# Grand Canonical Ensemble: N operator problem

1. Oct 19, 2014

### Xemnas92

I have a problem in understanding the quantum operators in grand canonical ensemble.
The grand partition function is the trace of the operator: $$e^{\beta(\mu N-H)}$$ (N is the operator Number of particle)
and the trace is taken on the extended phase space:
$$\Gamma_{es}= \Gamma_1 \times \Gamma_2 \times .... \Gamma_N ...$$
(product of all the phase spaces with arbitrary N).
But N is a constant of motion: $$[H,N]=0$$ (if it were not so, it would be impossible to define the grand partition function in this way).
So my question is: how fluctuations of the avarage value of N can be justified with the fact that N is a constant of motion?
I understand that if we fix the chemical potential (that is exactly the hypothesis of the grand canonical ensemble), there are many corresponding values of N, but their avarage is fixed . But if N (operator) is constant, the time evolution should be only in a phase space $$\Gamma_N$$ with fixed N and so no fluctuations should be allowed (the system cannot change its own phase space during time evolution).

Last edited: Oct 19, 2014
2. Oct 19, 2014

### vanhees71

First of all Fock space is not the direct (tensor) product of the n-particle Hilbert spaces but their orthogonal sum. It also includes the vacuum, which represents the case that no particles are present.

The grand canonical ensemble can be invisaged as a large subsytem of an even larger many-particle system, where both energy (heat) and particle exchange with the larger system is possible. E.g., for a gas, you can consider a smaller (but still macroscopic) subvolume within a big volume.

If you have the case of a small subsystem of fixed particle number in contact with a heat reservoir, allowing for exchange of heat, you are in the canonical ensemble.

If you have a closed system, you are at the level of the microcanonical ensemble.

The point is that, if the energy fluctuations in the canonical and the energy and particle-number fluctuations in the grand-canonical ensemble are small compared to the exepctation values of these quantities the descriptions are all macroscopically equivalent.

3. Oct 19, 2014

### Xemnas92

Thanks for the answer!
Sorry, my mistake...
Yeah, ok...but my doubt was another (probably, much more phylosophical than pragmatic): on a single subspace with fixed N, the ensemble is canonical ($e^{\mu N}$ is a constant). But N is even a constant of motion, so the evolution of a "microstate" in the phase space can happen only in the subspace with that precise value of N. So, how can an average over all the ensemble can be justified - if the microstate cannot reach the other subspaces?

It seems to me a sort of contradiction: the grand canonical ensemble is introduced for open systems, but N must be a constant of motion in the phase space.

4. Oct 19, 2014

### atyy

The number of particles is not fixed in the grand canonical ensemble. But it isn't any different than energy for the canonical ensemble. For a closed system, energy is conserved, but in the canonical ensemble we allow energy to fluctuate, because it is exchanging energy with its surroundings. The grand canonical ensemble is an extension of this concept where we allow the system to exchange particles with its surroundings.

5. Oct 19, 2014

### Xemnas92

Yes, but if the operator number of particles is not a constant of motion, the operator $e^{(\mu N-H)\beta}$ cannot be built. I know that from a macroscopic point of view the number of particles is not fixed, but I do not know how to interpret the fact that the operator N commutes with the hamiltonian.

Anyway my doubts come from Bugoliugov,Statistical Physics. He builds up all the statistical operator starting from the fact that H commutes with N.

My explanation is that the dynamic of the system cannot change the number of particles, that could be changed only through interactions with a larger system. So, a microstate evolves (in the stationary state) only in a single subspace, with fixed N, but we don't know a priori what N of the whole system is, so the integration must be performed all over the extended phase space.But I don't know if this is the correct interpretation, or there are some mathematical aspects hidden behind (the ergodic theorem, for instance...)

6. Oct 19, 2014

### atyy

In non-relativistic quantum mechanics, N and H operators commute, just like the operator for E and H commute.

I don't know anything about relativistic thermodynamics, so someone else will have to comment on that.

