Introduction: The Heart of Quantum Mechanics
The commutation relation \([\hat{x}, \hat{p}] = i\hbar\) is the defining equation of quantum mechanics. It encodes the wave-particle duality, the uncertainty principle, and the non-classical nature of the quantum world—all in one elegant formula.
Derivation
Compute \([\hat{x}, \hat{p}]\psi(x)\) using \(\hat{x} = x\) and \(\hat{p} = -i\hbar\frac{d}{dx}\):
\[[\hat{x}, \hat{p}]\psi = \hat{x}\hat{p}\psi - \hat{p}\hat{x}\psi\] \[= x\left(-i\hbar\frac{d\psi}{dx}\right) - \left(-i\hbar\frac{d}{dx}\right)(x\psi)\] \[= -i\hbar x\frac{d\psi}{dx} + i\hbar\left(\psi + x\frac{d\psi}{dx}\right)\] \[= i\hbar\psi\]Therefore: \([\hat{x}, \hat{p}] = i\hbar \hat{I}\)
The Canonical Commutation Relations
In three dimensions:
- \([\hat{x}_i, \hat{p}_j] = i\hbar\delta_{ij}\)
- \([\hat{x}_i, \hat{x}_j] = 0\)
- \([\hat{p}_i, \hat{p}_j] = 0\)
Position components commute with each other, momentum components commute with each other, but position and its conjugate momentum do not commute.
Worked Examples
Example 1: Alternative Representations
The commutation relation constrains but doesn't uniquely determine the operators. In momentum representation:
- \(\hat{p} = p\) (multiplication)
- \(\hat{x} = i\hbar\frac{d}{dp}\) (differentiation)
Check: \([\hat{x}, \hat{p}]\tilde{\psi}(p) = i\hbar\tilde{\psi}(p)\) ✓
Example 2: Powers of x and p
\[[\hat{x}, \hat{p}^2] = [\hat{x}, \hat{p}]\hat{p} + \hat{p}[\hat{x}, \hat{p}] = i\hbar\hat{p} + \hat{p}(i\hbar) = 2i\hbar\hat{p}\]More generally: \([\hat{x}, \hat{p}^n] = ni\hbar\hat{p}^{n-1}\)
Example 3: No Finite-Dimensional Representation
Taking the trace: \(\text{Tr}([\hat{x}, \hat{p}]) = \text{Tr}(i\hbar I) = i\hbar \cdot n\)
But \(\text{Tr}(AB - BA) = 0\) for finite matrices!
Contradiction → \(\hat{x}\) and \(\hat{p}\) must be infinite-dimensional.
The Quantum Connection
This commutation relation is the postulate that distinguishes quantum from classical mechanics. In classical physics, \(\{x, p\} = 1\) (Poisson bracket). Dirac's quantization rule replaces:
\[\{A, B\}_\text{classical} \to \frac{1}{i\hbar}[\hat{A}, \hat{B}]_\text{quantum}\]The non-zero commutator is why quantum mechanics is quantum.