# Misunderstanding with ladder operators: $(a^\dagger)^k a^k\overset{!}{=}N^k$

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?

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.
• Thank you. One can find $[N,a^j]=-ja^j$ by induction and show that everything works out fine. – TheoreticalMinimum Apr 30 at 8:35
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
$$\frac{(a^\dagger)^k}{\sqrt{k!}}\frac{a^k}{\sqrt{k!}}|k\rangle = \overbrace{\frac{a^\dagger a^\dagger\cdots a^\dagger}{\sqrt{k!}}}^{\text{k-times}}\overbrace{\frac{a a\cdots a}{\sqrt{k!}}}^{\text{k-times}}|k\rangle = \frac{a^\dagger a^\dagger\cdots a^\dagger Na a\cdots a}{k!}|k\rangle$$