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.
 A: 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 $x$.
A: I've tried to calculate the evolution operator in your case few years ago. I wanted to show that such a Hamiltonian never has a quantum state at any time. The solution I found is that such a model only have a coherent state at infinite time. I still do not know about intermediate times, even if the solution below allow you to calculate whatever you want at whichever time. A perturbation method can be found in the Landau (on quantum mechanics). Also, the result below can be found in Gardiner and Zoller book about quantum noise if I remember correctly. Finally, the oldest study of this question I found is an analysis by Carruthers and Nieto (1965) ; they use Green function to show that at infinite time, a quantum harmonic oscillator is described by a coherent state.
I've used the so-called Lie Algebraic method (see Wei and Norman 1963 for a good review ; my worksheet is self-contained I believe) because it a general method which allows you to calculate the evolution operator quite easily for simple Hamiltonian. This method is not so known 
Before going further into the derivation, let me give the references (if one is lacking please let me know) I used


*

*Carruthers, P., & Nieto, M. (1965). Coherent States and the Forced Quantum Oscillator. American Journal of Physics, 33(7), 537. doi:10.1119/1.1971895

*Wei, J., & Norman, E. (1963). Lie Algebraic Solution of Linear Differential Equations. Journal of Mathematical Physics, 4(4), 575. doi:10.1063/1.1703993

*Lo, C. F. (1991). Generating displaced and squeezed number states by a general driven time-dependent oscillator. Physical Review A, 43(1), 404–409. doi:10.1103/PhysRevA.43.404
Now my LaTeX file (I didn't try to edit it for the present display, so it might be the case that some remarks are totally stupid): 
Then, suppose we treat the following Hamiltonian 
$$
H=\hslash \omega \hat{a}^{+}\hat{a}+f\left( t\right) \hat{a}^{+}+f^{\ast
}\left( t\right) \hat{a}
$$
for any given function $f$ and $f^{\ast }$ depending on time, and with $\hat{
a}$ and $\hat{a}^{+}$ the annihilation and creation bosonic at frequency $
\omega $ mode operator. The unitary time evolution of this Hamiltonian can
be written, according to the Lie algebraic resolution method $H\left( t_{
\text{f}}\right) =\hat{U}^{+}\left( t_{\text{f}},t_{\text{i}}\right)
.H\left( t_{\text{i}}\right) .\hat{U}\left( t_{\text{f}},t_{\text{i}}\right) 
$ with the propagator (see \cite{Wei1963} for the first mathematical
treatment, and \cite{Lo1993} for the example of the quantum oscillator) : 
$$
\hat{U}\left( t_{\text{f}},t_{\text{i}}\right) =e^{-\mathbf{i}\hat{a}^{+}
\hat{a}\omega \left( t_{\text{f}}-t_{\text{i}}\right) }e^{\hat{a}^{+}\alpha
\left( t_{\text{f}},t_{\text{i}}\right) }e^{-\hat{a}\alpha ^{\ast }\left( t_{
\text{f}},t_{\text{i}}\right) }e^{-\left\vert \alpha \left( t_{\text{f}},t_{
\text{i}}\right) \right\vert ^{2}/2}e^{\gamma \left( t_{\text{f}},t_{\text{i}
}\right) }
$$
with 
$$
\alpha \left( t_{\text{f}},t_{\text{i}}\right) =\int_{t_{\text{i}}}^{t_{
\text{f}}}f\left( t\right) e^{\mathbf{i}\omega t}\dfrac{dt}{\mathbf{i}
\hslash }\text{ and }\gamma \left( t_{\text{f}},t_{\text{i}}\right) =-
\mathbf{i}\int_{t_{i}}^{t_{f}}\text{Im}\left\{ \alpha \left( \tau \right) 
\dfrac{\partial }{\partial \tau }\alpha ^{\ast }\left( \tau \right) \right\}
d\tau
$$
acting as two parameters for the unitary transformation.
Now, note the particular case when both $t_{\text{i}}$ and $t_{\text{f}}$
tend to infinity, $\gamma \left( t_{\text{f}},t_{\text{i}}\right) $ averages
to zero and : 
$$
\fbox{$\hat{U}\left( +\infty ,-\infty \right) =e^{\hat{a}^{+}\alpha }e^{-
\hat{a}\alpha ^{\ast }}e^{-\left\vert \alpha \right\vert ^{2}/2}=e^{\hat{a}
^{+}\alpha -\hat{a}\alpha ^{\ast }}=\hat{D}\left( \alpha \right) $}
$$
where $\alpha =f\left( \omega \right) /\mathbf{i}\hslash $ is the $\omega $
component of the Fourier transform of $f\left( t\right) $, divided by the
quantum box size $\hslash $. The unitary evolution of a quantum system
drived by a classical force thus corresponds to the displacement operator of
coherent states, as found in \cite{Carruthers1965}. Then, for a sufficiently
long interaction time, the monomode quantum harmonic oscillator transforms
to the quasi-classic oscillator.
Proof using the Lie algebraic method
Frist, rewritte the Hamiltonian $H$ as 
$$
H\left( t\right) =a_{0}\hat{a}^{+}\hat{a}+a_{+}\left( t\right) \hat{a}
^{+}+a_{-}\left( t\right) \hat{a}
$$
where $a_{0}$ does not depend on time. Then, remark that the operators $\hat{
1}$, $\hat{a}^{+}\hat{a}$, $\hat{a}^{+}$ and $\hat{a}$ form an closed algebra in the
Lie sens, \emph{i.e.} with respect to their commutators : 
$$
\left[ \hat{a}^{+}\hat{a},\hat{a}^{+}\right] =\hat{a}^{+}~;~\left[ \hat{a}
^{+}\hat{a},\hat{a}\right] =-\hat{a}~;~\left[ \hat{a},\hat{a}^{+}\right] =
\hat{1}
$$
Then, remark that the Schr\"{o}dinger equation 
$$
H\left( t\right) \left\vert \Psi \left( t\right) \right\rangle =\mathbf{i}
\hslash \dfrac{\partial }{\partial t}\left\vert \Psi \left( t\right)
\right\rangle \Rightarrow \left\vert \Psi \left( t\right) \right\rangle =
\hat{U}\left( t\right) \left\vert \Psi \left( 0\right) \right\rangle
$$
accept the unitary operator $\hat{U}\left( t\right) $ as the time dependent
solution with $\hat{U}\left( 0\right) =\hat{1}$. We thus have the following
equation : 
$$
H\left( t\right) \hat{U}\left( t\right) =\mathbf{i}\hslash \dfrac{\partial }{
\partial t}\hat{U}\left( t\right) \Rightarrow H \left(t\right)=\mathbf{i}
\hslash\frac{\partial U}{\partial t}U^{+}\left(t\right)
$$
corresponding to the Schr\"{o}dinger one. The second expression is obtained
from the first one by multiplication by $U^{+}$ on the right. Now, because
all the operators appearing in $H\left( t\right) $ form a Lie algebra, the
general form of $\hat{U}\left( t\right) $ is : 
$$
\hat{U}\left( t\right) =e^{\alpha _{0}\left( t\right) \hat{a}^{+}\hat{a}
}e^{\alpha _{+}\left( t\right) \hat{a}^{+}}e^{\alpha _{-}\left( t\right) 
\hat{a}}e^{\alpha \left( t\right) \hat{1}}
$$
where the functions $\alpha \left( t\right) $, $\alpha _{0}\left( t\right) $
, $\alpha _{+}\left( t\right) $ and $\alpha _{-}\left( t\right) $ have to be
find. Be carfeull, these $\alpha $'s have nothing to do with the $\alpha $
