This was a problem on my final exam that has been really bugging me.
Consider the quantum Harmonic oscillator prepared in an energy eigenstate, $\psi_n$(x). Calculate the expectation value of the potential energy, using the recurrence relation between the Hermite polynomials, $H_n(\alpha x)$ together with the orthogonality relation for the energy eigenfunctions.
It is given that the normalized harmonic oscillator eigenstates:
$$\Psi_n(x)=(\frac {\alpha} {\sqrt \pi 2^nn!})^{\frac 1 2} e^{\frac {-\alpha ^2x^2} 2}H_n(\alpha x)$$
My attempt:
$$V= \frac 1 2 kx^2$$
$$\langle V \rangle = \int^\infty_{-\infty}\Psi_n^*(x)V\Psi_n(x)\space dx$$
$$=\frac {\alpha}{\sqrt \pi2^nn!} \frac k2\int x^2e^{- \alpha^2 x^2}H_n(\alpha x)^2 dx \space\space \space \space \space \space \space \space \space \space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$$
The recursion is given as
$$H_{n+1}(x)=2xH_n(x)-H'_n(x)$$
This is where I get lost. I tried parts where
$$dv = xe^{-\alpha^2x^2}dx$$
$$u=xH_n(\alpha x)^2$$
$$v = -\frac 1 {2\alpha^2}e^{-\alpha^2 x^2}$$
$$du = H_n(\alpha x)^2+2xH_n(\alpha x)H'_n(\alpha x)$$
From equation (1)
$$\int x^2e^{- \alpha^2 x^2}H_n(\alpha x)^2dx=\frac 1 {2\alpha^2} \int \left( H_n(\alpha x)^2e^{-\alpha^2x^2}+2xH_n(\alpha x)H'_n(\alpha x)e^{-\alpha^2 x^2}\right)dx$$
So then I use the recursion:
$$=\frac 1 {2\alpha^2}\int\left( (\frac {H_{n+1}(\alpha x)+H'(\alpha x)}{2x})^2+2xH_n'(\alpha x)e^{-\alpha^2 x^2}H_n(\alpha x) \right) dx$$
This appears to be a dead end. I cannot simplify at all. We need to use the orthogonality principle, but I do not see where any $\int H_nH_m$ pops up.
This is just a problem that has been stuck in my head for a while. So any help is appreciated. Thanks!
Best Answer
You are on the right track. To start off the orthogonality principle for the Hermite polynomials are weighted: $$ \int_{-\infty}^\infty H_n(x)H_m(x)e^{-x^2}dx=\delta_{nm}2^nn!\sqrt{\pi}, $$ where $\delta_{nm}$ is the Dirac delta function. There are also two relations for Hermite polynomials that are needed: $$ H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)\\ H'_n(x)=2nH_{n-1}(x). $$ Starting from your Equation (1), and lettings $y=\alpha x$: $$ \langle V\rangle=\dfrac{1}{\sqrt{\pi}2^nn!}\dfrac{k}{2\alpha^2}\int_{-\infty}^\infty y^2 e^{-y^2}H_n(y)^2dy $$ Using $u=yH_n(y)^2$ and $dv=ye^{-y^2}dy$ for integrating by parts, the integral becomes part of $\langle V\rangle$: $$ -\dfrac{y}{2}e^{-y^2}H_n(y)^2\big|_{-\infty}^\infty+\int_{-\infty}^\infty \dfrac{1}{2}e^{-y^2}(H_n(y)^2+4nyH_n(y)H_{n-1}(y))dy, $$ where I use the $H_n'(x)=2nH_{n-1}(x)$ property. The non-integral expression goes to zero because of $e^{-y^2}$, which leaves the integral: $$ \int_{-\infty}^\infty \dfrac{1}{2}e^{-y^2}H_n(y)^2dy+2n\int_{-\infty}^\infty ye^{-y^2}H_n(y)H_{n-1}(y)dy $$ Integrating the rightmost integral by parts with $u=H_n(y)H_{n-1}(y)$ and $dv=ye^{-y^2}dy$ gives: $$ 2n\int_{-\infty}^\infty ye^{-y^2}H_n(y)H_{n-1}(y)dy=-ye^{-y^2}nH_n(y)H_{n-1}(y)\big|_{-\infty}^\infty\\+n\int_{-\infty}^\infty e^{-y^2}(2nH_{n-1}(y)H_{n-1}(y)+2(n-1)H_n(y)H_{n-2}(y))dy $$ In this case $du=H'_n(y)H_{n-1}(y)+H_n(y)H'_{n-1}(y)=2nH_{n-1}(y)H_{n-1}(y)+2(n-1)H_n(y)H_{n-2}(y)$. Next, using the orthogonality condition: $$ 2n\int_{-\infty}^\infty ye^{-y^2}H_n(y)H_{n-1}(y)dy=2n^22^{n-1}(n-1)!\sqrt{\pi}=n2^{n}n!\sqrt{\pi}\\ \int_{-\infty}^\infty \dfrac{1}{2}e^{-y^2}H_n(y)^2dy=2^{n-1}n!\sqrt{\pi} $$ Putting the integral part of $\langle V\rangle$ is the sum of the above two integrals: $$ \langle V\rangle=\dfrac{1}{\sqrt{\pi}2^nn!}\dfrac{k}{2\alpha^2}(n2^{n}n!\sqrt{\pi}+2^{n-1}n!\sqrt{\pi})=\dfrac{k}{4\alpha^2}(2n+1) $$