Quantum thermodynamics view
on the Gibbs paradox
and work fluctuations
Theo M. Nieuwenhuizen
University of Amsterdam
Oldenburg 26-10-2006
Outline
Crash course in quantum thermodynamics
Maximal extractable work = ergotropy
What is the Gibbs Paradox?
On previous explanations: mixing entropy
Application of mixing ergotropy to the paradox
The Bochkov-Kuzovlev-Jarzinsky relation
In 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?
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
Second law for finite quantum systems
No 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 dQ  TdS may be violated
due to formation of cloud of bath modes
C
T
Consequence : S  0 dT '
T'
- 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.
Maximal 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
“ergotropy”
W  U (0)  U min  U (0)    i i
i 1
  work;
  turn,
energy   - 
 in - work
transforma tion
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
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.
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
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, 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.
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: W . TS
.
• Also in quantum situation??
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 relations
Hamiltonian 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 steps
Kurchan: 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
Subensembles are obtained from initial (Gibbs) state
by measurement + selection: preparation process
Within one subensemble, repeated measurements at time t
determine average work
Outcome fluctuates from subensemble to subensemble
Average[exp(w)] differs from exp[Average(w)]
Q-work fluctuation theorems are either impossible,
or are not operational (not about work)
Summary
Q-thermodynamics describes thermo of small (nano) systems
First law holds, various formulations of second law broken
Explanation Gibbs paradox by formulation in terms of work
Mixing ergotropy = loss of maximal extractable work due to mixing
Operational definition: less work from less good apparatus
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