We know that the rule for creating excited states for a Quantum Harmonic Oscillator is $|n\rangle=\frac{(a^\dagger)^n(|0\rangle)}{\sqrt{n!}}$. I wanted to derive from this the familiar rule in terms of wavefunctions, $\psi_{n}=\frac{(a^\dagger)^n(\psi_{0})}{\sqrt{n!}}$. I tried applying $\langle x|$ on both sides of the former equation, which gives $\langle x|n\rangle=\frac{\langle x|(a^\dagger)^n|0\rangle}{\sqrt{n!}}$. The LHS gives me $\psi_{n}$, but how to deal with the RHS? We know that $\langle x|(a^\dagger)^n=a^n\langle x|$, so this means that $\psi_{n}=\frac{(a)^n(\psi_{0})}{\sqrt{n!}}$ which is not correct. Could someone explain what is wrong with this method, and how can we perform this derivation correctly?
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2I believe you are having essentially the same issue as this question, although dressed up a bit differently – By Symmetry Nov 29 '23 at 15:55
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2Does this answer your question? On the two different solution approaches of the quantum harmonic oscillator – Tobias Fünke Nov 29 '23 at 15:59