7. Oct 19, 2014

### vanhees71

Bogoliubov is right of course. If you want to describe an equilibrium system the Statistical Operator must be a function of the conserved quantities of the system and it must be the state of maximum entropy with the constraints given by the initial conditions. In the grand-canonical ensemble you assume an open subsystem within a larger system, where both energy and particle exchange between the subsystem and the rest of the system is possible. Here you maximize the entropy under the constraint that the mean energy and the mean particle number of the subsystem is given. The Lagrange multipliers lead to the temperature and chemical potential. They are adjusted such that the mean energy and particle number is correctly described by the corresponding grand-canonical statistical operator:
$$\hat{R}=\frac{1}{Z} \exp \left (-\frac{\hat{H}-\mu \hat{N}}{T} \right ), \quad Z=\mathrm{Tr} \left (-\frac{\hat{H}-\mu \hat{N}}{T} \right ).$$
In non-relativistic quantum theory for many systems the particle number (of a closed system) is conserved, i.e., the corresponding operator commutes with the Hamiltonian. Then you can choose $\hat{N}$ as the particle-number operator.

In relativistic physics I'm not aware of any system of interacting particles where the particle number is conserved. The reason is that any interacting quantum-field theory allows for particle creation and destruction processes. Nevertheless there are many conserved "charges" in the Standard Model like electric charge, the baryon number (or better isospin), strangeness, charm, bottom under strong interactions, etc. Then, instead of the particle number $\hat{N}$ one uses one or more conserved charges in the grand canonical statistical operator. The meaning is the same: You adjust temperature and chemical potential such as to describe the correct average energy and charge in the subsystem.

8. Oct 19, 2014

### DrDu

What I find strange is that Bogoliubov also developed a theory of symmetry breaking, e.g. in superconductors, where he explicitly adds particle number non-conserving terms to the hamiltonian.
Also the reduced BCS hamiltonian does not conserve particle number. Nevertheless one uses the grand canonical ensemble in both cases.

9. Oct 20, 2014

### Demystifier

Xemnas, just because a quantity is conserved does not imply that it doesn't "fluctuate". In QM, a "fluctuation" is not a time dependent process. "Fluctuation" of a quantity in QM just means that the quantity is "uncertain", i.e. that the state of the system is not an eigenstate of the operator corresponding to this quantity.

10. Oct 20, 2014

### Xemnas92

Thank you all for the answers!
Yeah, probably this was the answer I was looking for.
Hence (please, correct me, if my reasoning is wrong) the fluctuation of N comes from the fact that the chemical potential is fixed (not N). And the state of the whole system cannot be considered just an eigenstate of N, but a superposition of ones with different N (but such that its avarage is constant).
So, seen in the phase space, each part of the function with fixed N evolves only in its own subspace with fixed N.

I think that this is coherent with what Bogoliubov says: he considers a function that represents a state (even with N not a constant of motion) and then he considers its projections on the subspaces with fixed N. So, from the point of view of Fock space, all the eigenstates with fixed N are involved in integrals or in traces.

Probably this is exactly what happens in canonical ensemble: the state is not in an energy eigenstate (because the temperature is fixed) - but energy is a constant of motion.

11. Oct 20, 2014

### Demystifier

Yes, that's correct.

12. Oct 20, 2014

### DrDu

When I learned statistics, my teacher used the Jaynes approach of ensembles. So you have an ensemble of equal systems and you ask about the maximal probability (or entropy) for the distribution of a total energy E and particle number N among the M members of the ensemble, starting from an a priori equal probability of all possible states. You find that the energy and particle numbers scatter around the mean value e=E/M and n=N/M.
Nothing fluctuates here, each member of the ensemble is always in an eigenstate with fixed E and N. The spread of energy and particle number is only present when we compare different members of the ensemble.

13. Oct 20, 2014

### vanhees71

That's all correct, but it's also right that for a closed system conserved quantities cannot fluctuate if you start with a state, where these quantities are determined, i.e., if the state is in an eigenstate of the corresponding self-adjoint operator.

As an example, where this is important, let's consider relativistic heavy-ion collisions, as performed at the Relativistic Heavy Ion Collider (RHIC) at the Brookhaven National Lab (Upton, NY) or at the Large Hadron Collider (LHC) at CERN (Geneva, Switzerland). There you collide two heavy nuclei like gold or lead at very high energies (200 GeV or 2.7 TeV per nucleon, respectively). It's pretty clear that in this process a hot state of matter, where quarks and gluons become the relevant degrees of freedom are produced, and this system becomes astonishingly quickly to local thermal equilibrium, i.e., it can be described as a collectively expanding (flowing) fireball of liquid, well described by (nearly) ideal hydrodynamics, as is measured by looking at the transverse-momentum spectra and the abundances of hadrons in the detectors.

