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

Derive the normalization constant \( c_n \) for the Harmonic oscillator solution

\begin{equation}\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},

\end{equation}

by deriving the orthogonality relationship using generating functions

\begin{equation}\label{eqn:hermiteOrtho:40}

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

\end{equation}

Start by working out the integral

\begin{equation}\label{eqn:hermiteOrtho:60}

I = \int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx,

\end{equation}

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

## Answer

First using the exponential definition of the generating function

\begin{equation}\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}

\end{equation}

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

\begin{equation}\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}

\end{equation}

Let

\begin{equation}\label{eqn:hermiteOrtho:120}

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

\end{equation}

and equate the two expansions of this integral

\begin{equation}\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},

\end{equation}

or, after equating powers of \( t^n \)

\begin{equation}\label{eqn:hermiteOrtho:160}

\sqrt{\pi} (2 s)^n =

\sum_{m=0}^\infty \frac{s^m}{m!} \alpha_{n m}.

\end{equation}

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

\begin{equation}\label{eqn:hermiteOrtho:180}

\sqrt{\pi} 2^n = \frac{1}{n!} \alpha_{n n},

\end{equation}

and

\begin{equation}\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!.

\end{equation}

The SHO normalization is fixed by

\begin{equation}\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!,

\end{equation}

or

\begin{equation}\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}

\end{equation}

# References

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