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 $[x,p] = -[p, x]$,
$$ [x,p] = -[p,x] = \i \hbar $$
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$(3), (4)$
$$ \begin{align*} [x^{2},p]\psi &= x^{2}p\psi-px^{2}\psi \\ &= x^{2}(-\i\hbar D_{x}\psi) + \i\hbar D_{x}(x^{2}\psi) \\ &= -\i\hbar x^{2}\psi_{x} + \i\hbar2x\psi + \i\hbar x^{2}\psi_{x} \\ &= 2\i\hbar x \psi \end{align*} $$
Therefore,
$$ [x^{2},p] = 2 \i\hbar x $$
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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.
$$ \begin{align*} [x^{2}, p] &= x[x,p] + [x,p]x \\ &= x \i \hbar +\i \hbar x \\ &= 2 \i \hbar x \end{align*} $$
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