# Finite temperature spontanous symmetry breaking and Goldstone bosons

I recently asked (and then attemped to answer) a question about spontaneous symmetry breaking in the Heisenberg model: Spontanous symmetry breaking in the Heisenberg model?

The question and then the conclusion I came to in the answer can be summarized as follows*:

Spontaneous symmetry breaking is when the a ground state $|GS\rangle$ of the Hamiltonian does not posses the same symmetry as the Hamiltonian itself and the reason we see spontaneously broken systems is due to imperfections (e.g. symmetry breaking fields).

Looking at the 1D Ising model the grounds states have either all spins up or all spins down. Thus we have a spontnous symmetry breaking of the $Z_2$ symmetry of the Hamiltonian.

That said at any finite temperature: $$\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty}\frac{1}{N} \sum_i\mathrm{Tr}(\rho_e \sigma_i)=0$$ where $\rho_e=e^{-\beta H}/T$. I.e. the thermal average of the magnetization is zero in the limit of the symmetry breaking field $h$ going to zero and the volume $V$ going to infinity. This means the symmetry breaking does not show at finite temperature.

I.e. we appear to have the following:

1. The 1D Ising model does have Spontaneous symmetry breaking.
2. At any finite temperature the symmetry breaking is not manifest.

I have seen several sources (e.g. here; pg1) state that (exact quote from linked source):

Ising model cannot have spontaneous symmetry breaking at finite temperature,...

I assume that this is an abuse of terminology and what is meant is that the spontaneous symmetry breaking does not manifest itself.

I am not sure the way I am using the terminology in the above is correct and as such want to ask the following clarifying question:

If, for a system, there is spontaneous symmetry breaking of a continuous symmetry which is not manifest at that temperature $T$. Will the system have Goldstone modes at temperature $T$?

*If this is wrong please feel free to answer that question correctly.

**Sorry for the long rambling before the actual question - I am trying to prevent it being an XY problem.

• It makes no sense to say that there is spontaneous symmetry breaking without specifying the conditions. What I would say is that the system exhibits spontaneous symmetry breaking at temperature $T$ if the corresponding Gibbs state at temperature $T$ does not possess all the symmetries of the Hamiltonian. By extension, one might indeed say that the system displays spontaneous symmetry breaking in the ground state if this happen at $T=0$ (although the Gibbs state at $T=0$ is somewhat degenerate, at least for classical systems). – Yvan Velenik Mar 29 '18 at 12:15
• I cannot parse the sentence : If a system at finite temperature the spontaneous symmetry breaking of a continuous symmetry which is not manifest at that temperature will have have Goldstone modes?, so I am not able to answer it. – Yvan Velenik Mar 29 '18 at 12:15
• @YvanVelenik Thanks for your comments (both on this and the linked question) I have edited that sentence so hopefully it should make more sense now. – Quantum spaghettification Mar 29 '18 at 12:26
• As per the Mermin-Wagner theorem, the reason a continuous symmetry remains unbroken at any temperature (say, in a two-dimensional system), is the presence of massless spin waves. – Yvan Velenik Mar 29 '18 at 12:29
• Well, I don't know. As I said, this is just jargon. The existence of such spin waves (in a massless phase) is however a mathematical fact, no matter what you call them. – Yvan Velenik Mar 29 '18 at 14:35

• @RyanThorngren I'm not sure if you're right. The OP seems to be thinking about thermal phase transitions, and the QTIM's $T=0$ phase transition comes from tuning Hamiltonian parameters rather than temperature. But I agree that the OP's question in ambiguous and they should edit it to clarify. – tparker Mar 29 '18 at 17:34