Mehler kernel

Remarks

Proof idea: insert the integral formula $H_n(x) = \frac{2^n}{\sqrt{\pi}} \int_{\R} (x + it)^n e^{-t^2} \, dt$ for both factors. The sum over $n$ becomes an exponential, and the remaining double integral is Gaussian and can be computed in closed form. Check at $x = y = 0$: $H_0(0) = 1$, $H_2(0) = -2$ and $H_4(0) = 12$ give $1 + \frac{\rho^2}{2} + \frac{3\rho^4}{8} + \cdots$, which is the expansion of $(1 - \rho^2)^{-1/2}$. In terms of the Hermite functions $\psi_n$ the formula reads $$\sum_{n=0}^{\infty} \rho^n \psi_n(x) \psi_n(y) = \frac{1}{\sqrt{\pi (1 - \rho^2)}} \exp\left( -\frac{(1 + \rho^2)(x^2 + y^2) - 4\rho x y}{2 (1 - \rho^2)} \right).$$ So the Mehler kernel is the integral kernel of the operator that multiplies the $n$-th Hermite function by $\rho^n$. The hypothesis $\abs{\rho} < 1$ is needed: at $\rho = 1$ the right side is undefined and the series does not converge to a function.

Going deeper

For a student

Write $N$ for the operator with $N \psi_n = n \psi_n$, the number operator of the harmonic oscillator. Then the Mehler kernel with $\rho = e^{-t}$, $t > 0$, is the kernel of $e^{-tN}$, a smoothing semigroup on $L^2(\R)$. In probability, the probabilists' version of the same formula gives the transition density of the Ornstein–Uhlenbeck process. Letting $\rho$ tend to a point $e^{-i\alpha}$ of the unit circle leads to the kernel of the fractional Fourier transform. The formula is a model for bilinear generating functions of other orthogonal polynomials, such as the Hille–Hardy formula for Laguerre polynomials.