Finding $\langle \psi_n | x | \psi_m \rangle$ for the harmonic oscillator

Question

I need to prove the following for the quantum harmonic oscillator, whose potential is given by $V(x) = \frac{1}{2}m\omega^2 x^2$:

$$\langle x\rangle_{n m}=\left\langle\psi_{n}|x| \psi_{m}\right\rangle=\left\{\begin{array}{cl} \frac{1}{\alpha}\left(\frac{n+1}{2}\right)^{1 / 2} & m=n+1 \\ \frac{1}{\alpha}\left(\frac{n}{2}\right)^{1 / 2} & m=n-1 \\ 0 & \text { otherwise } \end{array}\right.$$

And I think I am on the right path but I am not very sure how to proceed. I know that the wavefunction obtained as a solution for Schrödinger's equation and the mentioned potential is:

$$\psi_n(x) = \sqrt{\frac{\alpha}{\sqrt{\pi}2^n n!}} H_n(\alpha x)e^{-a^2x^2/2}$$

Where $H_n(x)$ is the $n$-th order Hermite polynomial and $\alpha = \sqrt{m\omega / \hbar}$. I know that the following holds for Hermite polynomials: $$H_{n+1} = 2 x H_n - 2 n H_{n-1}$$ And therefore: $$x H_n = \frac{1}{2}H_{n+1} + n H_{n-1}$$

I also know that these are a set of orthogonal polynomials, since:

$$\int_{-\infty}^\infty e^{-x^2}H_n(x) H_m(x) dx = \sqrt{\pi} 2^n n! \delta_{nm}$$ Where $\delta_{nm}$ is the Kronecker delta. With these things in mind, this is what I have tried to do: $$\langle x \rangle_{nm} = \int_{-\infty}^\infty \psi_n(x)\cdot x \cdot \psi_m(x) dx$$

Now some constants come out of the integral, and the integral which I am having trouble solving is: $$I = \int_{-\infty}^\infty H_n(\alpha x) H_m(\alpha x) e^{-\alpha^2 x^2} \cdot x \cdot dx$$

And taking $\mu = \alpha x$ we get:

$$I = \frac{1}{\alpha^2}\int_{-\infty}^\infty H_n(\mu) H_m(\mu) e^{-\mu^2} \cdot \mu \cdot d\mu$$

It is clear that the orthogonality of the Hermite polynomials is going to be useful somehow, but I don't know how to keep going because the simpler integrals I obtain always diverge or turn out to be more difficult to solve. Any tips would be appreciated.

Answer 1

You already wrote down everything you need.

First you wrote: $$x H_n = \frac{1}{2}H_{n+1} + n H_{n-1}$$

Then you wrote: $$I = \frac{1}{\alpha^2}\int_{-\infty}^\infty H_n(\mu) H_m(\mu) e^{-\mu^2} \cdot \mu \cdot d\mu$$

So.. now you can substitute $\mu H_n$: $$I = \frac{1}{\alpha^2}\int_{-\infty}^\infty e^{-\mu^2} H_m(\mu) \left[\frac{1}{2}H_{n+1}(\mu) + n H_{n-1}(\mu)\right] d\mu$$

Where you can use now something you wrote already: $$\int_{-\infty}^\infty e^{-x^2}H_n(x) H_m(x) dx = \sqrt{\pi} 2^n n! \delta_{nm}$$

Does this make it easier?