Explanation of the Gibbs Paradox within the Framework of Quantum Thermodynamics Theo M. Nieuwenhuizen Physikalisches Kolloquium Johann Wolfgang Goete Universitaet Frankfurt am Main 31-01, 2007

Outline Crash course in Quantum Thermodynamics Maximal work = ergotropy Application of mixing ergotropy to the paradox Who was Josiah Willard Gibbs? What is the Gibbs Paradox? On previous explanations: mixing entropy

Josiah Willard Gibbs 1839 – 1903 Carreer in Yale : Travel to Paris, Berlin, Heidelberg Gustav Kirchhoff, Hermann von Helmholtz Gibbs free energy Gibbs entropy Gibbs ensembles Gibbs Duhem relation Gibbs distribution Gibbs state Gibbs paradox Copley Medal 1901

The Gibbs Paradox (mixing of two gases) Josiah Willard Gibbs 1876 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. mixing entropy

Gibbs = N log 2

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.

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 due to non-commutivity, the paradox is still unexplained.

Quantum mixing entropy argument ranges continuously from 2N ln 2 (orthogonal) to 0 (identical). Many scholars believe this solves the paradox. Von Neuman entropy After mixing Mixing entropy 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.

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, certain measurements would not expose a difference between the gasses. So 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 operational, the Gibbs paradox should be formulated in terms of work. Classically:.. Also in quantum situation??

Quantum Thermodynamics = Thermodynamics applying to: System finite (small, non-extensive) Bath extensive Work source extensive (e.g. laser) No thermodynamic limit Bath has to be described explicitly Non-negligible interaction energy

Caldeira-Leggett model: particle + harmonic bath Langevin equation (if initially no correlation between S and B)

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) dynamics in the small Hilbert space of the system. Work: Energy-without-entropy added to the system by a macroscopic source. Energy related to uncontrollable degrees of freedom 1) Just energy increase of work source 2) Gibbs-Planck: energy of macroscopic degree of freedom. Picture developed by Allahverdyan, Balian, Nieuwenhuizen ’00 -’04

Roger Balian (1933-) CEA Saclay; Academie des Sciences B phase = Balian –Werthamer phase (p-wave pairing) - Eigenfrequencies of Schroedinger operators in finite domain - Casimir effect: Balian-Duplantier sum rule - Book: From microphysics to macrophysics - Quantum measurement process

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 Experiments proposed for mesoscopic circuits and quantum optics. A+N: PRL 00 ; PRE 02, PRB 02, J. Phys A 02

Armen Allahverdyan Yerevan, Armenia statistical mechanics quantum thermodynamics quantum measurement process astrophysics, cosmology, arrow of time adiabatic theorems quantum optics quantum work fluctuations Gibbs paradox > 35 common papers

Work extraction from finite Q-systems Thermodynamics: minimize final energy at fixed entropy Assume 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) Couple to work source and do all possible work extractions Optimal final situation: eigenvectors of become those of H

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. (divine action, Aristotle)

Aspects of ergotropy - Optimal unitary transformations U(t) do yield, in examples, explicit Hamiltonians for achieving optimal work extraction -Comparison of activities: 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 -non-gibbsian states can be passive

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 well behaved: 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.

Luca Leuzzi, Rome PhD in Amsterdam 2002, cum laude L+N book: Thermodynamics of the glassy state

Summary Explanation by formulation in terms of work 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 versa Many details in Allahverdyan + N, Phys. Rev. E 73, (2006) Gibbs paradox not solved up to now Mixing entropy argument has its own drawbacks

Are adiabatic processes always optimal? One of the formulations of the second law: Adiabatic thermally isolated processes done on an equilibrium system are optimal (cost least work or yield most work) In finite Q-systems: Work larger or equal to free energy difference But adiabatic work is not free energy difference. A+N, PRE 2003: -No level crossing : adiabatic theorem holds -Level crossing: solve using adiabatic perturbation theory. Diabatic processes are less costly than adiabatic. Work = new tool to test level crossing. Level crossing possible if two or more parameters are changed. Review expts on level crossing: Yarkony, Rev Mod Phys 1996