# Goldstone bosons 1st order phase transition

Hi!

I read in Zinn-Justin, that first order phase transitions always have a finite correlation length. Since correlation length is the inverse of the smallest physical mass within a model, this would mean that there can be no Goldstone bosons for a 1st order phase transition. How can that be? I mean massless Goldstone modes always apper, if a global continous symmetry is broken spontaneously??

Best regards

## Answers and Replies

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blechman
spontaneous breaking of the symmetry is 2nd order. If it's a 1st order transition, the effective potential has a tilt, which gives explicit mass for the goldstone modes, if you wish. This is the memory/hysteresis effect that always goes with 1st order transitions.

Example: The Heisenberg ferromagnet has a goldstone mode as the orientation of the spins of the sample. However, in the ferromagnetic phase, as you vary the external magnetic field, this is no longer a flat direction, since the spins will want to align with the field (except, of course, at zero external field). Furthermore, if you vary the field from +H to -H, say, then there is a hysteresis effect (1st order PT) where the field will change suddenly rather than in a continuous way, as you would have expected if the goldstone modes were still massless.

Hope that helps.

Hi blechman!

Is there a general proof, that spontaneous symmetry breaking cannot take place in 1st order phase transitions?

It isn't obvious for me. Consider for example the evolution of the effective potential (plotted against condensate p) with temperature. Or in the case of a magnet, the evolution of the free energy (plottet against magnetization) with temperature (or volume).
I attached a little bitmap. Imagine we start with a Z_2-symmetric T=0 potential U=p^2+p^4 . At T=0 global minimum (denoted as a dot) is at p=0. As temperature grows, it might be possible that the global minimum breaks the symmetry spontaneously (it isn't symmetric any more under Z_2). In the picture it jumps discontinously, we have a 1st order phase transition.

Maybe such a behaviour of V is not allowed, but I guess from graphical considerations alone, one cannot prove it.

Best regards, Martin

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blechman
Well, first of all, Z_2 breaking does not lead to goldstone bosons anyway, but never mind...

But take a look at your thumbnail. Does that potential look symmetric to you?! The symmetric vacuum has different (free) energy than the asymmetric vacuum, so you have EXPLICITLY (as opposed to "spontaneously") broken the symmetry. This EXPLICIT symmetry can be thought of as a mass-term for the goldstone bosons, and corresponds physically to the fact that it now costs energy to excite these modes (hysteresis). To see it explicitly, go ahead and Taylor-expand the free energy about the symmetry-breaking vacuum rather than at the origin, and you will get a quadratic term for the goldstone mode (phase of the order parameter). There's your proof.

In the case of a second-order PT, this would not happen; the free energy is STILL symmetric, it's only that the vacuum has broken the symmetry, but it costs no energy to go from one vacuum state to another, and so there's a goldstone mode. However, if it's first-order (see your thumbnail), the vacuum is unique and there are no more flat directions. Does that make sense?

>Well, first of all, Z_2 breaking does not lead to goldstone bosons anyway, but never mind...
I admit it's a bad example , Z_2 is not continous, so better consider O(2). At T=0 we have the potential
$$V= (\vec{p})^2 + ((\vec{p})^2)^2$$

>Does that potential look symmetric to you?!
Sorry, I corrected the thumbnail

Ok here's my corrected argument:
We have a symmetric potential resp. free energy but at a critical temperature the vacuum jumps from p=0 to finite p. Therefore 1st order phase transition. The O(2) symmetry is spontaneously broken. The vacua now lie on a circle. One can rotate on the circle without energy loss, 2nd derivative vanishes, i.e. no mass (p_2 is our Goldstone boson). By choosing one vacuum (real vacuum), we spontaneously break the O(2) symmetry.

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blechman
Hmmm.... a good counterexample. There is definitely a goldstone mode here, and it is first order. I guess I have to take back my previous statements; sorry about that!

Well, you're saying that the correlation length is finite in this case, since it's first order. And this violates Goldstone's theorem....why? It is not true that the goldstone mass is $\propto\xi^{-1}$. For example, consider the ferromagnetic phase below the critical temperature. There's a goldstone mode there, but the correlation length is finite (hence the formation of domains) - it only diverges at the critical temperature itself. So I don't see a contradiction.

>It is not true that the goldstone mass is $$\propto \xi^{-1}$$
But in Field Theory sense, the inverse correlation length (dependant of temperature T) is the smallest mass appearing within the model. (Zinn Justin - Quantum Field Theory and Critical Phenomena, chapter 20.1). So, if a massless particle appears at critical temperature Tc (where phase transition takes place), the correlation length diverges for T>Tc.
If there are Goldstone particles for 1st order phase transition, as in the given example, the correlation length diverges. But Zinn-Justin says it should be finite.

blechman
I have to take a look at Zinn-Justin's book; I have a copy in my office, but I won't be in until Monday.

But I take some issue with the relation of the goldstone mass to the correlation length. The fact that a theory has a mass gap (finite inverse correlation length) says nothing about the goldstone modes themselves, but rather about the modes above the goldstone modes (modes of finite energy).

Example: in the ferromagnetic transition again (H=0) there is a vanishing mass gap at the critical point. If you think of the potential at this point, you see that there is a continuum of modes that relate one vacuum to the other in the RADIAL direction, since there is no barrier.

Compare this to your middle thumbnail. Now there are still goldstone modes, but it is no longer true that there is a continuum of states that take you radially from one vacuum to the other, since you must now cross a barrier. This is just the usual QM problem of the double well. I believe that THIS is what Zinn-Justin is referring to.

So in summary: the diverging correlation length tells you that there are a continuum of modes above the vacuum, while the finite correlation length tells you that there is a discrete spectrum of modes, separated by the inverse correlation length in energy. But none of this has to do with the goldstone modes themselves: that just tells you how the vacuum behaves.

I hope this makes sense. I cannot say more without looking at Zinn-Justin so I can comment more directly on your question. I'll try to stop by on Monday. In the meantime, if anyone else has a better way to say this, plz chime in!

blechman
Let me put the last post another way:

The mass gap tells you about the EXCITATIONS of the system. So saying that the correlation length is finite (finite mass gap) means that it takes a finite amount of energy to excite the system. However, if the correlation length is infinite (vanishing mass gap) it means that it takes infinitesimally small amount of energy to excite the system.

However, none of this has anything to do with the vacuum structure of the theory. This is where the goldstone modes live.

Perhaps that makes more sense. Hope it helps!

Hi!

I haven't been in office yesterday, too. So I wasn't able to reply until now.

I googled a bit and found two papers related to the topic.

The ginsparg-file (page 60) says that "when the correlation length diverges, the theroy becomes massless"

The other one says (in chapter 4) that a system develops Goldstone bosons at a 1st order transition.

I haven't read the whole papers, just the above remarks.

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blechman
Hi Sunset.
Again, when Ginsparg says "the theory is massless" he is referring to the excitations above the ground state, NOT the goldstone modes. What it means is that it no longer costs a finite, discrete amount of energy to create a spin-wave (or phonon in a solid, or photon in a conductor, etc). That has nothing to do with the goldstone modes.

This looks like a nice set of lectures. I look forward to reading them!

Hi!

I had a discussion with someone from the institute here, he told me mass M in the formula
$$\xi=1/M$$ relates only to the particle p_1 and not to the Goldstone mode p_2. I guess, this is what you said earlier.

Ok, thanks for your help!

Best regards, Martin

blechman