Grand Canonical Ensemble: N operator problem

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  • #26
atyy
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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.
I think there is a global symmetry breaking in the BCS theory also. Greiter does says that the BCS has a global symmetry breaking, and http://arxiv.org/abs/0704.3703 by van Wezel and van den Brink says that there is a nearly degenerate thin spectrum that reflects global symmetry breaking in finite superconductors.

I think the theory you are thinking about is the Abelian Higgs model which is related to the description of the Meissner effect by Anderson in superconductors, eg. http://arxiv.org/abs/cond-mat/0404327 by Hansson, Oganesyan and Sondhi says that in 3+1 dimensions the Abelian Higgs model "is a plausible description of a gapped BCS superconductor with particle-hole symmetry but it has the topological features of interest even if the choice of a Lorentz invariant dynamics is non-generic."


But what is the ground state degeneracy in the Abelian Higgs model? If I read Hansson et al correctly, they say that the 2+1D Abelian Higgs ground state degeneracy depends on the boundary conditions or the topology of the manifold. The Scholarpedia Higgs article by Kibble http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism does agree that it is misleading but conventional to talk about "spontaneous gauge symmetry breaking" when he refers to an Abelian model, and that it is better to say that there is an explicit breaking of the gauge symmetry by some gauge choices, but the state is gauge invariant. However, he also does say "the resulting theory does retain a global phase symmetry that is broken spontaneously by the choice of the phase of ##\langle\Phi\rangle##."
 
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  • #27
vanhees71
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That's of course also true. There's always a global symmetry for the gauge fixed gauge symmetry in the Faddeev-Popov formalism, the socalled BRST symmetry (named after Becchi, Rouet, Stora, Tyutin) but that doesn't imply the degeneracy of the ground state, because it's a state in physical Hilbert space and thus invariant under BRST transformations, i.e., the corresponding charge for these states is 0. This is made explicit in the operator treatment of the quantization of gauge symmetries, which is a bit cumbersome compared to the path-integral formalism, but it helps to understand such issues. Very detailed papers are the following, written by the inventors of the "covariant operator quantization" of gauge theories:

Kugo, T., Ojima, I.: Manifestly Covariant Canonical Formulation of the Yang-Mills Field Theories. I, Progress of Theoretical Physics 60(6), 1869–1889, 1978

Kugo, T., Ojima, O.: Manifestly Covariant Canonical Formulation of Yang-Mills Field Theories. II: SU (2) Higgs-Kibble Model with Spontaneous Symmetry Breaking, Progress of theoretical physics 61(1), 294–314, 1979

Kugo, T., Ojima, I.: Manifestly Covariant Canonical Formulation of Yang-Mills Field Theories. III—Pure Yang-Mills Theories without Spontaneous Symmetry Breaking, Progress of Theoretical Physics 61, 644–655, 1979

Of course, also here "spontaneous symmetry breaking" has to be silently substituted by "Higgsed gauge symmetry" ;-)). It's really colloquial and sometimes misleading for beginners in the field.
 
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  • #28
DrDu
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I think the present discussion has drifted quite far from answering the question of the OP whether the macro canonical ensemble treatment justifies to talk about particle number fluctuations despite [H,N]=0.
I think there are two answers:
1. In most systems the variation of N refers not to a single system but to the ensemble it is taken from. It is then often argued that this construction can also applied to systems which aren't completely isolated but can exchange particles with some particle bath. To justify this one has certainly to make some assumptions about the coupling of the system to the bath, which must be weak in some sense to be made more precise.
2. In systems where a symmetry is spontaneously broken, it may prove convenient to work in a representation where particle number or some other conserved quantity is unsharp. The article by Haag discusses this in detail for a model Hamiltonian for a superconductor:
http://link.springer.com/article/10.1007/BF02731446
However, as van Hees pointed out, this is not a realistic model of a real superconductor.
Another example is Bose Einstein condensation. There, Galilean invariance is broken. This is a real case of broken symmetry, as Goldstone modes, namely second sound, is also found experimentally.
It is again convenient to work in a representation where particle number isn't fixed.
Considering other conserved quantities than N, e.g. angular momentum, more familiar systems can be discussed. E.g. it is convenient in the case of ferromagnetics to use ensembles where the B field (which corresponds to ##\mu##is used in place of the magnetization which is proportional to angular momentum and corresponds to the N variable.
 
