How can work be generated from quantum correlations?

The maximum average amount of work that can be extracted from a quantum system in state (i.e., density operator) $X_S$, with Hamiltonian $H_S$, using an arbitrary thermal environment at temperature $T$, is given by $kT \log Z_S$ plus the non-equilibrium quantum free energy: \begin{equation} F(X_S,H_S) = \mathrm{Tr}[H_S X_S] - kT S(X_S), \end{equation} with $S$ the von Neumann entropy: $S(X_S) = - \mathrm{Tr}[ X_S \log X_S]$, and $\mathrm{Tr}$ the trace. Here $Z_S = \mathrm{Tr}[e^{H_S/kT}]$ is the partition function and k is Boltzmann's constant. Note that F has the standard form "average energy" $- kT$ "entropy", but it is defined out of equilibrum. The above has been known for a long time; for a simple proof that the average extracted work cannot be larger than $F(X_S,H_S) + kT \log Z_S$ see the appendix of arXiv:1705.05397. Saturation is more complicated, and is related e.g. to the results in New J. Phys. 16, 103011 (2014).

Now, suppose S is a bipartite system AB, i.e. $X_S = X_{AB}$, with Hamiltonian $H_S = H_A \otimes I_B + I_A \otimes H_B$ ($I_X$ is the identity). $F$ then admits a decomposition into "local parts plus correlations" \begin{equation} F(X_{AB}) = F(X_A) + F(X_B) + kT I(A:B) \end{equation} where $I(A:B)$ is the quantum mutual information: $I(A:B) = S(X_A) + S(X_B) - S(X_{AB})$ (this can be verified by summing and subtracting the local entropy terms $S(X_A)$, $S(X_B)$ to the expression for $F(X_{AB})$. Here $X_A$ ($X_B$) is the partial trace over $B$ ($A$) of $X_{AB}$. Hence correlations, as measured by $I(A:B)$, contribute directly to the non-equilibrium free energy of the state, and hence increase the extractable work. For a Bell state, $I(A:B) = 2 \log 2$. Note, interestingly, that $I(A:B) \leq \log 2$ for qubit states that are not entangled: the fact that you can go above $\log 2$ is a quantum feature.

An explanation that I found useful for understanding the general idea here is from the community blog lesswrong: "The Second Law of Thermodynamics, and Engines of Cognition".

So (again ignoring quantum effects for the moment), if you know the states of all the molecules in a glass of hot water, it is cold in a genuinely thermodynamic sense: you can take electricity out of it and leave behind an ice cube.

This doesn't violate Liouville's Theorem, because if Y is the water, and you are Maxwell's Demon (denoted M), the physical process behaves as:

M1,Y1 -> M1,Y1

M2,Y2 -> M2,Y1

M3,Y3 -> M3,Y1

M4,Y4 -> M4,Y1

Because Maxwell's demon knows the exact state of Y, this is mutual information between M and Y. The mutual information decreases the joint entropy of (M,Y): H(M,Y) = H(M) + H(Y) - I(M;Y). M has 2 bits of entropy, Y has two bits of entropy, and their mutual information is 2 bits, so (M,Y) has a total of 2 + 2 - 2 = 2 bits of entropy. The physical process just transforms the "coldness" (negentropy) of the mutual information to make the actual water cold - afterward, M has 2 bits of entropy, Y has 0 bits of entropy, and the mutual information is 0.

See also: Szilard's engine on wikipedia