In the following problems, a unit system so that $\hbar =1$ and $e =1$ is used.

a. The equation of motion is

$$\begin{eqnarray*} \frac{d a (0)} {dt} = -i [a, H] \end{eqnarray*}$$

Substitute

$$\begin{eqnarray*} a = \frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{m\omega}} \hat{p}- i \sqrt{m\omega} \hat{x}\right] \end{eqnarray*}$$

$$\begin{eqnarray*} a^\dagger = \frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{m\omega}} \hat{p}+ i \sqrt{m\omega} \hat{x}\right] \end{eqnarray*}$$

into

$$\begin{eqnarray*} H = \omega ( a^\dagger a +\frac{1}{2}) \end{eqnarray*}$$

to get

$$\begin{eqnarray*} H = \frac{1}{2m}\hat{p}^2 +\frac{1}{2}m \omega^2 \hat{x}^2 \end{eqnarray*}$$

using $[\hat{x}_j,\hat{p}_k]=i\delta_{jk}$. This shows that that $a$ and $a^\dagger$ can be considered as an operator factorization of H. Note that these so called creation and annihilation operators satisfy the following commutation relationship,

$$\begin{eqnarray*}[a,a^\dagger]=1 \end{eqnarray*}$$

Write out the commutator explicitly and commute the $a$ and $a^\dagger$ in the first term.

$$\begin{eqnarray*} \frac{d a (0)} {dt} = -i \omega (a a^\dagger a + \frac{1}{2}a ... ...\\ = -i\omega (a^\dagger a a - a^\dagger a a + a) = -i \omega a \end{eqnarray*}$$

Upon integration

$$\begin{eqnarray*} a(t) = a(0) e^{-i \omega t} \end{eqnarray*}$$

b. Given

$$\begin{eqnarray*} a^\dagger(t) = a^\dagger(0) e^{i\omega t} \end{eqnarray*}$$

And from part a

$$\begin{eqnarray*} a^\dagger (0) = \frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{m\omega}} \hat{p}(0) + i \sqrt{m\omega} \hat{x}(0) \right] \end{eqnarray*}$$

so

$$\begin{eqnarray*} a(t) = \frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{m\omega}} \hat{... ...ga}} \hat{p}(t) - i \sqrt{\frac{1}{2}m\omega} \hat{x}(t) \right] \end{eqnarray*}$$

$$\begin{eqnarray*} a^\dagger(t) = \frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{m\omega... ...ga}} \hat{p}(t) + i \sqrt{\frac{1}{2}m\omega} \hat{x}(t) \right] \end{eqnarray*}$$

Add these

$$\begin{eqnarray*} a(t)+a^\dagger(t)= \sqrt{\frac{2}{m\omega}} \hat{p}(t) = \frac... ...\omega}{2}} \hat{x}(0) \left[e^{i\omega t}-e^{-i\omega t}\right] \end{eqnarray*}$$

Solve for $\hat{p}$

$$\begin{eqnarray*} \hat{p}(t) = \hat{p}(0) \cos \omega t - m \omega \hat{x}(0) \sin \omega t \end{eqnarray*}$$

To get $\hat{x}$, subtract

$$\begin{eqnarray*} a^\dagger(t)-a(t)= \sqrt{2 m\omega} \hat{x}(t) = \frac{1}{\sqr... ...\omega}{2}} \hat{x}(0) \left[e^{i\omega t}+e^{-i\omega t}\right] \end{eqnarray*}$$

Solve for $\hat{x}$

$$\begin{eqnarray*} \hat{x}(t) = \hat{x}(0) \cos \omega t + \frac{1}{m \omega} \hat{p}(0) \sin \omega t \end{eqnarray*}$$

c. We know that ${\langle }q {\rangle }$ and ${\langle }p {\rangle }$ obey the Erenfest theorem and we can write:

$$\begin{eqnarray*} \frac{ {\langle }q {\rangle }}{dt} &=&\frac{{\langle }p {\rang... ...ngle }p {\rangle }}{dt}&=&-{\langle }V' {\rangle }. \\ \nonumber \end{eqnarray*}$$

And they are equivalent to classical equations when ${\langle }V' {\rangle }$ = ${\langle }V'_{cl} {\rangle }$. If dispersion $(\Delta q)^2={\langle }q^2 {\rangle }-{\langle }q {\rangle }^2$ of wave packet is small then we can write:

$$V'(q)=V'_{cl}({\langle }q {\rangle })+(q-{\langle }q {\rangle })V''_{cl}+(q-{\langle }q {\rangle })^2 V'''_{cl}+ ...$$

We see that $V'=V'_{cl}$ if all term higher that $V''$ are equal to zero. Harmonic oscillator potential has quadratic form and hence quantum and classical equation of motion look similar.