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 $F(t)$


\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?

  • $\begingroup$ Your <E(t)> and $<E^2(t)>$ are independent of time: no matter how long the force acts, your expressions give the same energy and dispersion. This doesn't sound right, does it? The problem is exactly solvable and you should take the averages <E(t)> and $<E^2(t)>$ using the unperturbed Hamiltonian (after the force ceases) wrt the evolved state at time t, not wrt the initial (ground) state. $\endgroup$
    – udrv
    Sep 12, 2015 at 19:39
  • $\begingroup$ This is exactly what I did. But I don't like the $\Delta E=0$. I think it is wrong but I don't know why. $\endgroup$
    – Jon Snow
    Sep 13, 2015 at 12:09
  • $\begingroup$ See my answer below, was too long for comment. Hope it helps. $\endgroup$
    – udrv
    Sep 13, 2015 at 17:49

1 Answer 1


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 $\langle \Delta E^2 \rangle \neq 0$.


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