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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!

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    $\begingroup$ To which older version are you referring? $\endgroup$
    – bmf
    Commented Mar 20, 2022 at 6:17
  • $\begingroup$ 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 $$ \frac{2^{n/2} e^{\left(\frac{1}{2}-i\right) n+\left(-1-\frac{i}{2}\right)} n^{-\frac{n}{2}-\frac{1}{2}} \cos \left(\frac{\pi n}{2}\right)}{\sqrt{\pi }}.$$ $\endgroup$
    – user64494
    Commented Mar 20, 2022 at 8:00
  • $\begingroup$ I am not sure, but I think it was Mathematica version 6 in 2007, where the equation gives a solution. $\endgroup$
    – Klaus
    Commented Mar 20, 2022 at 9:06
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    $\begingroup$ 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). $\endgroup$
    – Mariusz Iwaniuk
    Commented Mar 20, 2022 at 10:49
  • $\begingroup$ Yes, this was the result I got with the sum. $\endgroup$
    – Klaus
    Commented Mar 20, 2022 at 12:11

1 Answer 1

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By factoring the exponential and pulling the factors not dependent on the running index n out of the sum Mathematica still calculates

Exp[-(1/2) (x^2) - \[Lambda]^2 - I (1/2) t] 
Sum[HermiteH[n, x](\[Lambda] Exp[-I t])^n/n!,{n,0,Infinity}]
(* E^(-((I*t)/2)-x^2/2+(2*x*\[Lambda])/E^(I*t)-
\[Lambda]^2 - \[Lambda]^2/E^(2*I*t)) *)
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