When deriving the analytic representation of Fock states , why doesn't the $\hat{P}$ operator act on the ground state?

Question

Starting with \begin{align*} |n \rangle=\frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^{n}|0\rangle \end{align*} and \begin{equation} \psi_{0}(x)=\left(\frac{m\omega}{\pi\hbar}\right) ^{\frac{1}{4}}\mathrm{exp}\left(-\frac{m\omega x^{2}}{2\hbar}\right) \end{equation} for ladder functions defined as \begin{equation} \hat{a}=\sqrt{\frac{m\omega}{2\hbar}}\hat{X}+i\frac{1}{\sqrt{2m\hbar\omega}}\hat{P}\text{, } \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}\hat{X}-i\frac{1}{\sqrt{2m\hbar\omega}}\hat{P} \end{equation} I derive that the analytic wavefunction for the general number state $n$ is given by: \begin{align} \begin{split} \psi_{n}(x)&=\langle x|n\rangle\cr &=\frac{1}{\sqrt{n!}}\langle x|\left(\hat{a}^{\dagger}\right)^{n}|0\rangle\cr &=\frac{1}{\sqrt{n!}}\langle x|\left(\sqrt{\frac{m\omega}{2\hbar}}\hat{X}-i\frac{1}{\sqrt{2m\hbar\omega}}\hat{P}\right)^{n}|0\rangle\cr &=\frac{1}{\sqrt{n!}}\left(\sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right)^{n}\psi_{0}(x)\cr &=\frac{1}{\sqrt{n!}}\left(\frac{m\omega}{\pi\hbar}\right) ^{\frac{1}{4}}\left(\sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right)^{n}\mathrm{exp}\left(-\frac{m\omega x^{2}}{2\hbar}\right) \end{split} \end{align} I will call the final line of my working above expression (1). At this point, textbooks (for example lectures on quantum mechanics gordon baym) use the identity $$H_{n}(y)=\left( 2y -\frac{d}{dy} \right)^n\cdot 1$$ (which can be proven by induction, and is only true when we have the multiplication $\cdot1$) to remove the terms in the brackets and obtain $$ \psi_{n}(x)=\frac{1}{\sqrt{2^{n}n!}}\left(\frac{m\omega}{\pi\hbar}\right) ^{\frac{1}{4}}H_{n}(\sqrt{\frac{m\omega}{\hbar}})\mathrm{exp}\left(-\frac{m\omega x^{2}}{2\hbar}\right) $$ however this implies that the differentiation isn't acting on $\psi_{0}(x)$, which doesn't make sense, since the point is we're using the ladder operator to raise the state. any ideas?

Answer 1

You are overwhelmed by superfluous symbols which prevent you from seeing the obvious. First, non-dimensionalize junk constants by setting them equal to one, $\sqrt{m\omega/\hbar}=1$ as practicing physicists normally do.

Then, understand the n =1 case, \begin{align} \begin{split} \psi_{1}(x)&=\langle x|1\rangle\cr &= \left(\frac{ 1}{4\pi }\right) ^{\frac{1}{4}}\left( x- \frac{d}{dx}\right) \mathrm{exp}\left(-\frac{ x^{2}}{2 }\right) . \end{split} \tag{1} \end{align} and $$H_{1}(y)= 2y, ~~~~~~ H_0(y)=1. $$
It is then evident that $$ \psi_{1}(x)=\frac{1}{\sqrt{2}}\left(\frac{ 1}{\pi }\right) ^{\frac{1}{4}} \left (2x -x -\frac{d}{dx}\right ) \exp\left(-\frac{ x^{2}}{2 }\right)= {H_1(x)\over \sqrt 2 } \exp\left(-\frac{ x^{2}}{2 }\right), $$ where the last two terms in the parenthesis cancel each other.

Now proceed to handle the expression for generic n. (Hint: associatively, $e^{x^2}\left (-\frac{d}{dx}\right )= \left (2x-\frac{d}{dx}\right ) e^{x^2}$.)