📂Functions

Orthogonality of Hermite Polynomials

Theorem

The Hermite polynomials $\left\{ H_{n} \right\}_{n=0}^{\infty}$ are orthogonal with respect to the weight function $w(x)=e^{-x^{2}}$ over the interval $(-\infty, \infty)$.

$$\braket{ H_{n} | H_{m} }_{e^{-x^{2}}} =\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}(x)H_{m}(x)dx=\sqrt{\pi}2^{n}n!\delta_{nm}$$

Here, $\delta_{nm}$ is the Kronecker delta.

Proof

Case 1: $n=m$

Let’s denote the differential operator as $D = \dfrac{d}{dx}$.

$$\int_{-\infty}^{\infty} e^{-x^{2}}H_{n}(x)H_{n}(x)dx$$

Hermite polynomials

$$H_{n}(x) = (-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}} = (-1)^{n}e^{x^{2}}D^{n}e^{-x^{2}}$$

If we solve the front part $H_{n}(x)$ of the above equation, it looks like this.

$$\begin{align*} \int_{-\infty}^{\infty} e^{-x^{2}}(-1)^{n}e^{x^{2}}\left[D^{n}e^{-x^{2}} \right]H_{n}(x)dx &= \int_{-\infty}^{\infty} (-1)^{n}\left[D^{n}e^{-x^{2}} \right]H_{n}(x)dx \end{align*}$$

Solving the equation partially, we get the following equation.

$$\begin{align*} &\int_{-\infty}^{\infty} (-1)^{n}\left[D^{n}e^{-x^{2}} \right]H_{n}(x)dx \\ &=\left[ (-1)^{n}\left(D^{n-1}e^{-x^{2}}\right)H_{n}(x) \right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}(-1)^{n}\left[ D^{n-1}e^{-x^{2}}\right]H^{\prime}_{n}(x)dx \tag{1} \end{align*}$$

Here, since the first term is $\lim \limits_{x\rightarrow \pm\infty}D^{n-1}e^{-x^{2}}=0$, it equals $0$. The reason why this limit converges to $0$ is because for any $n$, it is expressed as

$$D^{n}e^{-x^{2}}=p(x)e^{-x^{2}}$$.

Here, $p(x)$ is an arbitrary polynomial. When $x \rightarrow \pm \infty$, the speed of convergence to $0$ is much faster than the divergence rate of the polynomial, thus the limit converges to $0$.

Recurrence relation of Hermite polynomials $$H_{n}^{\prime}(x) =2nH_{n-1}(x)$$

By the recurrence relation of the Hermite polynomials, $(1)$ is as follows.

$$-2n\int_{-\infty}^{\infty}(-1)^{n}\left[ D^{n-1}e^{-x^{2}} \right]H_{n-1}(x)dx$$

By the same logic as just now, performing partial integration once more, we obtain the following.

$$(-1)^{2}2^{2}n(n-1)\int_{-\infty}^{\infty}(-1)^{n}\left[ D^{n-2}e^{-x^{2}} \right]H_{n-2}(x)dx$$

Therefore, integrating $n$ times, we obtain the following equation.

$$(-1)^{n}2^{n}n!\int_{-\infty}^{\infty}(-1)^{n}e^{-x^{2}}H_{0}(x)dx$$

Since it is $H_{0}(x)=1$, it is summarized as follows.

$$2^{n}n!\int_{-\infty}^{\infty}e^{-x^{2}}dx$$

The above integral is the Gaussian integral, and its value is $\sqrt{\pi}$. Thus, we get the following.

$$\int_{-\infty}^{\infty} e^{-x^{2}}H_{n}(x)H_{n}(x)dx=\sqrt{\pi}2^{n}n!$$

Case 2: $n\ne m$

Without loss of generality, let’s assume $n \gt m$.

$$\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}(x)H_{m}(x)dx$$

Expressing only $H_{n}(x)$ from here looks as follows.

$$\int_{-\infty}^{\infty}e^{-x^{2}}(-1)^{n}e^{x^{2}}\left[D^{n}e^{-x^{2}}\right]H_{m}(x)dx=\int_{-\infty}^{\infty}(-1)^{n}\left[D^{n}e^{-x^{2}}\right]H_{m}(x)dx$$

Using the same method as in the case of $n=m$, performing $m$ times partial integration gives us the following equation.

$$(-1)^{n+m}2^{m}m!\int_{-\infty}^{\infty}\left[ D^{n-m}e^{-x^{2}} \right]\cdot 1 dx$$

Integrating once more gives the following.

$$\begin{align*} &(-1)^{n+m}2^{m}m!\int_{-\infty}^{\infty}\left[ D^{n-m}e^{-x^{2}} \right]\cdot 1 dx \\ &= (-1)^{n+m}2^{m}m!\left(\left[D^{n-m-1}e^{-x^{2}} \right]_{-\infty}^{\infty}+2(m+1)\int_{-\infty}^{\infty}\left[ D^{n-m-1}e^{-x^{2}}\right]\cdot 0 dx \right) \end{align*}$$

The first term, as explained in the above proof, equals $0$, and the second term also equals $0$. Therefore, we obtain the following.

$$\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}(x)H_{m}(x)dx=0, \quad n \ne m$$

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