# Broken symmetry, superconductivity and all that

1. Dec 11, 2009

### DrDu

I am struggling for some time to understand the concept of broken symmetry. As I come more from the solid state side than from high energy physics. My problem is the following: I understand, how e.g. the rotational symmetry in a ferromagnet is broken. The magnetic moment is observable and I can imagine it to break rotational symmetry by pointing in an arbitrary direction.
But what does it mean that global gauge symmetry (which is unobservable) is broken in a superconductor?
These days I came across the interesting classic article by Rudolph Haag, "The Mathematical Structure of the Bardeen-Cooper-Schrieffer Model", Nuovo Cimento, Vol 25(2), pp.287 (1962).
There Haag finds that the states of broken symmetry, which do not conserve particle number and lead to a diagonalization of the BCS Hamiltonian via the Bogoliubov transformation, correspond to irreducible representations of the algebra of the local field operators. There are different irreducible representations which are labeled by a parameter alpha which ranges from 0 to 2 pi. The ground states of all these irreducible representations have the same energy, but each one lives in a Hilbert space on its own.
He also constructs representations where the ground state(s) have a fixed particle number, however, these states are not irreducible, but, I think physically more relevant.

What is the physical relevance of a representation being reducible or irreducible? Why do states which unsharp particle number become so important in a problem which in principle conserves particle number?

2. Dec 11, 2009

### DrDu

Why has my post be moved here? I am mainly interested in the general principles behind symmetry breaking. It is of relevance not only in solid state physics but also in QFT that is why I posted it in the general QM forum.

3. Dec 16, 2009

### hiyok

（1）Whether a representation is reducible or irreducible depends on the system's symmetry. And I dont think, there is peculiarity with either choice.
（2）It should be noted that, the particle number is not conserved actually. The superconducting electrons are usually within a thin layer of Fermi surface. In addition to superconduting electrons, there are also normal electrons. So, the number of total superconducting electrons fluctuate due to exchange with normal electrons. But such exchange is not explicitly included in the Hamiltonian, because the normal electrons can be taken a part of heat bath. What is peculiar to supercondutivity is not only electron pair-up, but alslo condensation of these pairs. Condensation means long-range phase correlation and loss of particle conservation.

4. Dec 17, 2009

### DrDu

If you split the electrons into a superconducting part and a normal part, each of the two is not conserved, but their sum still is (would be asonishing if cooling a lump of metal were to lead to the production of electrons). I also don't see that the BCS Hamiltonian refers only to one type of these electrons, so also the fluctuation of either particle number should be included. Actually, the wavefunction with a fixed number of particles can explicitly be written down (see, e.g., the book "Superconductivity" by Schrieffer). There are also derivations of the superconductive ground state, Tc etc. which respect particle number conservation (Richardson in the 1960s) and are therefore of interest especially in nuclear theory, where the number of particles is rather small. However, the Richardson derivation is not very intuitive and I wonder what can be learned from the infinite system limit.

5. Dec 17, 2009

### hiyok

Of course, the number of normal and superconducting electrons together is conserved. But superconducting electrons can not be conserved, which is reflected in the fact that, an excitation in a superconducting system consists of a partial hole and a partial electron. If you inspect on the BCS Hamiltonian, you'll see the attraction between electrons is confined to within a very thin sheet about Fermi surface. Thus, superconducting electrons comprise only a small portion of total electrons. You are right that BCＳ Hamiltonian is written in number-conserving form. Nevertheless, in dealing with spontaneous symmetry breaking, interaction between superconducting electrons and heat bath (normal electrons and the rest) is implicit. One would not have a BCS-type wave function if particle number were conserved. This is also the case with BEC. The heat bath always has a subtle role in understanding symmetry breaking.

Would you show where I can find Rechardson, 1960. ?

Thanks.

Last edited: Dec 17, 2009
6. Dec 17, 2009

### DrDu

7. Dec 18, 2009

### hiyok

Dear DrDu,
Thanks for the paper.
I dont see any contraditions of their work with what i said.
Pls notice that, they are talking about paring correlations, which of course should be completely determined by the original number-conserving Hamiltonian. Such correlations are inherent to the complete Hamiltonian. However, they are not the sole cause of symmetry breaking. Symmetry breaking implies a particular phase. This phase cannot be picked out by the number-conserving Hamiltonian. It must be due to random interaction with heat bath. However small this interaction may be, it eventually leads to a symmetry breaking zero temperature state, in the presence of divergent paring correlation length at Tc.

In a word, there are two distinct things involved in symmetry breaking: (1)divergent correlations, which are solely determined by the Hamiltonian that respects symmetry and (2)symmetry breaking interaction with heat bath, which can be neglected above Tc but crucial at Tc.