Now a very interesting question is, whether in this way one can learn something about the phase diagram of strongly interacting matter. E.g., there must be a phase transition from a Quark-Gluon Plasma (QGP) to a hot and dense hadron gas (confinement-deconfinement phase transition) and from a chirally symmetric state (QGP) to a state, where this symmetry is spontaneously broken (hadronic phase). One way to find the phase transition is to look at the fluctuations of conserved quantities (like the baryon number or the electric charge) in the sense of grand-canonical ensembles.

Of course, when you look at all particles in the final state after the collision, there are no fluctuations in these numbers, because you start with the two nuclei. As a whole these two nuclei have well-defined baryon number and electric charges, and there cannot be any fluctuations in these numbers, because they are conserved exactly. This is described by the microcanonical ensemble. In fact, of course one measures only a part of the system, because the detectors only observe the hadrons in a certain rapdity window.

At the same time the system is finite and the thermodynamic limit is not exactly realized. The particle numbers are also pretty small (some thousand particles are produced in each event). So it is important to make this rapidity window not too large, because then the constraints from overall charge conservation influence the measured fluctuations of these quantities, and you don't get the statistics that's described by the grand-canonical ensemble. On the other hand, if you make the window too small, you get all kinds of trivial fluctuations but not the thermal ones described by the grand-canonical ensemble.

Another question is, whether one can expect to find the fluctuations of the grand-canonical ensemble in such a system, which is not in full thermal equilibrium in each event, because it's a finite system and a rapidly expanding fireball. The buildup of the fluctuations, particularly the non-trivial (non-Gaussian) ones around a phase transition (or even at a critical endpoint, where the phase transition becomes 2nd order) takes time, and maybe the fireball expands fast enough such that these fluctuations are not developed at all. This is an open question of theoretical heavy-ion physics, which is quite exciting at the moment, because with the beam-energy scan at RHIC, where one looks at all kinds of observables for different collision energies and at the new Facility of Antiproton an Ion Research (FAIR) in Darmstadt, Germany, where one wants to look at states of high net-baryon densities but lower temperatures than at top RHIC and LHC energies. Here the search for the first-order phase-transition line (both deconfinement and chiral phase transition) or even the critical endpoint of this line is one of the main motivation to build this new accelerator and detectors!

14. Oct 20, 2014

### atyy

I think the standard view is the ensemble viewpoint for both classical and quantum statistical mechanics. In classical statistical mechanics, this is assumed to have a justification from some probability distribution over the initial conditions, coarse grained observables, and some sort of time averaging, so that the ensembles are not real, but some means to calculate probabilities for coarse grained time-averaged observables of single systems, which in the thermodynamic limit can be obtained with microcanonical, canonical, grand canonical ensembles equivalently (for all practical purposes).

The surprising thing is that the "single system view" may be possible in quantum mechanics without time averaging. I think the possibility of this goes back to von Neumann, but it's really only demonstrated in detail quite recently, eg. http://arxiv.org/abs/1003.5424, http://arxiv.org/abs/1007.3957, http://arxiv.org/abs/1112.0740, http://arxiv.org/abs/1302.3138

Last edited: Oct 20, 2014
15. Oct 20, 2014

### atyy

I looked up the explanation in Ketterson and Song's book, and they say one should find the variational wave function that minimizes the energy subject to a fixed average number of particles, and that putting in this Lagrange multiplier is equivalent to adding a chemical potential. So it seems like the chemical potential is used as a "cure" for the non-constant number of particles in the reduced BCS Hamiltonian? http://books.google.co.il/books/about/Superconductivity.html?hl=iw&id=PdEu62yEjg0C (p201)

Jenö Sólyom has a similar explanation that because the number of particles is not fixed, we have to use the grand canonial ensemble. http://books.google.com/books?id=rL5eGGiY1WUC&dq=bcs hamiltonian grand canonical&source=gbs_navlinks_s (p407)

I guess this is either a magnificent kludge that works, or it shows that it isn't really necessary to have the density matrix commute with the Hamiltonian. Traditionally, the density matrix and Hamiltonian should commute since the density matrix represents equilibrium and doesn't change with time. There are many ensembles that can be associated with a mixed density matrix but we assume that it is the mixture corresponding to diagonalization in the energy basis. However, the idea that many single pure states can be thermal, and that these can be reached by unitary evolution from states that are not energy eigenstates provides a way out of this.

