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Given that $a$ and $a^\dagger$ are bosonic annihilation and creation operators (in the language of second quantisation); are there any simpler ways to calculate the commutators of arbitrary products of these operators, without laboriously using the key bosonic identity $[a,a^\dagger]=1$ over and over again. For example, given the commutator

$$[a^\dagger a a^\dagger a^\dagger, a^\dagger a a^\dagger a],$$

are there any handy tricks for just reading off the result, or making the computations quick?

Qmechanic
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Tom
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1 Answers1

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Sure, there are exactly 137 tricks and 29.5 relevant questions on this site, and the linked questions to them.

For "mixed" expressions like yours, the very first thing you do is isolate the number operators $N=a^\dagger a$, for which $$ [N,a^{\dagger ~n}]=n a^{\dagger ~n}, \qquad [N,a^{n}]=-n a^{ n}, $$ with N commuting with itself, of course.

So you reduce your expression to commutators of powers of creators with powers of annihilators, for which @Qmechanic 's link steers you to elegant expressions.

For example, your expression is $$ [N a^{\dagger ~ 2}, N^2]= N\{ [a^{\dagger ~ 2},N],N\}= 4N(1-N)a^{\dagger ~ 2}, $$ etc. You should be able to hack many of them by inspection.

Also take a look at this.

Cosmas Zachos
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