For an isolated quantum system, one can study the time evolution of entanglement entropy after a quantum quench (always a pure state), which has a rich behaviour in various different models.

However, one can study the classical limit of the quantum system, which also has non-trivial time evolution of other physical observables, such as magnetisation in a spin chain. Sometimes one can find a classical-quantum correspondence of the dynamics of some physical observables after the quench.

I'm wondering if there exists a similar classical concept that links to the quantum entanglement entropy. It seems to me that it cannot be classical thermodynamic entropy, since the classical counterpart doesn't have a well defined temperature.

  • $\begingroup$ well, if you take the "classical limit" of a state you get some totally mixed state, and therefore its entanglement entropy becomes the regular Shannon entropy of the classical state/probability distribution. Or am I missing something? $\endgroup$ – glS May 29 at 9:24
  • $\begingroup$ Actually I don't get mixed state at all! As I mentioned in the question, there is no "thermodynamics" after taking the classical limit. For instance, I take a ferromagnetic state of a 1D spin chain, the classical limit is still a ferromagnetic "configuration" in my classical theory (say ferromagnetic spin wave theory). (of course it is almost impossible to describe a highly entangled quantum state à la this). In the classical field theory, it is just an initial value problem, which does NOT have a thermodynamic entropy for instance. $\endgroup$ – Exhaustive May 29 at 13:01
  • $\begingroup$ well, how do you define "classical limit" in general then? $\endgroup$ – glS May 29 at 13:09
  • $\begingroup$ For instance, section 4 of arxiv.org/abs/1708.07192 in another word, take $\hbar$ to 0. Of course this doesn't say anything about the state, merely the hamiltonian. But the states (at least the product states) can be obtained similarly. $\endgroup$ – Exhaustive May 29 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.