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## Derivative recurrence relation for Hermite polynomials

Posted by peeterjoot on January 3, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# 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 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.