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The standard Damped one-dimensional Harmonic Oscillator with sinusoidal driving force has equation

$$\frac{d^2}{dt^2}x(t)+2\zeta\omega_0\frac{d}{dt}x(t)+\omega_0^2x(t)=\frac{1}{m}F_0\sin(\omega t).$$

Here is $\zeta>0$ for damping.

The solutions of this system are linear combinations of a transient function (that goes to 0 with time) and a steady-state solution that is periodic with frequency $\omega$ (see Wikipedia).

I would like to quantize this to a Quantum Harmonic Oscillator. In the case without damping and driving force we get

$$i\hbar \frac{\partial}{\partial t}\psi = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac12 m\omega_0^2x^2\right)\psi$$

To add the driving force, we can create a potential energy field $U(x,t)$ such that the driving force is the gradient of this. Then add $U$ to the quantum hamiltonian:

$$i\hbar \frac{\partial}{\partial t}\psi = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac12 m\omega_0^2x^2+F_0\sin(\omega t)x\right)\psi$$

Now we have to add the damping term. But I am not sure how to do that. My question is then: Are the steady-state solutions of this system also periodic with frequency $\omega$? Is there an easy way to prove that?

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  • $\begingroup$ What do you call “stationary solutions” in this case? It is not at all obvious that there will be solutions proportional to $\exp(-i\Omega t)$. $\endgroup$
    – Gec
    Commented Jan 24 at 14:16
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    $\begingroup$ 1. Your second and third equations are not the Schrödinger equations of a harmonic oscillator (dimensional check!). $\partial^2/\partial t^2$ on the LHS should be replaced by $\partial / \partial t$ and on the RHS, a $-\hbar^2$ is missing in the kinetic term of the Hamilton operator. 2. Regarding your first equation as the equation of motion for the position operator in the Heisenberg picture, the problem can, of course, be solved quantum mechanically. However, solving the einenvalue problem of the time-dependent Hamiltonian is not extremely useful in this case. $\endgroup$
    – Hyperon
    Commented Jan 24 at 14:21
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    $\begingroup$ I took the liberty of correcting typos $\endgroup$
    – Gec
    Commented Jan 24 at 14:24
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    $\begingroup$ By the confusing term "steady state" Wikipedia means just a special solution of the inhomogeneous differential equation. As your first equation is a linear differential equation, the solution of the QM Heisenberg equation of motion is exactly the same as in classical mechanis, just replacing $x_0$ and $p_0$ by operators. Does this answer your question? $\endgroup$
    – Hyperon
    Commented Jan 24 at 14:34
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    $\begingroup$ There is no such thing as stationary states of Hamiltonians such as above. In order to study such systems, you need to use the "Floquet formalism" for driven systems. There, you will recover something called Floquet states and Floquet quasienergies which describe how solutions to the time dependent Schrödinger equation evolve in time. $\endgroup$
    – Physiker
    Commented Jan 24 at 15:12

3 Answers 3

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It is usually convenient to write problems about a quantum harmonic oscillator in terms of the creation-annihilation operators $a^\dagger$, $a$. The Schrödinger equation now looks like $$ i\hbar\frac{\partial}{\partial t}|\psi, t\rangle = \left(\hbar\omega_0\ a^\dagger a + \alpha\sin(\omega t)(a+a^\dagger)\right)|\psi,t\rangle $$ Let's look for a solution to this equation in the form of a coherent state: $$ |\psi,t\rangle = A(t)\exp(\phi(t)\ a^\dagger)|0\rangle $$ The functions $A(t)$, $\phi(t)$ must satisfy the following equations $$ i\hbar\dot{A} = A\alpha\sin(\omega t)\phi,\quad i\hbar\dot{\phi} = \hbar\omega_0\phi + \alpha\sin(\omega t) $$ These equations posses periodic solution of the form $$ \phi(t) = -\frac{\alpha}{2i\hbar}\left(\frac{e^{i\omega t}}{\omega+\omega_0} + \frac{e^{-i\omega t}}{\omega-\omega_0}\right), $$ $$ A(t) = A(0)\exp\left(-i\frac{\alpha^2}{2\hbar^2}\frac{\omega_0}{\omega^2-\omega_0^2}t \right)\exp\left(-\frac{\alpha^2}{8\hbar^2\omega}\left( \frac{e^{i2\omega t}}{\omega+\omega_0} + \frac{e^{-i2\omega t}}{\omega-\omega_0} \right)\right) $$ Due to the phase factor in $A(t)$, the vector $|\psi,t\rangle$ is not strictly periodic with frequency $\omega$. However, the average value of the coordinate, for example, demonstrates the expected time dependence $$ \langle x\rangle \sim \langle a + a^\dagger\rangle = \frac{\alpha}{\hbar} \frac{2\omega_0}{\omega^2-\omega_0^2}\sin(\omega t) $$ I suppose there may be other solutions with properties similar to those of an explicitly constructed solution.

