Problem 3. Find the unitary operator U^ which diagonalizes one of the Pauli matrices σ^x=(0110) Obtain the eigenvalues and eigenvectors of σ^x

Step 1 Given Pauli matrices σ^x=[0110] According to question U^σ^x=[λ100λ2] Where U^ is the unitary operator which diagonalizes the the Pauli matrices. Step 2 Now Eigenvalues of Pauli matrix. |σ^x−λI|=[−λ11−λ]=0 ⇒[−λ11−λ]=0⇒λ2−1=0 ⇒λ=1,−1 Now Eigenvector, λ=1 [−111−1][xy]=0 ⇒x=y Hence Eigenvector for λ=1 is X=[11] Eigenvector for λ=−1 [1111][xy]=0 ⇒x=−y Hence Eigenvector for λ=−1 is Y=[1−1] Let for unitary operator, U^=[abcd] then [abcd][0110]=[100−1] ⇒[abcd]=[100−1][0110]−1 ⇒[abcd]=[0−110] ⇒U^=[0−110]

Best solutions to - "Problem 3. Find the unitary operator U^ which..."

Lewis Harvey

Answered question

2021-01-04

Problem 3. Find the unitary operator

\hat{U}

which diagonalizes one of the Pauli matrices

{\hat{σ}}_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

Obtain the eigenvalues and eigenvectors of

{\hat{σ}}_{x}

Answer & Explanation

bahaistag

Skilled2021-01-05Added 100 answers

Step 1
Given
Pauli matrices
${\hat{σ}}_{x} = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$
According to question
$\frac{\hat{U}}{{\hat{σ}}_{x}} = [\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}]$
Where $\hat{U}$ is the unitary operator which diagonalizes the the Pauli matrices.
Step 2
Now Eigenvalues of Pauli matrix.
$| {\hat{σ}}_{x} - λ I | = [\begin{matrix} - λ & 1 \\ 1 & - λ \end{matrix}] = 0$
$\Rightarrow [\begin{matrix} - λ & 1 \\ 1 & - λ \end{matrix}] = 0 \Rightarrow λ^{2} - 1 = 0$
$\Rightarrow λ = 1, - 1$
Now Eigenvector,
$λ = 1$
$[\begin{matrix} - 1 & 1 \\ 1 & - 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = 0$
$\Rightarrow x = y$
Hence Eigenvector for $λ = 1$ is
$X = [\begin{matrix} 1 \\ 1 \end{matrix}]$
Eigenvector for
$λ = - 1$
$[\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = 0$
$\Rightarrow x = - y$
Hence Eigenvector for $λ = - 1$ is
$Y = [\begin{matrix} 1 \\ - 1 \end{matrix}]$
Let for unitary operator,
$\hat{U} = [\begin{matrix} a & b \\ c & d \end{matrix}] then$
$[\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}]$
$\Rightarrow [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] {[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]}^{- 1}$
$\Rightarrow [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$
$\Rightarrow \hat{U} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$

Jeffrey Jordon

Expert2022-01-24Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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