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:


the Lagrangian:


and the correlation operator:


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:


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?


2 Answers 2


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}

  • $\begingroup$ Thank you very much. Can you say why such an assumption is made? Or a better question, what assumption one must have in order to claim that the position and momentum operators are not explicitly time dependent and the only explicit time dependence for the Hamiltonian is due to the time dependent frequency of the oscillations? $\endgroup$
    – Ali Pedram
    Commented Oct 14, 2021 at 11:22
  • 1
    $\begingroup$ What this assumption is effectively saying is that a particle with no momentum and with no forces acting on it will not start moving to the left because that is somehow "just the fundamental nature of the position operator". The $\frac{\partial}{\partial t}$ term represents some sort of external influence not generated by the system dynamics, so here $H$ has an explicit time dependence because we are reaching in and changing the fields being applied to the system. An explicitly time dependent $Q$ or $P$ could imply some sort of moving or non-inertial reference frame. $\endgroup$ Commented Oct 14, 2021 at 11:30

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 $a^\dagger$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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