Pauli Matrix Commutator Calculation

To find the value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$, we first calculate the commutator $[\sigma_x, \sigma_y]$.

• Compute the product $\sigma_x \sigma_y$: $ \sigma_x \sigma_y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} (0)(0) + (1)(i) & (0)(-i) + (1)(0) \\ (1)(0) + (0)(i) & (1)(-i) + (0)(0) \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} $
• Compute the product $\sigma_y \sigma_x$: $ \sigma_y \sigma_x = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0) + (-i)(1) & (0)(1) + (-i)(0) \\ (i)(0) + (0)(1) & (i)(1) + (0)(0) \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} $
• Calculate the commutator $[\sigma_x, \sigma_y]$: $ [\sigma_x, \sigma_y] = \sigma_x \sigma_y - \sigma_y \sigma_x = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} - \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} i - (-i) & 0 - 0 \\ 0 - 0 & -i - i \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} $
• Note that $[\sigma_x, \sigma_y] = 2i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = 2i \sigma_z$.

Trace Calculation for $\sigma_z [\sigma_x, \sigma_y]$

Next, we compute the matrix product $\sigma_z [\sigma_x, \sigma_y]$.

• Substitute the result for the commutator: $ \sigma_z [\sigma_x, \sigma_y] = \sigma_z (2i \sigma_z) $
• Use the property that $\sigma_z^2 = I$ (the $2 \times 2$ identity matrix): $ \sigma_z (2i \sigma_z) = 2i \sigma_z^2 = 2i I = 2i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & 2i \end{pmatrix} $
• Calculate the trace, which is the sum of the diagonal elements: $ \text{Tr}(\sigma_z [\sigma_x, \sigma_y]) = \text{Tr}\begin{pmatrix} 2i & 0 \\ 0 & 2i \end{pmatrix} = 2i + 2i = 4i $

Therefore, the value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$ is $4i$.