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  Dissipation in quantum systems

+ 3 like - 0 dislike

I'm confused by what it means to calculate the work done in a quantum system (refer to [Tong's notes on kinetic theory, Sec 4.3.1]) in linear response theory. The set-up is as follows:

Assume we initially have a closed system with static Hamiltonian $H$, so that we can define energy eigenstates with energy eigenvalues etc.

 Then I perturb it by a source so that the Hamiltonian is now:
H(t) = H + \mathcal{O}\phi(t),
so that in the Schrodinger picture the density matrix evolves as 
$\rho(t) = U(t) \rho_0 U(t)^\dagger$ where the unitary time operator satisfies $i \partial_t U(t) = H(t) U(t)$ with appropriate initial condition.

Given that the Hamiltonian is now time-dependent, the concept of energy is lost. So what does it mean to calculate the 'work done' on the system? How are we measuring the 'energy' pumped into the system? What is the operator that measures this 'energy'?

I would have argued as follows: it makes sense to measure energy with respect to the original, unperturbed Hamiltonian $H$. This is because it is $H$ that has the originally the concept of equilibrium, energy eigenstates and eigenvalues etc. So, assuming we start from an equilibrium state  $\rho_0 = e^{-\beta H}$ of $H$, the drive will perturb it out of equilibrium ($\rho(t) \neq \rho_0$) and cause it to populate different energy eigenstates -- and now what we want to do is measure this non-equilibrium state's energy i.e.
E(t) = \text{Tr}(\rho(t) H),
so that the work done is $dE(t)/dt = \text{Tr}(\dot\rho(t) H)$.

However, in Tong's notes, he seems to define 'energy' as measured by $H(t)$, i.e. the 'instantaneous energy', so that
E(t) = \text{Tr}(\rho(t) H(t) )
and so the work done is
\frac{dE}{dt} = \text{Tr}(\rho(t) \mathcal{O}) \dot\phi(t)
and in linear response theory the average work done is 
\frac{\overline{dE}}{dt} = 2 \Omega \chi''(\Omega)|\phi_0|^2
which is the imaginary part of the response function. 

But my question is why define 'energy' or work done by the instantaneous Hamiltonian $H(t)$? Why is that the physical energy? How should I think about it? My interpretation makes sense to me: $H$ is static and thus has a time-invariant spectrum, and what the perturbation $\mathcal{O}$ does is cause the population over this spectrum (i.e. the decomposition of the state over this spectrum) to move in time, and correspondingly with it there is an associated energy change. But such an interpretation is apparently incorrect.


asked Apr 22, 2017 in Theoretical Physics by nervxxx [ no revision ]

1 Answer

+ 1 like - 0 dislike

In one word: dissipation.

Tong's approach is correct, and to see why, it is helpful to consider an analogous classical system, which he does in sections 4.2.1 and 4.2.2.

Basically, the reason for considering the time-dependent Hamiltonian is that the time-evolution of the system is not energy conserving. The system is dissipating energy (perhaps to a heat reservoir of some kind), and the way the Hamiltonian is set up (both in the classical and quantum case) encapsulates this. He goes into further detail in this in the next section, where he explicitly derives the form that the dissipation takes.

Imagine you have a pendulum that has friction, and is initially stationary, and you give it a push. You come back an hour later and the pendulum is again stationary. It would be incorrect to say you performed 0 work on the system. Instead, you have to say that you performed a positive amount of work during some period, and then that energy was dissipated.

answered Sep 10, 2018 by AlNejati (10 points) [ revision history ]

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