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I wish to prove that the Hermite polynomials defined as $$H_n(x) := (-1)^n e^{x^2} D^n(e^{-x^2})$$ are orthogonal wrt the inner product $$\langle f,g\rangle = \int_{\mathbb{R}} e^{-x^2} {f(x)}\overline{g(x)} dx $$ and that $||H_n||^2=n!2^n\sqrt{\pi}$.

Not much else to do than compute $\langle H_m, H_n \rangle$.

The integral becomes $$(-1)^{n+m} \int_{\mathbb{R}} e^{x^2}D^m(e^{-x^2})D^{n}(e ^{-x^2}) dx $$ Let's look at the derivatives of $e^{-x^2}$: $$ \begin{array}{|} \hline n &&D^n(e^{-x^2}) \\ \hline 1 &&-2xe^{-x^2} \\ \hline2 &&-2e^{-x^2}+4x^2e^{-x^2} \\\hline3 &&4xe^{-x^2}+8xe^{-x^2}-8x^3e^{-x^2}\\ \vdots && \vdots\end{array}$$ It is obvious $D^{n}(e^{-x^2}) = P_n(x)e^{-x^2}$ for a polynomial $P_n$ of degree $n$ where the leading coefficient in $P_n$ is $(-1)^n2^n$. It should also be apparant that $P_n$ only contains terms $\{c_kx^k |k \equiv n \mod 2, 0\leq k\leq n, c_k\neq 0\}$. However, I am struggling to find the rest of the coefficients of $P_n$ under closed form. So, I am looking for a way to compute this integral without having to find the coeffiecients. Alas, I still haven't found what I am looking for. Any ideas?

Lozansky
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  • Orthogonality just follows from integration by parts. This is a shared phenomenon: Rodrigues formula $\Rightarrow$ orthogonality with respect to some inner product. – Jack D'Aurizio Mar 21 '18 at 12:34

1 Answers1

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Orthogonality

Since $m\ne n$, we may assume $m < n$. Rewrite $H_n$ according to the definition, but keep $H_m$ as is: $$ \int_\mathbb{R} H_m(x)H_n(x) e^{-x^2}\,dx = (-1)^n \int_\mathbb{R} H_m(x) D^n(e^{-x^2})\,dx $$ Integrate by parts $n$ times, throwing the derivative onto $H_m$ each time. No boundary terms appear because of super exponential decay of $e^{-x^2}$ at infinity, which beats any polynomial. Integration by parts results in $$ (-1)^{m+n} \int_\mathbb{R} D^n(H_m(x)) e^{-x^2}\,dx = 0 $$ since the $n$-th derivative of $H_m$ is identically zero.

Norm

When $m=n$, we end up with $$ D^n(H_n(x)) \int_\mathbb{R} e^{-x^2}\,dx = \sqrt{\pi} D^n(H_n(x)) $$ where $D^n(H_n(x))$ is $n!$ times the leading coefficient of $H_n$. That coefficient is $2^n$, as one can see directly from the definition $(-1)^n e^{x^2} D^n(e^{-x^2})$: to get $x^n$ here, one has to apply the derivative to the exponential part every time, getting the factor of $(-2x)$ each time.