# Explanation of massive Goldstone modes

I'm solving this exercise with a Heisenberg Hamiltonean in linear spin-wave theory and at some point we are asked to compute the dispersion relation at $$k=0$$, which leads me to finding two different stationary modes:

$$\epsilon_{+} = 2 E_0$$

$$\epsilon_{-} = 0$$ Where $$E_0$$ is a constant in the problem.

As I understand, the gapless energy mode corresponds to the Goldstone mode - which are modes that can be excited without any energy cost and are associated with slow, long-wavelength fluctuations of the order parameter (in this case, the magnetization of the ferromagnet). In the case of a ferromagnet, the Goldstone modes correspond to oscillating spin waves. When it comes to symmetry-breaking states where the order parameter (the magnetization) is not conserved, the Nambu-Goldstone modes are usually massless (thus leading us to a gapless, zero excitation energy state which can be described as a condensation of spin-wave of momentum 0). This would mean the 0 energy mode corresponds to a spontaneous symmetry breaking of the system at $$k=0$$

In this lens, the other mode (the "gapped" mode) should be a massive Goldstone mode. Now this is where I feel my knowledge on this topic starts to fail. From what I know, the massive Goldstone modes appear when a symmetry is explicitly broken weakly (say, for example, a small external magnetic field is added to the Heisenberg model creating an anisotropy, the existing Heiseiberg Goldstone modes that originate from the spontaneous symmetry breaking at $$k=0$$ become massive, attaining a finite energy, and it becomes harder to excite the system because the external potential "picks" a preferential direction for the order parameter, aka the magnetization).

What I don't understand is how this system can have both. I feel like there is gap (pun intended) in my understanding of these two definitions that is stopping me from seeing the bigger picture. Can someone help clarify these concepts and the meaning of having these two modes?

PS: In case it helps, the initial Heisenberg Hamiltonean was a Ferromagnetic model for a honeycomb lattice.

• Maybe I am not getting the point, but in general there is both a Goldstone mode and a massive mode in a spontaneous symmetry breaking, right? I am more familiar with superconductivity, where the Goldstone mode is an excitation of the phase of the order parameter, and the massive mode is an excitation of the amplitude Commented Jun 23, 2023 at 8:18
• Could you point me to some resources on that topic maybe? I am trying to understand what kind of symmetry breaking I can associate to this model given it does show those two modes at the boson condensation limit (k=0).
– Rye
Commented Jun 23, 2023 at 10:38
• The first reference that comes into my mind is Altland and Simons' book. Talking about your case, I think the symmetry which is broken is $SU(2)$ spin invariance, and the residual symmetry of the ground state is $U(1)$ (any direction of magnetization is degenerate to the other). Maybe the massive mode corresponds to an increment of magnetization in your case, while the massless mode is the one you have described Commented Jun 23, 2023 at 13:28
• @Matteo yes this makes a lot of sense - the massive mode translating an increase in the value of remnant magnetization and the goldstone, "zero energy" mode just being the general rotation of the system. If you could elaborate this into an answer I can accept it
– Rye
Commented Jul 5, 2023 at 20:32
• Could you write the Hamiltonian at play here? Commented Jan 4 at 11:24

I'm not familiar with the system you're describing, but are you sure that the massive mode is a Goldstone mode? In general, when a symmetry is spontaneously broken, part of the symmetry remains after the breaking. If we view the symmetry as being generated by a set of infinitesimal transformations, then each of the generators that remains unbroken corresponds to a Goldstone mode, while the remaining broken generators correspond to massive (not Goldstone) modes.

• I was looking for information about this and found some papers (like arxiv.org/pdf/1306.1240.pdf) that mention massive/gapped goldstone modes: "for relativistic systems at finite density, Goldstone modes associated with certain internal symmetries can become gapped". For example, if we take $H'=H+\mu Q$, if the new, approximate symmetries of $H'$ "are also spontaneously broken with a symmetry breaking scale much bigger than $\mu$, the corresponding Goldstones will not be exactly gapless, but will have a small gap proportional to the symmetry."
– Rye
Commented Jun 23, 2023 at 7:50
• So in this scenario we are assuming there is an external potential that gaps the Goldstone mode. But in my problem, I start off with a Heisenberg Hamiltonean in a honeycomb lattice, so I don't know how to make sense of all this.
– Rye
Commented Jun 23, 2023 at 7:52

I try to elaborate my comment on an answer.

Spontaneous symmetry breaking occurs when the ground state of the system in the thermodynamic limit (infinitely large system) is not invariant under the action of the symmetry group of the Hamiltonian. Such state can be described by an order parameter, which is the expectation value of some operator that vanishes in the normal state and becomes non zero in the symmetry broken state. For a classical ferromagnet, it is the magnetization vector, while in the case of a quantum Heisenberg ferromagnet, it is the average spin $$\vec{M} = \frac{1}{N} \sum_i \langle \vec{S}_i \rangle$$, where $$N$$ is the number of sites and $$\vec{S}_i$$ the spin operator on site $$i$$.

Typically the ground state is still invariant under some transformations of the original group that form a subgroup. For example, in the classical ferromagnet, the symmetry group is the rotation group $$O(3)$$, and a symmetry-broken state is, for instance, a state with all the spins polarized on the $$z$$-direction. Such state is left invariant by all the rotations about the $$z$$ axis, which form a subgroup $$O(2)$$ of the original $$O(3)$$. On the other hand, all the other rotations bring the state into another ground state, but with a different magnetization. The same happens in a quantum ferromagnet, the symmetry group being $$SU(2)$$, and the subgroup that leaves the ground state invariant being $$U(1)$$, which corresponds to multiplying a state like $$|\uparrow \uparrow \uparrow ... \rangle$$ by a phase factor.

The Goldstone soft modes arise from the fact that there are many degenerate states with a different direction of the magnetization: if a very small amount of energy is provided to the system, the direction of the magnetization can change over long distances, creating a spin wave with long wavelength and small momentum. The number of independent Goldstone modes is given by the difference between the dimension of the symmetry group (3 in our example) and the dimension of the subgroup (1 in our example), so this system has 2 independent Goldstone modes (you need two polar angles $$\theta_i$$ and $$\phi_i$$ to describe a 3d spin wave).

Besides Goldstone modes, the system also supports a "massive" or hard mode, which is excited only if a minimum amount of energy is provided. In this example, the massive mode corresponds to a change in the magnitude of the magnetization. This requires a lot of energy simply because states with different magnetizations are not degenerate ground states in the first place. The number of independent massive modes is given by the number of components of the order parameters (3 for a 3d magnetization) minus the number of Goldstone modes (2 in our case), so we just have 1 massive mode here, which we have already described.

This beautiful picture completely breaks down if an external magnetic field is provided, for example in the $$z$$ direction. The reason is very simple: now the Hamiltonian is different, the symmetry group is $$O(2)$$ (or $$U(1)$$ for the quantum version) and the ground state is invariant under this group, so there is no symmetry breaking and thus no Goldstone modes. Of course you still have massive modes, because changing the magnetization is still pretty hard to do, but everything is much more trivial here.

Hope this helps!