thermodynamics - Mathematical proof of non-negative change of entropy $Delta Sgeq0$

I understand that we can prove that for any process that occurs in an isolated and closed system it must hold that

$$\Delta S\geq0$$

via Clausius' theorem. My question is, how can I prove this in a mathematical way?

Answer

In the context of quantum mechanics, the entropy of a system whose initial state is given by a density matrix $\rho(0)$ is given by the so-called von Neumann entropy;

$$S_\mathrm{vn}(\rho) = -k\,\mathrm{tr}(\rho\ln\rho)$$ For an isolated system, quantum mechanical time evolution is unitary; for each time $t$, there is a unitary operator $U(t)$ such that the state of the system at time $t$ is given by $$\rho(t) = U(t) \rho(0)\, U^\dagger (t)$$ It can be shown that the von Neumann entropy is invariant under unitary similarity transformation of $\rho$; in other words $$S_\mathrm{vn}(U\rho U^\dagger) = S_\mathrm{vn}(\rho)$$ and it immediately follows that $$S_\mathrm{vn}(\rho(0)) = S_\mathrm{vn}(\rho(t))$$ In other words, the entropy of an isolated quantum system does not change with time in accordance with the second law of thermodynamics.

Author's Admission. I have always been somewhat bothered by the argument I just gave you, not because I think it's incorrect, but rather because in light of the conclusion we draw from it regarding isolated systems, why don't people say that the stronger statement $dS=0$ for isolated systems as opposed to $dS\geq 0$. It's not that these are inconsistent statements; one is just stronger than than the other, so I would think one should simply assert the stronger one in the context of isolated systems.

Addendum. In response to my "admission," I should note that there is a cute argument I have seen for the non-negativity of a change in total (von-Neumann) entropy of an isolated system provided one defines total entropy properly. Here it is.

Suppose that we have an isolated system, let's call it the universe, described by a Hilbert space $\mathcal H$. Suppose that this system can be divided into two subsystems $a$ and $b$ so that the combined Hilbert space can be written $\mathcal H = \mathcal H_a\otimes\mathcal H_b$. If the density matrix of the universe is $\rho$, then the density matrices of the subsystems $a$ and $b$ are defined as partial traces over $\rho$;

$$\rho_a = \mathrm{tr}_{\mathcal H_a}\rho, \qquad \rho_b = \mathrm{tr}_{\mathcal H_b}\rho$$ Now we can prove the following:

If systems $a$ and $b$ are initially uncorrelated, then then the total entropy $S(\rho_a) + S(\rho_b)$ will never be lower than at the initial time.

Proof. If the systems are initially uncorrelated, then by definition the total density operator at the initial time is a tensor product $\rho(0) = \rho_a^0\otimes \rho_b^0$. It follows from taking partial traces and using the fact that the density operator is unit trace that the density matrices of the subsystems $a$ and $b$ at the initial time are

$$\rho_a(0) = \rho_a^0, \qquad \rho_b(0) = \rho_b^0$$ Now, at any later time, the total density matrix evolves unitarily, so that $$S(\rho(0)) = S(\rho(t))$$ On the other hand, entropy is subadditive which means that $$S(\rho(t)) \leq S(\rho_a(t))+S(\rho_b(t))$$ and is additive for uncorrelated systems which gives $$S(\rho(0)) = S(\rho_a(0)) + S(\rho_b(0))$$ Putting this all together yields $$S(\rho_a(0)) + S(\rho_b(0)) \leq S(\rho_a(t))+S(\rho_b(t))$$

I've always been somewhat unsatisfied with this argument, however, because (i) it assumes that the subsystems are originally uncorrelated and (ii) it's not clear to me that the definition of total entropy as the sum of the entropies of the reduced density operators of the subsystems is what we should be calling $S$ when we write $\Delta S \geq 0$.

By the way, this argument was stolen from lectures I took: Eric D'Hoker's quantum lecture notes.