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What is Goldstone's theorem for classical, purely thermodynamic systems?

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In the context of nonrelativistic physics (e.g. condensed matter), Goldstone's theorem tells us that spontaneous symmetry breaking leads to gapless excitations, i.e. excitations with arbitrarily low energy above the ground state. Since $E \sim \omega$ in quantum mechanics, this tells us classically that there should exist modes with vanishing frequency.

However, Goldstone's theorem is also applied to 'purely thermodynamic' systems, such as the classical XY model, which have no dynamics of their own! That is, if you time-evolve with the XY model Hamiltonian, absolutely nothing happens, because there is no canonical momentum anywhere in sight. The system just sits there.

In practice, time evolution would happen due to coupling to a thermal reservoir, but that's not written into the Hamiltonian and certainly doesn't lead to a unique $\omega$ for a mode, or even oscillations at all. So it's hard to define any sort of "mode" for such a system.

In this case, what is the formal statement of Goldstone's theorem for such systems, and how does it relate to the usual statement of Goldstone's theorem?

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user knzhou
asked Dec 2, 2016 in Theoretical Physics by knzhou (45 points) [ no revision ]
Most voted comments show all comments
Your second paragraph is vague. About the third paragraph : time evolution exists in all of quantum systems, regardless of the existence of any thermal reservoirs. Another point is that the Goldstone's theorem holds for all of physical systems with a global continuous symmetry(classical or quantum systems)

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user Hosein
@Hosein I'm working classically here. Classically, if you have a Hamiltonian $H(q)$ that doesn't depend on the canonical momenta $p$, then the equation of motion is just $\dot{q} = 0$, and nothing happens whatsoever.

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user knzhou
How do you define a classical spin degree of freedom?

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user Hosein
@Hosein See the classical XY model I linked in the question.

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user knzhou
The XY model mentioned in the link is just a defined partition function. There is not such a thing as classical XY model which is really classical. The quantum XY model in the limit of large spin can be approximated by a classical XY model (yet the equations of motion for spins should be derived from quantum commutation relations)

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user Hosein
@Hosein I don't understand your first statement; the very first equation is a definition of a Hamiltonian. And Goldstone's theorem seems to be applied just fine to the classical XY model, without any reference to quantum mechanics, even if it is unrealistic as you say. I'm asking exactly how that works.

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user knzhou

What I wanted to say was that the classical XY model actually has dynamics which is derived from quantum mechanical relations. So this model isn't static without the existence of a thermal bath. However your main question still remains.

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user Hosein

1 Answer

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The connection is easiest to see in terms of the quantum effective potential in the presence of a classical background, which is analogous to a free energy functional in the context of statistical field theory. You can show using arguments equivalent to ones in most QFT textbooks that as long as interactions are short-ranged or local, the free energy cost associated with long-wavelength deformations perpendicular to the direction of symmetry breaking vanishes with the spatial frequency of the applied external field. Usually this implies a divergence in a susceptibility function transverse to the 'VEV' (or direction of spontaneous symmetry breaking). The 'modes' in question are analogous to the low frequency normal modes of coupled harmonic oscillators.

This post imported from StackExchange Physics at 2016-12-09 18:26 (UTC), posted by SE-user fs137
answered Dec 5, 2016 by fs137 (10 points) [ no revision ]

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