Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Dissipation in quantum systems

+ 3 like - 0 dislike
766 views

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:
\begin{align}
H(t) = H + \mathcal{O}\phi(t),
\end{align}
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.
\begin{align}
E(t) = \text{Tr}(\rho(t) H),
\end{align}
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
\begin{align}
E(t) = \text{Tr}(\rho(t) H(t) )
\end{align}
and so the work done is
\begin{align}
\frac{dE}{dt} = \text{Tr}(\rho(t) \mathcal{O}) \dot\phi(t)
\end{align}
and in linear response theory the average work done is 
\begin{align}
\frac{\overline{dE}}{dt} = 2 \Omega \chi''(\Omega)|\phi_0|^2
\end{align}
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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...