## Question: Hermite polynomial normalization constant ([1] pr. 2.21)

Derive the normalization constant $$c_n$$ for the Harmonic oscillator solution

\label{eqn:hermiteOrtho:20}
u_n(x) = c_n H_n\lr{ x \sqrt{\frac{m\omega}{\Hbar}} } e^{-m \omega x^2/2 \Hbar},

by deriving the orthogonality relationship using generating functions

\label{eqn:hermiteOrtho:40}
g(x,t) = e^{-t^2 + 2 t x} = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}.

Start by working out the integral

\label{eqn:hermiteOrtho:60}
I = \int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx,

consider the integral twice with each side definition of the generating function.

First using the exponential definition of the generating function

\label{eqn:hermiteOrtho:80}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
e^{-t^2 + 2 t x}
e^{-s^2 + 2 s x} e^{-x^2} dx \\
&=
e^{-t^2 -s^2}
\int_{-\infty}^\infty
e^{-(x^2- 2 t x – 2 s x)} dx \\
&=
e^{-t^2 -s^2 + (s + t)^2}
\int_{-\infty}^\infty
e^{-(x – t – s)^2} dx \\
&=
e^{2 st}
\int_{-\infty}^\infty
e^{-u^2} du \\
&= \sqrt{\pi} e^{2 st}.
\end{aligned}

With the Hermite polynomial definition of the generating function, this integral is

\label{eqn:hermiteOrtho:100}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}
\sum_{m=0}^\infty H_m(x) \frac{s^m}{m!}
e^{-x^2} dx \\
&=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx.
\end{aligned}

Let

\label{eqn:hermiteOrtho:120}
\alpha_{n m} = \int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx,

and equate the two expansions of this integral

\label{eqn:hermiteOrtho:140}
\sqrt{\pi} \sum_{n=0}^\infty \frac{(2st)^n}{n!}
=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\alpha_{n m},

or, after equating powers of $$t^n$$

\label{eqn:hermiteOrtho:160}
\sqrt{\pi} (2 s)^n =
\sum_{m=0}^\infty \frac{s^m}{m!} \alpha_{n m}.

This requires $$\alpha_{n m}$$ to be zero for $$n \ne m$$, so

\label{eqn:hermiteOrtho:180}
\sqrt{\pi} 2^n = \frac{1}{n!} \alpha_{n n},

and

\label{eqn:hermiteOrtho:200}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx = \delta_{n m} \sqrt{\pi} 2^n n!.

The SHO normalization is fixed by

\label{eqn:hermiteOrtho:220}
\int_{-\infty}^\infty u_n^2(x) dx
= c_n^2
\int_{-\infty}^\infty H_n^2(x/x_0) e^{-(x/x_0)^2} dx
= c_n^2 x_0 \sqrt{\pi} 2^n n!,

or

\label{eqn:hermiteOrtho:240}
\begin{aligned}
c_n
&= \inv{\sqrt{ \sqrt{\pi} 2^n n! \sqrt{\frac{\Hbar}{m \omega}}}} \\
&= \lr{ \frac{m \omega}{\Hbar \pi} }^{1/4} 2^{-n/2} \inv{\sqrt{n!}}
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.