schrödinger equation of a free particle:
+
+$$i \hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}. $$
+
+To attempt a solution, we will make a *separation ansatz*,
+
+$$\psi(x,t)=\phi(x) f(t).$$
+
+!!! info "Separation ansatz"
+    The separation ansatz is a restrictive ansatz, not a fully general one. In
+    general, for such a treatment to be valid, an equation and the boundary 
+    conditions given with it have to fulfill certain properties. In this course
+    however, you will only be asked to use this technique when it is suitable.
+    
+!!! info "General procedure for the separation of variables:"
+
+    1. Substituting the separation ansatz into the PDE,
+
+        $$i \hbar \frac{\partial \phi(x)f(t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \phi(x)f(t)}{\partial x^2} $$
+        $$i \hbar \dot{f}(t) \phi(x) = - \frac{\hbar^2}{2m} \phi''(x)f(t). $$
+
+        Notice that in the above equation the derivatives on $f$ and $\phi$ can each be
+        written as ordinary derivatives, $\dot{f}=\frac{df(t)}{dt}$, 
+        $\phi''(x)=\frac{d^2 \phi}{dx^2}$. This is so because each one is a function of 
+        only one variable. 
+
+    2. Next, divide both sides of the equation by $\psi(x,t)=\phi(x) f(t)$,
+
+        $$i \hbar \frac{\dot{f}(t)}{f(t)} = - \frac{\hbar^2}{2m} \frac{\phi''(x)}{\phi(x)} = constant := \lambda. $$
+
+        In the previous line we concluded that each part of the equation must be equal 
+        to a constant, which we defined as $\lambda$. This follows because the left hand
+        side of the equation only has a dependence on the spatial coordinate $x$, whereas 
+        the right hand side only has dependence on the time coordinate $t$. If we have 
+        two functions $a(x)$ and $b(t)$ such that 
+        $a(x)=b(t) \quad \forall x, \quad t  \in \mathbb{R}$, then $a(x)=b(t)=const$.
+    3. The constant we defined, $\lambda$, is called a *separation constant*. With it, 
+        we can break the spatial and time dependent parts of the equation into two separate equations,
+
+        $$i \hbar \dot{f}(t) = \lambda f(t)$$
+
+        $$-\frac{\hbar^2}{2m} \phi''(x) = \lambda \phi(x) .$$
+
+    To summarize, this process has broken one partial differential equation into two
+    ordinary differential equations of different variables. In order to do this, we 
+    needed to introduce a separation constant, which remains to be determined.
+
+###8.2.3. Boundary and eigenvalue problems
+
+Continuing on with the Schrödinger equation example from the previous 
+section, let us focus on the spatial part
+
+$$-\frac{\hbar^2}{2m} \phi''(x) = \lambda \phi(x),$$
+$$\phi(0)=\phi(L)=0.$$
+
+This has the form of an eigenvalue equation, in which $\lambda$ is the 
+eigenvalue, $- \frac{\hbar^2}{2m} \frac{d^2}{dx^2}[\cdot]$ is the linear 
+operator and $\phi(x)$ is the eigenfunction. 
+
+Notice that this ordinary differential equation is specified 
+along with its boundary conditions. Note that in contrast to an initial value
+problem, a boundary value problem does not always have a solution. For example, 
+in the figure below, regardless of the initial slope, the curves never reach $0$
+when $x=L$. 
+
+![image](figures/DE2_1.png)
+
+For boundary value problems like this, there are only solutions for particular 
+eigenvalues $\lambda$. Coming back to the example, it turns out that solutions
+only exist for $\lambda>0$. 
+
+*This can be shown quickly, feel free to try it!*
+
+For simplicity, define $k^2:= \frac{2m \lambda}{\hbar^2}$. The equation then 
+reads
+
+$$\phi''(x)+k^2 \phi(x)=0.$$
+
+Two linearly independent solutions to this equation are 
+
+$$\phi_{1}(x)=\sin(k x), \ \phi_{2}(x) = \cos(k x).$$
+
+The solution to this homogeneous equation is then 
+
+$$\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x).$$
+
+The eigenvalue, $\lambda$, as well as one of the constant coefficients, can be 
+determined using the boundary conditions. 
+
+$$
+\begin{align}\phi(0) &=0 \ \Rightarrow \ \phi(x)=c_1 \sin(k x), \ c_2=0. \\
+\phi(L) &=0 \ \Rightarrow \ 0=c_1 \sin(k L) 
+\end{align} \, .
+$$
+
+In turn, using the properties of the $\sin(\cdot)$ function, it is now possible
+to find the allowed values of $k$ and hence also $\lambda$. The previous 
+equation implies, 
+
+$$k L = n \pi, \, n  \in  \mathbb{N}$$
+
+$$\lambda_n = \big{(}\frac{n \pi \hbar}{L} \big{)}^2.$$
+
+The values $\lambda_n$ are the eigenvalues. Now that we have determined 
+$\lambda$, it enters into the time equation, $i \hbar \dot{f}(t) = \lambda f(t)$
+only as a constant. We can therefore simply solve,
+
+$$\dot{f}(t) = -i \frac{\lambda}{\hbar} f(t)$$
+
+$$f(t) = A e^{\frac{-i \lambda t}{\hbar}}.$$
+
+In the previous equation, the coefficient $A$ can be determined if the original
+PDE is supplied with an initial condition. 
