Commutators

Given two operators $\hat{A}$ and $\hat{B}$, their commutator is defined as

$[\hat{A},\,\hat{B}] \;=\; \hat{A}\hat{B} - \hat{B}\hat{A}.$

The result is itself an operator. If $[\hat{A},\hat{B}] = 0$ the operators commute; if the commutator is non-zero they do not.

Why commutators matter

In quantum mechanics an observable is represented by a Hermitian operator. Two observables can be measured simultaneously with arbitrary precision if and only if their operators commute. When $[\hat{A},\hat{B}] \ne 0$ the Robertson uncertainty relation imposes a lower bound on the product of their standard deviations:

$\sigma_A\,\sigma_B \;\ge\; \frac{1}{2}\bigl|\langle[\hat{A},\hat{B}]\rangle\bigr|.$

The most famous example is position and momentum. In one dimension,

$[\hat{x},\,\hat{p}] = i\hbar,$

which leads directly to Heisenberg's uncertainty principle $\sigma_x\,\sigma_p \ge \hbar/2$.

Computing a commutator by hand

Take two $2\times 2$ matrices. The commutator is computed by ordinary matrix multiplication, then subtraction. Consider the Pauli operators

$X = \begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad Z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$

First multiply in the order $XZ$:

$XZ = \begin{pmatrix}0\cdot1+1\cdot0 & 0\cdot0+1\cdot(-1)\\1\cdot1+0\cdot0 & 1\cdot0+0\cdot(-1)\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}.$

Then multiply in the reverse order $ZX$:

$ZX = \begin{pmatrix}1\cdot0+0\cdot1 & 1\cdot1+0\cdot0\\0\cdot0+(-1)\cdot1 & 0\cdot1+(-1)\cdot0\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix}.$

Subtracting,

$[X,Z] = XZ - ZX = \begin{pmatrix}0-0 & -1-1\\ 1-(-1) & 0-0\end{pmatrix} = \begin{pmatrix}0&-2\\2&0\end{pmatrix}.$

This equals $-2iY$ where $Y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$ is the Pauli-$Y$ operator (you can verify: $-2i \cdot \begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}0&-2\\2&0\end{pmatrix}$). In general the three Pauli operators satisfy

$[X,Y]=2iZ,\qquad [Y,Z]=2iX,\qquad [Z,X]=2iY,$

a cyclic pattern that mirrors the cross-product relations for angular momentum. Since no two Pauli operators commute, no two of the corresponding spin components can be simultaneously sharp — a concrete physical consequence of non-commutativity.

Try it

This is a numerical exercise — your code should return a number. Using the matrices above, $[X,Z] = \begin{pmatrix}0&-2\\2&0\end{pmatrix}$. The $(1,0)$ entry (row 1, column 0) of this commutator is $2$. Return that entry divided by $2$.

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