Is this Hermite polynomial identity known?

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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?

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    up vote
    4
    down vote

    favorite

    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?

    share|cite|improve this question

      up vote
      4
      down vote

      favorite

      up vote
      4
      down vote

      favorite

      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?

      share|cite|improve this question

      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

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      asked Nov 29 at 5:31

      Zurab Silagadze

      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 $H_n(x)H_n(y)$.

          share|cite|improve this answer

          • 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.
            – Zurab Silagadze
            Nov 29 at 7:48

          • 1

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

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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 $H_n(x)H_n(y)$.

          share|cite|improve this answer

          • 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.
            – Zurab Silagadze
            Nov 29 at 7:48

          • 1

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

          up vote
          12
          down vote

          accepted

          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 $H_n(x)H_n(y)$.

          share|cite|improve this answer

          • 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.
            – Zurab Silagadze
            Nov 29 at 7:48

          • 1

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

          up vote
          12
          down vote

          accepted

          up vote
          12
          down vote

          accepted

          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 $H_n(x)H_n(y)$.

          share|cite|improve this answer

          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 $H_n(x)H_n(y)$.

          share|cite|improve this answer

          share|cite|improve this answer

          share|cite|improve this answer

          answered Nov 29 at 7:03

          Gjergji Zaimi

          61.5k4161306

          61.5k4161306

          • 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.
            – Zurab Silagadze
            Nov 29 at 7:48

          • 1

            There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
            – Zurab Silagadze
            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.
            – Zurab Silagadze
            Nov 29 at 7:48

          • 1

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

          1

          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.
          – Zurab Silagadze
          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.
          – Zurab Silagadze
          Nov 29 at 7:48

          1

          1

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

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

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