  • #29
vanhees71
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Why is Galilei invariance (or Poincare invariance in the relativistic case) broken for a BEC? That doesn't make sense to me. How can a Galilei- (Poincare-) covariant theory break Galilei (Poincare) invariance?

It's true that a heat bath defines a preferred reference frame, namely the one where the medium is at rest, but that doesn't mean that Galilei or Poincare invariance are broken. You can define everything in a covariant way, and then you see that there is no breaking of these fundamental symmetries, you just need the fluid-flow vector as an additional ingredient to characterize the state.

Haag's model, of course, is not a gauge-invariant model, and he does not discuss the "spontaneous breaking of a local gauge symmetry" (meant in the colloquial sense of the high-energy physicists; as I stressed several times before, it's better named "Higgsed local gauge symmetry"). Then the ground state of the quantum field theory is of course degenerate as usual for a spontaneous breaking of a global symmetry (which here is the invariance under multiplication of the fields with space-time-independent phase factors). As Haag also stresses, all observable quantities do not depend on the so introduced arbitrarily chosen phase [itex]\alpha[/itex]. Anyway, it's a very well readable paper with concise statements about the meaning of each limit (especially the meaning of the grand-canonical ensemble and the finite quantization volume introduced for proper definition of the states and the appropriate infinite-volume limit).
 
  • #30
atyy
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Haag's model is the BCS model, which I think this is usually considered gauge invariant, but the electromagnetic field is not dynamical.
 
  • #31
atyy
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The Scholarpedia Higgs article by Kibble http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism does agree that it is misleading but conventional to talk about "spontaneous gauge symmetry breaking" when he refers to an Abelian model, and that it is better to say that there is an explicit breaking of the gauge symmetry by some gauge choices, but the state is gauge invariant. However, he also does say "the resulting theory does retain a global phase symmetry that is broken spontaneously by the choice of the phase of ##\langle\Phi\rangle##."
I guess the interesting point about Kibble's remark here is that if the Abelian Higgs model does break a "global symmetry" because there is a phase with non-zero order parameter, why then are there no Goldstone bosons? The commentary given by Hansson et al is that the gauge invariant order parameter in this type of "symmetry breaking" is nonlocal, and doesn't give rise to Goldstone bosons. Hansson et al say that ground state degeneracy in the Abelian Higgs case depends on the topology of the manifold.

http://arxiv.org/abs/cond-mat/0404327 (p5) "With a local order parameter this would be a signature of Goldstone bosons. In fact, the Anderson-Higgs mechanism forbids any such bosons in the actual spectrum, which shows that a description based on ##\phi_{D}## does not have the character of the standard sigma model."
 
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  • #32
DrDu
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Why is Galilei invariance (or Poincare invariance in the relativistic case) broken for a BEC? That doesn't make sense to me. How can a Galilei- (Poincare-) covariant theory break Galilei (Poincare) invariance?
At the moment, I find no better source than wikipedia http://en.wikipedia.org/wiki/Goldstone_boson
There is quite simple an argument working backward: In a BEC ##\langle a_0\rangle\neq 0##. Hence particle number is unsharp. However, in a nonrelativistic context particle number conservation is a superselection rule following from the invariance under boosts from Galilei group and time reversal. So invariance under the subgroup generated by the boosts must be broken.
 
  • #33
DrDu
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I guess the interesting point about Kibble's remark here is that if the Abelian Higgs model does break a "global symmetry" because there is a phase with non-zero order parameter, why then are there no Goldstone bosons? The commentary given by Hansson et al is that the gauge invariant order parameter in this type of "symmetry breaking" is nonlocal, and doesn't give rise to Goldstone bosons. Hansson et al say that ground state degeneracy in the Abelian Higgs case depends on the topology of the manifold.

http://arxiv.org/abs/cond-mat/0404327 (p5)
The following article by Kennedy and King is interesting (Hansson cites the PRL short version of it)
http://projecteuclid.org/download/pdf_1/euclid.cmp/1104115008
They show that there is a Goldstone boson in Landau gauge and suspect that it doesn't couple to physical degrees of freedom.
For the relativistic Higgs models it is known that the Goldstone theorem holds in Lorentz gauge, but that the Goldstone bosons aren't in the physical sector and hence unobservable.
 
