Question:
Consider the Pauli matrices \(\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), \(\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\), \(\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\).
The value of \(\text{Tr}\big(\sigma_z [\sigma_x, \sigma_y]\big)\) is
Consider the Pauli matrices \(\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), \(\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\), \(\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\).
The value of \(\text{Tr}\big(\sigma_z [\sigma_x, \sigma_y]\big)\) is
The value of \(\text{Tr}\big(\sigma_z [\sigma_x, \sigma_y]\big)\) is
Show Hint
Multiply out \(\sigma_x\sigma_y\) and \(\sigma_y\sigma_x\) as \(2\times2\) matrices first, then subtract to get the commutator before multiplying by \(\sigma_z\).
Updated On: Jul 28, 2026
- \(2i\)
- \(i\)
- \(4i\)
- \(\dfrac{i}{2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Multiply \(\sigma_x\sigma_y\).
\[ \sigma_x \sigma_y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \]
Each entry comes from the usual row times column rule: top-left is \(0\cdot 0 + 1\cdot i = i\), top-right is \(0\cdot(-i) + 1\cdot 0 = 0\), and so on for the rest.
Step 2: Multiply \(\sigma_y\sigma_x\).
\[ \sigma_y \sigma_x = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \]
Step 3: Form the commutator \([\sigma_x, \sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x\).
\[ [\sigma_x, \sigma_y] = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} - \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} = 2i\,\sigma_z \]
Step 4: Multiply by \(\sigma_z\) and take the trace.
\[ \sigma_z [\sigma_x,\sigma_y] = \sigma_z (2i\sigma_z) = 2i\,\sigma_z^2 \]
Since \(\sigma_z^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I\), we get \(\sigma_z[\sigma_x,\sigma_y] = 2iI\). The trace of \(2iI\) is \(2i\) times the trace of the \(2\times2\) identity, which is \(2i \times 2 = 4i\).
Step 5: Why the other options are wrong.
Option (A), \(2i\), is only the coefficient in \([\sigma_x,\sigma_y]=2i\sigma_z\) before taking the trace with \(\sigma_z\); it skips the last multiplication and trace step. Options (B) and (D) do not match this direct computation at all and would only appear from an arithmetic slip in the matrix multiplication.
Final Answer:
\(\text{Tr}(\sigma_z[\sigma_x,\sigma_y]) = 4i\), which is option (C).\[ \boxed{4i} \]
\[ \sigma_x \sigma_y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \]
Each entry comes from the usual row times column rule: top-left is \(0\cdot 0 + 1\cdot i = i\), top-right is \(0\cdot(-i) + 1\cdot 0 = 0\), and so on for the rest.
Step 2: Multiply \(\sigma_y\sigma_x\).
\[ \sigma_y \sigma_x = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \]
Step 3: Form the commutator \([\sigma_x, \sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x\).
\[ [\sigma_x, \sigma_y] = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} - \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} = 2i\,\sigma_z \]
Step 4: Multiply by \(\sigma_z\) and take the trace.
\[ \sigma_z [\sigma_x,\sigma_y] = \sigma_z (2i\sigma_z) = 2i\,\sigma_z^2 \]
Since \(\sigma_z^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I\), we get \(\sigma_z[\sigma_x,\sigma_y] = 2iI\). The trace of \(2iI\) is \(2i\) times the trace of the \(2\times2\) identity, which is \(2i \times 2 = 4i\).
Step 5: Why the other options are wrong.
Option (A), \(2i\), is only the coefficient in \([\sigma_x,\sigma_y]=2i\sigma_z\) before taking the trace with \(\sigma_z\); it skips the last multiplication and trace step. Options (B) and (D) do not match this direct computation at all and would only appear from an arithmetic slip in the matrix multiplication.
Final Answer:
\(\text{Tr}(\sigma_z[\sigma_x,\sigma_y]) = 4i\), which is option (C).\[ \boxed{4i} \]
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