Canonical Commutation Relation
- Quantum Mechanics 1.10
· 6 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
The position and momentum operators.
Translation operator
$ \global\def\T{\hat{\mathscr{T}}} $
Let’s define the spatial displacement, or translation operator $\hat{\mathscr{T}}(\Delta\b{x})$ as follows:
\[ \T\ket{\b{x}} \coloneqq \ket{\b{x}+\Delta\b{x}} \]
Then, the translation operator $\T$ is unitary:
\[ \brktop{\b{x}}{ \T^\dagger\T }{\b{x}} = \brkt{\b{x}}{\b{x}} \]
\[ \therefore \T^\dagger\T = \mathbb{I} \]
Also, infinitesimal translation operator should be additional:
\[ \T(\d\b{x}_2) \T(\d\b{x}_1) = \T(\d\b{x}_1+\d\b{x}_2) \]
Similarly, we expect the opposite direction of translation operator is the inverse of the original one:
\[ \T(-\d\b{x}) = \T^{-1}(\d\b{x}) \]
Finally, we demand that as $\d\b{x}\to 0$, $\T(\d\b{x})\to \mathbb{I}$.
\[ \lim_{\d\b{x}\to 0} \T(\d\b{x}) = \mathbb{I} \]
These properties are satisfied if the difference between $\T(\d\b{x})$ and $\mathbb{I}$ be of the first order in \d\b{x}. Therefore, we can demonstrate the infinitesimal translation operator in such form,
\[ \T(\d\b{x}) = \mathbb{I} - i\b{K}\cdot\d\b{x} \]
where the components of $\b{K}$ are Hermitian operators. The formula shows that $\b{K}$ is the generator of translation. By a simple calculation, we can show the following commutation relation:
\[ \comm{ \b{x} }{ \T(\d\b{x}) } = \d\b{x} \]
Therefore, we obtain the commutation relation between $\b{x}$ and $\b{K}$.
\[ \comm{ x_i }{ K_j } = i\delta_{ij} \]
Momentum operator
$ \global\def\p{\hat{\b{p}}} $
The momentum operator $\p$ is defined as the generator of translation, or equivalently as the gradient of the translation operator:
\[ \p \coloneqq \hbar \left( \nabla\T(\b{x}) \middle)\right\vert_{\b{x}=\b{0}} \]
We actually get familiar form when we expand the gradient:
\[ \p = -i\hbar\nabla \]
The definition of momentum operator then results some commutation relations trivially,
- $ \comm{p_i}{p_j} = 0 $
- $ \comm{x_i}{p_j} = i\hbar\delta_{ij} $
Also, by the fact that the momentum operator is a generator of translation, we get several properties. First, the momentum operator is Hermitian since the translation operator is unitary. Second, arbitrary translation operator can be expressed as follows:
\[ \begin{align*} \T(\Delta\b{x}) &= \lim_{N\to\infty} \left[ 1-\frac{ i\b{p}\cdot\Delta\b{x} }{ N\hbar } \right]^N \nl &= \exp\left( -\frac{ i\b{p}\cdot\Delta\b{x} }{ \hbar } \right) \end{align*} \]
Canonical commutation relation
We summarize the commutation relations we have obtained so far:
\[ \begin{align*} \comm{x_i}{x_j} &= 0 \nl \comm{p_i}{p_j} &= 0 \nl \comm{x_i}{p_j} &= i\hbar\delta_{ij} \end{align*} \]
