# Photon number representation of a position eigenstate

I wanted to calculate the photon number representation of a position eigenstate, so I developed as follows.

\begin{align} \vert x\rangle =\sum_{n}\vert n\rangle\langle n\vert x\rangle =\sum_{n}\psi_{n}(x)\vert n\rangle, \end{align} where \begin{align} \langle n\vert x\rangle=\psi_{n}(x)=\frac{1}{\sqrt{2^{n}n!}}\left(\frac{1}{\pi^{1/4}}\right)e^{-\frac{1}{2}x^{2}}H_{n}(x) \end{align} with a Hermite polynomial $$H_{n}(x)$$. From the normalization condition, i.e., $$\langle x\vert x\rangle=1$$, I would expect to see \begin{align} \sum_{n}\vert \psi_{n}(x)\vert^{2}=1, \end{align} but I failed.

Q. Did I make any mistake above? or am I completely wrong?

• The expression $\langle x|x\rangle$ cannot be equal to 1. $|x\rangle$ is not normalizable a la Born. Instead, it is more customary to assume $\langle x|x'\rangle=\delta(x-x')$. – Ján Lalinský Jan 22 at 16:34
• I don't understand this question. Photons do not have position eigenstates - they are relativistic objects without a good position operator. What exactly is the set up here that you think you have a photonic position state, and why are the "photon number" states Hermite polynomials? – ACuriousMind Jan 22 at 17:22
• @JánLalinský oh, yes. You’re right. By the way, is the above expansion valid? – Veteran Jan 22 at 18:30
• @ACuriousMind Sorry that The question was not so clear. I was asking more about things that come out of the canonical quantization, in which there are an excitation number state (corresponding to photons in optics) and a canonical quadrature operator x. I meant the projection of the eigen state of the quadrature operator x to the excitation number state. I hope this is now clear. – Veteran Jan 22 at 18:34
• @JánLalinský : “position eigenstate” is a perfectly standard terminology in quantum optics, even if the “position” is not the position of anything real, but the position of the particle in the harmonical oscillator canonically equivalent to the mode of the electromagnetic field currently considered. – Frédéric Grosshans Jan 23 at 11:45

\begin{align} \vert x\rangle =\sum_{n}\vert n\rangle\langle n\vert x\rangle =\sum_{n}\psi_{n}(x)\vert n\rangle, \end{align} where \begin{align} \langle n\vert x\rangle=\psi_{n}(x)=\frac{1}{\sqrt{2^{n}n!}}\left(\frac{1}{\pi^{1/4}}\right)e^{-\frac{1}{2}x^{2}}H_{n}(x) \end{align} with a Hermite polynomial $$H_{n}(x)$$
However, the position eigenstate $$|x\rangle$$ is not normalizable, so attempting to set $$\langle x| x \rangle = 1$$ will only end in disaster. The normalization for position states is instead normally chosen via a delta-function normalization $$\langle x| x' \rangle = \delta(x-x')$$ (which seems to be the normalization that's required for your result to be true, if I remember the formulas correctly). As such, what you expect is to have something like $$\sum_{n=0}^\infty \frac{1}{\sqrt{\pi}2^{n}n!} e^{-\frac{1}{2}x^{2}-\frac{1}{2}x'^{2}} H_{n}(x)H_{n}(x') = \delta(x-x') ,$$ where the product of orthogonal polynomials inside the sum is normally called the Christoffel-Darboux kernel for the polynomial family.