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The most common definition of the (physicists') Hermite polynomials that I have found in the literature is the following: $$ H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} $$

Now, Wikipedia also gives the equivalent definition

$$ H_n(x) = \left( 2x - \frac{d}{dx} \right)^n(1) $$ where the operator above is being applied to the constant function $g(x) \equiv 1$. So for instance for $n = 1$, we have

$$ H_1(x) = \left( 2x - \frac{d}{dx} \right)(1) = 2x\cdot 1 - \frac{d}{dx}(1) = 2x $$

My quesion is the following.

Question:

Assuming that I only know one of these two definitions, is there a way to derive the other expression for the Hermite polynomials directly from the given definition, say without having to guess by induction the form that such an expression should take?

Thank you very much for any help.

Adrián Barquero
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  • Well... Physically speaking Hermite Polynomials are found while solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oszillator hence you derive both expression quite easily while solving that particular situation. I personally prefer the first expression simply because there is no (physical) operator involved (I'm a mathematician) although there is (obviously) no difference between the two :) – b00n heT Feb 17 '14 at 20:32
  • Dear @b00nheT, do you know a reference that someone whose only knowledge of physics amounts to what little he can remember from three General Physics courses taken in college about 8 years ago could understand? (I really hope that there's another way to pass from one definition to the other without having to solve the Schrodinger equation that you mention). Thanks a lot for your comment =) – Adrián Barquero Feb 17 '14 at 20:42
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    Unfortunately the only reference I can give you is the one I used while studying which is in German, but you can still give it a try: (from 10.2.3 till 10.2.5) http://qudev.ethz.ch/lecture-notes/PhysikIII-HTMLch10 And a small addendum at the end of the script: http://qudev.ethz.ch/lecture-notes/PhysikIII-HTMLap5.html#x111-283000E Good luck! – b00n heT Feb 17 '14 at 21:26

1 Answers1

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It's not easy to derive one definition from the other, but here's one possible way: the Hermite polynomials can also be expressed in terms of their generating function, that is a function $g(t,x)$ such that $$ g(t,x) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ It follows that $$ H_n(x) = \left.\frac{\partial^n}{\partial t^n}g(t,x)\right|_{t=0} $$ Now, if we take the first definition $$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}, $$ we see that it looks quite similar. We need to find a way to change the derivatives to $x$ into derivatives to $t$, and we can do that by noting that $$ \frac{\partial}{\partial t}e^{-(t-x)^2} = -\frac{\partial}{\partial x}e^{-(t-x)^2} $$ Thanks to this trick, the function $g(t,x)$ follows immediately: $$ g(t,x) = e^{x^2}e^{-(t-x)^2} = e^{2xt-t^2}. $$ This allows us to derive yet another form for the Hermite polynomials, by equating the terms in $$ e^{2xt-t^2} = \sum_{n=0}^\infty\frac{(2xt-t^2)^n}{n!} = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ Let us examine the $(n-k)$th term in the first series: $$ (2x-t)^{n-k}\frac{t^{n-k}}{(n-k)!} = \sum_{l=0}^{n-k}\binom{n-k}{l}(2x)^{n-k-l}(-1)^l\frac{t^{n-k+l}}{(n-k)!}. $$ The $l=k$ term in this sum contains $t^n$: $$ (-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ provided that $2k\leqslant n$. In other words, all the terms that contribute to the $t^n$ term are $$ H_n(x)\frac{t^n}{n!} = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ so that we've found another expression for $H_n(x)$: $$ H_n(x) = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{n!}{k!\,(n-2k)!}. $$ We can write this as $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}\frac{(n-k)!}{(n-2k)!}(-1)^k(2x)^{n-2k}, $$ which is $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}(-1)^k\frac{d^k}{dx^k}(2x)^{n-k}, $$ and this is a more explicit notation of the other definition $$ \left(2x - \frac{d}{dx}\right)^n (1) $$ In principle, you can follow the steps in the reverse order to derive the first definition from the second, but it's certainly not intuitive. Alternatively, one might use the recurrence relations $$ H_{n+1}(x) = 2xH_n(x) - 2n H_{n-1}(x) $$ $$ H'_n(x) = 2nH_{n-1}(x). $$ It's easy to verify that both definitions satisfy these relations.

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