# Entanglement decay and the second law of thermodynamics

It is known that entanglement is rather fragile. Due to interactions with the environment, an entangled state can easily loose its entanglement and evolve to become a separable state. This is the much studied effect of entanglement decay.

Now if entangled states are numerous and separable states are so few. Then we see here a natural tendency for a system to move from the more numerous to the few. This is opposite to the situation described by the second law of thermodynamics where systems tend to move from high order (few states) to low order (numerous state).

So here's the question: does entanglement decay violate the second law of thermodynamics?

Quantum entropy is not a good entanglement measure, hence it is not so much a measure of available states with given entanglement content, but of pure states in a statistically equivalent ensemble representing a given quantum state $\rho$. So a decreasing number of available states under decay of entanglement need not imply a decrease in entropy at all, to the contrary, and it is easy to construct a simple counterexample.

At the very least we can say that it all depends on the dynamics: decay of entanglement may well comply with the Second Law.

For the counterexample, take two systems in local states $\rho_A$, $\rho_B$ and some reasonable entanglement measure E. In general there exists a continuum of entangled states $\rho$ with identical entanglement content E and local states $\rho_A = Tr_B\rho$, $\rho_B = Tr_A\rho$. As usual, entropy reads $S(\rho) = -Tr\rho\ln\rho$. Let the two systems undergo maximal disentanglement from some such $\rho$ into the completely uncorrelated product state, $\rho \rightarrow \rho_A \otimes \rho_B$. In terms of entanglement content the number of available states decreases from many to 1, but the entropy increases because $0 \le S(\rho) \le S(\rho_A) + S(\rho_B)$.

Clarification on entropy vs. equivalent statistical ensembles (following request in comments): Maximum entropy occurs for the maximally mixed state (microcanonical state), which is proportional to the identity operator and so it is unique as a density matrix. But in terms of equivalent statistical ensembles, it is maximally undetermined or disordered:

• an equivalent ensemble can be generated using any orthonormal basis set and even non-orthogonal overcomplete sets;

• the number of ways in which the elements (system copies) of any given ensemble can be distributed on available pure states is maximal (equivalently, each element has the same probability of being in any ensemble pure state).

No other density matrix has this property, nor maximal entropy.

In fact, for any other mixed state the pure state sets that can realize equivalent ensembles are much more limited, although they generally still form a continuum (for example, in absence of degeneracies there is a unique orthonormal set that is not necessarily a basis) and/or there are fewer ways to distribute elements of an ensemble on available pure states.

• I'm still digesting your answer. Seems to me there is something missing, or perhaps contradictory, at least in terms of my understanding. If I take a completely mixed state, which is represented by the identity matrix, then we get for the von Neuman entropy $\log d$ (in $d$-dimensions). However, the identity is a unique state. So the order should be maximal (disorder should be minimal). Can you perhaps clear up my confusion here? Commented Aug 25, 2016 at 4:15
• The maximally mixed state is unique as a density matrix, or equivalently, the identity as an operator. Now think in terms of statistical ensembles equivalent to this state: how many ensembles corresponding to orthonormal pure state bases are compatible with the maximally mixed state? All of them, since any orthonormal basis can generate an ensemble representing the maximally mixed state. This is what the maximal disorder of the maximally mixed state is about. No other density matrix has this property, nor a maximal entropy.
– udrv
Commented Aug 25, 2016 at 12:27
• I think you've answered my question. Don't you want to include this comment into your answer? Then I can accept is as the write answer. Commented Aug 25, 2016 at 12:55
• Sorry, got carried away with the ensembles idea. There is something to it, but it's not the relevant part. I corrected the issue when I added the clarification in the answer. Doesn't change the rest of the argument.
– udrv
Commented Aug 25, 2016 at 17:23

No, entanglement decay does not violate the second law of thermodynamics.

To understand what is going on, consider a quantum system which is initially in some pure state which might or might not be entangled.

Let us now couple this system to a environment, which e.g. forms a thermal reservior. Then, after a while, the system will be in a thermal state $$\rho \propto e^{-\beta H}\ .$$ This is, the system will be in a mixture of energy eigenstates $|\psi_i\rangle$ of the system with a corresponding weight $e^{-\beta E_i}$.

The scenario here is therefore very different to the question you linked to (which talks about the entanglement of a randomly drawn pure state): Here, we don't have a known pure state drawn from some distribution -- which indeed would be entangled with high probability -- but an ensemble. Now the point is that even if the individual pure states in an ensemble are entangled (which is very likely following the linked question), this does not at all have to be true for their mixture!

To understand why this is the case, consider e.g. the four Bell states \begin{align} |\Phi^\pm\rangle &= \tfrac{1}{\sqrt{2}}(|00\rangle\pm|11\rangle)\\ |\Psi^\pm\rangle &= \tfrac{1}{\sqrt{2}}(|01\rangle\pm|01\rangle)\ . \end{align} They are all maximally entangled. However, an equal weight mixture of the four Bell states is the maximally mixed state, which can equally well be understood as a mixture of four product states $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$ -- this is, the mixed state is unentangled (as it can be constructed without using any entanglement). (More generally, any mixture where none of the four Bell states has a weight $>1/2$ is unentangled.)

Thus, there is no contradiction to the second law: While thermalization indeed goes "from one to many", and almost all pure states are entangled, many pure states together (i.e., an ensemble of many pure states) do not need to be entangled (and indeed such a mixed state is unentangled with a significant probability).

Having thought about my question, I believe a found an answer which is different from the other answers.

If one wants to consider the entropy of a system and check that it satisfies the second law of thermodynamics, then such a system must be a closed system. The notion of the decay of entanglement is usually found in the case of open quantum systems. The basic quantum system is in contact with, and interacting with, an environment. So for an analysis of the total entropy, one needs to consider both the basic system and the environment together as a complete closed system.

As the initial complete system, consider a finite dimensional maximally entangled pure state $|\psi\rangle$, together with an infinite dimensional separable pure state for the environment $|{\cal E}\rangle$. The complete state $|\psi\rangle \otimes |{\cal E}\rangle$ now experiences some big unitary operation $U(|\psi\rangle \otimes |{\cal E}\rangle)$, which entangles the basic system with the environment. If one wants to look at the resulting state of the basic system alone, after the unitary operation, one needs to trace out the environmental degrees of freedom $${\rm tr}_{\cal E} \{U(|\psi\rangle \otimes |{\cal E}\rangle) (\langle\psi| \otimes \langle{\cal E}|)U^{\dagger}\} = \rho_{\psi} .$$ The resulting state of the basic system $\rho_{\psi}$ would be mixed and therefore less entangled (even completely separable), depending on the amount of entanglement that the unitary operation created between the basic system and the environment.

So now we can look at the entropy of the initial and final systems. In the initial system the entanglement is restricted to the basic system, which is finite dimensional. The entropy of this system would be given by the size (cardinality) of the space of entangled states for this finite dimensional system.

For the entropy of the final system we need to look at it in its pure form $U(|\psi\rangle \otimes |{\cal E}\rangle)$. Here we find entanglement that is spread over an infinite dimensional system. So the entropy of this system, which involves the cardinality of an infinite dimensional space, would be much larger that of the basic system in the initial state. Hence, entropy did indeed increase, so the second law of thermodynamics is safe.