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I am struggling to understand how the following is true for the fermionic creation/annihilation operators $a^\dagger, a$: $$[a^\dagger a, a]=-a$$

If someone could walk me through the math derivation of this, I would greatly appreciate it!

I see that $[a^\dagger a, a]= a^\dagger aa - a a^\dagger a = (1-a a^\dagger)a - a a ^\dagger a = a - 2 a a^\dagger a$ but I'm not sure how to simplify this further to get $-a$ at the end.

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Your first step was already good. And then you can exploit $aa=0$.

$$\begin{align} [a^\dagger a, a] &=a^\dagger \underbrace{aa}_{=0}-aa^\dagger a \\ &=-aa^\dagger a \\ &=-a(1-aa^\dagger) \\ &=-a+\underbrace{aa}_{=0}a^\dagger \\ &=-a \end{align}$$

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  • $\begingroup$ $aa=0$ is only true for fermions because the max occupation number in a state is 1, right? $\endgroup$
    – photonica
    Commented Dec 20, 2023 at 2:19
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    $\begingroup$ @photonica yes, it is only for fermions. $a^\dagger a^\dagger=0$ is because max occupation number is 1, and $aa=0$ is because min occupation number is 0. $\endgroup$
    – Thomas Fritsch
    Commented Dec 20, 2023 at 2:26
  • $\begingroup$ @photonica This follows from the anti-commutation relations. $\endgroup$
    – Tobias Fünke
    Commented Dec 20, 2023 at 7:38

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