Coherence Is a Charged Battery, Not a Fuel: Two Accounting Questions in Quantum Thermodynamics — Epoche C2
Thesis: coherence really does raise the extractable work The claim under examination is that quantum coherence and entanglement act as additional fuels, allowing a quantum engine to exceed the Carnot limit. Half of this is correct, and the correct half should be stated precisely before the mistaken half is dismantled. Units are SI, with energies given in units of $k_BT$ and level splittings as $\hbar\omega$, as in the quantum-thermodynamics literature cited. Let a system have Hamiltonian $H$, be in state $\rho$, and have access to a bath at temperature $T$. Define the non-equilibrium free energy $F(\rho) = \mathrm{tr}(\rho H) - TS(\rho)$, where $S(\rho) = -k_B\,\mathrm{tr}(\rho\ln\rho)$ is the von Neumann entropy. Writing the Gibbs state as $\rho_{th} = e^{-H/k_BT}/Z$ with partition function $Z$ and substituting $\ln\rho_{th} = -H/k_BT - \ln Z$ into the quantum relative entropy $D(\rho\|\rho_{th}) = \mathrm{tr}(\rho\ln\rho - \rho\ln\rho_{th})$ gives an identity in one line: $$W_{ext} \;\le\; F(\rho) - F(\rho_{th}) \;=\; k_BT\,D(\rho\|\rho_{th}).$$ The bound is saturated by a reversible protocol. Because relative entropy is non-negative and vanishes only at $\rho = \rho_{th}$, no work can be drawn from a state already in equilibrium — which is the second law in this setting. Now split the state's departure from equilibrium into two parts. Let $\Delta(\rho)$ be the state obtained by deleting the off-diagonal elements of $\rho$ in the energy eigenbasis. Since $\rho_{th}$ is diagonal in that basis, the relative entropy decomposes exactly: $D(\rho\|\rho_{th}) = D(\rho\|\Delta\rho) + D(\Delta\rho\|\rho_{th})$. The first term is the relative entropy of coherence, $C(\rho) = [S(\Delta\rho) - S(\rho)]/k_B$; the second is the contribution of the populations alone. A qubit makes this concrete. Take $H = \hbar\omega|1\rangle\langle1|$ with the level splitting tuned to the bath, $\hbar\omega = k_BT$, so $Z = 1 + e^{-1} = 1.3679$ and $F(\rho_{th}) = -k_BT\ln Z = -0.3133\,k_BT$. Prepare the pure superposition $|+\rangle = (|0\rangle + |1\rangle)/\sqrt2$, for which $\mathrm{tr}(\rho H) = 0.5\,k_BT$ and $S = 0$, so $F = 0.5\,k_BT$ and the extractable work is $0.5 + 0.3133 = 0.8133\,k_BT$. Dephase the same state to $\Delta\rho = \mathrm{diag}(1/2,1/2)$: the energy is unchanged but the entropy rises to $k_B\ln2$, so $F = 0.5 - 0.6931 = -0.1931\,k_BT$ and the extractable work falls to $0.1201\,k_BT$. The coherence is worth the difference, $0.6931\,k_BT = k_BT\ln2$, which is $0.6931/0.8133 = 85\%$ of the total. There is nothing metaphorical about calling that a resource. Antithesis: the resource has to be bought, and the price is the same The step that the "extra fuel" reading omits is preparation. The bound above is an inequality on a single discharge, from $\rho$ to $\rho_{th}$. An engine is a cycle, and a cycle must return the working medium to $\rho$. Apply the same inequality in reverse: the work required to drive the system from $\rho_{th}$ back to $|+\rangle$ is at least $F(\rho) - F(\rho_{th}) = 0.8133\,k_BT$, with equality only for a reversible protocol. Over the closed cycle the net work is therefore at most zero. Coal is dug out of the ground; coherence is not. This is not merely an accounting convention, and there is a structural reason it cannot be evaded. Contact with a thermal bath, modelled as an energy-conserving unitary between system and a Gibbs state — the class known as thermal operations — commutes with free evolution generated by $H$. Any such map is therefore covariant under time translation, and a time-translation-covariant map cannot create coherence between energy eigenspaces from a state that has none. Lostaglio, Jennings and Rudolph showed in 2015 that this makes coherence a genuinely independent constraint: the free-energy inequality alone does not capture it, and a whole family of additional restrictions applies. The practical statement is blunt. Coherence in the energy basis has to be imported from outside the thermal machinery, and importing it costs at least what it later yields. The two standard counterexamples, examined Two celebrated results are routinely cited as engines beating Carnot, and both survive scrutiny only with a redefinition of terms. The phaseonium engine. Scully and colleagues showed in 2003 that a photo-Carnot engine whose hot reservoir consists of atoms with a small coherence between two nearly degenerate ground states can, in principle, extract work at an efficiency above $1 - T_c/T_h$. The reservoir, however, is not a thermal state at $T_h$. It has an effective temperature that depends on the coherence phase, and referring the efficiency to that effective temperature restores the ordinary bound. The coherence must also be maintained by an external field, whose work cost is outside the quoted cycle. The squeezed-bath engine. Roßnagel and colleagues showed in 2014 that a cycle driven by a squeezed thermal reservoir exceeds $1 - T_c/T_h$. Again the reservoir is not thermal: squeezing gives it a non-thermal energy distribution with more energy at fixed entropy. A generalised bound involving the squeezing parameter is obeyed, and the work of squeezing the reservoir has been paid elsewhere. The pattern is identical in both cases. A non-thermal reservoir is described by a temperature, the temperature is then used in Carnot's formula, and the formula is violated because its hypothesis was violated first. Carnot's bound is a theorem about two thermal reservoirs. No experiment has contradicted it, and no derivation in this field claims to. Synthesis: the real quantum effects are on power and on fluctuation Rejecting the fuel metaphor does not leave quantum thermodynamics with nothing to say. It relocates the interesting question, and the relocation is quantitative. Carnot efficiency is attained only in the quasi-static limit, where the output power is zero. The classical benchmark for finite power is the Curzon–Ahlborn efficiency $\eta_{CA} = 1 - \sqrt{T_c/T_h}$,