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.

  • 1
    $\begingroup$ The evolution operator is the time-ordered exponential of the integral, $T\exp\int H\, dt/i\hbar$. For this Hamiltonian that only depends quadratically on $x,p$, I think that the value may be calculated explicitly, I mean analytically. The resonance near $\Omega=\omega_0$ should be seen in the result, too. I don't know how to write the result on the top of my head now. $\endgroup$ Dec 7, 2012 at 17:58

2 Answers 2


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 $x$.

  • $\begingroup$ I think that in the first equation you have a mistake. In the Heisenberg equation there is the Hamiltonian in the Heisenberg picture. In fact Hamiltonian depend explicitly on time, so $H_S\neq H_H$. And we don't know the Hamiltonian $H_H$, so we can't use Heisenberg equation. $\endgroup$
    – Oiale
    Dec 9, 2012 at 17:39
  • 1
    $\begingroup$ Because $H$ is quadratic its Heisenberg picture is given by simply replacing $p\rightarrow p_H(t)$ and $x\rightarrow x_H(t)$. The commutation relations between $x$ and $p$ do not change for equal time. $\endgroup$
    – Olaf
    Dec 9, 2012 at 17:46
  • $\begingroup$ Thank you for comment, but this statement doesn't clear for me.Could you explain it in details? Namely I don't understand the second equality in your post. $\endgroup$
    – Oiale
    Dec 9, 2012 at 18:00
  • $\begingroup$ So what part is it exactly that you do not understand? $\endgroup$
    – Olaf
    Dec 12, 2012 at 14:01
  • $\begingroup$ @Olaf I would question the replacement procedure. The solution to the Heisenberg equation of $\hat{x}_H(t)$ is the sum of the homogeneous and inhomogeneous solution. The square of $\hat{x}_H(t)$, i.e., $\hat{x}_H(t)^2$ then contains a term quadratic in the inhomogeneous part $\endgroup$
    – bodokaiser
    Jun 15, 2021 at 15:11

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.

  • $\begingroup$ Thank you for such a wonderful answer, but Olaf's answer was closer to what I needed $\endgroup$
    – Oiale
    Dec 12, 2012 at 15:57
  • $\begingroup$ @Caloric No problem. The full unitary evolution can be of help for someone else. Thanks for your interesting question. $\endgroup$
    – FraSchelle
    Dec 12, 2012 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.