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I am trying to understand how Boltzmann's statistical entropy relates to Von Neumann's quantum entropy. For example, the canonical ensemble describes a system in contact with a thermal bath at fixed temperature $T$, where the probability to be in a state $n$ is

\begin{equation} p_n=e^{-E_n /{k_b T}} \end{equation}

we can compute the entropy of this system as

\begin{equation} S=-\sum_n p_n \ln p_n \end{equation}

Conceptually, what's happening is that our actual system is rather complicated in its interactions with the thermal bath, so we instead make $N$ copies of the system with different definite energies such that the probability distribution follows $p_n$ and then make predictions about the ensemble.

The important thing is that even if we don't know what's happening with the system, our physical real system is very much in some state. However, at the quantum level, if we think of the system as trully being in one state, then the Von Neumann entropy automatically gives as zero entropy. I think this would be solved if we notice that the system is in contact with a thermal bath and thus is entangled with the enviroment. Local measurements on the system would then look as if the system is in a mixed state, thus giving entropy.

My question is: Is classical statistical entropy caused at the quantum level by tracing out the degrees of freedom of the thermal bath? If not, how can we explain that a macroscopical classical system in a definite state has some entropy but if we describe it quantum mechanically it does not?

Related to this, if entanglement with the enviroment is what creates statistical entropy by tracing out the degrees of freedom in the thermal bath, then how can we explain entropy in the microcanonical ensemble?

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According to the Eigenstate Thermalization Hypothesis, the answer is yes. The thermodynamic entropy of an open system is the bipartite (von Neumann) entanglement entropy. Entropy hides information about the initial state throughout the system + environment, so that it is no longer accessible to local operators. Thermal systems, and generic classical systems, thus have an entropy of entanglement (with their environments) that scales with the volume of the system. In the microcanonical ensemble, every individual energy eigenstate (of the full, closed system) is a valid statistical ensemble, where every subsystem has volume-law entanglement with the (larger) rest of the system, which plays the role of the environment. See the Wiki page. I can look for other references if desired.

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The important thing is that even if we don't know what's happening with the system, our physical real system is very much in some state. However, at the quantum level, if we think of the system as trully being in one state, then the Von Neumann entropy automatically gives as zero entropy.

This does not happen in quantum theory/quantum statistical physics for interacting system, because that state is not pure. In case the system interacts with the bath, it is not an isolated system and thus cannot be assigned single "ket" state, or pure "density matrix" state. In other words, its state is mixed, not pure, and von Neumann entropy can't be zero.

It is useful to realize that von Neumann entropy is a function. A function of all density matrix elements, diagonal and off-diagonal. In special basis where the matrix is diagonal, it reduces to the usual information-theoretic entropy formula in terms of probabilities only.

I think this would be solved if we notice that the system is in contact with a thermal bath and thus is entangled with the enviroment. Local measurements on the system would then look as if the system is in a mixed state, thus giving entropy.

This is the standard view, the interacting system does not have its own pure quantum state, only the hypothetical isolated supersystem including the bath does. Thus there is nothing to solve.

My question is: Is classical statistical entropy caused at the quantum level by tracing out the degrees of freedom of the thermal bath?

Not in classical statistical physics. But in quantum statistical physics, density matrix of the system can be got as result of "tracing out" the environment out of the supersystem density matrix, and its typical lack of purity is necessary consequence of allowing interaction.

If not, how can we explain that a macroscopical classical system in a definite state has some entropy but if we describe it quantum mechanically it does not?

You have to differentiate between the von Neumann and thermodynamic entropy. Isolated macroscopic system can have definite pure state and thus zero von Neumann entropy, but non-zero thermodynamic entropy. Just as in classical statistical physics, the Gibbs entropy of delta distribution is zero, but thermodynamic entropy of the corresponding state is not.

Related to this, if entanglement with the enviroment is what creates statistical entropy by tracing out the degrees of freedom in the thermal bath, then how can we explain entropy in the microcanonical ensemble?

In microcanonical ensemble we do not have single pure state either, only set of constraints like $E,V,N$, which define set of compatible states that has many members and thus non-zero information entropy. Thus the Gibbs (in classical statistical physics) or von Neumann (in quantum statistical physics) entropy is not zero.

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  • $\begingroup$ The notion of a mixed state is always subjective. For all we know the complete system with the entanglement with the bath included could be a pure state. It is our ignorance that is being represent in terms of the mixed state. $\endgroup$ Commented Sep 17, 2023 at 3:06

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