Lesson 270: Commutators: When Order Matters

Introduction: The Failure of Commutativity

Unlike ordinary numbers, operators don't always commute: \(\hat{A}\hat{B} \neq \hat{B}\hat{A}\) in general. The commutator measures this non-commutativity and lies at the heart of quantum mechanics.

Definition

The commutator of operators \(\hat{A}\) and \(\hat{B}\) is:

\[[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\]

If \([\hat{A}, \hat{B}] = 0\), the operators commute.

The anticommutator is: \(\{\hat{A}, \hat{B}\} = \hat{A}\hat{B} + \hat{B}\hat{A}\)

Key Properties

Worked Examples

Example 1: Pauli Matrices

\[[\sigma_x, \sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x = i\sigma_z - (-i\sigma_z) = 2i\sigma_z\]

The Pauli matrices satisfy: \([\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k\)

Example 2: Position and Momentum

In position representation, \(\hat{x} = x\) and \(\hat{p} = -i\hbar\frac{d}{dx}\):

\[[\hat{x}, \hat{p}]\psi = \hat{x}\hat{p}\psi - \hat{p}\hat{x}\psi = x(-i\hbar\psi') - (-i\hbar)(x\psi)'\] \[= -i\hbar x\psi' + i\hbar(\psi + x\psi') = i\hbar\psi\]

Therefore: \([\hat{x}, \hat{p}] = i\hbar\)

Example 3: Commuting Operators

\([\hat{x}, \hat{y}] = 0\) (position components commute)

\([\hat{p}_x, \hat{p}_y] = 0\) (momentum components commute)

The Quantum Connection

The commutator is the quantum analog of the Poisson bracket from classical mechanics: \([\hat{A}, \hat{B}] \leftrightarrow i\hbar\{A, B\}\). Non-zero commutators lead to the uncertainty principle:

\[\Delta A \cdot \Delta B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|\]

Position and momentum don't commute → they can't both be precisely known.