# Many-body harmonic oscillator with time-dependent modes

Consider the following problem of $N$ noninteracting particles described by the harmonic oscillator Hamiltonian $$\hat{H}(t) = \hbar \sum\limits_{k=0} \epsilon_k(t) \hat{a}^{\dagger}_k(t) \hat{a}_k(t),$$ where $\hat{a}^{\dagger}_k(t)$ creates a particle in single-particle mode wavefunction $\phi_k(x,t)$ which is time-dependent. I do not specify what is the form of the single-particle states, but one can imagine a situation where $N$ particles are placed in a harmonic potential with time-dependent frequency or periodic potential with time dependent amplitude. In other words I consider just any time-dependent potential $V(x,t)$.

We start evolution at time $t=0$ with all particles in the lowest energy state $k=0$: $|\psi(t=0)\rangle = |N_{\phi_0(t=0)}\rangle$ or $$|\psi(t=0)\rangle = \frac{\left[ \hat{a}^{\dagger}_0(t=0)\right]^N}{\sqrt{N!}}|0\rangle.$$ During the time interval $t \in [0,t_f]$ we change the potential $V(x,t)$ in some way (general case) and we would like the study the efficiency of a transfer from state $|\psi(t=0)\rangle$ to $$|\psi(t)\rangle = \frac{\left[ \hat{a}^{\dagger}_0(t)\right]^N}{\sqrt{N!}}|0\rangle.$$ In other words, how efficient is the transfer of the condensate from state $\phi(x,t=0)$ to the state $\phi(x,t)$. Of course, transfer of all the particles is possible only when the process is fully adiabatic. For finite times $t_f$ condensate fraction will be depleted because some fraction of particles will get excited to other higher-energy modes. I would like study occupation of the instantaneous lowest energy mode $\langle \hat{n}_0(t)\rangle$ and thus have the fraction of particles that are excited (or depletion fraction).

I wanted to do this in the Heisenberg picture by writing the equation of motion for the creation operator $$\frac{d}{dt}\hat{a}^{\dagger}_k(t) = \frac{i}{\hbar}[\hat{H}(t),\hat{a}^{\dagger}_k(t)] + \hat{U}^{\dagger}(t) \frac{\partial \hat{a}^{\dagger}_k(t)}{\partial t} \hat{U}(t),$$ where $$\hat{U}(t) = \mathcal{T} \exp\left( -\frac{i}{\hbar}\int\limits_{0}^{t} dt'\ \hat{H}(t')\right).$$

I have the problem with the second term on the right-hand side. I don't have a clue how to relate it somehow to modes $\hat{a}^{\dagger}_k(t)$ so that I can solve the coupled differential equation.

The specific question is: How to solve the Heisenberg equation of motion if the operators are time dependent (the second term on the right-hand side remains)? Is there any general method? How do I interpret $\partial \hat{a}^{\dagger}_k(t)/\partial t$?

I am posting this as an example, but I want to apply the method to some other problem.

### UPDATE

I may have some idea how to bring the equation to a form that can be subsequently analyzed. Let's change a little bit notation and introduce operator $\hat{A}_k(t)$ which is the equivalent of $\hat{a}_k(t)$, but in the Heisenberg picture. So the previous equation of motion should look like this

$$\frac{d}{dt}\hat{A}^{\dagger}_k(t) = \frac{i}{\hbar}\epsilon_k(t) \hat{A}^{\dagger}_k(t) + \hat{U}^{\dagger}(t) \frac{\partial \hat{a}^{\dagger}_k(t)}{\partial t} \hat{U}(t).$$

Now, operator $\partial\hat{a}^{\dagger}_k(t)/\partial t$ creates a particle in the state $\partial \phi_k(x,t)/\partial t$ which can be decomposed into a linear combination of states $\phi_k(x,t)$. Thus, we can write $$\frac{\partial \hat{a}^{\dagger}_k(t)}{\partial t} = \sum\limits_{k'} C_{k'}(t) \hat{a}^{\dagger}_{k'}(t)$$ $$C_{k'}(t) = \int dx \phi_{k'}^{*}(x,t)\frac{\partial \phi_{k}(x,t)}{\partial t}$$

