Explanation ofthe Gibbs Paradox

within the Framework ofQuantum Thermodynamics

Theo M. Nieuwenhuizen

Physikalisches KolloquiumJohann Wolfgang Goete UniversitaetFrankfurt am Main31-01, 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 Gibbs1839 – 1903

Carreer in Yale

1866-69: Travel to Paris, Berlin, HeidelbergGustav Kirchhoff, Hermann von Helmholtz

Gibbs free energyGibbs entropyGibbs ensemblesGibbs Duhem relationGibbs distributionGibbs stateGibbs paradox

Copley Medal 1901

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

Gibbs --------------------------------------------------------------------1876

= 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??S TW

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 explicitlyNon-negligible interaction energy

Caldeira-Leggett model: particle + harmonic bath

ii bath, linear ninteractio system

ixi im

i imip

ixi

icxxb

m

ptotH

)2222

2(2

22

2

Langevin equation (if initially no correlation between S and B)

)'()'()()()( ttKt t ,t xx bxm

ba xa

m

pH :nHamiltonia System ,2

22

2

22

2

2

) ( 2

) ( :

iiii

i

m

cJbathOhmicquasi

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

Roger Balian (1933-)

CEA Saclay; Academie des Sciences

B phase =Balian –Werthamer phase(p-wave pairing)

He3

- 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 systemsNo 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

TdSdQ

T

T

CdTS 0 '

' :eConsequenc

ninformatio of erasurefor inequalityLandauer ofBreakdown - 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

0dm if 0 m

dm

2)0( :modelLeggett Caldeira

2

TdQ

Armen AllahverdyanYerevan, Armenia

statistical mechanicsquantum thermodynamicsquantum measurement processastrophysics, cosmology, arrow of timeadiabatic theoremsquantum opticsquantum work fluctuations

Gibbs paradox

> 35 common papers

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)

Couple to work source and do all possible work extractions

Optimal final situation: eigenvectors of become those of H

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:

dd ... ,... 2121

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: )()();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 than Then can be more active than

-Thermodynamic regime reduced to states that majorize one another

1...nk for k

1jj

k

1jj if , majorizes sr

,

-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, RomePhD in Amsterdam 2002, cum laudeL+N book: Thermodynamics of the glassy state

Summary

Explanation by formulation in terms of workMixing 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, 056120 (2006)

Gibbs paradox not solved up to nowMixing 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