4

In theoretical studies of Bose-Einstein condensates, it's common to look at the one-body density matrix: $$ n_{ab}=\langle\hat{c}_a^\dagger\hat{c}_b\rangle $$ where the $\hat{c}_j$ are annihilation operators for some single-particle modes. (Often, it's written in the position basis as $n(\mathbf{x},\mathbf{y})=\langle\hat{\psi}^\dagger(\mathbf{x})\hat{\psi}(\mathbf{y})\rangle$, but I'm working in a discrete basis.)

The von Neumann entropy of a system is defined as: $$S=-\text{Tr}\left\{\hat{\rho}\log\left(\hat{\rho}\right)\right\}$$ where $\hat{\rho}$ is the full many-body density matrix. I'm working with a method in which I can't tractably obtain the full many-body density matrix, but frequently use the one-body density matrix. Consider the quantity: $$S^{(n)}=-\text{Tr}\left\{\mathbf{n}\log\left(\mathbf{n}\right)\right\}$$ where $\mathbf{n}$ is the one-body density matrix, with matrix elements $n_{ab}$ as defined in my first equation. In the case of a single particle, the one-body density matrix is the full density matrix, and this would give us the von Neumann entropy. What about for a many-body system? Is it meaningful in any way then? Intuition makes me suspect that it's something like a lowest-order approximation to the full von Neumann entropy that discards many-body effects (and my needs make me hope so!), but I'm not sure how I'd go about formally showing this, or deducing its meaning.

user502382
  • 593
  • 2
  • 12
  • 1
    What precisely do you want to know? Why are you computing the entropy, which is the property you actually want to learn? This is probably rather relevant in assessing whether the single-particle density matrix bears any relevance. -- Note also that if you have Gaussian states, $n_{ab}$ contains all the information about the state. There is also some ways in which Gaussian states are extremal among all states with the same $n_{ab}$ (would have to check in which way), so your entropy might be some upper/lower bound on the true entropy. – Norbert Schuch Apr 07 '20 at 12:33
  • Bogolubov transformation is a mean field theory that is reducing the interacting many-particle system to non-interacting one, and solving it self-consistently. If your system is effectively a non-interacting one, you treat it in single-particle terms. – Roger V. Jun 21 '20 at 15:40
  • Great question! It has a lot of sense to look both at single particle and many body quantities, such as the entropy. By looking at the difference you can quantify the many body effects - for example entanglement entropy, validity of independent electron approximation and so on. Slight different example is the single particle gap in superconductors, while the many body state is gapless. Many examples in condensed matter physics and quantum optics. – Alexander Jun 23 '20 at 18:55
  • 1
    A possible reference Phys. Rev. Lett., 115, 046603 (2015) – Adam Jun 23 '20 at 18:55

2 Answers2

1

The one-body entropy of a system at equilibrium is equivalent to the thermodynamic entropy per particle. It is ubiquitous in discussions of cooling ultracold gases to reach low-entropy ordered states (here is one example of many), and is often referred to as the "entropy per particle" ($S/N$) or just the "entropy" of the gas. This equivalence is a consequence of the Eigenstate Thermalization Hypothesis.

This is often applied to categorize phases of matter by the entropy required. For example, in a homogeneous Bose gas condensation occurs when $S/N=1.3 k_B$. Granted, this is calculated for a noninteracting gas, but it remains a good approximation for a gas with weak repulsive interactions in which the state no longer factorizes.

Edit: Since the bounty asks for sources, here is an experimental work that explores the correspondence between single-body Von Neumann entropy and thermodynamic entropy at equilibrium. They look at the entropy of a single site, rather than a single particle, but since they consider a uniformly filled system these are interchangeable as $S/V=nS/N$. More generally, the Eigenstate Thermalization Hypothesis that is studied applies to all few-body observables. I have discussed this paper more in a related previous question.

Rococo
  • 7,671
  • 1
  • 23
  • 54
0

For a system of identical particles, this paper https://journals.aps.org/pre/abstract/10.1103/PhysRevE.81.021119 defines the quantum entropy (Eq. (6)) in terms of the one-body density matrix.

Chaos
  • 903