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$\def\textdagger{\dagger}\def\ket#1{\left|#1\right\rangle}$I'm reading through the Schwartz text on QFT & the Standard Model. I've found myself bogged down already in section 1.3! In a derivation of the spontaneous emission coefficient, he notes:

$$\hat{N}a^\textdagger \ket{n} = a^\textdagger a a^\textdagger \ket{n} = a^\textdagger \ket{n} + a^\textdagger a^\textdagger a \ket{n}=(n+1)a^\textdagger\ket{n}$$

where he previously notes that $a$, the annihilation operator, and $a^\textdagger$, the creation operator, have commutator 1:

$$[a, a^\textdagger]=1$$

I understand these lines to mean that the number of modes in a quantum state goes up by one after the creation operator is applied to that state. Makes sense. But I'm getting caught up on that first line.

Now, I'm no physics guy, but I am a little bit of a math guy, and if I'm not horribly mistaken... when the commutator of two operators is equal to the group identity, doesn't that normally just mean they commute? If that's the case, wouldn't it simply be:

$$a^\textdagger a a^\textdagger \ket{n} = a^\textdagger a^\textdagger a \ket{n}$$

Where does the additional term come from, which is clearly very physically important?

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  • $\begingroup$ with due respect: you should start with elementary quantum mechanics and gain experience with operators before launching into Schwartz and QFT. $\endgroup$ Sep 16 at 12:58
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Simply note that $[a,a^\dagger]=1 \implies aa^\dagger = 1+a^\dagger a$.

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  • $\begingroup$ Does the commutator in a Hilbert space have the ring-like definition $[a, b]=ab-ba$ or the group-like definition $[a,b] = a^{-1}b^{-1}ab$? I though Hilbert spaces only had the one binary operation (vector addition) formally defined? Of course there's the inner product, but that results in a scalar, not an element of the Hilbert space... am I way off the mark here? $\endgroup$
    – neph
    Sep 16 at 4:51
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    $\begingroup$ I thought about it some more & decided a) I'm really stupid & b) it's been too long since my last algebra class. Thanks for clarifying! Of course an inner product space has a ring-like commutator. It's a lot harder to learn this stuff out of school $\endgroup$
    – neph
    Sep 16 at 4:55
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    $\begingroup$ @neph No worries, we all have those days. For clarity, $a$ and $a^\dagger$ are (unbounded) operators on the Hilbert space $\mathcal H$, so the commutator between them is (a) ring-like as you describe, and (b) defined on the domain $\mathrm{dom}(aa^\dagger) \cap \mathrm{dom}(a^\dagger a)\subsetneq \mathcal H$. $\endgroup$
    – J. Murray
    Sep 16 at 12:07

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