# Hermitian Matrix. Comutators (1-325)

Specify the condition that must be specified by a matrix so that it is both unitary and Hermitian.

Consider the matrices  $\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$ , $\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}$ and $\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$

Do they satisfy this condition?

A matrix is Hermitic if $A=(A^T)^*$ (matrix is hermitic if it is equal to the conjugate of its transpose)

A matrix is unitary if $A^{-1}=(A^T)^*$ ( therefore $I=A(A^T)^*$ ) (matrix is unitary it its inverse is equal to the conjugate of its transpose)

Hence matrix is both unitary and Hermitic if and only if: $A=A^{-1}$ or equivalent $A=[1/det⁡(A)]*adj(A)$ where $adj(A)$ is the adjunct matrix of A.

For a 2×2 matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ one has  $A^{-1}=\frac{1}{(ad-bc)}*\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

In our case:

$\begin{pmatrix}0&1\\1&0\end{pmatrix}^{-1}=[1/(0-1)]*\begin{pmatrix}0&-1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ so that it is unitary and Hermitic

$\begin{pmatrix}0&-i\\i &0\end{pmatrix}^{-1}=[1/(0-1)]*\begin{pmatrix}0&i\\-i&0\end{pmatrix}=\begin{pmatrix}0&-i\\i &0\end{pmatrix}$ so that it is unitary and Hermitic

$\begin{pmatrix}1&0\\0&-1\end{pmatrix}^{-1}=[1/(-1-0)]*\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ so that it is unitary and Hermitic

Let $\hat A$ and $\hat B$ be two operators. Assuming that $\hat B^{-1}$ exists show that $[A,B^{-1}] =-B^{-1}[A,B]B^{-1}$

$[A,B^{-1} ]=AB^{-1}-B^{-1} A$

$B^{-1} [A,B] B^{-1}=B^{-1} (AB-BA) B^{-1}=(B^{-1} AB-B^{-1} BA) B^{-1}=$

$=(B^{-1} AB-A) B^{-1}=B^{-1} ABB^{-1}-AB^{-1}=B^{-1} A(BB^{-1} )-AB^{-1}=B^{-1} A-AB^{-1}$

Therefore

$[A,B^{-1} ]=-B^{-1} [A,B] B^{-1}$