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
Who was Josiah Willard Gibbs?
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
Josiah Willard Gibbs
1839 – 1903
Carreer in Yale
1866-69: 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
SA , S B
S A B  S A  SB  ( N A  NB ) k log 2
mixing entropy
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.
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
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.
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: W  T
. S. 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
2
p
mi 2 2
b 2
i
H tot 
 x  x  ci xi   (
  i xi )
2m 2
i
i 2mi 2
system  interaction  linear bath,  i  i 
p2
 ci2
2
quasi  Ohmic bath : J ( )  i
 ( i   )   2
2mi i
 2
Langevin equation (if initially no correlation between S and B)
m x  (b   ) x   x   (t ) ,
  (t ) (t ' )  K (t  t ' )
p2 a 2
System Hamiltonian : H 
 x ,
2m 2
a  b  
First law: is there a thermodynamic description,
though the system is finite?
dU  dQ  dW
U  H 
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.
dW
Work: Energy-without-entropy added to the system by
a macroscopic source.
1) Just energy increase of work source
2) Gibbs-Planck: energy of macroscopic degree of freedom.
dQ
Energy related to uncontrollable degrees of freedom
Picture developed by Allahverdyan, Balian, Nieuwenhuizen ’00 -’04
Roger Balian (1933-)
CEA Saclay; Academie des Sciences
He
3
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 dQ  TdS may be violated
due to formation of cloud of bath modes
 dm
Caldeira Leggett model : dQ(T  0) 
 0 if dm  0
2
2 m
- Rate of energy dispersion may be negative
Breakdown= of
for erasure of
informatio n
Classically:
T *Landauer
( rate ofinequality
entropy production
): non-negative
C
Consequenc
S J.0Phys
dT '
A+N: PRL 00 ; PRE
02, PRBe :02,
A 02
T'
T
Experiments proposed for mesoscopic circuits and quantum optics.
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
Couple to work source and do all possible work extractions
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,
 (t )  U (t )  (0)U (t )
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
Maximal work = ergotropy
Lowest final energy:
highest occupation in ground state,
one-but-highest in first excited state, etc
(ordering 1   2  ...   d , 1  2  ...  d )
n
Umin    i  i
i 1
n
Maximal work:
W  U (0)  U min  U (0)    i i
i 1
  work;
  turn,
energy   - 
 in - work
transforma tion
(divine action, Aristotle)
entropy   -   in - transforma tion (Clausius)
ergotropy   -   work - transforma tion
Allahverdyan, Balian, Nieuwenhuizen, EPL 03.
Aspects of ergotropy
-non-gibbsian states can be passive
-Comparison of activities:
U (0;  )  U (0; ) but S (  )  S ( )
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
k
k
j 1
j 1
 majorizes  ,    , if  r j   s j for k  1...n
- Optimal unitary transformations U(t) do yield, in examples,
explicit Hamiltonians for achieving optimal work extraction
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
Gibbs paradox not solved up to now
Mixing entropy argument has its own drawbacks
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, 056120 (2006)
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