+
+Putting the solutions to the two ODEs together and redefining 
+$\tilde{A}=A \cdot c_1$, we arrive at the solutions for the PDE,
+
+$$\psi_n(x,t) = \tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} \sin(\frac{n \pi x}{L}).$$
+
+Notice that there is one solution $\psi_{n}(x,t)$ for each natural number $n$. 
+These are also very special solutions that are important in the context of physics. We will next discuss how to 
+obtain the general solution in our example. 
+
+##8.3. Self-adjoint differential operators
+
+<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/p4MHW0yMMvY?rel=0" frameborder="0" allow="accelerometer; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
+
+###8.3.1. Connection to Hilbert spaces
+
+As hinted earlier, it is possible to re-write the previous equation by 
+defining a linear operator, $L$, acting on the space of functions which satisfy
+$\phi(0)=\phi(L)=0$:
+
+$$L[\cdot]:= \frac{- \hbar^2}{2m} \frac{d^2}{dx^2}[\cdot]. $$
+
+Then, the ODE can be written as 
+
+$$L[\phi]=\lambda \phi.$$
+
+This equation looks exactly like, and it turns out to be, an eigenvalue equation!
+
+!!! info "Connecting function spaces to Hilbert spaces"
+    
+    Recall that a space of functions can be transformed into a Hilbert space by 
+    equipping it with a inner product,
+    
+    $$\langle f, g \rangle = \int^{L}_{0} dx f^*(x) g(x) $$
+    
+    Use of this inner product also has utility in demonstrating that particular 
+    operators are *Hermitian*. The term "Hermitian" is precisely defined below.
+    Of considerable interest is that Hermitian operators have a set of convenient 
+    properties including all real eigenvalues and orthonormal eigenfunctions. 
+    
+The *nicest* type of operators for many practical purposes are Hermitian 
+operators. In quantum physics, for example, all physical operators must be 
+Hermitian. 
+
+!!! info "Hermiticity of an operator"
+    Denote a Hilbert space $\mathcal{H}$. An operator $H: \mathcal{H} \mapsto \mathcal{H}$ is said to be Hermitian if it satisfies
+    $$\langle f, H g \rangle = \langle H f, g \rangle \ \forall \ f, \ g \ \epsilon \ \mathcal{H}.$$
+
+Now, we would like to investigate whether the operator we have been working with,
+$L$, satisfies the criterion of being Hermitian over the function space 
+$\phi(0)=\phi(L)=0$ equipped with the inner product defined above (i.e. it is a
+Hilbert space).
+
+1.  First, denote this Hilbert space $\mathcal{H}_{0}$ and consider $f, \ g \ \in \ \mathcal{H}_0$ which are two functions from the Hilbert space. Then, we can investigate
+    $$\langle f, L g \rangle = \frac{- \hbar^2}{2m} \int^{L}_{0} dx f^*(x) \frac{d^2}{dx^2}g(x).$$
+
+2. In the next step, use the fact that it is possible to do integration by parts in the integral,
+    $$
+    \langle f, L g \rangle = \frac{+ \hbar^2}{2m} ( \int^{L}_{0} dx \frac{d f^*}{dx} \frac{d g}{dx} - [f^*(x)\frac{d g}{dx}] \big{|}^{L}_{0} )
+    $$
+    The boundary term vanishes due to the boundary conditions $f(0)=f(L)=0$, which directly imply $f^*(0)=f^*(L)=0$. 
+4. Now, integrate by parts a second time 
+    $$\langle f, L g \rangle = \frac{- \hbar^2}{2m} (\int^{L}_{0} dx \frac{d^2 f^*}{dx^2} g(x) - [\frac{d f^*}{dx} g(x)] \big{|}^{L}_{0} ).$$
+    As before, the boundary term vanishes, due to the boundary conditions $g(0)=g(L)=0$. 
+    After canceling the boundary term, the expression on the right hand side contained in the integral simplifies to $\langle L f, g \rangle$. 
+5. Therefore,
+    $$\langle f, L g \rangle=\langle L f, g \rangle. $$
+
+
+Thus, we demonstrated that $L$ is a Hermitian operator on the space $\mathcal{H}_0$. As a hermitian operator, $L$ has the property that its eigenfunctions form an orthonormal basis for the space $\mathcal{H}_0$. Hence, it is possible to expand any function $f \in \mathcal{H}_0$ in terms of the eigenfunctions of $L$.
+
+!!! info "Connection to quantum states"
+    
+    Recall that a quantum state $|\phi\rangle$ can be written in an orthonormal 
+    basis $\{ |u_n\rangle \}$ as 
+    $$|\phi\rangle = \underset{n}{\Sigma} \langle u_n | \phi \rangle\, |u_n\rangle.$$ 
+    
+    In the case of Hermitian operators, their eigenfunctions play the role of the orthonormal basis. In the context of our running example,
+    the 1D Schrödinger equation of a free particle, the eigenfunctions 
+    $\sin(\frac{n \pi x}{L})$ play the role of the basis functions $|u_n\rangle$.
