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

Starting with \begin{align*} |n \rangle=\frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^{n}|0\rangle \end{align*} and $$$$\psi_{0}(x)=\left(\frac{m\omega}{\pi\hbar}\right) ^{\frac{1}{4}}\mathrm{exp}\left(-\frac{m\omega x^{2}}{2\hbar}\right)$$$$ for ladder functions defined as $$$$\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}\hat{X}+i\frac{1}{\sqrt{2m\hbar\omega}}\hat{P}\text{, } \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}\hat{X}-i\frac{1}{\sqrt{2m\hbar\omega}}\hat{P}$$$$ I derive that the analytic wavefunction for the general number state $$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?

• What do you mean with the differentiation isn't acting on $\psi_0(x)$? Can you elaborate? Does this help? Commented Oct 5, 2022 at 17:50
• So you see the expression which has $\left(\sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right)^{n}\mathrm{exp}\left(-\frac{m\omega x^{2}}{2\hbar}\right)$. The $\frac{d}{dx}$ in the brackets should act on the exponential outside the brackets. However the substitution assumes that the the operator acts on $1$ outside the brackets. Commented Oct 5, 2022 at 17:53
• Sorry, I really don't understand. What expression 'has' the formula you mention above? You should be as clear as possible and edit the question accordingly.... Perhaps it is just me not understanding the question, tho. Commented Oct 5, 2022 at 17:55
• sorry, I meant expression (1), is that clearer now? Commented Oct 5, 2022 at 18:04
• Your formulas are fine, but you are lost in superfluous symbols so you are misreading them! Try n=1. Commented Oct 5, 2022 at 18:20

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, $$e^{x^2}\left (-\frac{d}{dx}\right )= \left (2x-\frac{d}{dx}\right ) e^{x^2}$$.)