Bivariate Hermite Polynomials

Question

I know I can get the Hermite polynomials in a single variable with: HermiteH[n, x]

Now I need the bivariate Hermite polynomials. I thought about building them with this procedure:

basis = Flatten[TensorProduct[x^Range[0, n], y^Range[0, n]]]
orthbasis = Orthogonalize[basis, Integrate[Exp[-x^2 - y^2]*#1*#2, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] &]

Is this procedure correct? Or is there a better way with internal Mathematica functions like HermiteH, without involving Gram-Schmidt?

@DanielHuber does it provide a complete orthogonal basis? If yes, I think it should do. — Enrico Detoma, Jun 25 '22 at 18:33
As Hermite polynomials are a orthogonal basis on R, so is the product H[n,x] H[m,y] on R^2. — Daniel Huber, Jun 25 '22 at 18:44
"bivariate Hermite polynomials" - do you have a literature reference for how these are defined? — J. M.'s missing motivation, Jun 25 '22 at 20:10

yarchik · Accepted Answer · 2022-06-26T07:53:50.843

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Normal Hermite polynomial can be derived from the generating function $$ \sum_{n=0}^\infty \frac{t^n}{n!}H_n(x)=e^{2tx-t^2}. $$ Analogically, bivariate Hermite polynomial are obtained from$$ \sum_{m=0}^\infty\sum_{n=0}^\infty \frac{s^m}{m!}\frac{t^n}{n!}H_{m,n}(x,y) =e^{sx+ty-s t}. $$ Thus, in order to obtain $H_{m,n}(x,y)$ just do series expansion

H[m_, n_] := 
 SeriesCoefficient[
  SeriesCoefficient[Exp[s x + t y - s t], {s, 0, m}], {t, 0, n}]
H[3, 4]
(*1/144 (-24 y + 36 x y^2 - 12 x^2 y^3 + x^3 y^4)*)

Notice that $H_{m,n}(x,y)$ is not a direct product of two independent single-variable Hermite polynomials.

Bivariate Hermite polynomials also satisfy orthogonality relation, which can be found in Ismail, M., 2016. Analytic properties of complex Hermite polynomials. Transactions of the American Mathematical Society, 368(2), pp.1189-1210: $$ \frac1\pi\frac{1}{m!n!}\int_{\mathbb{R}^2}H_{m,n}(x+iy,x-iy)H^*_{p,q}(x+iy,x-iy)e^{-x^2-y^2}dxdy=\delta_{m,p}\delta_{n,q} $$

edited Jun 26 '22 at 07:53

answered Jun 25 '22 at 21:41

yarchik

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More completely, H[m_Integer?NonNegative, n_Integer?NonNegative][x_, y_] := SeriesCoefficient[Exp[s x + t y - s t], {s, 0, m}, {t, 0, n}] – Bob Hanlon Jun 25 '22 at 23:25
One remark: it seems to me that only some of those polynomials are orthogonal (I expected them to be all orthogonal). E.g. Integrate[H[4, 0]*H[1, 1] Exp[-x^2 - y^2], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] gives -(Pi/32). – Enrico Detoma Jun 26 '22 at 06:21
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@EnricoDetoma I extended my answer to provide source of information on the orthogonality relation. – yarchik Jun 26 '22 at 07:47
@yarchik Thank you very much for the complete reference! – Enrico Detoma Jun 26 '22 at 09:22

Bivariate Hermite Polynomials

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