Question

If sigma x, sigma y, and sigma z are three components of a pauli spin matrix sigma, then show that [sigma x, sigma y]=2i sigma z; [sigma y, sigma z]= 2i sigma x

Answer 1

Answer on Question #79316- Physics- Quantum Mechanics

Question: 1. If sigma $x$, sigma $y$, and sigma $z$ are three components of a pauli spin matrix sigma, then show that [sigma $x$, sigma $y$] = 2i sigma $z$; [sigma $y$, sigma $z$] = 2i sigma $x$

Answer:

In the so-called $\sigma_z$-representation the Pauli matrices can be written in the form as follows [1]:

$$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

One should also recall the definition of the commutator in quantum mechanics:

$$[\mathrm{A}, \mathrm{B}] = \mathrm{A B} - \mathrm{B A},$$

where A and B are two arbitrary quantum operators.

Substituting the matrices (1) into the definition (2), we obtain:

$$\begin{aligned} [\sigma_x, \sigma_y] &= \sigma_x \sigma_y - \sigma_y \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} - \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} = \\ &= 2i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = 2i \sigma_z \end{aligned}$$

In a similar fashion we get:

$$\begin{aligned} [\sigma_y, \sigma_z] &= \sigma_y \sigma_z - \sigma_z \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2i \\ 2i & 0 \end{pmatrix} = \\ &= 2i \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 2i \sigma_x \end{aligned}$$

[1] (Electronic resource) https://en.wikipedia.org/wiki/Pauli_matrices

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