Explanation of the Gibbs Paradox within the Framework of Quantum Thermodynamics

1. Explanation ofthe Gibbs Paradoxwithin the Framework ofQuantum Thermodynamics Armen E. Allahverdyan (Yerevan Physics Institute)Theo M. Nieuwenhuizen (University of Amsterdam)Phys. Rev. E 73, 056120 (2006) The Second Law of Thermodynamics,Foundations and Status.University of San Diego June 19-20, 2006

2. Outline What is the Gibbs Paradox?On previous explanations: mixing entropy Crash course in quantum thermodynamics Maximal work = ergotropy Application of mixing ergotropy to the paradox

3. The Gibbs ParadoxJosiah Willard-Gibbs 1876 • Gas A in container 1 with volume V. • Gas B in container 2 with volume V. • Mix them: each atom in volume 2V. • Entropy per atom increases by k ln 2. • But if A and B identical, no increase. • The paradox: There is a discontinuity, still k ln 2 for very similar but non-identical gases.

4. Proper setup for the limit B to A • Isotopes: too few to yield a good limit • Let gases A and B both have translational modes at equilibrium at temperature T,but their internal states (e.g. spin) be described by a different density matrix or Then the limit B to A can be taken continuously.

5. Current opinions: The paradox is solved within information theoretic approach to classical thermodynamics Solution has been achieved within quantum statistical physics due to feature of partial distinguishability Quantum physics is right starting point.But a specific peculiarity (induced by non-commutivity) has prevented a solution:The paradox is still unexplained.

6. Quantum mixing entropy argument Von Neuman entropy After mixing Mixing entropy ranges continuously from 2N ln 2 (orthogonal) to 0 (identical) .Many scholars believe this solves the paradox. Dieks+van Dijk ’88: thermodynamic inconsistency, because there is no way to close the cycle by unmixing.If nonorthogonal to any attempt to unmix (measurement) will alter the states.

7. Another objection: lack of operationality • The employed notion of ``difference between gases’’ does not have a clear operational meaning. • If the above explanation would hold, there could be situations where measurements would not expose a difference between the gasses. So in practice the ``solution’’ would depend on the quality of the apparatus. • There is something unsatisfactory with entropy itself. It is non-unique. Its definition depends on the formulation of the second law. • To be operatinal, the Gibbs paradox should be formulated in terms of work.Classically: . . Also in quantum situation??

8. Quantum Thermodynamics= thermodynamics applying to: • System finite (non-extensive) • Bath extensive, work source extensive

9. First law: is there a thermodynamic description,though the system is finite? where H is that part of the total Hamiltonian, that governs the unitary part of (Langevin) dynamicsin the small Hilbert space of the system. Work: Energy-without-entropy added to the system bya macroscopic source. 1) Just energy increase of work source2) Gibbs-Planck: energy of macroscopic degree of freedom. Energy related to uncontrollable degrees of freedom Picture developed by Allahverdyan,Balian, Nieuwenhuizen ’00 -’04

10. Second law for finite quantum systems No thermodynamic limit Thermodynamics endangered Different formulations are inequivalent -Generalized Thomson formulation is valid: Cyclic changes on system in Gibbs equilibrium cannot yield work (Pusz+Woronowicz ’78, Lenard’78, A+N ’02.) • Clausius inequality may be violated • due to formation of cloud of bath modes - Rate of energy dispersion may be negative Classically: = T*( rate of entropy production ): non-negative A+N, PRL 00 ; PRE 02, PRB 02, J. Phys A,02 Experiments proposed for mesoscopic circuits and quantum optics.

11. Work extraction from finite Q-systems Couple to work source and do all possible work extractions Thermodynamics: minimize final energy at fixed entropyAssume final state is gibbsian: fix final T from S = const.Extracted work W = U(0)-U(final) But: Quantum mechanics is unitary, So all n eigenvalues conserved: n-1 constraints, not 1. (Gibbs state typically unattainable for n>2) Optimal final situation: eigenvectors of become those of H

12. Maximal work = ergotropy Lowest final energy:highest occupation in ground state,one-but-highest in first excited state, etc(ordering ) Maximal work: Allahverdyan, Balian, Nieuwenhuizen, EPL 03.

13. -non-gibbsian states can be passive -Comparison of activities: Aspects of ergotropy Thermodynamic upper bounds: more work possible from But actual work may be largest from -Coupling to an auxiliary system : if is less active than Then can be more active than -Thermodynamic regime reduced to states that majorize one another - Optimal unitary transformations U(t) do yield, in examples, explicit Hamiltonians for achieving optimal work extraction

14. Resolution of Gibbs paradox • Formulate problem in terms of work:mixing ergotropy = maximal extractable work before mixing – ( idem, after mixing) • Consequence: limit B to A implies vanishing mixing ergotropy.Paradox explained. • Operationality: difference between A and B depends on apparatus: extracted work need not be maximal • More mixing does not imply more work, and vice versa.Counterexamples given in A+N, PRE 06.

15. Summary Gibbs paradox not solved up to nowMixing entropy argument has its own drawbacks Present approach: express the paradox in terms of extractable work. ABN ‘03 solution for maximally extractable work: ergotropy.Mixing ergotropy = loss of maximal extractable work due to mixing Operational definition: less work from less good apparatus More mixing does not imply more work and vice versaMany details in A+N, PRE 06.