Question on distinct phases through spontaneous symmetry breaking of finite groups.

Question

If I have a many-body Hamiltonian that is invariant under the representation of a certain group $G$ i.e [H,U(g)] = 0, how many phases can exist through spontaneous symmetry breaking (SSB) of this symmetry by the ground state? Specifically, if a certain subgroup $H \in G$ can occur in $G$ multiple times, does SSB to each of these represent different phases?

To make my question clear with an example, consider a Hamiltonian invariant under some representation of $\mathbb{Z}_2 \times \mathbb{Z}_2 =\{e, a, x, ax \}$. There are three ways  the ground state can break down to $\mathbb{Z}_2:$ 

$1) \mathbb{Z}_2^{A}: \{e,a\},\\2)~\mathbb{Z}_2^{B}: \{e,x\},\\3) ~\mathbb{Z}_2^{C}: \{e,ax\}$. 

Does the residual symmetry being the above three $\mathbb{Z}_2$ representations represent different phases of matter? Or are they the same?

Answer 1

Phases with different residual subgroups of $\mathbb{Z}_2\times\mathbb{Z}_2$ should be considered distinct, since they can be distinguished by order parameters.

Since you define the symmetry to be $\mathbb{Z}_2\times\mathbb{Z}_2$, the two generators must be represented by distinct physical operators, and very likely are associated with different physical degrees of freedoms (for example, one $\mathbb{Z}_2$ could correspond to conservation of particle numbers mod 2, the other may correspond to spin rotation by $\pi$). By measuring the order parameter it is clear that one can differentiate which $\mathbb{Z}_2$ subgroups are broken -- although viewed abstractly as groups they are isomorphic in the mathematical sense, they have physical meanings attached.

Comments

Thanks. Can you help me understand why this does not happen with an isotropic Ferromagnet, with Hamiltonian $H = -\sum_i \vec{\sigma}_i.\vec{\sigma}_{i+1}$? Here there are infinite ways to break the $SO(3)$ symmetry to $SO(2)$ (about the various possible axes). Do we consider each of the infinite axes of magnetization to represent different phases? 

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I think one needs to be careful with non-Abelian groups. If the subgroup is not a normal subgroup, it means that under symmetry action (i.e. conjugation) it is not invariant. Physically, this is rather obvious: the magnetization axis for the residual SO(2) subgroup is rotated as you apply any out of plane rotation, and this is the Goldstone mode (for continuous groups). Therefore in this case you probably do not want to consider different magnetization axes as distinct phases. For Abelian groups like $\mathbb{Z}_2\times\mathbb{Z}_2$ this kind of complication does not show up.

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Another way I think to say what Meng is saying is that the "phases" with different axes can be connected to each other without going through a phase transition. This is a hallmark of 2nd order phase transitions. For an Ising magnet, for example, one can continuously connect the up "phase" with the down "phase" by going around the critical point. Likewise, one can continuously connected the liquid "phase" of water with the gaseous "phase".

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Thank you. However I am still a little confused. Is this a feature of continuous groups? Or non-abelian groups? Or normal subgroups? Even for a finite non-abelian group, what happens for different instances of the same non-normal subgroup?

Perhaps I can rephrase my question with a better example (closer to the actual problem I am thinking about). If I consider SSB of a global $A_4$ symmetry to the 4 possible $Z_3$ subgroups (which are not normal to $A_4$), are the corresponding phases distinct? Or do I think of them in the same way as the different magnetization axes for SSB of  $SO(3) \mapsto SO(2)$? @Meng @Ryan Thorngren. 

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This is indeed a very good way to put it. From this point of view, my earlier comment was a bit too limited to non-Abelian groups, since the same question could be asked for the spontaneous breaking of ${U}(1)$ symmetry. In these cases changes in the order parameters (i.e. magnetization axes) are very "soft", and infinitesimal changes basically cost no energy and thus should not be thought as distinct phases.

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@sawd Good question. I think in the example (three $Z_3$ subgroups of $A_4$) the three SSB should really be thought as the same, by a very similar argument. To be precise, let me define SSB using the long-range correlation function of order parameters (so one does not need to worry about the problem of symmetric ground states in a finite volume system). For two order parameters $O_1$ and $O_2$, if they can be related to each other by conjugation, then they must have the same correlation functions on a symmetric ground state, and therefore by definition SSB does not distinguish these two.

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@sawd, 

 Here there are infinite ways to break the SO(3) symmetry to SO(2) (about the various possible axes). Do we consider each of the infinite axes of magnetization to represent different phases? 

It depends. Do you imagine yourself a creature living inside of the magnet having no way of interacting with outside world or an outsider? In the former case there's no way to distinguish different phases, although you could possibly infer there exist other phases; in the latter case there clearly is a way to distinguish phases.