Lewis Harvey

2021-01-04

Problem 3. Find the unitary operator $\stackrel{^}{U}$ which diagonalizes one of the Pauli matrices
${\stackrel{^}{\sigma }}_{x}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$
Obtain the eigenvalues and eigenvectors of ${\stackrel{^}{\sigma }}_{x}$

### Answer & Explanation

bahaistag

Step 1
Given
Pauli matrices
${\stackrel{^}{\sigma }}_{x}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
According to question
$\frac{\stackrel{^}{U}}{{\stackrel{^}{\sigma }}_{x}}=\left[\begin{array}{cc}{\lambda }_{1}& 0\\ 0& {\lambda }_{2}\end{array}\right]$
Where $\stackrel{^}{U}$ is the unitary operator which diagonalizes the the Pauli matrices.
Step 2
Now Eigenvalues of Pauli matrix.
$|{\stackrel{^}{\sigma }}_{x}-\lambda I|=\left[\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right]=0$
$⇒\left[\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right]=0\phantom{\rule{0ex}{0ex}}⇒{\lambda }^{2}-1=0$
$⇒\lambda =1,-1$
Now Eigenvector,
$\lambda =1$
$\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0$
$⇒x=y$
Hence Eigenvector for $\lambda =1$ is
$X=\left[\begin{array}{c}1\\ 1\end{array}\right]$
Eigenvector for
$\lambda =-1$
$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0$
$⇒x=-y$
Hence Eigenvector for $\lambda =-1$ is
$Y=\left[\begin{array}{c}1\\ -1\end{array}\right]$
Let for unitary operator,

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$
$⇒\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^{-1}$
$⇒\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
$⇒\stackrel{^}{U}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon