Retarded Quantum Harmonic Oscillator Suppose there is a harmonic oscillator and at some time acts on him a force. The external force $F (t)$ being zero before $t = 0$ and after $t = T$ .
The oscillator was in its ground state for all time $t < 0$ until the force
starts acting on it.
Compute the dispersion ∆E of the energy at times $t > T$ , i.e. after the
force ceases its action.
How does the product ∆E · T behave, as a function of T ?
I solved it and i found out that 
\begin{equation*}
\begin{split}
\big< E(t)\big>&=\bigg< n=0\bigg|\hbar\omega\left(\alpha^{\dagger}\alpha+\dfrac{1}{2}\right)\bigg|n=0 \bigg>\\
&=\dfrac{1}{2}\hbar\omega+\hbar\omega \langle n=0|(a^{\dagger}+\mathcal{F}^*)(a+\mathcal{F})|n=0\rangle\\
&=\dfrac{1}{2}\hbar\omega+\hbar\omega|\mathcal{F}|^2
\end{split}
\end{equation*}
\begin{equation}
\big< E(t)\big>=\hbar\omega\left(\dfrac{1}{2}+\dfrac{\vert\tilde{F}(\omega)\vert^2}{2\hbar\omega} \right)
\end{equation}
where $\alpha^{\dagger}$ and $\alpha$ are the new creation and anihhilation operators and $a^{\dagger}$ and $a$ are the old ones (before the action of the force).
$\mathcal{F} = \frac{\vert\tilde{F}(\omega)\vert}{2\hbar\omega}$ and $\tilde{F}(\omega)$ is the fourier transform of the function $F(t)$
and
\begin{equation*}
\begin{split}
\big< E^2(t)\big>&=(\hbar \omega)^2\left(\dfrac{1}{4}+\dfrac{\vert\tilde{F}(\omega)\vert^2}{2\hbar\omega} +\dfrac{\vert\tilde{F}(\omega)\vert^4}{4\left(\hbar\omega\right)^2} \right)
\end{split}
\end{equation*}
And so the uncertainty in the energy is $\Delta E=0$. Is it right?
 A: Say the unperturbed Hamiltonian is $H_0 = \hbar \omega \left( a^\dagger a + \frac{1}{2} \right)$, the perturbed one is $H_1 = \hbar \omega \left( \alpha^\dagger \alpha + \frac{1}{2} \right) = U_0^\dagger H_0 U_0$ and the state at time $t=T$ reads 
$$
|T\rangle = e^{-\frac{i}{\hbar}H_1 T} |0\rangle = U_0^\dagger e^{-\frac{i}{\hbar}H_0 T} U_0 |0\rangle
$$
The formula you use for the energy at time $t>T$ is
$$
\langle E(t) \rangle = \langle 0 | H_1 | 0 \rangle = \langle 0 | U_0^\dagger H_0 U_0 | 0 \rangle
$$ 
But for $t>T$ we have $|t\rangle = e^{-\frac{i}{\hbar}H_0 (t-T)} |T\rangle$, so the corresponding energy is 
$$
\langle E(t) \rangle = \langle t | H_0 | t \rangle = \langle T | H_0 | T \rangle = \langle 0|U_0^\dagger e^{\frac{i}{\hbar}H_0 T} U_0 H_0 U_0^\dagger e^{-\frac{i}{\hbar}H_0 T} U_0 |0\rangle \neq \langle 0 | U_0^\dagger H_0 U_0 | 0 \rangle = \langle 0 | H_1 | 0 \rangle 
$$
since 
$$
e^{\frac{i}{\hbar}H_0 T} U_0 H_0 U_0^\dagger e^{-\frac{i}{\hbar}H_0 T} \neq H_0
$$
Caution:
Eventually it occurred to me that you mentioned a time-dependent force over the interval $[0, T]$, $F = F(t)$, while I considered simply $F = const \neq 0$. But when we consider $F = F(t)$, it turns out that a time independent result may actually be applicable in the limit of $\omega T >> 1$. Here is why:
If $F = F(t)$ and $H_1 = H_1(t)$, then the state at $t=T$ is no longer given by $e^{-\frac{i}{\hbar}H_1 T} |0\rangle$, but reads
$$
|T \rangle = {\mathcal T} e^{-\frac{i}{\hbar}\int_0^T{d\tau H_1(\tau)}} |0 \rangle = e^{-\frac{i}{\hbar}H_0 T} {\mathcal T} e^{-\frac{i}{\hbar}\int_0^T{d\tau V_I(\tau)}} |0 \rangle
$$
where the last expression uses the interaction picture of $H_0$, and 
$$
V_I(t) = e^{\frac{i}{\hbar}H_0 t} V(t) e^{-\frac{i}{\hbar}H_0 t}
$$
for $V(t) = H_1(t) - H_0 = F^*(t)a + F(t) a^\dagger$. This gives 
$$
\langle E(t>T) \rangle = \langle T| H_0 | T \rangle = \langle 0| {\mathcal T} e^{\frac{i}{\hbar}\int_0^T{d\tau V_I(\tau)}} H_0 {\mathcal T} e^{-\frac{i}{\hbar}\int_0^T{d\tau V_I(\tau)}} |0\rangle
$$
But given the simple form of $V(t)$, in the limit of very large $T$, $\omega T >> 1$, we can approximate 
$$
{\mathcal T} e^{-\frac{i}{\hbar}\int_0^T{d\tau V_I(\tau)}} \approx e^{-\frac{i}{\hbar}\int_0^\infty{d\tau V_I(\tau)}} = e^{-\frac{i}{\hbar}\left( {\tilde F}^*(\omega)a + {\tilde F}(\omega)a^\dagger \right) }
$$
and so 
$$
\langle E(t>T) \rangle = \langle 0| e^{\frac{i}{\hbar}\left( {\tilde F}^*(\omega)a + {\tilde F}(\omega)a^\dagger \right) } H_0 e^{-\frac{i}{\hbar}\left( {\tilde F}^*(\omega)a + {\tilde F}(\omega)a^\dagger \right) } |0\rangle = \hbar \omega\left( \frac{1}{2} + \frac{|{\tilde F}(\omega)|^2}{\omega} \right)
$$
is time independent. Similarly $\langle E^2(t>T) \rangle$ and $\langle \Delta E^2 \rangle$ are time-independent, but please note that
$$
\langle E^2(t>T) \rangle = \langle 0| e^{\frac{i}{\hbar}\left( {\tilde F}^*(\omega)a + {\tilde F}(\omega)a^\dagger \right) } H_0^2 e^{-\frac{i}{\hbar}\left( {\tilde F}^*(\omega)a + {\tilde F}(\omega)a^\dagger \right) } |0\rangle = \\
(\hbar \omega)^2 \left[ \frac{1}{4} + |{\tilde F}(\omega)|^2 + \langle 0| \left( a^\dagger - {\tilde F}^*(\omega) \right) \left(a - {\tilde F}(\omega) \right) \left( a^\dagger - {\tilde F}^*(\omega) \right) \left(a - {\tilde F}(\omega) \right)|0\rangle\right] \neq \langle E(t>T) \rangle^2
$$
and so $\langle \Delta E^2 \rangle \neq 0$.
