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!


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: $$ \langle V\rangle=\dfrac{1}{\sqrt{\pi}2^nn!}\dfrac{k}{2\alpha^2}(n2^{n}n!\sqrt{\pi}+2^{n-1}n!\sqrt{\pi})=\dfrac{k}{4\alpha^2}(2n+1) $$

  • $\begingroup$ Very nice! The only thing that you have incorrect is that $\delta_{mn}$ is the Kronecker delta, but I knew what you meant $\endgroup$ – yankeefan11 Jul 30 '13 at 17:18

Actually, the more convenient form of relation is

$$H_{n+1}=2x H_{n}-2n H_{n-1},$$

thus we find: $x H_{n} = \frac{1}{2} H_{n+1}+n H_{n-1}$. Therefore the integral you need to calculate is:

$$ I_{n} =\frac{1}{\sqrt{\pi}2^n n!}\int \left(\frac{1}{2} H_{n+1}+n H_{n-1}\right)^2 e^{-x^2} dx.$$

Using orthogonality of the Hermite polynomials, we find: $$ I_{n} =\frac{1}{\sqrt{\pi}2^n n!}\int \left(\frac{1}{4} H_{n+1}^2+n^2 H_{n-1}^2\right)e^{-x^2} dx =\frac{n+1}{2}+\frac{n}{2}=n+\frac{1}{2},$$ as it should be.

  • $\begingroup$ Where does the k go? Does that not affect anything' $\endgroup$ – yankeefan11 Jul 30 '13 at 17:09
  • $\begingroup$ I used dimensionless integration variable $\alpha\, x\to x$ and omitted the overall factor $k/(2\alpha^2)$. Hence the expectation value of the potential energy is $I_{n} k/(2\alpha^2) $. $\endgroup$ – Grisha Kirilin Jul 30 '13 at 17:12

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