Question

# 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.

Solution

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(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}

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