Hermite function ($H_n(x)=\frac{e^{-x^2/2}}{{\pi }^{1/4}\sqrt{2^{n}n!}}(-1{)}^n{e}^{x^2}\frac{d^n}{d{x}^n}{e}^{-x^2}$)

The normalized Hermite functions multiply the physicists' Hermite polynomials by a Gaussian. They form an orthonormal basis of $L^2(\mathbb{R})$ and satisfy $(-d^2/d{x}^2+x^2)H_n=(2n+1)H_n$. With $a=(x+d/dx)/\sqrt{2}$ and $a^{\ast }=(x-d/dx)/\sqrt{2}$, they are $(a^{\ast }{)}^n{H}_0/\sqrt{n!}$ and the quantum harmonic oscillator is $2a^{\ast }a+1$. For completeness, if $u$ is orthogonal to all of them, all moments of $u(x)e^{-x^2/2}$ vanish. Its transform $\int u(x)e^{-x^2/2}{e}^{zx}dx$ is entire by Cauchy-Schwarz inequality, has every derivative zero at zero, and hence vanishes identically. Uniqueness of the Fourier transform gives $u=0$.

Use the unitary Fourier transform with phase $e^{-i\lambda x}$. It commutes with $-d^2/d{x}^2+x^2$, because differentiation transforms to multiplication by $i\lambda$ and multiplication by $x$ transforms to $id/d\lambda$. The Hermite differential equation has a unique polynomial solution of each degree up to scale, and multiplying by the Gaussian gives a Hermite function. Transforming a polynomial times this Gaussian again gives a polynomial times the Gaussian of the same degree; comparison of leading coefficients gives the eigenvalue $(-i{)}^n$.

For $|t|<1$, this identity sums the Hermite function expansion into a Gaussian. One proof takes the Bargmann-Fock space inner product of the generating functions ${\pi }^{-1/4}\exp (-x^2/2+\sqrt{2}xz-z^2/2)$ at $\sqrt{t}z$. Writing $z=u+iv$, the two resulting real Gaussian integrals have quadratic coefficients $1+t$ and $1-t$ and linear coefficients $\sqrt{2t}(x+y)$ and $i\sqrt{2t}(x-y)$. Evaluation gives the formula. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ gives the harmonic oscillator transition kernel of $e^{-s(-d^2/d{x}^2+x^2)}$.