Last edited: Oct 20, 2014
16. Oct 21, 2014

### DrDu

Yes, one of the interesting aspects of the BCS hamiltonian is the fact that N is not a function of E, only, although H does not commute with N. For other particles whose hamiltonian does not conserve particle number, like photons or phonons, this is not the case. The physical reason is that the Cooper pair condensate has always energy 0, independently of the number of pairs in the condensate. The same argument holds also for other systems containing a Bose Einstein condensate.
I am also not sure whether the matematical formulation is water tight. The BCS hamiltonian is usually applied in the thermodynamic limit where neither H nor N are well defined for the total infinite volume. What one is really fixing introducing the chemical potential, is not the particle number, which is infinite, but the number density, lim V-> infinity of N/V. The number density commutes with H restricted to any finite volume V.

17. Oct 21, 2014

### vanhees71

If you have spontaneous symmetry breaking of a global symmetry, the ground state is degenerate, and you have a "condensate", which serves as an order parameter of the corresponding phase transition. E.g. in the case of a Bose-Einstein Condensate for the non-relativistic case you have (in the thermodynamic limit) $\mu=0$ and a non-zero value of the condensate. For a QFT treatment of the free gas in the relativistic case, see p. 96 ff in

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

In the case of the Bardeen-Cooper-Shriefer theory of superconductivity you don't have spontaneous symmetry breaking, because the "broken symmetry" is em. gauge theory. An effective model is scalar QED with the scalar field building a condensate, and the "would-be Goldstone boson" can be absorbed into the photon field by choosing a special gauge ("unitary gauge"). Then the photon becomes massive, and the additional degree of freedom serves as the third polarization state of the massive photon (a massless vector particle has only two polarization states). For details, see

Weinberg, The Quantum Theory of Fields, Vol. 2

Note that the ground state is not degenerate, because the degeneracity you'd have for a global symmetry is just a gauge transformation in the case of the local gauge symmetry, and thus all these configurations represent the same state.

On the microscopic level, perturbation theory breaks down at the BCS transition. Thus one has to treat this system differently. Bogoliubov has invented the technique of the Bogoliubov transformation to handle such cases, which is a canonical transformation. See the very clearly written paper, where also the conservation of electron number via the grand-canonical formalism is discussed in detail:

N. N. Bogoliuobov, Nuov. Cim. 7, 794 (1958)

Last edited: Oct 21, 2014
18. Oct 21, 2014

### DrDu

I wouldn't say so. The global gauge symmetry is broken, too. That the breaking does not lead to the appearance of Goldstone bosons is another story. Furthermore, there is a subtlety here:
The reduced BCS hamiltonian I was referring to is not invariant w.r.t. local gauge transformations but only to global ones. Hence there is also no local gauge symmetry breaking. That nevertheless no Goldstone modes appear in the reduced BCS model is due to the Hamiltonian being non-local. Locality is one of the prerequisites of the Goldstone theorem.
This is quite nicely discussed in Strocchi's book on symmetry breaking.
I also find quite enlightening the article by Rudolph Haag: The mathematical structure of the BCS model,
Haag was probably the first one to demystify the meaning of the degenerate ground states you are also referring to. I think it is one of the best articles on gauge symmetry breaking ever written.

You also mentioned that the chemical potential vanishes for a gas of non-interacting bosons. However, for interacting bosons, mu is usually different from zero and you can use it to control the condensate density.

19. Oct 21, 2014

### vanhees71

Hm, non-gauge invariant QED is, however, a contradiction in itself. It doesn't make sense at all, and it's not true for BCS theory of course. See the very nice pedagogical explanation in (also concerning the issue of global "gauge invariance")

Greiter, Martin: Is electromagnetic gauge invariance spontaneously violated in superconductors?, Annals of Physics 319(1), Elsevier, 217–249, 2005
http://dx.doi.org/10.1016/j.aop.2005.03.008 [Broken]
http://arxiv.org/abs/cond-mat/0503400

Last edited by a moderator: May 7, 2017
20. Oct 21, 2014

### DrDu

I was not talking about QED not being gauge invariant, but the reduced BCS hamiltonian as introduced in their classic paper. This was viewed at as a big problem by many people and Anderson was one of the first to work how to set up a locally gauge invariant theory. However much more readable is the analysis by Nambu, for which he finally got the Nobel prize.

The article by Greiter I know very well, however, I doubt it to be correct. I laid this out already in this forum:
https://www.physicsforums.com/threa...eally-a-spontaneous-symmetry-breaking.623786/