# Projecting energy eigenstates of quantum harmonic oscillator into the coordinate basis

I am trying formally derive the projection of the energy eiegenstates of the 1D quantum harmonic oscillator into the $x$ basis $$\phi_n(x) = \langle x | n \rangle = \langle x | \frac{{a^{\dagger}}^{n}}{\sqrt{n!}} | 0 \rangle = \frac{{a^{\dagger}}^{n}}{\sqrt{n!}} \, \phi_0(x)$$

I'm not sure how to make the last step

$$\langle x |\frac{{a^{\dagger}}^{n}}{\sqrt{n!}} | 0 \rangle = \frac{{a^{\dagger}}^{n}(x)}{\sqrt{n!}} \langle x | 0 \rangle$$

where I use ${a^{\dagger}}^{n}(x)$ to clarify that the operator is in the $x$ basis.

I feel I need to use the resolution of identity

$$\int \langle x | \frac{{a^{\dagger}}^{n}}{\sqrt{n!}}| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x'$$

I can do this integral for a single creation operator

$$\int \langle x | a^{\dagger}| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x' = \int \langle x | \gamma X - i \epsilon P| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x'$$

where $\gamma = \sqrt{\frac{m \omega}{2 \hbar}}$ and $\epsilon = \sqrt{\frac{1}{2m \omega \hbar}}$. Using the matrix representations of the $X$ and $P$ operators in the $x$ basis

$$\langle x |\gamma X - i \epsilon P| x' \rangle = x' \gamma \delta(x - x') + \hbar \epsilon\delta^{(1)}(x - x')$$

where $\delta^{(1)}(x)$ is the first derivative of the Dirac delta function. So the integral is

$$\phi_1(x) = \int \left[ x' \gamma \delta(x - x') + \hbar \epsilon\delta^{(1)}(x - x') \right] \phi_0 (x') \: \mathrm{d} x' = \left[ \gamma x - \hbar \epsilon\frac{\mathrm{d}}{\mathrm{d} x} \right]\phi_0 (x)$$

So how do I continue with this derivation? Do I make some sort of inductive step? Is this even the right path to go down, is there a much simpler method?

• You've already done it! What do you even want help with? – Prahar Apr 20 '17 at 13:15
• To check the recursive generation of Hermite polynomials, consult WP. – Cosmas Zachos Apr 20 '17 at 14:35

I have not checked the details and this is too long for a comment but it looks right and yes it is induction. Instead of starting with $\phi_n(x)=\langle x\vert \frac{(\hat a^\dagger)^n}{\sqrt{n!}}\vert 0\rangle$, start with \begin{align} \phi_{n+1}(x)&=\frac{1}{\sqrt{n+1}}\langle x\vert \hat a^\dagger\vert n\rangle\, ,\\ &= \frac{1}{\sqrt{n+1}}\left[\gamma x-\hbar \epsilon \frac{d}{dx}\right] \phi_n(x) \tag{1} \end{align} where you obtain (1) by repeating your steps starting with $\vert n\rangle$ rather than $\vert 0\rangle$. You can then start the induction.