Peeter Joot's (OLD) Blog.

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Derivative recurrence relation for Hermite polynomials

Posted by peeterjoot on January 3, 2012

[This post is included in my phy456 quantum mechanics notes available here.]

Motivation.

For a QM problem I had need of a recurrence relation for Hermite polynomials. I found it in [1], but thought I’d try to derive the relation myself.

Guts

The starting point I’ll use is the Rodgigues’ formula definition of the Hermite polynomials

\begin{aligned}H_n = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2}\end{aligned} \hspace{\stretch{1}}(2.1)

Let’s write D = d/dx, and take the derivative of H_n

\begin{aligned}(-1)^n D H_n &= D \left( e^{x^2} D^n e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} D^n \left( - 2 x e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^n \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2} \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^1 \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2} \\ &= \not{{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(-\not{{2 x D^{n} e^{-x^2} }}- 2 n D^{n-1} e^{-x^2} \right) \\ &= - 2 n e^{x^2} D^{n-1} e^{-x^2} .\end{aligned}

So we have the rather simple end result

\begin{aligned}\frac{d}{dx} H_n(x)=2 n H_{n-1}(x).\end{aligned} \hspace{\stretch{1}}(2.2)

References

[1] M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.

This entry was posted on January 3, 2012 at 11:18 pm and is filed under Math and Physics Learning.. Tagged: , , , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Derivative recurrence relation for Hermite polynomials”

  1. Mike said

    November 5, 2015 at 1:08 pm

    Very nice derivation! For completeness, I think you’re missing a factor of (-1)^n from the start of the main calculation.

    Reply

  2. Daniel Pires said

    April 11, 2020 at 9:52 am

    Very well explained !! For completeness, I think there’s a e^{x^2} missing in the last equality. (it’s suppose to be – e^{x^2} 2n D^{n-1} e^{-x^2}).

    Reply

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