Commutator
- Quantum Mechanics 1.4

09 Feb 2023 in Physics on Quantum Mechanics · 2 min

from 「Modern Quantum Mechanics」: Sakurai, J. J.

Commutators not only imply the non-commutativity of operators. You should look further.

• Commutator
  • Properties
• Anticommutator

Commutator

For operators $\hat{A}$ and $\hat{B}$ acting on wavefunctions, the following operator is called the commutator of $\hat{A}$ and $\hat{B}$.

\[ \comm{\hat{A}}{\hat{B}} \coloneqq \hat{A}\hat{B}-\hat{B}\hat{A} \]

If $\comm{\hat{A}}{\hat{B}}=0$, we say that, $\hat{A}$ and $\hat{B}$ commute. In other words, there is no difference between the result of acting $\hat{A}$ after acting $\hat{B}$, and the result of acting $\hat{B}$ after acting $\hat{A}$.

Properties

1. Bilinearity
   • $ \comm{\hat{A}+\hat{B}}{\hat{C}} = \comm{\hat{A}}{\hat{C}}+\comm{\hat{B}}{\hat{C}} $
   • $ \comm{\hat{A}}{\hat{B}+\hat{C}} = \comm{\hat{A}}{\hat{B}}+\comm{\hat{A}}{\hat{C}} $
2. Alternativity
   • $ \comm{\hat{A}}{\hat{B}} = -\comm{\hat{B}}{\hat{A}} $
3. Jacobi identity
   • $ \comm{\hat{A}}{\comm{\hat{B}}{\hat{C}}} + \comm{\hat{B}}{\comm{\hat{C}}{\hat{A}}} + \comm{\hat{C}}{\comm{\hat{A}}{\hat{B}}} = 0 $
4. Additional Property
   • $ \comm{\hat{A}}{\hat{B}\hat{C}} = \comm{\hat{A}}{\hat{B}}\hat{C}+\hat{B}\comm{\hat{A}}{\hat{C}} $
   • $ \comm{\hat{A}\hat{B}}{\hat{C}} = \hat{A}\comm{\hat{B}}{\hat{C}}+\comm{\hat{A}}{\hat{C}}\hat{B} $

Read Commutator (ring theory) for further details.

Also, if a common eigenfunction of any two different operators exists, they commute.

\[ \begin{cases} A\psi=a\psi \nl B\psi=b\psi \end{cases} \Rightarrow \comm{A}{B}=0 \]

It’s very easy to show.

Anticommutator

We can also define anticommutators:

• $ \acomm{\hat{A}}{\hat{B}} \coloneqq \hat{A}\hat{B}+\hat{B}\hat{A} $