quantum mechanics - Expectation value of time-dependent Hamiltonian

I'm trying to solve a problem in QM with a forced quantum oscillator. In this problem I have a quantum oscillator, which is in the ground state initially. At $t=0$, the force $F(t)=F_0 \sin(\Omega t)$ is switched on and after time $T$ turned off again. I need to find $\langle \hat{H} \rangle$ at time $T$.

I started out with this Hamiltonian:

$$\hat H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega_0^2\hat{x}^2-\hat{x}F_0\sin(\Omega t)$$

And I want to solve this problem in the Heisenberg picture. Then

$$\langle \hat{H}\rangle=\langle\psi(T)|\hat{H}_S|\psi(T)\rangle=\langle\psi(0)|\hat{H}_H|\psi(0)\rangle\quad,$$

where $\hat{H}_H$ is the Hamiltonian in the Heisenberg picture and $|\psi(0)\rangle$ is the ground state of the harmonic oscillator.

Since $\hat{H}_H=U^{\dagger}(T)\hat{H}_SU(T)$, I need to find the time evolution operator $U$. I previously asked a question regarding this operator, but don't see how to apply it to this problem.

Answer

My approach would be: first determine the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. For $\hat{x}$ you have $$\frac{d}{dt}\hat{x}_H(t) = i[H_H,\hat{x}_H(t)] = \frac{i}{2m} [\hat{p}_H(t)^2,\hat{x}_H(t)] = \frac{\hat{p_H(t)}}{m}$$ and for $p$ you have (assuming $0\leq t \leq T$) $$\frac{d}{dt}\hat{p}_H(t) = i[H_H(t),\hat{p}_H(t)] = -m\omega_0^2 \hat{x}_H(t) + F_0\sin(\Omega t)$$ These are coupled differential equations, which you can decouple by differentiating them once more with respect to time and performing a substitution. For instance,

$$\frac{d^2}{dt^2} \hat{x}_H(t) = \frac{1}{m} \frac{d}{dt} \hat{p}_H = -\omega_0^2 \hat{x}_H(t)+\frac{F_0}{m}\sin(\Omega t)$$ where I substituted $\frac{d}{dt} \hat{p}_H(t)$ by its equation of motion found earlier. You can also get an equation like this for $\hat{p}_H(t)(t)$, which I leave for you..

Now, these equations can be solved using your favorite method, provided you give them suitable boundary conditions. Note that you only need one boundary condition for $x$ and $p$ (which is $x_H(0)=\hat{x}_S$ and $p_H(0)=\hat{p}_S$ It will give you some expression for $\hat{x}_H(t)$ and $\hat{p}_H(t)$ in terms of $\hat{x}_S$ and $\hat{p}_S$. The Heisenberg Hamiltonian is then easily determined by substituting $\hat{x}_H(t)$ and $\hat{p}_H(t)$.

With that expression in hand you should be able to find $\langle H(t)\rangle$ (note that you should consider the cases where $t<0$ and $t>T$ separately).

EDIT: The proof regarding my statement below: In the Schroedinger picture the Hamiltonian is

$$\hat H_S=\frac{\hat{p}_S^2}{2m}+\frac{1}{2}m\omega_0^2\hat{x}_S^2-\hat{x}_SF_0\sin(\Omega t)$$

and the Heisenberg picture is given by $H_H = U^\dagger(t) H_S U(t)$. So if you take for instance the first term you get:

$$U^\dagger(t) \frac{\hat{p}_S^2}{2m}U(t) =\frac{1}{2m} (U^\dagger(t) \hat{p}_S U^\dagger(t))(U(t)\hat{p}_SU(t)) =\frac{1}{2m} \hat{p}_H(t)^2$$

You can do the same for the other terms. In the end you just effectively replace $p_S\rightarrow p_H(t)$ and the same for $x$.