# Is this Hermite polynomial identity known?

Clash Royale CLAN TAG#URR8PPP

In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $$H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$$ are Hermite polynomials and
$$r=sqrt{x^2+y^2+z^2}$$. Is this identity known?

In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $$H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$$ are Hermite polynomials and
$$r=sqrt{x^2+y^2+z^2}$$. Is this identity known?

In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $$H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$$ are Hermite polynomials and
$$r=sqrt{x^2+y^2+z^2}$$. Is this identity known?

In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $$H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$$ are Hermite polynomials and
$$r=sqrt{x^2+y^2+z^2}$$. Is this identity known?

mp.mathematical-physics orthogonal-polynomials

10.7k2569

10.7k2569

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If we define the generating functions $$F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$$ and $$G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $$H_n(x)H_n(y)$$.

• Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

• There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

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oldest

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If we define the generating functions $$F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$$ and $$G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $$H_n(x)H_n(y)$$.

• Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

• There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

If we define the generating functions $$F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$$ and $$G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $$H_n(x)H_n(y)$$.

• Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

• There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

If we define the generating functions $$F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$$ and $$G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $$H_n(x)H_n(y)$$.

If we define the generating functions $$F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$$ and $$G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $$H_n(x)H_n(y)$$.

Gjergji Zaimi

61.5k4161306

61.5k4161306

• Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

• There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

• Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

• There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

1

Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
Nov 29 at 7:48

1

There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
Nov 29 at 8:24

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