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  • $\begingroup$ Thank you, 'periodic up to a phase factor' seems to be the perfect description $\endgroup$
    – Riemann
    Commented Jan 24 at 16:25
  • $\begingroup$ Now we could use the fact that coherent states form an overcomplete basis $\endgroup$
    – Riemann
    Commented Jan 24 at 18:33
  • $\begingroup$ @Riemann I do not see here an infinite system of solutions in the form of coherent states. At the moment, only one solution has been found. But I have a suspicion that all solutions can be found for this non-stationary Schrödinger equation. And it is possible that even in the case of an arbitrary dependence of force, $F(t)$, on time. $\endgroup$
    – Gec
    Commented Jan 24 at 18:42
  • $\begingroup$ You're right, only one coherent state works (up to a phase factor) $\endgroup$
    – Riemann
    Commented Jan 24 at 20:38
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    $\begingroup$ @Riemann Yes, in the case where the ratio $\omega_0/\omega$ is not a rational number, the new solution will not be periodic. The general solution of the classical equation is also not periodic. $\endgroup$
    – Gec
    Commented Jan 24 at 21:03
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Almost. The only exception being when you hit resonance $\omega = \omega_0$. Then you don't have any steady state solutions at all.

To see this quantum-mechanically, you can go to a rotating frame by a unitary transformation

$\hat{U} = e^{-i \, a^{\dagger} \!a \,\omega t}$

on your Hamiltonian

$ H = \hbar \omega_0 ( a^{\dagger} \!a +1/2) + \lambda ( a^{\dagger} + a) \cos(\omega t)$

(here I took liberty to redefine your force as $F(t) \propto \cos(\omega t)$ rather than $\sin (\omega t)$, which doesn't change anything conceptually, but is more convenient to make my point; and also for simplicity defined a constant for force $\lambda$ which is equal to the product of $F_0$ and some constants of the oscillator).

Now, within rotating frame approximation your Hamiltonian can be written as

$\hat{H}_{rot}= \hat{U}^{\dagger} H \hat{U} - i\hbar \hat{U}^{\dagger} \frac{\partial \hat{U}}{\partial t} = \hbar \omega_0 ( a^{\dagger} \!a +1/2) -\hbar \omega \, a^{\dagger} \!a + \lambda (a + a^{\dagger}) + \{\mathrm{quickly\,oscillating \, terms}\}$

As you see, as long as $\omega \neq \omega_0$, one deals with the standard problem of a QHE with constant field, which upon returning to the original reference frame, produces a solution oscillating at $\omega$, as one would expect.

Conversely at $\omega = \omega_0$ one looses the restoring potential and there is no well-defined solution anymore, even in the rotating frame. Physically this corresponds to resonance with no stationary state solution.

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  • $\begingroup$ According to my calculations, the quickly oscillating terms are equal to $\frac{\lambda}{i}\sin(\omega t)(a^{\dagger}-a)$. Why can we ignore these? $\endgroup$
    – Riemann
    Commented Jan 24 at 17:01
  • $\begingroup$ The quickly oscillating terms must be (in my $\cos$ case) $\propto a \exp \{ -2i\omega t\} + a^{\dagger} \exp \{ 2i\omega t \}$. If we assume that $\omega \approx \omega$ (but not equal!), then these terms are too far away from resonance and hence their effects average out over time. $\endgroup$
    – John
    Commented Jan 24 at 17:20
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    $\begingroup$ For reference, this is also called the rotating wave approximation. $\endgroup$
    – march
    Commented Jan 24 at 17:38
  • $\begingroup$ @John I would be interested in a proof that these oscillating terms average out over time $\endgroup$
    – Riemann
    Commented Jan 24 at 18:06
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The problem of a Damped and Forced Quantum Harmonic Oscillator has been studied in this article. It turns out that adding the damping term is quite tricky. You can't just add a term to the Hamiltonian because there is nontrivial interacting with the environment. The system becomes entangled with the environment, and therefore we have to use the density operator instead of the wavefunction. The Schrödinger Equation gets replaced by a Lindbladian (Master Equation).

In the article, they study (among others) the case of harmonic driving force $f(t)=f_0\cos(\Omega t)$. In the limit $t\rightarrow\infty$, the density operator $\rho(t)$ converges to a limit cycle with frequency $\Omega$ (Equations 99 and 101). So we can write $\rho(t)$ as the sum of a transient state and a periodic state, just as in the classical case.

The section containing Equations 99 and 101 is called 'A special set of solutions'. So it seems that they apply only to certain solutions. In section 6 (Concluding remarks) they say:

In particular we have derived an analytic expression for the asymptotic density operator for harmonic driving, the limit cycle.(...) The present analysis covers, however, only a special class of solutions and the derivation of a general solution remains a challenging problem for future studies as well as a deeper study of the interrelation between the Lindblad description and non-Hermitian Hamiltonians.

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  • $\begingroup$ Expectation values are obviously going to obey these identities. Ehrenfest's theorem is just Newton's laws in disguise. $\endgroup$
    – Physiker
    Commented Jan 24 at 15:11

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