8. Dec 18, 2009

### DrDu

I was not referring to the pairing correlations in small grains, which is the main topic of that paper. In that article, an exact method for the determination of the fixed particle number ground state of the reduced BCS Hamiltonian is reviewed, which is due originally to Richardson (all the references to the articles of Richardson are contained and the derivation is a little bit stream lined in comparison with the original papers). In the thermodynamic limit, the usual BCS gap equation is recovered. This method is well known in nuclear physics, in fact, I learned about it first from the book by Ring and Schuck on nuclear theory.

I also don't think that an interaction with a heat bath (or better to say a particle bath) will lead to an observable choice of absolute phase, at least as long as the bath itself is not a superconductor. The reason is that only phase differences are observable at all (like in the Josephson effect, which measures the phase difference between two superconductors).

9. Dec 18, 2009

### hiyok

(1)The energy gap is no more than a manifestation of paring correlations. This gap is of course determined by the complete Hamiltonian,which however is inadequate for symmetry breaking;
(2)The fact that only phase difference is observable does not pose any inconsistency. If you dont think it is the heat bath that makes the choice, whatelse candidate would you have in mind ?

Last edited: Dec 18, 2009
10. Dec 18, 2009

### hiyok

It is expected rather than a surprise that the exact wave function gives the same energy gap and paring information as BCS wave function. But it is impossible to get the phase coherence from the exact wave function. This information is unique to BCS resovant. This is clear if you look at paring amplitude rather than paring correlation function.

Last edited: Dec 18, 2009
11. Dec 19, 2009

### wbwb

This is a rather frequent issue: the grand-canonical nature of most BCS-like approaches is not really related to any problems with concepts of broken symmetry. There are several truly canonical 'projected' versions of BCS available - which are eg. relevant in the context of small SC grains. Check out eg. Phys. Rev. Lett. 81, 4712 (1998) and refs. therein. Its a 'nice read'.

wbwb

12. Dec 21, 2009

### DrDu

That's by the same authors and the same year as the article I cited in post #6 and I mentioned the particle number projection of the BCS wavefunction in post #4. The point is, that, at least in the limit of an infinitely large system, the particle number conserving state is a reducible state (on the algebra of local field operators), while the gc states are irreducible, hence, they are more fundamental in a way whose meaning I would like to understand better.

13. Dec 22, 2009

### DrDu

I just found a very nice article on that topic: Anthony J. Leggett and Fernando Sols, On the concept of spontaneously broken gauge symmetry in condensed matter physics, Foundations of Physics, Vol. 21, No. 3, 1991

14. Dec 22, 2009

### jensa

Dear Dr Dru,

First of all let me thank you for bringing up the interesting article by Haag, I have been looking for something like this. I too have been struggling with the same issue for some time, namely the question:

Q: In which sense is SSB of a "gauge-symmetry" the same as the SSB of a "real" symmetry? More importantly, are the different ground states parametrized by $\alpha$ (superconducting phase) physically inequivalent states?

I have had several spikes of interest in this but never had the time to really go deep and get a completely satisfactory understanding. Nevertheless I will try to break down what I think about this issue in hopes that some discussion will deepen my understanding.

Let me first discuss the issue of (global) gauge invariance in general. A couple of months ago I started a thread about problems I had with this issue (https://www.physicsforums.com/showthread.php?t=283979). In summary my question was the following: Why would we assume that the global U(1) rotations of the form

$$\hat{\psi}\rightarrow \hat{\psi}e^{i\varphi}$$

should correspond to a gauge degree of freedom? I understand that a global phase is unobservable, but a U(1) rotation as defined above corresponds to a change in the relative phase between sectors of different particle numbers, so there is no obvious reason that such a rotation should not be observable. Now, if the number of particles of our system is truly fixed, i.e. our Hilbert space is constrained to within a sector of fixed number of particles, then the U(1) rotation would be truly unobservable (i.e. corresponds to a gauge degree of freedom).

This question led me to superselection rules and eventually to the work of Haag. Basically, any observable quantity corresponds to an element of a particular subset of the local operator algebra (R in Haags paper), called the algebra of observables. Now it is conjectured that this set consists of operators created from equal numbers of $\hat{\psi}$ and $\hat{\psi}^\dag$. If this is so, then all observables are invariant under the U(1) rotations. This means that a pure state which is a coherent superposition of states with different particle numbers (or rather different charge sectors) is physically equivalent to a mixed state of states with different particle numbers (different charge). Here, by physically equivalent, I mean physically indistinguishable.

Now, what this means for the ground states of the BCS superconductors is that $|\Psi_0(\alpha)\rangle$ is physically indistinguishable from the state $|\Psi_0(\alpha')\rangle$.

