This is an upshot of the question here. The up-aligned and the down-aligned spin configurations are assumed to be two distinct phases in case of an Ising ferromagnet. But for Heisenberg ferromagnet, there are infinitely many spin configurations distinguished by different orientations of the spins in full 3-dimensional space.

  1. Does it mean a Heisenberg ferromagnet has an infinite number of phases? If not, why?

  2. If yes, what is the nature of phase transition between any two such phases below $T_c$? Is it a first order or a continuous transition? Note that, in the case of Ising ferromagnet, the transition between up-aligned ferromagnetic phase and the down-aligned phase is first order.

  • $\begingroup$ Yes they form different phases (in the sense that they correspond to distinct extremal Gibbs measures). As for the transition between them, you'd have to specify which external parameter you are varying. For example, if you apply a magnetic field $\vec{h}=(h,0,0)$, and vary $h$ from positive values to negative ones, then there will be a first-order phase transition at $h=0$, with a discontinuity of the magnetization $\vec{m}(h)=(m(h),0,0)$, $m(h)$ jumping from a positive value to a negative one. (Of course, I am assuming that the lattice dimension is at least $3$.) $\endgroup$ Jul 29, 2017 at 7:44

1 Answer 1

  1. Yes, but only in the same sense that there are infinitely many directions for the hydrogen m=1 state to point it's angular momentum in. This degeneracy is from global rotational symmetry. They are not distinct phases, simply rotating the entire magnet moves you from one to the other.

  2. All of these states have the same energy, so there is no concept of phase transitions between them. It costs zero energy to go between them. This is related to the fact that symmetry breaking of the ferromagnet comes with a (gapless) Goldstone mode

Edit: So OP is adamant about definining these different states as different phases, fine its up to you what you define as "phase", but it disagrees with many classic texts of phase transitions (cf. Phase transitions by Goldenfeld). In that case, yes there are as many "phases" as there are on the surface of a sphere.

As I mentioned in the comments, there is only a first order transition if you pass through $H=0$ going from one of these states to another. This means the "phase" boundary between these states is a single point in 3D configurational $H$ space. This is why I would say that there is no distinction between these phases, you can get from one to the other without passing through any transition (1st, 2nd order or otherwise).

You then ask why can't it be that there is a no line or plane in $H$ that separates states. I return this question with: why would there be? Surely you agree that all of these states are degenerate when $H$ points in the appropriate direction. Because the energy goes smoothly as $\mathbf{M} \cdot \mathbf{H}= M H \cos(\theta)$, you then can see there is no energy barrier going from one state to the other if $\theta$ is changed continuously.

  • $\begingroup$ How does it answer my question? Up-aligned magnetization and down-aligned magnetization are also energetically degenerate. So if you can talk about distinct phases in case of Ising ferromagnet (which is common in the books of phase transition) why not Heisenberg ferromagnet? What does Goldstone mode have to do with my question (even though what you have said is correct)? $\endgroup$
    – SRS
    Jul 29, 2017 at 6:03
  • $\begingroup$ Because it is meaningless to consider them separate phases. In the Ising case sure you have a 1st order transition from all spin up to all spin down, but this only because you have to pass H=0 (where you get directional singularities). But in the Heisenberg case you can smoothly move from one to the other without going through H=0. That's why there is noo phase transition. Moreover, most books do not consider the two Ising states separate phases, having a 1st order line between the two isn't the proper way to distinguish phases (cf liquid to gas) $\endgroup$
    – KF Gauss
    Jul 29, 2017 at 6:28
  • $\begingroup$ @user157879 : which books do not consider them different phases? Note that you can even have spatial coexistence of the $+$ and $-$ states. Of course, it is possible to go from one to the other without undergoing a phase transition (as you can go from liquid to gas). This does not make them the same phase (or we are using very different notions of phases). $\endgroup$ Jul 29, 2017 at 7:50
  • $\begingroup$ @user157879 Even though it costs zero energy to move between degenerate states, this will not happen spontaneously in a thermodynamic system. Once the phase transition has occurred, that phase with a given orientation of magnetization remains locked where it is. It's true that for first order phase transition in the Ising case, one has to cross the $H=0$ line. I'm not sure that in the Heisenberg case, the line of transition is still $H=0$? Can't it be that the transition line is at some non-zero H? Recall in liquid-gas transition the pressure (analog of H) is nonzero on the coexistence line. $\endgroup$
    – SRS
    Jul 29, 2017 at 8:48
  • $\begingroup$ @SRS : As I explain in my comment to your question above, the magnetic field is now a vector, not a scalar. If you fix its direction, but not its (signed) magnitude, as I do in the comment, then a first-order phase transition does occur as the magnetic field "changes sign". However, if you fix the magnitude of the magnetic field and smoothly change its direction, the magnetization changes continuously. $\endgroup$ Jul 29, 2017 at 9:30

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