A question of the norm calculation of Hermite polynomials

Question

Define the (physicist's) Hermite polynomial $H_n (x)$ by $$H_n (x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} $$ then prove that $$ \int_{\mathbb R} |H_n (x) |^2 e^{-x^2} dx = 2^n n! \sqrt{\pi}$$.

Answer 1

We have the recursive relation $$H'_n(x)=(-1)^ne^{x^2}\left(2x\frac{d^n}{dx^n}e^{—x^2}+\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)=-H_{n+1}(x)+2xH_n(x).$$ We use this in the integral: \begin{align} \int_{\Bbb R}H_n(x)^2e^{-x^2}dx&=\int_{\Bbb R}(2xH_{n-1}(x)-H'_{n-1}(x))H_n(x)e^{-x^2}dx\\ &=\int_{\Bbb R}2xe^{-x^2}H_{n-1}(x)H_n(x)dx-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\ &=\left[-e^{-x^2}H_{n-1}(x)H_n(x)\right]_{—\infty}^{+\infty}+\int_{\Bbb R}e^{-x^2}(H'_{n-1}(x)H_n(x)+H_{n-1}(x))H'_n(x))\\ &-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\ &=\int_{\Bbb R}H_{n-1}H'_n(x)e^{-x^2}dx. \end{align} By induction, we get $$\int_{\Bbb R}H_n(x)^2e^{-x^2}dx=\int_{\Bbb R}H_0(x)H^{(n)}_n(x)e^{-x^2}dx=\int_{\Bbb R}H^{(n)}_n(x)e^{-x^2}dx.$$ Indeed, denote $\langle P,Q\rangle:=\int_{\Bbb R}P(x)Q(x)e^{-x^2}dx$ for two polynomials $P$ and $Q$. We can show by an analogous way that $\langle H_n,H_{n+k}\rangle=\langle H_{n-1},H'_{n+k+}\rangle$ for integers $k$ and $n$, then the induction relation.

Now, to get the result, we need the following facts:

• $H_n$ is a polynomial of degree $n$, whose leading terms is $2^n$ (show it by induction);
• $\int_{\Bbb R}e^{-x^2}=\sqrt \pi$.

By the way, this shows that the Hermite polynomials are orthogonal.

Answer 2

Due to orthogonality,

$$\int_{-\infty}^\infty x^k H_n(x) \exp(-x^2)\mathrm dx=0\qquad k<n$$

we can simplify the integral like so:

$$\int_{-\infty}^\infty H_n(x)^2 \exp(-x^2)\mathrm dx=2^n\int_{-\infty}^\infty x^n H_n(x) \exp(-x^2)\mathrm dx$$

where we used the fact that the leading coefficient of $H_n(x)$ is $2^n$.

If we substitute the Rodrigues formula for the Hermite polynomial into the integral,

$$2^n\int_{-\infty}^\infty x^n H_n(x) \exp(-x^2)\mathrm dx=(-2)^n\int_{-\infty}^\infty x^n \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2)\mathrm dx$$

One can then do $n$-fold integration by parts to yield

$$\begin{align*} (-2)^n\int_{-\infty}^\infty x^n \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2)\mathrm dx&=(-2)^n (-1)^n n! \int_{-\infty}^\infty \exp(-x^2)\mathrm dx\\ &=2^n n! \sqrt{\pi} \end{align*}$$

where we used the known value of the Gaussian integral.