Consider a one-dimensional simple harmonic oscillator. (a) Using

$$\left.\begin{array}{l}{a} \\ {a^{\dagger}}\end{array}\right\}=\sqrt{\frac{m \omega}{2 \hbar}}\left(x \pm \frac{i p}{m \omega}\right), \left.\begin{array}{l}{a|n\rangle} \\ { a^{\dagger}|n\rangle}\end{array}\right\}=\left\{\begin{array}{l}{\sqrt{n}|n-1\rangle} \\ {\sqrt{n+1}|n+1\rangle}\end{array}\right.$$

evaluate $\langle m|x| n\rangle,\langle m|p| n\rangle,\langle m|\{x, p\}| n\rangle,\left\langle m\left|x^{2}\right| n\right\rangle$ and $\left\langle m\left|p^{2}\right| n\right\rangle$. (b) Check that the virial theorem holds for the expectation values of the kinetic energy and the potential energy taken with respect to an energy eigenstate.

(a)

Using relations given in the problem $x$ and $p$ can be expressed in terms of $a$ and $a^\dagger$.

$$\begin{align} x &= \sqrt{\frac{\hbar}{2 m\omega}}\left( a + a^\dagger \right )\\ p &= i \sqrt{\frac{\hbar m\omega }{2}}\left( a^\dagger - a \right) \end{align}$$

Now matrix elements can be computed

$$\begin{align} \bra{m} x \ket{n} &= \sqrt{\frac{\hbar}{2 m\omega}}\left(\bra{m} a \ket{n} + \bra{m} a^\dagger \ket{n} \right ) \\ \bra{m} x \ket{n} &= \sqrt{\frac{\hbar}{2 m\omega}}\left(\sqrt{n}\;\delta_{m , n-1} + \sqrt{n+1}\;\delta_{m , n+1} \right ) \\ \bra{m} p \ket{n} &= i \sqrt{\frac{\hbar m\omega }{2}}\left(\bra{m} a^\dagger \ket{n} - \bra{m} a \ket{n} \right ) \\ \bra{m} p \ket{n} &= i \sqrt{\frac{\hbar m\omega }{2}}\left(\sqrt{n+1}\;\delta_{m , n+1} - \sqrt{n}\;\delta_{m , n-1} \right ) \end{align}$$

Now, we can compute matrix element of $x^2$. We will make use of $\left[a, a^\dagger\right] = 1$

$$\begin{align} x^2 &= \frac{\hbar}{2 m \omega}\left( a + a^\dagger \right)^2 = \frac{\hbar}{2 m \omega}\left( a^2 + 2 a^\dagger a+ 1+ \left(a^\dagger\right)^2 \right) \\ \bra{m}x^2\ket{n} &= \frac{\hbar}{2 m \omega}\left( \sqrt{n(n-1)}\;\delta_{m,n-2} + (2 n+1)\; \delta_{mn} +\sqrt{(n+1)(n+2)}\;\delta_{m,n+2} \right) \end{align}$$

To compute the rest of values consider following relations

$$\begin{align} \hbar\omega a^\dagger a + \frac{1}{2}\hbar\omega &= \frac{1}{2m}p^2 +\frac{1}{2}m\omega^2 x^2 \\ a^2 - \left(a^\dagger\right)^2 &= \frac{i}{\hbar}\left( xp + px \right) = \frac{i}{\hbar}\left\{x, p\right\} \end{align}$$