Apologies for the long question. This problem has been mentioned multiple times on this platform, with no real resolution, in my opinion. So I'd like to gather all my thoughts below with hopes of finding a satisfactory answer.
Consider a quantum system with a state described by a density operator $\rho$. The (Von Neumann) entropy of this system is defined as $$S\left(\rho\right) \equiv - \mathrm{Tr}[\rho\log\rho] = - \sum_n p_n \log\left(p_n\right),$$ where $p_n$ is the $n$th eigenvalue of $\rho$. This is equivalent to the Shannon entropy typically used in information theory.
Now for an isolated system, the time evolution of the state is described by a unitary operator $U(t)$. It's a straightforward exercise to show that under such unitary evolution, the entropy stays constant:
\begin{align} S\left(\rho(t)\right) &= - \mathrm{Tr}[\rho(t)\log\rho(t)]\\ &= - \mathrm{Tr}[U(t)\rho(0)U^\dagger(t)\log\left(U(t)\rho(0)U^\dagger(t)\right)]\\ &= - \mathrm{Tr}[U(t)\rho(0)\log\rho(0)U^\dagger(t)]\\ &= S\left(\rho(0)\right). \end{align}
This makes sense so far. Intuitively too, I would expect the "information content" of the state to be conserved if it's undergoing unitary evolution, which is reversible. However, the 2nd law of thermodynamics states that
The total entropy of an isolated system never decreases.
which seems to leave out the possibility for the entropy of the system to increase, even if it's completely isolated. Admittedly, the entropy change for a reversible process is taken to be zero in thermodynamics, but the confusing thing is that there's again a possibility left out for an irreversible process, even though the system is isolated. In what way can an isolated system undergo an irreversible process, even though we know its time evolution, which is governed by its internal Hamiltonian, has to be unitary and therefore reversible.
There have been several attempts at this question over the years on this website, e.g. this question from 2021, this one from 2020, this from 2018 and this from 2015, non of which are given a fully satisfactory answer, in my opinion.
One interesting explanation I came across is this insightful answer by @joshphysics. The argument, which he attributes to these notes by Eric D'Hoker, goes like this: If I imagine the isolated system to be composed of subsystems $A$ and $B$, then by subadditivity of the entropy, we know that $$S\left(\rho(t)\right) \leq S\left(\rho_A(t)\right) + S\left(\rho_B(t)\right),$$ where $\rho_A(t)$ and $\rho_B(t)$ are the reduced density operators for $A$ and $B$, respectively. If we assume that the two subsystems are uncorrelated at $t=0$, $S\left(\rho(t)\right) = S\left(\rho(0)\right) = S\left(\rho_A(0)\right) + S\left(\rho_B(0)\right)$, which gives $$S\left(\rho_A(0)\right) + S\left(\rho_B(0)\right) \leq S\left(\rho_A(t)\right) + S\left(\rho_B(t)\right),$$ i.e. the sum of the entropies of the subsystems can increase assuming that the subsystems are uncorrelated to begin with. Intuitively, this makes sense. The equality in the subadditivity property is only satisfied when the subsystems are uncorrelated. If I think of the quantity $C_{AB} = S(\rho_A) + S(\rho_B) - S(\rho) \geq 0$ as an indicator of correlations between $A$ and $B$, correlations can form between the subsystems as time goes on, which would increase $C_{AB}$, and therefore $S(\rho_A) + S(\rho_B)$ (since $S(\rho)$ is constant).
While this approach seems consistent, it has some major problems:
- It assumes that the subsystems are uncorrelated to begin with.
- It suggests that we need to interpret "total entropy" in the 2nd law as the sum of the entropy of the constituents, rather than the entropy of the overall system. But this is ambiguous, since that notion depends on how you partition the system into subsystems, which is not unique. Furthermore, the notion of entropy discussed in a lot of modern treatments of statistical mechanics (e.g. the famous papers by Jaynes or these notes) does seem to be $S\left(\rho\right)$, rather than something like $S\left(\rho_A\right) + S\left(\rho_B\right)$.
I also found this answer by @oleg interesting, but I'd like something more quantitative, if possible.
In short: how exactly does one resolve the fact that according to quantum mechanics, the entropy of an isolated system has to stay constant, while the 2nd law of thermodynamics allows for increases in entropy, even in full isolation.