defined in the main text. To do that, let us first calculate 
\begin{eqnarray*}
\dfrac{\partial \hat{U}}{\partial t}\hat{U}^{+}&=&\left[ \dfrac{\partial
\alpha _{0}}{\partial t}\hat{a}^{+}\hat{a}+\dfrac{\partial \alpha }{\partial
t}\right] +\dfrac{\partial \alpha _{0}}{\partial t}e^{\alpha _{0}\left(
t\right) \hat{a}^{+}\hat{a}}\hat{a}^{+}e^{-\alpha _{0}\left( t\right) \hat{a}
^{+}\hat{a}} \\
&&+\dfrac{\partial \alpha _{-}}{\partial t}e^{\alpha _{0}\left( t\right) 
\hat{a}^{+}\hat{a}}e^{\alpha _{+}\left( t\right) \hat{a}^{+}}\hat{a}
e^{-\alpha _{+}\left( t\right) \hat{a}^{+}}e^{-\alpha _{0}\left( t\right) 
\hat{a}^{+}\hat{a}} \\
&=&\left[ \hat{a}^{+}\hat{a}\dfrac{\partial \alpha _{0}}{\partial t}+\dfrac{
\partial \alpha }{\partial t}+\hat{a}^{+}\dfrac{\partial \alpha _{+}}{
\partial t}e^{\alpha _{0}\left( t\right) }-\alpha _{+}\left( t\right) \dfrac{
\partial \alpha _{-}}{\partial t}+\hat{a}\dfrac{\partial \alpha _{-}}{
\partial t}e^{-\alpha _{0}\left( t\right) }\right]
\end{eqnarray*}
with the help of the following relations : 
$$
e^{\lambda \hat{a}^{+}\hat{a}}\hat{a}=\hat{a}e^{\lambda \hat{a}^{+}\hat{a}
}e^{-\lambda }~;~e^{\lambda \hat{a}^{+}\hat{a}}\hat{a}^{+}=\hat{a}
^{+}e^{\lambda \hat{a}^{+}\hat{a}}e^{\lambda }\text{ and }e^{\lambda \hat{a}
^{+}}\hat{a}=\left( \hat{a}-\lambda \right) e^{\lambda \hat{a}^{+}}
$$
plus the definition of unitary operator $UU^{+}=\hat{1}$. We can
equivalently use the well-known Hadamard lemma, which states that 
$$
e^{\hat{S}}\hat{A}e^{-\hat{S}}=\hat{A}+\left[\hat{S},\hat{A}\right]+\frac{1}{
2!}\left[\hat{S},\left[\hat{S},\hat{A}\right]\right]+\frac{1}{3!}\left[\hat{S
},\left[\hat{S},\left[\hat{S},\hat{A}\right]\right]\right]+\ldots
$$
and the commutation relations. Usually, for simple groups, this methods is
faster because the series has only null components after few iterations.
Injecting the above $\left(\partial \hat{U} /\partial t\right)U^{+}$ into
the Schr\"{o}dinger equation $H\left( t\right)=\mathbf{i}\hslash
\left(\partial \hat{U}/\partial t\right)U^{+}$, we have : 
$$
\left\{ 
\begin{array}{l}
\mathbf{i}\hslash \dot{\alpha}_{0}=a_{0} \\ 
\dot{\alpha}-\alpha _{+}\left( t\right) \dot{\alpha}_{-}=0 \\ 
\mathbf{i}\hslash \dot{\alpha}_{-}e^{-\alpha _{0}\left( t\right)
}=a_{-}\left( t\right) \\ 
\mathbf{i}\hslash \dot{\alpha}_{+}e^{\alpha _{0}\left( t\right)
}=a_{+}\left( t\right)
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{l}
\alpha _{0}=a_{0}\left( t-t_{\text{i}}\right) /\mathbf{i}\hslash \\ 
\\ 
\alpha _{-}\left( t\right) =\displaystyle\int_{t_{\text{i}}}^{t}a_{-}\left(
\tau \right) e^{a_{0}\tau }\dfrac{d\tau }{\mathbf{i}\hslash } \\ 
\\ 
\alpha _{+}\left( t\right) =\displaystyle\int_{t_{\text{i}}}^{t}a_{+}\left(
\tau \right) e^{-a_{0}\tau }\dfrac{d\tau }{\mathbf{i}\hslash }
\end{array}
\right.
