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I was reading this StackExchange answer, which I will briefly summarize here. There is a room that is at a fixed temperature and is isolated from the environment. There is a block of ice inside the room. As the block melts, the temperature of the room will slightly decrease. The net entropy change of the room + ice will be positive (e.g. proportional to $1/T_{\rm ice} - 1/T_{\rm room}$).

Now my confusion. The room + ice is a mixed quantum state; somehow or another the universe created a pure state such that the corresponding reduced density operator describes the room + ice before the ice begins to melt. Then the room + ice becomes isolated from the rest of the universe. There is an interaction between the room and the ice that causes the ice to melt and the temperature of the room to decrease. If the interaction was unitary, then the entropy of the room + ice must remain constant throughout the melting process, so clearly the interaction must not be unitary. But we are considering that the room + ice is isolated from the rest of the universe, so how can it be that the interaction is not unitary?

I thought the only way in which evolution in nature could be non-unitary is when they are Completely Positive Trace Preserving maps on a subsystem, meaning that they are unitary on the whole system. But in this situation, the entire interaction is solely between the room and ice, so it seems like it should described without any reference to a bigger system.

To be a little more clear, the entropy of a subsystem is a measure of how much that subsystem is entangled to the environment. It is not a measure of how much that subsystem is entangled with itself. So the entropy of the room + ice does not characterize how much the room and ice are entangled. So indeed the entropy of the ice will change, and the entropy of the room will change. But when looking at the room and the ice as a single isolated system, I don’t see how that changes.

Does the Von Neumann entropy not reproduce the thermodynamic entropy in this case? If so, when people say that the entropy of an isolated system can increase, is that just a classical approximation? If the Earth, for example, were to somehow be isolated from space such that it received no interaction from anything outside the atmosphere, would its thermodynamic entropy increase while its Von Neumann entropy remained the same? What does that even mean?

Any thoughts would be much appreciated, thanks!

Joe
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    This is not unlike the problem that the classical mechanics describing the particles in e.g. two mixing gases is deterministic. – Norbert Schuch Mar 02 '21 at 19:46
  • @NorbertSchuch Can you please elaborate? I don’t quite follow what you’re saying. – Joe Mar 02 '21 at 23:32
  • If you describe a classical gas on the level of particles, there is also no increase in entropy, as they behave deterministically. Yet, e.g. the mixing of two gases clearly increases entropy. – Norbert Schuch Mar 03 '21 at 00:00
  • @NorbertSchuch Do you have a reference for this? I guess I'm surprised that physicists talk about different types of entropy that are in some sense incompatible. It seems to me that an isolated thermodynamic system is still an isolated quantum system, and therefore it seems odd to say that its entropy will increase even though a quantum system's entropy will not. – Joe Mar 03 '21 at 00:27
  • Even classically, the concept of entropy comes from averaging all microscopic configurations compatible with the macroscopic thermodynamic observables, i.e. neglecting all microscopic degrees of freedom. This should be discussed at the beginning of every statistical mechanics textbook. – Norbert Schuch Mar 03 '21 at 00:35
  • @NorbertSchuch Yes I'm comfortable with that. If we were to make the assertion that the universe is a closed system (not expanding, etc?), then it seems like the statement is that the statistical mechanical entropy of the universe is always increasing while the Von Neumann entropy is constant. Would we see a "heat death" of the universe? It seems we would always be a coherent quantum state, never approaching such an end. But the thermodynamic statement that the entropy is indeed increasing seems to suggest that we would see that end. Apologies for the potentially strange discussion. – Joe Mar 03 '21 at 01:44

1 Answers1

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Copenhagen interpretation: At time $t_1$ wave-funtion w is such that it will most probably collapse into simple measurements.

At time $t_2$ said wave-function has evolved so that it most probably collapses into complicated measurements.

Always the situation can be described by the same wave-equation and one number - the time. Except not after the measurement.

Many worlds interpretation: At time $t_1$ wave-funtion w is such that it causes a small amount of decohering of an observer that interacts with it.

At time $t_2$ said wave-function has evolved so that it causes much more decohering of an observer that interacts with it.

Always the situation can be described by the same wave-equation and one number - the time. Except not after the interaction.

stuffu
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