Furthermore, as Haag shows in the article you referred to, the Hilbert spaces, $\mathcal{H}_\alpha$ constructed from the ground state $|\Psi_0(\alpha)\rangle$ by repeated application of the bogoliubov operator $\gamma^\dag(\alpha)$ for different $\alpha$, are inequivalent irreps of the operator algebra. This means that we cannot produce a unitary operator from the operator algebra (corresponding to a physical transformation) which takes $\mathcal{H}_\alpha$ to $\mathcal{H}_{\alpha'}$. This is distinctly different from ordinary SSB when, although the given Hamiltonian is invariant wrt the symmetry, we still can write down a unitary operator corresponding to a physical transformation of one ground state to another.

What all this means to me is that there is no true (physical) space of degenerate ground states (or rather, this space only contains one element), and that the "label" $\alpha$ is completely meaningless (redundant).

I believe it is this redundancy (absence of true physical space of degenerate ground states) that leads to the absence of goldstone modes through the Higgs mechanism associated with the local gauge invariance.

(Note that the superselection rule I am talking about is actually the "charge superselection rule" so it only applies to charged superconductors and not to neutral superfluids)

15. Dec 23, 2009

### DrDu

Dear jensa,

I agree with you that the physical relevance of the different irreps parameterized by alpha is not clear, and, like you, I would like to understand it better. I also had a look at the very interesting discussion about global gauge invariance you iniciated some months ago. I just want to remark that in theories with several particles (e.g. electrons and positrons) the conservation of charge and that of particle number are not equivalent so that even in a gauge invariant theory (unbroken) the particle number need not be fixed. However, this is not the case in the example of supersymmetry where we only consider electrons and the gauge symmetry is broken. The operator of total charge arises from Noethers theorem from the invariance of the Laplacian under global gauge transformations, so it is no wonder that in the irreps in which gauge invariance is broken, also particle number is not fixed (this I understood only recently).

I cannnot agree on your statement:
"This is distinctly different from ordinary SSB when, although the given Hamiltonian is invariant wrt the symmetry, we still can write down a unitary operator corresponding to a physical transformation of one ground state to another."
Consider e.g. a ferromagentic system of an infinite number of spins. The point is that the elements from the local algebra can only change a finite number of spins in some (eventually arbitrary large) volume but not the spins at "infinity". A specific ground state is one where all spins point in the same direction. It is clear that another ground state,
where the spins point in another direction cannot be reached by an element from the local algebra, especially, it cannot be implemented unitarily within that algebra.

Well, maybe later more.

16. Dec 28, 2009

### jensa

Yes I am a aware of this. Indeed it should be obvious that in such systems the individual U(1) rotations of fields of different particle species is not a symmetry of the action, and thus particle number of individual species is not a conserved quantity. There is no reason what so ever to call such transformations "gauge symmetries".

In my previous post I tried to slip in that it is actually charge (not particle number) that is conjectured to always be a globally conserved quantity. If this is a fundamental principle, then I would agree that the associated U(1) rotations (which will depend on the charge of the individual particle species) would indeed qualify as gauge symmetries.

On this note I would like to add that I find it extremely misguiding (although I see it sometimes in literature) to refer to the U(1) rotations of uncharged fields as gauge symmetries, and the corresponding breaking of such symmetries as "breaking of gauge symmetry" (e.g. neutral superfluids). The meaning of the phase parametrizing the degenerate ground states in this particular case should be clear to be the conjugate variable of the particle number. The fact that the particle number is not conserved can be explained by the exchange of particles with a heat bath, as has been suggested earlier. This exchange, however should be random so I am not sure if the global phase is a meaningful parameter even in this case. Either way, I suspect that the only observable effect is due to phase difference between two interacting subsystems so this should not be of any major importance.

I am sorry, I don't know enough about supersymmetry to know if what you wrote makes sense in that context but did you mean to write superconductivity ? If so, then I agree with what you wrote. My point, however, is that a true gauge symmetry can not be broken (in the usual sense of the word) as a matter of principle. Gauge transformations do not correspond to physical transformations so it does not make sense to distinguish ground states that are related by such a transformation.

Yes, I believe you are correct about this. I probably put too much importance on the local operator algebra. However, your argument suggests that this mathematical feature (degenerate ground states are members of different hilbert spaces) is a general feature of spontaneously broken symmetries, right?
Furthermore, I assumed that the fact that all local observables were invariant under global U(1) rotations, implied that such rotations cannot be considered physical. I believe the question whether these transformations are physical or not depend on whether the conjectured charge superselection rule has a fundamental origin or if it is rather dynamical in origin (see decoherence induced superselection rules). There is a very interesting article on the subject: Symmetries, superselection rules, and decoherence, Giulini et al, Phys. Lett. A 199 291-298. Here is a link if you are interested and have access: http://www.sciencedirect.com/scienc...erid=10&md5=c344cfcee9d7f5516c11ae57bd88114a". They also discuss briefly the specific case of superconductors.

