Ferromagnetism is sometimes given as an example of broken symmetry. However, it should be stressed that is the different from spontaneous symmetry breaking which:
- is a strictly quantum effect
- only exists in the thermodynamic limit
In ferromagnetism, specifically, the ground state is an eigenstate of the relevant continuous symmetry, and as a result the symmetry is unbroken and the low-energy excitations have no new properties. Broken symmetry proper occurs when the ground state is not an eigenstate of the original group, as in antiferromagnetism or superconductivity; only then does one have the concepts of quasidegeneracy and of Goldstone bosons and the "Higgs" phenomenon.It is also worth reading the ensuing exchange between Anderson and Rudolf Peierls on the subject. I thank John Fjaerestad for making me aware of this correspondence.
Note added: Peierls discussed his point of view in more detail in his 1992 Dirac Memorial lecture at Cambridge, published here.
Also, Piers Coleman told me that in the early 90's Peierls came to Princeton specifically to have a public debate with Anderson about the issue. The large audience, including some famous theorists, was divided about who was correct.
Nice post, I had been told that ferromagnetism was an example of spontaneous broken symmetry myself (and unfortunately not looked into it to find out whether it was true).
ReplyDeleteHmmm... I don't get it. And I didn't get it either when I happened across this exchange when I was a graduate student. I suspect that I am not the only one. I'd appreciate if someone could expand on this.
ReplyDeletePWA wrote, "Broken symmetry proper occurs when the ground state is not an eigenstate of the original group" But isn't this trivially the case with a ferromagnet that spontaneously chooses a net magnetization along (say) z? The high T state has continuous rotation symmetry along all three axes and the low T state now only has it along z.
In his followup, PWA gives examples of states that DO meet his definition of broken symmetry, superconductivity and AFism, but I fail to see how they are different i.e. "This is analogous to BCS theory, where the Hamiltonian is charge conserving but the quasiparticles do not create charge eigenstates. In the ferromagnetic case the excitations - spin waves - can be chosen to create states with a definite spin quantum number, so the analogy to ferromagnetism is flawed." But here, the spin waves are only states of definite Sz, not Sx and Sy. So I don't see how it is any different than the BCS example.
In his followup, PWA implies that the distinction is one about excitations, but the original letter was all about ground states...
Any insight would be appreciated.
PS Hi John!
Peter,
ReplyDeleteWhen we say "an eigenstate of the continuous group" what is meant is an eigenstate of the *generators* of the group in a given representation. For the spin-rotation symmetry group, these are labeled by S (total spin) and S_z (spin in some direction). So, the eigenstate for the group only needs to have definite S_z and S^2. The ground state of the ferromagnet is an eigenstate of the rotation and satisfies this property. That of the AFM does not. (When S<>0 there is no state with definite S_z as well as S_x and S_y.)
This is reflected in the excitation spectrum as well. In the case of the FM, excitations carry definite spin (i.e. belong to a single representation of the spin-rotation group). In the case of the AFM, they don't.
The point PWA makes is to draw a distinction between these two case ("FM" vs. "AFM"). In particle physics, he says, only the AFM type of symmetry breaking is important. In condensed matter, I think both are important, and I think we generally count both as spontaneous symmetry breaking. There is, however, a semantic aspect to all of this, which I do not find interesting.
Oops, I should correct my original statement, ... When we say "an eigenstate of the continuous group" what is meant is an eigenstate in a given representation (not the *generators* of the group as written above -- S_x, S_y and S_z are the generators of the spin-rotation group).
ReplyDeleteThanks so much! My PhD adviser did his doctorate with Anderson and mentioned this exchange but I couldn't find it until now.
ReplyDelete