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I am getting confused about the number of Goldstone modes present in the Heisenberg model. After a Holstein-Primakoff transformation the energy can be written a: $$H=-JS^2 Nz+ \sum_{\vec k} \varepsilon(\vec k) a^\dagger(\vec k) a(\vec k)+\text{higher order terms}$$ where $a^\dagger(\vec k)$ and $a(\vec k)$ are Bosonic creation and annihilation operators and: $$\varepsilon (\vec k)=2J S(3- \cos( k_x a) -\cos(k_y a)-\cos(k_z a))$$ To me this indicates $3$ independent Goldstone modes. But I have also read that we should have one Goldstone mode per continuous symmetry generator broken - from this I would expect $2$ Goldstone modes. This answer on a related question also indicates that the Heisenberg model is an exception.

My question is therefore: How many Goldstone modes does the Heisenberg model have and why?

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  • $\begingroup$ It is definitely not 3. Why would you say that ? Also, is the model with ferromagnetic or antiferromagnetic interactions ? That changes everything in that case. $\endgroup$ – Adam May 11 '18 at 13:29
  • $\begingroup$ @Adam Ferromagnetic. And $3$ modes as you appear to have a mode corresponding to $\vec k$ in the $x$, $y$ or $z$ direction. $\endgroup$ – Quantum spaghettification May 11 '18 at 14:41
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    $\begingroup$ You are making a huge confusion here : the dimension of space (related to the number of $k$ components) has nothing to do with the number of Goldstone modes (which is related to the number of internal symmetries not broken by the ground-state). For example, the O(N) model has N-1 Goldstone mode in any dimension d>2 (which is the lower-critical dimension). $\endgroup$ – Adam May 11 '18 at 14:45
  • $\begingroup$ @Adam Ok fair, If you expand on where this confusion lies - it may form a good answer. $\endgroup$ – Quantum spaghettification May 11 '18 at 16:45
  • $\begingroup$ @Quantumspaghettification Maybe helpful physics.stackexchange.com/questions/113773/… $\endgroup$ – SRS Jun 30 '18 at 14:18
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Having a Goldstone mode at momentum $\boldsymbol k = (k_x,k_y,k_z)$ requires that the energy vanishes there, i.e. $\varepsilon(\boldsymbol k) = 0$. In the periodic Brillouin zone $\left[ -\frac{\pi}{a} , \frac{\pi}{a} \right] \times \left[ -\frac{\pi}{a} , \frac{\pi}{a} \right] \times \left[ -\frac{\pi}{a} , \frac{\pi}{a} \right]$, this only happens at the zone center, $\boldsymbol k = (0,0,0)$. More precisely, for small momenta, we have that $$\varepsilon(\boldsymbol k) \approx JSa^2 (k_x^2 + k_y^2 + k_z^2).$$

So we seemingly have only one Goldstone mode. How does this rhyme with the number of Goldstone modes being equal to the number of broken symmetry-generators (which is indeed two in this case)?

The answer that the above rule of thumb for counting Goldstone modes is true for relativistic theories, where the Goldstone modes have a low-energy dispersion $\varepsilon \sim |\boldsymbol k|$. However, in the above case, our low-energy dispersion is quadratic. Indeed, the more general formula is given in this 1976 article "On how to count Goldstone bosons" by Nielsen and Chadha: the modes where $\varepsilon_{\boldsymbol k} \sim |\boldsymbol k|^z$ count double if $z$ is even. Hence, $$ \# \left(\textrm{broken generators}\right) = \# \left(\textrm{Goldstone modes with $z$ odd} \right) + 2 \# \left( \textrm{Goldstone modes with $z$ even} \right). $$

Example: the ferromagnetic Heisenberg model has one Goldstone mode with a quadratic dispersion, whereas the antiferromagnetic Heisenberg model has two Goldstone modes with a linear dispersion. In both cases, this agrees with the number of broken generators being two.

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  • $\begingroup$ Related reference: Unified Description of Nambu-Goldstone Bosons without Lorentz Invariance arxiv.org/abs/1203.0609 @RubenVerresen $\endgroup$ – SRS Jun 30 '18 at 14:21

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