# How to calculate the sum of the series of Hermite polynomial?

I want to calculate the infinite sum of Hermite polynomials, which works fine with older versions of Mathematica, but with version 13 it doesn't. The infinite sum is:

Sum[HermiteH[n, x] λ^n/ n! Exp[-(1/2) (x^2) - λ^2 - I ( n + 1/2) t], {n,0,Infinity},Assumptions -> Element[{λ,x,t}, Reals]]


Can someone help? Thank you very much!

• To which older version are you referring?
– bmf
Commented Mar 20, 2022 at 6:17
• That series may diverge: DiscreteAsymptotic[ HermiteH[n, x] \[Lambda]^n/ n! Exp[-(1/2) (x^2) - \[Lambda]^2 - I (n + 1/2) t] /. {x -> 0, t -> 1, \[Lambda] -> 1}, {n, Infinity, 1}] results in $$\frac{2^{n/2} e^{\left(\frac{1}{2}-i\right) n+\left(-1-\frac{i}{2}\right)} n^{-\frac{n}{2}-\frac{1}{2}} \cos \left(\frac{\pi n}{2}\right)}{\sqrt{\pi }}.$$ Commented Mar 20, 2022 at 8:00
• I am not sure, but I think it was Mathematica version 6 in 2007, where the equation gives a solution. Commented Mar 20, 2022 at 9:06
• Looks like answer is: E^((-1/2*I)*t - x^2/2 + (2*x*\[Lambda])/E^(I*t) + (-1 - E^((-2*I)*t))*\[Lambda]^2). Commented Mar 20, 2022 at 10:49
• Yes, this was the result I got with the sum. Commented Mar 20, 2022 at 12:11

Exp[-(1/2) (x^2) - \[Lambda]^2 - I (1/2) t]