+    
+To close our running example, consider the initial condition 
+$\psi(x,0) = \psi_{0}(x)$. Since the eigenfunctions $\sin(\frac{n \pi x}{L})$ 
+form a basis, we can now write the general solution to the problem as 
+
+$$\psi(x,t)  = \overset{\infty}{\underset{n}{\Sigma}} c_n e^{-i \frac{\lambda_n t}{\hbar}} \sin(\frac{n \pi x}{L}),$$
+
+where in the above we have defined the coefficients as a Fourier 
+coefficient,
+
+$$c_n:= \int^{L}_{0} dx \sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
+    
+###8.3.2. General recipe for separable PDEs
+
+!!! tip "General recipe for separable PDEs" 
+
+    1. Make the separation ansatz to obtain separate ordinary differential 
+        equations.
+    2. Choose which equation to treat as the eigenvalue equation. This will depend 
+        upon the boundary conditions. Additionally, verify that the linear 
+        differential operator $L$ in the eigenvalue equation is Hermitian.
+    3. Solve the eigenvalue equation. Substitute the eigenvalues into the other 
+        equations and solve those too. 
+    4. Use the orthonormal basis functions to write down the solution corresponding 
+        to the specified initial and boundary conditions. 
+
+One natural question is: *"What if the operator $L$ from step 2 is not Hermitian?"*
+
+- It is possible to try and make it Hermitian by working on a Hilbert space equipped with a different inner product. This means that one can consider modifications to the definition of $\langle \cdot, \cdot \rangle$ such that $L$ is Hermitian with respect to the modified inner product. This type of technique falls under the umbrella of *Sturm-Liouville Theory*, which forms the foundation for a lot of the analysis that can be done analytically on PDEs.
+
+Another question is of course: *"What if the equation is not separable?"*
+
+- One possible approach is to try working in a different coordinate system. There are a few more analytic techniques available. However, in many situations, it becomes necessary to work with numerical methods of solution.
+
+*** 
+
+##8.4. Problems
+
+1.  [:grinning:] Which of the following equations for $y(x)$ is linear?
+
+    1. $y''' - y'' + x \cos(x) y' + y - 1 = 0$
+
+    2. $y''' + 4 x y' - \cos(x) y = 0$
+
+    3. $y'' + y y' = 0$
+
+    4. $y'' + e^x y' - x y = 0$
+
+2.  [:grinning:] Find the general solution to the equation 
+
+    $$y'' - 4 y' + 4 y = 0. $$
+
+    Show explicitly by computing the Wronski determinant that the 
+    basis for the solution space is actually linearly independent. 
+
+3.  [:grinning:] Find the general solution to the equation 
+
+    $$y''' - y'' + y' - y = 0.$$
+
+    Then find the solution to the initial conditions $y''(0) =0$, $y'(0)=1$, $y(0)=0$. 
+
+4.  [:smirk:] Take the Laplace equation in 2D:
+
+    $$\frac{\partial^2 \phi(x,y)}{\partial x^2} + \frac{\partial^2 \phi(x,y)}{\partial y^2} = 0.$$
+
+    1.  Make a separation ansatz $\phi(x,y) = f(x)g(y)$ and write 
+        down the resulting ordinary differential equations.
+
+    2.  Now assume that the boundary conditions $\phi(0,y) = \phi(L,y) =0$ for all y, i.e.  $f(0)=f(L)=0$. Find all solutions $f(x)$ and the corresponding eigenvalues.
+    3.  Finally, for each eigenvalue, find the general solution $g(y)$ for this eigenvalue. Combine this with all solutions $f(x)$ to write down the general solution (we know from the lecture that the operator $\frac{d^2}{dx^2}$ is Hermitian - you can thus directly assume that the solutions form an orthogonal basis). 
+
+
+5.  [:smirk:] Consider the following partial differential equations, and try to make a separation ansatz $h(x,y)=f(x)g(y)$. What do you observe in each case? (Only attempt the separation, do not solve the problem fully)
+    
+    1.  $\frac{\partial h(x,y)}{\partial x} + x \frac{\partial h(x,y)}{\partial y} = 0. $
+
+    2.  $\frac{\partial h(x,y)}{\partial x} + \frac{\partial h(x,y)}{\partial y} + xy\,h(x,y) = 0$
+
+6.  [:sweat:] We consider the Hilbert space of functions $f(x)$ defined for $x \ \epsilon \ [0,L]$ with $f(0)=f(L)=0$. 
+
+    Which of the following operators $\mathcal{L}$ on this space is Hermitian?
+
+    1.  $\mathcal{L}_1 f(x) = A(x) \frac{d^2 f}{dx^2}$
+
+    2.  $\mathcal{L}_2 f(x) = \frac{d}{dx} \big{(} A(x) \frac{df}{dx} \big{)}$

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