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When assuming that the ground state is not degenerate one can show using $$ N=a^\dagger a \quad [a,a^\dagger]=1 $$ that there is exactly one eigenstate with norm 1 for every eigenvalue in $\textrm{Spec}(N)=\mathbb{N_0}$ given by $$|k\rangle=\frac{(a^\dagger)^k}{\sqrt{k!}}|0\rangle$$ But I'm confused because by applying k-times the annihilation operator on this $|k\rangle$ I can construct the normed state $$ \frac{a^k}{\sqrt{k!}}|k\rangle\overset{!}{=}|0\rangle$$ with eigenvalue 0. Because the ground state is non-degenerate this state has to be equal to $|0\rangle$. Combining both equations $$|k\rangle=\frac{(a^\dagger)^k}{\sqrt{k!}}\frac{a^k}{\sqrt{k!}}|k\rangle=\frac{N^k}{k!}|k\rangle=\frac{k^k}{k!}|k\rangle.$$ Clearly something is wrong here. Can someone point me to my mistake?

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Your mistake is by assuming $(a^{\dagger})^k a^k = N^k$, which is not true. Once you have $(a^{\dagger})^k a^k = (a^{\dagger})^{k-1} N a^{k-1}$ you need to "pull" $N$ to the right or to the left, and each movement through a ladder operator results in a numerical factor due to the commutation relations.

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    $\begingroup$ Thank you. One can find $[N,a^j]=-ja^j$ by induction and show that everything works out fine. $\endgroup$ – TheoreticalMinimum Apr 30 at 8:35
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The notation of the exponentiation is not to be taken by the letter. The $a^k$, as well as $a^{\dagger k}$, means that you have to apply $k$-times the operator. So

$$ \frac{(a^\dagger)^k}{\sqrt{k!}}\frac{a^k}{\sqrt{k!}}|k\rangle = \overbrace{\frac{a^\dagger a^\dagger\cdots a^\dagger}{\sqrt{k!}}}^{\text{k-times}}\overbrace{\frac{a a\cdots a}{\sqrt{k!}}}^{\text{k-times}}|k\rangle = \frac{a^\dagger a^\dagger\cdots a^\dagger Na a\cdots a}{k!}|k\rangle $$

and so on.

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