$$
where $\dot{\alpha}_{0}=\partial \alpha _{0}/\partial t$, $\dot{\alpha}=$...
and equating each operator to itself on each side of the Schr\"{o}dinger
equation. We supposed that $t_{\text{i}}$ is the initial time, where $\alpha
_{0}\left( t=t_{\text{i}}\right) =0$, $\alpha \left( t_{\text{i}}\right) =0$
, ... because the unitary evolution operator must verify $\hat{U}\left(
t=0\right) =\hat{U}\left( t,t\right) =\hat{1}$, for the single time $\hat{U}
\left( t\right) $ or the two times $\hat{U}\left( t_{\text{i}},t_{\text{f}
}\right) $ unitary evolution operator. Now let us find the phase factor $
\alpha \left( \tau \right) $. By virtue of an integration by parts, we have
: 
\begin{eqnarray*}
\alpha \left( t\right) &=&\int_{t_{\text{i}}}^{t}\alpha _{+}\left( \tau
\right) \dot{\alpha}_{-}\left( \tau \right) d\tau =\left[ \alpha _{+}\left(
t\right) \alpha _{-}\left( t\right) \right] _{t_{\text{i}}}^{t}-\int_{t_{
\text{i}}}^{t}\dot{\alpha}_{+}\left( \tau \right) \alpha _{-}\left( \tau
\right) d\tau \\
&=&\int_{t_{\text{i}}}^{t}\dfrac{\alpha _{+}\left( t\right) \dot{\alpha}
_{-}\left( t\right) -\dot{\alpha}_{+}\left( t\right) \alpha _{-}\left(
t\right) }{2}dt+\dfrac{1}{2}\left[ \alpha _{+}\left( t\right) \alpha
_{-}\left( t\right) \right] _{t_{\text{i}}}^{t}
\end{eqnarray*}
for a more convenient form. If this last form might be more complex than the
previous ones, it appears to be easier to evaluate in our case, because $
\alpha _{-}\left( t\right) =-\alpha _{+}^{\ast }\left( t\right) $ (this is
true as long as the choosen Hamiltonien is Hermitic), so that : 
$$
\alpha \left( t\right) =-\mathbf{i}\int_{t_{\text{i}}}^{t}\text{Im}\left\{
\alpha _{+}\left( \tau \right) \dot{\alpha}_{+}^{\ast }\left( \tau \right)
\right\} d\tau -\dfrac{1}{2}\left[ \left\vert \alpha _{+}\left( \tau \right)
\right\vert ^{2}\right] _{t_{\text{i}}}^{t}
$$
which is the form used in the main text. Now, because $t_{\text{i}}$ is the
initial time, $\alpha _{+}\left( t_{\text{i}}\right) =0$, and thus $\left[
\left\vert \alpha _{+}\left( \tau \right) \right\vert ^{2}\right] _{t_{\text{
i}}}^{t}=$ $\left\vert \alpha _{+}\left( t\right) \right\vert ^{2}$. Remark
that the previous expression is just one of the several possibilities to
express $\alpha\left(t\right)$. We have in general 
\begin{eqnarray*}
\alpha\left(t\right)&=&-\int\alpha_{+}d\alpha_{+}^{*}=-\int\alpha_{-}^{*}d
\alpha_{-} \\
&=&-\int\alpha_{-}d\alpha_{-}^{*}+\left\vert\alpha_{-}\right\vert^{2}=-\int
\alpha_{+}^{*}d\alpha_{-}+\left\vert\alpha_{+}\right\vert^{2}
\end{eqnarray*}
for this phase term.