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  • #34
vanhees71
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Haag's model is the BCS model, which I think this is usually considered gauge invariant, but the electromagnetic field is not dynamical.
Hm, but (3) is just the effective two-body potential of fermions, not the electromagnetic interaction. You get necessarily Cooper pairs as effective degrees of freedom, when this becomes attractive (in condensed matter physics via the electron-phonon interactions), as it is very nicely described in Haag's paper, but as I said, I'm not an expert in condensed matter physics, so that perhaps I misunderstand all this. As far as I can see all this is solved for a long time. The paper by Nambu has been already cited. A very clear description can also be found in the book by Schrieffer:

J. R. Schrieffer, Theory of Superconductivity, Westview Press (1999)
 
  • #35
DrDu
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Hm, but (3) is just the effective two-body potential of fermions, not the electromagnetic interaction. You get necessarily Cooper pairs as effective degrees of freedom, when this becomes attractive (in condensed matter physics via the electron-phonon interactions), as it is very nicely described in Haag's paper, but as I said, I'm not an expert in condensed matter physics, so that perhaps I misunderstand all this. As far as I can see all this is solved for a long time. The paper by Nambu has been already cited. A very clear description can also be found in the book by Schrieffer:

J. R. Schrieffer, Theory of Superconductivity, Westview Press (1999)
 
  • #36
DrDu
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Yes, Schrieffers book is nice. Also the following article by S. Cremer, M. Sapir and D. Lurie, Collective Modes Coupling Constants and Dynamical-Symmetry Rearrangement in Superconductivity, Il Nuovo Cimento, Vol 6(2), pp. 179 much more accessible than Nambu's paper. I found it especially interesting, how they derive the appearance of bound composite modes, something I find rather difficult to grasp in field theory.

Could you comment on my doubts about Greiters paper?
 
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  • #37
vanhees71
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Hm, I get more and more confused. On the one hand I know for sure that there cannot be spontaneous symmetry breaking of a local gauge symmetry and Higgsing such a local gauge symmetry does not lead to Goldstone modes, which is very important for the electroweak standard model since there shouldn't be a massless scalar or pseudoscalar particle, because it's not observed after all.

On the other hand many people state that in superconductivity there is a Goldstonde mode present, including the just cited paper in #36. However, there the Hamiltonian is not gauge invariant, and they only consider the spontaneous breakdown of the usual U(1) symmetry of the effective Hamiltonian.

I always thought that superconductivity is just Higgsing the electromagnetic local U(1) symmetry due to a condensate of Cooper pairs. Then there shouldn't be any Goldstone modes in a superconductor.

Is there any experimental hint for gapless excitations in a superconductor?
 
  • #38
DrDu
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You mean the paper by Cremer et al? They show that the Goldstone mode is no longer massless once the electromagnetic field (i.e. local gauge invariance) is taken into account. Their hamiltonian is invariant wrt local gauge trafos.
 
  • #39
atyy
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Hm, but (3) is just the effective two-body potential of fermions, not the electromagnetic interaction. You get necessarily Cooper pairs as effective degrees of freedom, when this becomes attractive (in condensed matter physics via the electron-phonon interactions), as it is very nicely described in Haag's paper, but as I said, I'm not an expert in condensed matter physics, so that perhaps I misunderstand all this.

I'm hardly an expert either, but if I have correctly understood Greiter http://arxiv.org/abs/cond-mat/0503400, the BCS theory is gauge invariant even though it omits the electromagnetic gauge field, in the sense that since the gauge field is ignored, the theory only has to be invariant under the part of the gauge transformation in Eq 14, which is a rewrite of the gauge transformations in Eq 9 and 13. If the electromagnetic field is included, then one has to add Eq 11 to Eq 9 and 13 for the complete gauge transformation. (But I think DrDu doesn't agree that Greiter's gauge transformation is correct.)
 

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