Time Derivative of the Hamiltonian for a Quantum Simple Harmonic Oscillator

Question

I am reading an article on quantum refrigerator. Here is the link of the article. The arXiv version is available here. The working medium is an ensemble of non-interacting particles in a harmonic potential. The authors argue that under certain assumptions we can describe the state of the system along the cycle (which is a reversed Otto cycle), using 3 operators. These operators are the Hamiltonian:

$\hat{H(t)}=\frac{1}{2m}\hat{P}^2+\frac{1}{2m}[\omega(t)]^2\hat{Q}^2$

the Lagrangian:

$\hat{L(t)}=\frac{1}{2m}\hat{P}^2-\frac{1}{2m}[\omega(t)]^2\hat{Q}^2$

and the correlation operator:

$\hat{C(t)}=\frac{1}{2}\omega(t)(\hat{Q}\hat{P}+\hat{P}\hat{Q})$

This is due to the fact that this set of operators form a closed Lie algebra. Hence:

$\hat{\rho} = \hat{\rho}(\hat{H},\hat{L},\hat{C})$

In the adiabatic strokes of the cycle, the system doesn't interact with the environment (closed system) therefore the time evolution of any operator can be given by:

$\frac{d\hat{O}(t)}{dt}=\frac{i}{\hbar}[\hat{H}(t),\hat{O}(t)]+\frac{\partial \hat{O}(t)}{\partial t}$

Then, they claim that in the adiabatic stroke the time evolution of the Hamiltonian can be given as:

$\frac{d}{dt}\hat{H}=\frac{\dot{\omega}}{\omega}(\hat{H}-\hat{L})$

I don't understand how they derived this. In my attempt I write the Hamiltonian in terms of the ladder operators:

$\hat{H}=\hbar \omega(t) (a^{\dagger}a+\frac{1}{2})$

The Hamiltonian commutes with itself. Therefore we only need to calculate the explicit time derivative:

$\frac{\partial \hat{H}}{\partial t} = \hbar \dot{\omega}(a^{\dagger}a+\frac{1}{2})+\hbar \omega (\dot{a}^{\dagger}a+a^{\dagger}\dot{a})$

The first term on the right hand side is indeed $\frac{\dot{\omega}}{\omega}\hat{H}$. However, in my calculations I couldn't verify that the second term is $-\frac{\dot{\omega}}{\omega}\hat{L}$. In order to express the time derivatives of the ladder operators I wrote them in terms of the position and the momentum operators and used the fact that for this particular system:

$\dot{\hat{Q}}=\frac{\hat{P}}{m}, \quad \dot{\hat{P}}= -m\omega^2\hat{Q}$

Which, upon calculation didn't give me the desired result. What am I doing wrong?

Answer 1

As you correctly state, the Hamiltonian commutes with itself for all times, so we only need to consider the explicit time dependence. The operators $P$ and $Q$ are assumed to have no explicit time dependence (They may have a time dependence due to the dynamics of the system, but this is taken into account in the vanishing commutator term). We therefor have \begin{align} \frac{\partial}{\partial t} H &= \frac{\partial}{\partial t}\left[\frac{1}{m} P^2 + \frac{1}{2}m\omega(t)^2 Q^2\right]\\ &= \frac{1}{2}mQ^2 \frac{\partial}{\partial t} \omega^2\\ &= m \omega \dot{\omega}Q^2\\ &= \frac{\dot\omega}{\omega}\left(H-L\right) \end{align}

Answer 2

You can't use $a$, $a^\dagger$ when the frequency $\omega$ is time dependent. $H(t)$ is not simply $\omega(t) (a^\dagger a+1/2)$ becaue the frequency change chnges the definition of $a$ in terms of $x$ and $p$, and so mixes $a$ with $a^\dagger$.