# Entropy of subsystem maximized?

Let $A$ and $B$ be 2 subsystems of a quantum mechanical system, so a state of the whole system is a vector in $A \otimes B$. As far as I understand, a density operator $\rho$ in general can't be written as a tensor product of the density operators of its subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independent systems, then you can write: $$\rho = \rho_A \otimes \rho_B.$$ Otherwise (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$: $$\tilde{\rho}_A = \mathrm{tr}_B \rho.$$

Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced density matrix $\tilde{\rho}_A$ is also at its maximum?

In the case of no entanglement, $$S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B]$$ holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore for an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?

Edit: To give a reason, why I'm asking this. The question I was thinking about originally was: If a quantum system is in thermodynamic equilibrium, are the sub systems also in thermodynamic equilibrium? My naive answer to that is "Yes, they should be", but I'm not sure about that, and I can't give a proper reason, why they should.

• What you are interested in is known as entanglement entropy. Nov 3, 2016 at 13:50
• Well, yes, especially in the behaviour of the entanglement entropy for thermodynamic equilibrium. Nov 4, 2016 at 10:08
• What do you mean by "thermodynamic equilibrium"? An isolated system in equilibrium, in which case it is is a maximum entropy state (in which case trivially all subsystems are also in a maximal entropy state), or a system coupled to a thermal bath in equilibrium? Nov 5, 2016 at 17:22
• What do you mean "in which case trivially all subsystems are also in a maximal entropy state"? Why is this trivial? This is the core of my question. If this is trivial, then it is trival, that $S[\mathrm{tr}_B \rho]$ is maximized when $S[\rho]$ is maximized. I know that you could treat the subsystems as canonical ensembles and the whole system as a microcanical system, and because of that, the answer to my question SHOULD be "Yes, of course". But is it? Nov 6, 2016 at 14:25
• Well, the maximum entropy state is the maximally mixed state $\rho\propto 1\!\!1$, which clearly a product state, and has maximally mixed marginals. If you want the maximally mixed state with a certain energy, then you first have to be more specific about the Hamiltonian. P.S.: Please use @NorbertSchuch in your responses, otherwise I do not get notified of your responses. Nov 6, 2016 at 16:01

Assuming you are considering non-interacting subsystems, let us take two of them, $S_1$ and $S_2$, at given total energy $E = E_1 + E_2$, and seek a maximum entropy state.

As you already noticed, subadditivity of entropy in the presence of entanglement, $S ≤ S_1 + S_2$, rules out entangled states, and means that entropy necessarily attains its maximum on the set of unentangled states compatible with the given $E$.

Now sort the latter set according to the energy of one subsystem, say $E_1 = \epsilon$. For each $E_1 = \epsilon$, the state of maximum entropy will be the direct product $\rho_1^0(\epsilon) \otimes \rho_2^0(E- \epsilon)$ of maximum entropy subsystem states $\rho_1^0(\epsilon)$, $\rho_2^0(E- \epsilon)$ corresponding to energies $\epsilon$, $E-\epsilon$. The total entropy is $S^0(\epsilon) = S_1^0(\epsilon)+S_2^0(E-\epsilon)$, and the problem is reduced to maximizing $S^0(\epsilon)$. That is, we need $\epsilon$ such that $$\frac{dS^0}{d\epsilon} = \frac{dS_1^0}{d\epsilon}(\epsilon) - \frac{dS_2^0}{d\epsilon}(E- \epsilon) = 0$$ From here a standard argument yields that the desired value of $\epsilon$ is that for which $S_1$ and $S_2$ are in mutual thermal equilibrium (common temperature) for the total energy $E$.

Does this mean that each subsystem is in its own maximum entropy state?

Relative to its other states of identical energy $E_i$, yes. Across the whole set $\{\rho_i^0(E_i)\}_{E_i}$ compatible with given $E$, no.

The reason is that the entropy of equilibrium states $\rho_i^0(E_i)$ increases with the energy $E_i$, and so for each subsystem the entropy $S_i^0(E_i)$ attains its maximum for maximum $E_i$. But when this happens the complementary subsystem has minimum energy $E-E_i$, hence minimum entropy, qed.

• The key to the answer to my question would be the part where you say that the subsystems won't be entangled in the maximum entropy case, and that the density operator then is a tensor product of the density operators of the subsystems. Why is that? You say, subadditivity rules out entangled states. This seems counter-intuitive to me, I would have expected the systems to become more and more entangled, as entropy increases. Nov 8, 2016 at 8:35
• @Quantumwhisp Sorry for the delay. Subadditivity basically says that the total entropy of a bipartite state $\rho_{12}$ is at most equal to the sum of the entropies of the "local" states $\rho_1 = Tr_2\rho_{12}$, $\rho_2 = Tr_1\rho_{12}$, or $S(\rho_{12}) ≤ S(\rho_1) + S(\rho_2)$. Equality occurs for $\rho_{12}=\rho_1\otimes\rho_2$. In other words, total entropy reaches a maximum on unentangled direct product states.
– udrv
Nov 9, 2016 at 5:04
• @Quantumwhisp To form an intuition, think in terms of correlations: maximum entropy = "absence of correlations", while entanglement = "(quantum) correlations present". Hence entangled, correlated states are lower in entropy than unentangled states.
– udrv
Nov 9, 2016 at 5:04
• I get the intuitive part, and I also get what you mean by the subadditivity ruling out the entanglement: Consider the maximum entropy operator $\hat{\rho}_{max}$. In a vicinity of it, consider all the $\hat{\rho}$ with the same local states $\hat{\rho}_1$ and $\hat{\rho}_2$. Then $\hat{\rho}_{max}$, out of all these operators, will be the one with $\rho = \rho_1 \otimes \rho_2$. I think this math is important :) Nov 10, 2016 at 0:22
• Yes, it is. If you are interested in a proof, see Sec. 11.3.4, pg.515-16, in Nielsen&Chuang, books.google.com/…
– udrv
Nov 10, 2016 at 4:39