📂Quantum Mechanics

Commutator of Momentum and Position

Formula

The commutator of position and momentum operator is given by the equation below.

$$\begin{align} [p, x] &= -\i \hbar \\ [x, p] &= \i \hbar \end{align}$$

This equation is termed the canonical commutation relation. The commutator of the square of position and momentum is as follows.

$$\begin{align} [x^{2}, p] &= 2 \i \hbar x \\ [p, x^{2}] &= -2 \i \hbar x \end{align}$$

Explanation

Since the momentum operator $p = \i\hbar \dfrac{d}{dx}$ is a differential operator, it commutes for different coordinates.

$$[x,p_{y}]=[x,p_{z}]=[y,p_{x}]=[y,p_{z}]=[z,p_{x}]=[z,p_{y}] = 0$$

This can be summarized as follows.

$$[x_{k}, p_{x_{\ell}}] = \i \hbar \delta_{k\ell} \\[1em] [p_{x_{k}}, x_{\ell}] = - \i \hbar \delta_{k\ell}$$

Here, $\delta_{k\ell}$ is the Kronecker delta.

Proof

Let $D_{x}$ be the differential operator.

$$D_{x} := \frac{d}{dx}$$

And, we denote the derivative $\dfrac{df}{dx}$ simply as follows.

$$f_{x} = \dfrac{df}{dx} = D_{x}f$$

$(1), (2)$

Since the momentum operator is $p=-\i \hbar \dfrac{d}{dx} = -\i \hbar D_{x}$, the following is obtained.

$$\begin{align*} [p, x] \psi &= px\psi - xp\psi \\ &= -\i \hbar D_{x} (x\psi) + \i \hbar x D_{x}\psi \\ &= -\i \hbar\psi - \i \hbar x \psi_{x} + \i \hbar x\psi_{x} \\ &= -\i \hbar \psi \end{align*}$$

Therefore,

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

Also, since $[x,p] = -[p, x]$,

$$[x,p] = -[p,x] = \i \hbar$$

■

$(3), (4)$

$$\begin{align*} [x^{2},p]\psi &= x^{2}p\psi-px^{2}\psi \\ &= x^{2}(-\i\hbar D_{x}\psi) + \i\hbar D_{x}(x^{2}\psi) \\ &= -\i\hbar x^{2}\psi_{x} + \i\hbar2x\psi + \i\hbar x^{2}\psi_{x} \\ &= 2\i\hbar x \psi \end{align*}$$

Therefore,

$$[x^{2},p] = 2 \i\hbar x$$

■

---

Due to the properties of commutators (4),

$$[ AB, C ] = A [ B, C ] + [ A, C ] B$$

It can be computed as follows due to the properties of commutators.

$$\begin{align*} [x^{2}, p] &= x[x,p] + [x,p]x \\ &= x \i \hbar +\i \hbar x \\ &= 2 \i \hbar x \end{align*}$$

■