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=−JS2Nz+∑→kε(→k)a†(→k)a(→k)+higher order terms where a†(→k) and a(→k) are Bosonic creation and annihilation operators and: ε(→k)=2JS(3−cos(kxa)−cos(kya)−cos(kza)) 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?
Answer
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.
No comments:
Post a Comment