Update 4_vector_spaces_QM.md

src/4_vector_spaces_QM.md

@@ -10,6 +10,7 @@ The lecture on vector spaces in quantum mechanics consists of four parts:
 - [Bra vectors](#bra-vectors)
 - [Inner product of state vectors](#inner-product-of-state-vectors)
 - [Matrix representation of ket and bra vectors](#matrix-representation-ket-bra-vectors)
+- [A two-dimensional Hilbert space](#two-dimensional-hilbert-space)
 
 **Total video length: xxx **
 
@@ -21,65 +22,29 @@ 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 is known as the *state space* of the system.
 Such 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: a *ket* is also refereed to as a state vector, *ket* vector, or just state. 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.
+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 refereed to as a state vector, *ket* vector, or just state. 
+
+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:
     
 - 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
      $$|{\Psi}\rangle=c_1|{\psi_1}\rangle+c_2|{\psi_1}\rangle+c_3|{\psi_3}\rangle+...+c_n|{\psi_n}\rangle$$
 where $c_1, c_2, c_3, ...$ are general complex numbers will also be a physically allowed state vector of the quantum system.
      
-- 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.
-
-As we will discuss below, in quantum mechanics it is advantageous to work with  *normalised vectors*, that is, whose *length* is one.
+- 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. 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.
 
 - 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 of the system 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 said to *span* the space.
+for some specific choice of coefficients $c_i$. The set of vector \{$|{\psi_i}\rangle$\} are then said to *span* the space.
     
 - 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
+Linear independence can also be expressed as the requirement that if one has that
 $$\sum_{i=1}^n c_i |{\psi_i}\rangle=0\;\text{then}\; c_i=0\;\text{for all}\; i$$
-The minimum number of vectors needed to form a complete set of basis states is known as the *dimension* of the state space (recall that this is
-called usually the Hilbert space).
-
-
-### Example: a two-dimensional Hilbert space
-
-As an example we can consider a Hilbert space in two dimensions.
-
-The elements of this space are of the form
-
-$$ |\Psi\rangle = \left( \begin{array}{c} a_1 \\ a_2 \end{array} \right)$$
-
-where $a_1$ and $a_2$ are general complex coefficients.
 
-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 minimum number of vectors needed to form a complete set of basis states is known as the *dimensionality* of the state space. In quantum mechanis 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).
 
-To identify a suitable basis, we need two vectors that span the whole space and are linearly independent of each other. A possible choice is
-$$ |{\psi_1}\rangle = \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \, ,\quad
-|{\psi_2}\rangle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$
-      
-You can check that any element of this Hilbert space can then be expressed as
-$$ |{\Psi}\rangle=\sum_{i=1}^2 c_i|{\psi_i}\rangle \, ,$$
 
-For instance, the values of the coefficients $c_1$ and $c_2$ for the examples above are, respectively,
-$$
-     (c_1,c_2) = (3,-2i) \, ,\qquad
-     (c_1,c_2) = (i,-4) \, ,\qquad
-      (c_1,c_2) = (2,-5) \, .\qquad 
-      $$
-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.
 
 ## Bra vectors
 We need now to extend a bit the Dirac notation
@@ -140,33 +105,7 @@ A set of orthonormal basis vectors $\{\ket{\psi_n}\text{;}\; n=1,2,3,...\}$ will
  All the vectors belonging to a Hilbert space have a finite norm, that is they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic
   interpretation described above.
 
- Again we can use our 2d quantum system as example. If
-  we have two state vectors
-  $$
-    \ket{\psi} = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
-    \ket{\phi} = \left( \begin{array}{c}0 \\1 \end{array} \right)
- $$
-  then their inner product is
-$$
-    \braket{\psi}{\phi} =
-    \frac{1}{\sqrt{2}} \left( 1 , i  \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
-\frac{i}{\sqrt{2}} 
-$$
-  and the associated probability will be given by
- $$
-    |\braket{\psi}{\phi}|^2 = \frac{1}{2}
- $$
-  meaning that if I measure the state $\ket{\phi}$, I will have a 50\% probability
-  of finding it in the state $\ket{\psi}$.
-
-  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
- $$
-    \braket{\psi}{\psi} =  \braket{\phi}{\phi} = 1 \, .
-  $$
-
-
+ 
 
 ## Matrix representation of ket and bra vectors
  
@@ -230,6 +169,9 @@ and ket vectors as column vector. The row vector can thus be treated as the {\it
 of the corresponding column vector.
 
 
+## A two-dimensional Hilbert space
+
+
 **The spin-1/2 case**: one very important system in quantum mechanics is that of the intrinsic angular
   momentum, known as {\it spin}, of the electron.
    
@@ -268,6 +210,64 @@ $$
   Note that one needs to take the complex conjugate when expressing a state vector
   as a bra vector.
 
+As an example we can consider a Hilbert space in two dimensions.
+
+The elements of this space are of the form
+
+$$ |\Psi\rangle = \left( \begin{array}{c} a_1 \\ a_2 \end{array} \right)$$
+
+where $a_1$ and $a_2$ are general complex coefficients.
+
+Examples of elements of this Hilbert space are the following:
+$$
+\left( \begin{array}{c}3 \\ -2i \end{array} \right) \, ,\quad
+\left( \begin{array}{c}i \\ -4 \end{array} \right) \, ,\quad
+\left( \begin{array}{c}2 \\ 5 \end{array} \right) \, .
+$$
+
+To identify a suitable basis, we need two vectors that span the whole space and are linearly independent of each other. A possible choice is
+$$ |{\psi_1}\rangle = \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \, ,\quad
+|{\psi_2}\rangle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$
+      
+You can check that any element of this Hilbert space can then be expressed as
+$$ |{\Psi}\rangle=\sum_{i=1}^2 c_i|{\psi_i}\rangle \, ,$$
+
+For instance, the values of the coefficients $c_1$ and $c_2$ for the examples above are, respectively,
+$$
+     (c_1,c_2) = (3,-2i) \, ,\qquad
+     (c_1,c_2) = (i,-4) \, ,\qquad
+      (c_1,c_2) = (2,-5) \, .\qquad 
+      $$
+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.
+
+Again we can use our 2d quantum system as example. If
+  we have two state vectors
+  $$
+    \ket{\psi} = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
+    \ket{\phi} = \left( \begin{array}{c}0 \\1 \end{array} \right)
+ $$
+  then their inner product is
+$$
+    \braket{\psi}{\phi} =
+    \frac{1}{\sqrt{2}} \left( 1 , i  \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
+\frac{i}{\sqrt{2}} 
+$$
+  and the associated probability will be given by
+ $$
+    |\braket{\psi}{\phi}|^2 = \frac{1}{2}
+ $$
+  meaning that if I measure the state $\ket{\phi}$, I will have a 50\% probability
+  of finding it in the state $\ket{\psi}$.
+
+  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
+ $$
+    \braket{\psi}{\psi} =  \braket{\phi}{\phi} = 1 \, .
+  $$
+
+
+
 Problems
 ========