How can entropy of a quantum subsystem increase?

Suppose that Hamiltonian of a global system is time-independent. Two subsystems encompass this global system. Adopt Heisenberg picture, and density matrix and state vector are constant. von Neumann entropy is defined on (reduced) density matrix, so in Heisenberg picture, entropy of a subsystem is constant. But this cannot be the right way of thinking.

So should one not use usual von Neumann entropy equation when in Heisenberg picture? Or is entropy of a subsystem indeed constant when Hamiltonian of a global system is time-independent? Is entangelment entropy different from von Neumann entropy?

You should not use the Heisenberg picture to compute entropy of a subsystem.

To explain this, let me focus on the case where the state of the whole system is pure. Then the entropy of a subsystem is basically telling you how much information about the state you lose, when you only have access to the subsystem. Or in the other words, when you can only measure observables which act only on that subsystem.

In the Heisenberg picture, the expectation value at time $$t$$ of an observable $$A$$ defined on a given subsystem is $$\langle \Psi | A(t) | \Psi \rangle$$, where $$|\Psi\rangle$$ is constant in time, but $$A(t)$$ evolves according to the Heisenberg equations of motion, with initial condition $$A(0) = A$$. But note that if $$A$$ acts only a given subsystem, then this is not going to be true of $$A(t)$$. By computing the subsystem entropy of $$|\Psi\rangle$$ (using the usual von Neumann formula) on the other hand, you are computing the information loss when you can only measure observables whose Heisenberg evolution $$A(t)$$ acts only on the subsystem, which is not what you want.

The von Neumann entanglement entropy is one of several different ways to quantify entanglement. Here I'll describe one that plays more nicely with the Heisenberg picture. I got this example from section 3 in the paper http://arxiv.org/abs/1702.04924, "Entanglement measures and their properties in quantum field theory".

This example quantifies the entanglement between two mutually commuting subalgebras of the algebra of observables, which is a different (more general) way of thinking about subsystems. This example is motivated by the idea that we can quantify entanglement in terms of the degree to which a particular Bell inequality (the CHSH inequality) can be violated. I'll use the notation $$\psi(\cdots)\equiv\frac{\langle\psi|\cdots|\psi\rangle}{\langle\psi|\psi\rangle}.$$ Let $$A$$ and $$B$$ be the two subalgebras whose mutual entanglement we want to quantify. The measure of entanglement is $$E(\psi)\equiv\frac{1}{2}\max\psi\big((a_1(b_1+b_2)+a_2(b_1-b_2)\big),$$ where the max is over all self-adjoint elements $$a_k\in A$$ and $$b_k\in B$$ with norms $$\|a_k\|\leq 1$$ and $$\|b_k\|\leq 1$$. This is equation (88) in the paper I cited. With the given restrictions on the norms of $$a_k$$ and $$b_k$$, it satisfies $$E(\psi)\geq 1.$$ Any state $$\psi$$ with $$E(\psi)>1$$ is entangled, but beware that some entangled states satisfy $$E(\psi)=1$$. The condition $$E(\psi)=1$$ is equivalent to the CHSH inequality. The maximum possible value of $$E(\psi)$$ is $$\sqrt{2}$$, regardless of the size of the Hilbert space.

I mention this example because it is expressed in terms of observables, so you can use it in the Heisenberg picture.

• How should one understand the functional dependence of $\psi$ on a and b in the definition of E? Any intuition for why that specific linear combination? Nov 21, 2018 at 13:59
• @user157879 Instead of thinking of $\psi$ as depending on $a$ and $b$, it's simpler to think of $\psi$ as an arbitrary state, and then $E(\psi)$ depends on both $\psi$ and on the observables $a$ and $b$. In other words, with this definition of "entanglement", the degree of entanglement depends on the observables as well as on the state. Regarding intuition for this specific linear combination, it's motivated by the fact that (when $a$ and $b$ are suitably chosen) it can violated the CHSH inequality, which is one of the simplest empirical manifestations of "entanglement" in the generic sense. Nov 21, 2018 at 15:47
• @user157879 The simplest model that accommodates CHSH inequality violations is a two-qubit model (or two-photon model in which only the photons' polarizations are accounted for), which can be expressed using a Hilbert space $H_2\otimes H_2$ where $H_2$ is a single-qubit Hilbert space. In this case, we can take the four observables in $E(\psi)$ to be single-qubit observables (two for each qubit). The related post physics.stackexchange.com/a/442240/206691 comments on the relationship between the two definitions of entanglement ($E(\psi)$ versus the traditional one) in this case. Nov 21, 2018 at 15:52