At the end we get $$\frac{d}{dt}\hat{A}^{\dagger}_k(t) = \left[ \frac{i}{\hbar}\epsilon_k(t) + C_k(t) \right]\hat{A}^{\dagger}_k(t) + \sum\limits_{k' \neq k}\hat{A}^{\dagger}_{k'}(t) C_{k'}(t) .$$

## Interpretation of $\partial \hat{a}^{\dagger}_k(t)/\partial t$

According to definition of derivative I can write $$\partial \hat{a}^{\dagger}_k(t)/\partial t = \lim\limits_{dt \rightarrow 0} \frac{\hat{a}^{\dagger}_k(t + dt) - \hat{a}^{\dagger}_k(t)}{dt}$$ Because at every moment of time I am dealing with complete basis of single-particle states I can write $$\hat{a}^{\dagger}_k(t + dt) = \sum\limits_{k' = 0} A_{k'}(t; dt)\hat{a}^{\dagger}_k(t)$$ where $$A_{k'}(t;dt) = \int\limits dx\ \phi^{*}_{k'}(x,t)\phi_k(x,t+dt)$$ Inserting this back in the equation and using identity $1 = \int\limits dx\ \phi^{*}_{k}(x,t)\phi_k(x,t)$, I arrive at $$\partial \hat{a}^{\dagger}_k(t)/\partial t = \hat{a}^{\dagger}_k(t)\lim\limits_{dt \rightarrow 0} \int dx\ \phi^{*}_k(x,t)\frac{\phi_k(x,t+dt) - \phi_k(x,t)}{dt} + \lim\limits_{dt \rightarrow 0} \sum\limits_{k' \neq k} \frac{A_{k'}(t;dt)}{dt} \hat{a}^{\dagger}_{k'}(t)$$ The second term is $0/0$, but using Taylor expansion $\phi_k(x,t+dt) \approx \phi_k(x,t) + \partial\phi_k(x,t)/\partial t dt$, we get the final answer $$\frac{\partial \hat{a}^{\dagger}_k(t)}{\partial t} = \sum\limits_{k' = 0} \hat{a}^{\dagger}_{k'}(t) \int dx\ \phi^{*}_{k'}(x,t) \frac{\phi_k(x,t)}{\partial t}$$ which is exactly what I used previously.

• -1. Not clear what you are asking. You appear to be asking for help with a calculation, which is off topic according to site policy. You must ask about a specific physics concept. – sammy gerbil Mar 18 '17 at 18:35
• @Nex_Friedrich: assuming your original time-dependent ${\hat H}(t)$ is in the Schroedinger picture, do you know the time-dependence of the Schroedinger picture ${\hat a}_k(t)$-s? Perhaps you can still write it as a unitary ${\hat V}(t) {\hat a}_k(t) {\hat V}^\dagger(t)$ and carry it over into the Heisenberg picture for $\partial {\hat a}_k(t)/\partial t$. – udrv Mar 19 '17 at 5:05
• @udrv I don't even know how to interpret $\partial \hat{a}^{\dagger}_k(t)/\partial t$ in the Schrodinger picture. I wonder if it means that I create a particle in state $\partial \phi_k(t)/\partial t$? – WoofDoggy Mar 20 '17 at 10:14
• @Nex_Friedrich $\partial a_k^\dagger(t)/\partial t$ concerns the explicit time-dependence of ${\hat a}^\dagger(t)$ stemming for instance from quantizing a time-dependent potential, as opposed to the time-dependence due to the unitary evolution corresponding to ${\hat H}(t)$. I don't think you should look at it as anything else more "physical". If you prefer, if ${\hat a}^\dagger(t)$ is defined w.r.t. some time-independent counterpart ${\hat a}^\dagger$, you should be able to interpret states created using ${\hat a}^\dagger(t)$ in terms of states created by ${\hat a}^\dagger$, but that's all. – udrv Mar 20 '17 at 18:29
• @udrv I updated my question and included interpretation of partial derivative of creation operator. – WoofDoggy Mar 21 '17 at 13:02