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At the beginning of Fetter, Walecka "Many body quantum mechanics" there is a statement, that every property of creation and annihilation operators comes from their commutation relation (I'm translating from my translation back to english, so it's not literal). I would like to know - how. Defining annihilation operator as: $$a\left|n\right\rangle=\sqrt n\left|n-1\right\rangle$$ I can derive the $[a,a^\dagger]=1$ equation. But in the opposite direction? I have no idea.

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    $\begingroup$ You need to use the commutation relation and the fact that the states $\lvert n\rangle$ are eigenvectors of the operator $a^\dagger a$. $\endgroup$ – Mark Mitchison Feb 21 '16 at 5:33
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Actually the commutation relation must be accompanied by the assumption of a vacuum state $|0\rangle$ for which $a|0\rangle = 0$ ($\langle 0| 0\rangle = 1$).

Then the commutation relation $aa^\dagger - a^\dagger a = 1$ implies also $$ a(a^\dagger)^2 - a^\dagger a a^\dagger = a^\dagger\\ a(a^\dagger)^2 - a^\dagger a a^\dagger = a(a^\dagger)^2 - (a^\dagger)^2 a - a^\dagger = a^\dagger\\ [a, (a^\dagger)^2] = 2a^\dagger $$ and from this one, again, $$ a(a^\dagger)^3 - (a^\dagger)^2 a a^\dagger = 2(a^\dagger)^2\\ a(a^\dagger)^3 - (a^\dagger)^2 a a^\dagger = a(a^\dagger)^2 - (a^\dagger)^3 a - (a^\dagger)^2 = 2(a^\dagger)^2\\ [a, (a^\dagger)^3] = 3(a^\dagger)^2 $$ and so on. By induction it follows that in general $$ [a, (a^\dagger)^n] = n(a^\dagger)^{n-1} $$ If we apply $[a, (a^\dagger)^n]$ to the vacuum state and assume $a|0\rangle = 0$, we obtain $$ [a, (a^\dagger)^n]|0\rangle = a (a^\dagger)^n|0\rangle = n (a^\dagger)^{n-1}|0\rangle $$ which is basically what we want. Indeed, let us denote $$ |n\rangle = A_n (a^\dagger)^n|0\rangle $$ where $A_n$ is a (real) normalization constant such that $\langle n| n\rangle = 1$. The above then means $$ a|n\rangle = n\frac{A_n}{A_{n-1}}|n-1\rangle $$ so all we have to do is figure out the $A_n$-s. But $$ 1 = \langle n| n\rangle = A_n^2 \langle 0|a^n (a^\dagger)^n |0\rangle = \\ = A_n^2 \langle 0| a^{n-1} \left[a (a^\dagger)^n\right] |0\rangle = A_n^2 \langle 0| a^{n-1} \left[(a^\dagger)^n a + n (a^\dagger)^{n-1}\right] |0\rangle = \\ = A_n^2 n \langle 0| a^{n-1} (a^\dagger)^{n-1} |0\rangle = \dots = A_n^2 n! \langle 0| 0\rangle = A_n^2 n! $$ and therefore $$ A_n = \frac{1}{\sqrt{n!}} \;\;\; \Rightarrow \;\;\; \frac{A_n}{A_{n-1}} = \frac{\sqrt{n!}}{\sqrt{(n-1)!}} =\sqrt{n} \;\;\; \Rightarrow a|n\rangle = \sqrt{n}|n-1\rangle $$ as expected. A similar argument, which you can do as an exercise, retrieves $a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$.

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