$5-23.$ To normalize the harmonic$-$oscillator wave functions and calculate various expectation values, we must be able to evaluate integrals of the form

$I_v(a)={\displaystyle \underset{-}{\overset{}{}}}{x}^{2v}{e}^{-a{x}^2}dx,v=0,1,2,$dots

We can simply either look them up in a table of integrals or continue this problem. First, show that

$I_v(a)=2{\displaystyle \underset{0}{\overset{}{}}}{x}^{2v}{e}^{-a{x}^2}dx$

Chapter $5/$ The Harmonic Oscillator and Vibrational Spectroscopy

The case $v=0$ can be handled by the following trick. Show that the square of $I_0(a)$ can be written in the form

$I_0^2(a)=4{\displaystyle \underset{0}{\overset{}{}}}{\displaystyle \underset{0}{\overset{}{}}}dxdy{e}^{-a(x^2+y^2)}$

Now convert to plane polar coordinates, letting

$r^2=x^2+y^2,$ and $,dxdy=rdrd$

Show that the appropriate limits of integration are $0r<mi></mi>$ and $0\frac{}{2}$ and that

$I_0^2(a)=4{\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}d{\displaystyle \underset{0}{\overset{}{}}}drr{e}^{-a{r}^2}$

which is elementary and gives

$I_0^2(a)=4*\frac{}{2}*\frac{1}{2a}=\frac{}{a}$

or

$I_0(a)=(\frac{}{a}{)}^{\frac{1}{2}}$

Now prove that the $I_v(a)$ may be obtained by repeated differentiation of $I_0(a)$ with respect to a and, in particular, that

$\frac{d^v{I}_0(a)}{d{a}^v}=(-1{)}^v{I}_v(a)$

Use this result and the fact that $I_0(a)=(\frac{}{a}{)}^{\frac{1}{2}}$ to generate $I_1(a),I_2(a),$ and so forth.