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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?

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    $\begingroup$ What about a product basis: H[n,x] H[m,y] $\endgroup$
    – Daniel Huber
    Jun 25 at 17:59
  • $\begingroup$ @DanielHuber does it provide a complete orthogonal basis? If yes, I think it should do. $\endgroup$
    – Enrico Detoma
    Jun 25 at 18:33
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    $\begingroup$ As Hermite polynomials are a orthogonal basis on R, so is the product H[n,x] H[m,y] on R^2. $\endgroup$
    – Daniel Huber
    Jun 25 at 18:44
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    $\begingroup$ The products are separable so yes, they will be orthogonal. $\endgroup$
    – Daniel Lichtblau
    Jun 25 at 18:45
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    $\begingroup$ "bivariate Hermite polynomials" - do you have a literature reference for how these are defined? $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jun 25 at 20:10

1 Answer 1

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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: $$ \frac1\pi\frac{1}{m!n!}\int_{\mathbb{R}^2}H_{m,n}(x+iy,x-iy)H^*_{p,q}(x+iy,x-iy)e^{-x^2-y^2}dxdy=\delta_{m,p}\delta_{n,q} $$

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    $\begingroup$ 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}] $\endgroup$
    – Bob Hanlon
    Jun 25 at 23:25
  • $\begingroup$ 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). $\endgroup$
    – Enrico Detoma
    Jun 26 at 6:21
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    $\begingroup$ @EnricoDetoma I extended my answer to provide source of information on the orthogonality relation. $\endgroup$
    – yarchik
    Jun 26 at 7:47
  • $\begingroup$ @yarchik Thank you very much for the complete reference! $\endgroup$
    – Enrico Detoma
    Jun 26 at 9:22

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