# Orthonormality of Hermite function

I was wondering if someone could tell me when the following relation holds? where $$H_{n}(x)$$ are Hermite polynomials and $$\delta(x-x')$$ is Dirac delta function: $$\sum_{n=0}^\infty \frac{1}{\sqrt{\pi}2^{n}n!} e^{-\frac{1}{2}x^{2}-\frac{1}{2}x'^{2}} H_{n}(x)H_{n}(x') = \delta(x-x') .$$

I asked this because I am trying to solve this:

https://physics.stackexchange.com/questions/455890/photon-number-representation-of-a-position-eigenstate

## migrated from physics.stackexchange.comJun 10 at 18:00

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• For all real $x$ and $x’$, I would assume. – G. Smith Jun 10 at 15:20
• how may i proceed to prove it? @G.Smith – Jason Jun 10 at 15:30
• – G. Smith Jun 10 at 15:36

Denote the left hand side of your identity by $$K(x,x’)$$. The identity $$K(x,x’)=\delta(x-x’)$$ is an identity between generalized functions. What it’s really saying is that $$\int_{-\infty}^{\infty} K(x,x’)f(x’)dx’ = f(x)$$ for any $$x$$ and any reasonably nice function $$f$$. In fact it’s enough to prove this for $$f(x)=e^{-x^2/2} H_m(x)$$ a Hermite function, since the Hermite functions are s complete orthogonal system in $$L_2(\mathbb{R})$$. For this $$f$$ the claim follows immediately from the orthogonality relation for the Hermite polynomials, since when we perform the integration for the $$n$$th term in the sum, we get 0 if $$n \neq m$$ and the correct thing for $$n=m$$.