Quantum thermodynamics view on the Gibbs paradoxand work fluctuations

Theo M. Nieuwenhuizen University of Amsterdam

Oldenburg 26-10-2006

Outline

Crash course in quantum thermodynamics

Maximal extractable work = ergotropy

Application of mixing ergotropy to the paradox

What is the Gibbs Paradox?On previous explanations: mixing entropy

The Bochkov-Kuzovlev-Jarzinsky relationIn the quantum domain?

Quantum Thermodynamics=

thermodynamics applying to:

• System finite (non-extensive) “nano”• Bath extensive, work source extensive

Toy models: - (An)harmonic oscillator coupled to harmonic bath (Caldeira-Leggett model)

- spin ½ coupled to harmonic bath (spin-boson model)

Complementary approach: Mahler, Gemmer, Michel: length scale at which temperature is well defined

First law: is there a thermodynamic description,though the system is finite?

dWdQdU HU where H is that part of the total Hamiltonian,

that governs the unitary part of (Langevin) dynamicsin the small Hilbert space of the system.

dW Work: Energy-without-entropy added to the system bya macroscopic source.

dQ Energy related to uncontrollable degrees of freedom

1) Just energy increase of work source2) Gibbs-Planck: energy of macroscopic degree of freedom.

Picture developed by Allahverdyan,Balian, Nieuwenhuizen ’00 -’04

Second law for finite quantum systemsNo thermodynamic limit Second law 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

TdSdQ

TTCdTS :eConsequenc 0 '

'

- 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

Maximal work extraction from finite Q-systems

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, )()0()()( tUtUt

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

Couple to work source and do all possible work extractions

Maximal work = ergotropy

tiontransforma-work- ergotropy (Clausius)tion transforma-in-entropy

work-in -energy

tiontransforma, work;

turn

n

iiiUUUW

1

min )0()0(

n

iiiU

1

min

Lowest final energy:highest occupation in ground state,one-but-highest in first excited state, etc(ordering )

Maximal work“ergotropy”

dd ... ,... 2121

Allahverdyan, Balian, Nieuwenhuizen, EPL 03.

Aspects of ergotropy

- Optimal unitary transformations U(t) do yield, in examples, explicit Hamiltonians for achieving optimal work extraction

-non-gibbsian states can be passive

-Comparison of activities: )()();0();0( SS but UU

Thermodynamic upper bounds: more work possible from But actual work may be largest from

-Coupling to an auxiliary system : if is less active thanThen can be more active than

-Thermodynamic regime reduced to states that majorize one another

1...nk for k

1jj

k

1jj if , majorizes sr

,

BSS ,A 2 logk ) ( BABABA NNSSS

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

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 and

Then the limit B to A can be taken continuously.

Current opinions:The paradox has been 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.

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

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??S TW

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 a measurement would not expose a difference between the gasses. So in practice the ``solution’’ would depend on the quality of the apparatus.

Resolution of Gibbs paradox

• Formulate problem in terms of work:mixing ergotropy = [maximal extractable work before mixing] – [max. extractable work 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.

Classical work fluctuation relationsHamiltonian changed in time. Work in trajectory starting with (x,p) :

Initial Gibbs state:

Bochkov + Kozovlev, 1977: cyclic change

Trajectories with negative work must exist

• Noncyclic process: Jarzinsky relation, defines free energy difference

Seifert: entropy of single trajectory

Average entropy:

Quantum situation

Bochkov + Kuzovlev: similar stepsKurchan: different approach Mukamel: other approach

Quantum work fluctuation theorem? A+N, PRE 2006

Work = average quantity Work fluctuation must be an average

over some quantum-subensemble

Average[exp(w)] differs from exp[Average(w)]

Q-work fluctuation theorems are either impossible,

or are not operational (not about work)

Subensembles are obtained from initial (Gibbs) stateby measurement + selection: preparation process

Within one subensemble, repeated measurements at time t determine average work Outcome fluctuates from subensemble to subensemble

Summary

Explanation Gibbs paradox by formulation in terms of workMixing ergotropy = loss of maximal extractable work due to mixing

Operational definition: less work from less good apparatus

Q-thermodynamics describes thermo of small (nano) systemsFirst law holds, various formulations of second law broken

Formulation of Q-work fluctuation theorem runs into principle difficulties

Q-theorems that have been derived, are non-operational

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