I Broken symmetry in ferromagnetism

[jean194]

Hello,

Today I found this paper and this one where P.W. Anderson says that there is no broken symmetry in ferromagnetism because the ground state is an eigenstate of the spin rotation operator. And so we don't have in this system Goldstone's mode for example.
But I thought spin waves were Goldstone's mode of ferromagnetic systems. So I'm a little confused, especially since in the second article Peierls and Kaplan do not seem to agree with Anderson.

 

[DrDu]

Maybe this article https://scipost.org/SciPostPhysLectNotes.11/pdf, especially exercise 2.7 and the following chapter clarifies the point Anderson is trying to make.

 

[jean194]

Thanks, the pdf is very interesting.
Tell me if I'm wrong but from what I understand, the uniqueness of the ground state and the notion of tower of states is fundamental in spontaneous symmetry breaking. The case of ferromagnetism is special in the sense that the ground state is degenerate and is an eigenvector of the Hamiltonian. So in this sense the symmetry is not broken.

Now I'm trying to make the connection with symmetry breaking in electroweak theory; I can't see where the notion of tower of states appears in the latter case.

 

[DrDu]

This tower of states is rather a special topic of one of the authors.
The ground state of macroscopic systems is always highly degenerate in the thermodynamic limit and by its very definition is an eigenvector of the hamiltonian.
My favourite article on the subject is by Rudolf Haag:

https://link.springer.com/article/10.1007/BF02731446

In the case of ferromagnetism, it is also an eigenstate of one of the generators of the broken symmetry, which makes it somewhat special. As is explained in the article, this leads, for example, to a reduced number of Goldstone bosons.

 

[jean194]

Thanks I will read that.

 

[atyy]

Here is another article (that has an author in common with the article above).
https://arxiv.org/abs/physics/0609177
Footnote 1`1: Notice that ferromagnetism is explicitly not included in this list. The ferromagnet has a large number of possible exact groundstates which are all precisely degenerate, and which all have a finite magnetization. The singling out of one of these eigenstates is in a sense more like classical symmetry breaking than like the quantum symmetry breaking discussed here. The quantum symmetry breaking causes a state which is not an eigenstate of the Hamiltonian to be realized, and thus goes much further than only singling out one particular eigenstate.

 

[DrDu]

I am not sure whether this statement about "not an eigenstate of the Hamiltonian" is correct. We are talking here about phases, which usually require the idealisation of systems of infinite extent. But in infinite systems, also broken symmetry states are eigenstates of a Hamiltonian.