Addressing #16

src/6_eigenvectors_QM.md

@@ -14,10 +14,16 @@ 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
 $$
 \hat{A}|\psi\rangle=|\varphi\rangle \, ,
@@ -65,7 +71,7 @@ the rules of matrix multiplication:
     $$
     where $\mathbb{I}$ is the identity matrix of dimensions $n\times n$, and ${\rm det}$ is the determinant.
 
-This relations follows from the eigenvalue equation in terms of components
+This relation follows from the eigenvalue equation in terms of components
 $$
 \begin{align}
 \sum_{j=1}^n A_{ij} v_j &= \lambda v_i \, , \\
@@ -73,12 +79,12 @@ $$
 \to \quad \sum_{j=1}^n\left( A_{ij} - \lambda \delta_{ij}\right) v_j &=0 \, .
 \end{align}
 $$
-Therefore the eigenvalue condition can be written as a set of coupled linear equations
+Therefore, the eigenvalue condition can be written as a set of coupled linear equations
 $$
 \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 condition), thus leading to the characteristic equation.
+(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 $\lambda_k$, $k=1,\ldots,n$.
   
@@ -98,7 +104,7 @@ $$
 = A_{11}A_{22} - A_{12}A_{21} \, ,
 $$
 while the corresponding expression for a matrix belonging to a vector
-space in $n=3$ dimensions will be given in terms of the previous expression
+space in $n=3$ dimensions in terms of the previous expression will be given as
 $$
 {\rm det}\left( A \right) = \left|  \begin{array}{ccc} A_{11}  & A_{12}  & A_{13} \\ A_{21}  &  A_{22}
 &  A_{23} \\ A_{31}  &  A_{32}
@@ -171,19 +177,19 @@ corresponding eigenvalue problem for this operator, what we called above as the
 This is most efficiently done in the matrix representation of this operation, where we have
 that the above operator equation can be expressed in terms of its components as
 $$
-\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} \, , \quad k=1,\ldots,n \, .
+\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
 $$
 \begin{pmatrix} A_{11}- \lambda_{\psi_k} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi_k} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi_k} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}
-\begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}=0 \, , \quad k=1,\ldots,n \, .
+\begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}=0 \, .
 $$
 
 !!! 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 condition:
+    the determinant of the matrix vanishes, as follows from the Cramer's condition:
     $$
     {\rm det} \begin{pmatrix} A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}=
     \left|  \begin{array}{cccc}A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{array} \right| = 0
@@ -211,36 +217,38 @@ $$
 
 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 $c$ some constant. In the context of quantum mechanics, we need to choose this overall rescaling constant
-to ensure that the eigenvectors are normalised, that is, that they satisfy
+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 set of eigenvectors
+With such a choice of normalisation, one says that the eigenvectors in a set
 are *orthogonal* among them.
 
-The set of all the eigenvalues of an operator is called *eigenvalue spectrum* of the operator. Note that different eigenvectors can also have the same eigenvalue. If this is the case the eigenvalue is said to be *degenerate*.
+!!! 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*.
 
 ***
 
-##Problems
+##6.3. Problems
 
-1. *Eigenvalues and Eigenvectors* 
+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. The Hamiltonian for a two-state system is given by 
- 
+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. Find the eigenvalues and eigenvectors of the matrices
+3. *Eigenvalues and eigenvectors II*
 
-    $$A=\begin{pmatrix} -2&-1&-1\\6&3&2\\0&0&1 \end{pmatrix}\,  \quad B=\begin{pmatrix} 1&1&2\\2&2&2\\-1&-1&-1 \end{pmatrix} $$.
+    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*
 
@@ -248,12 +256,14 @@ The set of all the eigenvalues of an operator is called *eigenvalue spectrum* of
     $$\hat{H}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix} \, .$$
     Determine the eigenvalues and eigenvectors of this operator.
 
-5. Show that the Hermitian matrix
+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. Confirm, by explicit calculation, that the eigenvalues of the real, symmetric matrix
+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.