Finally, even if the parametrization $\alpha$ does correspond to physically distinct states I do not believe it is directly observable (and due to the same arguments as above regarding neutral superfluids, probably is not meaningfull since you most likely have a random exchange of electrons with the environment). The relative phase between two interacting superconductors (Josephson Effect) on the other hand is both observable and physically meaningful, at least in the idealized case. There is no violation of the charge superselection rule if the total number of electrons is conserved. It is also known that incoherent tunneling due to high energy quasiparticles causes delocalization of the relative phase which is consistent with what I said about the effect of a "particle bath".

Last edited by a moderator: Apr 24, 2017
17. Jan 11, 2010

### DrDu

I had been out for vacation some time and used the time to think about the problem a bit. Especially, I read about the symmetry breaking in ferromagnets which is more intuitive than superconductivity and I think I understood some points.
Let's consider a typical isotropic ferromagnet: The symmetry is broken from SU(2) to U(1) (or from (SO(3) to SO(2)). However, I shall in the following single out the z-axis so that the symmetry broken is U(1), like in the case of superconductivity.

There is a spontaneous net magnetization, or equivalently, a net total spin [tex]\langle \mathbf{S}\rangle [\tex]. The vector S has two totally different functions in this context. Let me assume that of the three components of S, the average [tex]\langle S_x \rangle [\tex] is different from zero; this is the order parameter of the problem. What is special in the problem of ferromagnetism is that S_x or S_y commute with the hamiltonian and are therefore a constant of motion. They correspond to the macroscopic wavefunction of the condensate in superconductivity.
On the other hand, the z-component S_z is the generator of the rotational symmetry. It corresponds to the particle number operator in superconductivity. It is also a constant of motion, but it does not commute with S_x and S_y. Hence any state with a nonvanishing expectation value of S_x or S_y cannot be an eigenstate of S_z and vice versa. Be [tex]|\alpha \rangle [\tex] a ground state with [tex]\langle \mathbf{S}\rangle =S_{max} (\cos \alpha, \sin \alpha,0)^T[\tex]
, then a state with a sharp value m of S_z can be constructed as [tex] \int d\alpha |\alpha \rangle \exp(im\alpha)[\tex]. All the states with S_max and any alpha are degenerate as are all states with S_max and all values of m.
Now we may consider what happens when the model becomes anisotropic when the z-axis becomes a hard axis Like, e.g., in a XXZ model. Then only S_z will remain a constant of motion for a finite system but S_x and S_y will fail to be constants of motion. This is more like the situation in a finite superconductor where the states with sharp m correspond to states with a fixed particle number. For a finite system, there will be a unique ground state with m=0 while the states with m=+-1 will be very near energetically and the level spacing will decrease when the system size is increased. In a large system, we can then construct coherent superpositions of states with small |m| so that the expectation value of S is very well localized somewhere in the xy-plane. In the limit of infinite system size, these will become exact ground states.

I found an interesting article by Koma and Tasaki, on the transition from finite size systems to the infinite volume limit. They also give an easy example based on the Ising model.
it is also available on somewhere on ArXive.

Last edited: Jan 11, 2010
18. Jan 11, 2010

### Physics Monkey

Hi DrDu,

In think the most physical approach is to study the structure of finite but very large systems. The article by Koma and Tasaki is good for this purpose, and I can point you to some others if you're interested. In particular, this approach allows one to understand how the degenerate gound states appear from the spectrum of "extremely" low lying states, often called the thin spectrum. It also allows one to understand how these states become approximately orthogonal in the limit of large system size. One further advantage is that it permits one to understand the crucial role played by the surface. This is a topic of great current interest in condensed matter physics.

Of course, it is interesting to consider what one can learn from the structure of formally infinite systems a la Haag. The limit of infinite system size permits new phenomena that are not possible in finite systems. As you might imagine, these phenomena can be understood as limits of finite dimensional phenomena, but sometimes the infinite dimensional description is more elegant or suggestive.

As I think you've discovered, superconductivity brings in a whole extra layer of complexity since the "gauge symmetry" is really a redundancy. Even when the true global symmetry remains well defined, there are complications.

We can chat more if you're interested.

19. Jan 11, 2010

### DrDu

Dear Monkey,

thanks for your reply. Of course I am interested in these articles and to chat about this topic! I must say that it is quite hard to find the relevant articles on that subject, probably because so much has already been published on superconductivity and broken symmetry, in general.

20. Jan 13, 2010

### Physics Monkey

Hi DrDu,

Sorry I haven't replied yet, I've been a little busy. May I ask what sort of background you have? I feel like if you can handle the Haag paper, you can handle anything, but I just want to check.

And you're right, not only is the subject old, there is also a great deal of conflicting terminology and confusion. This makes finding useful discussions a bit challenging.