2
$\begingroup$

It is known within the Ising or Heisenberg model that the exchange constant $J$, combined with the dimensionality/connectivity of the system, sets critical temperature for a phase transition into a ferromagnetic or antiferromagnetic state.

For example, within the 2D Ising model, we then get $k_B T_c = \frac{2}{\mathrm{ln}(1+\sqrt{2})} J \approx 2.3J$. From a theory perspective, this exchange constant $J$ is assumed to be completely temperature-independent.

Now, let's say we cool a realistic material, like iron, below its ferromagnetic transition temperature. Will there be corrections to the exchange constant $J$ now that the phase is ordered? Experimentally speaking, my gut feeling is that there should be some modification of the exchange parameter $J$, maybe like how phonon energies are modified upon ordering in crystals. But I don't know the order of magnitude for this change.

Given that $J$ is not a term in the first-principles Hamiltonian, but is actually an effective parameter, I think it shouldn't generically be the same above and below the transition.

Finally, I'd also like to know if these changes (or lack thereof) extend to other phase transitions, for example liquid crystals or ferroelectricity.

$\endgroup$
1
$\begingroup$

I don't think I'm well-equipped to discuss any experiments, but I do think there are two important insights to be gained from the theoretical perspective:

  1. Prior to the modern understanding of phase transitions, it was wondered by physicists whether the same microscopic Hamiltonian could possibly describe the two phases separated by a phase transition (there's a nice discussion of this in Nigel Goldenfeld's Lectures on Phase Transitions and the Renormalization Group, at the end of chapter 2). Of course, we now know this to be the case: if $J$ corresponds to a coupling constant in your microscopic Hamiltonian, then it always takes the same value in every phase.

  2. However, as you say in your post, $J$ does not usually correspond to some constant in the microscopic Hamiltonian: it usually corresponds to some effective coupling. One way to think about this is from the perspective of the renormalization group: your microscopic Hamiltonian might have all sorts of crazy terms with different coupling constants, but upon course-graining and rescaling these coupling constants will generally change. In particular, the renormalization group "flow" of your coupling constants will depend on the phase of the system -- indeed, it's precisely the sudden change in RG fixed point as you tune across the transition that accounts for the sudden change in long-distance behavior of the system. For example, in a simple 2D Ising model, you expect the temperature to flow towards infinity upon starting with $T > T_c$, while it will flow to zero for $T < T_c$.

So in summary, to answer your question: yes, the effective couplings of your system will appear to change suddenly across a transition. One particularly nice example of this is in the KT transition, where the superfluid stiffness undergoes a universal "jump" upon tuning from ordered to disordered. This is because your effective coupling $K = \beta J$ immediately switches from flowing to zero to flowing to $2/\pi$ at long distances (a nice reference for this is Kardar's Statistical Physics of Fields, chapter 8). This has indeed been experimentally verified.

$\endgroup$
2
  • $\begingroup$ Thanks, this is the kind of thing I was interested in. Do you know of other examples besides the BKT transition? RG analysis of the Heisenberg Hamiltonian wouldn't tell you anything about a jump in $J$, is there any other method for figuring out the size of the jump? $\endgroup$ – KF Gauss Apr 7 at 10:54
  • $\begingroup$ Not to my knowledge -- I'm not an expert, but I think the KT transition is a fairly special example. For a classical Heisenberg Hamiltonian, you expect $K = \beta J$ to be a relevant coupling separating just two phases at one critical point, so you expect K to flow to zero below the critical point and infinity above the critical point. But that's certainly not what you would measure. $\endgroup$ – Zack Apr 7 at 14:23
-1
$\begingroup$

Yes, in principle a real material's Hamiltonian is not the Ising Hamiltonian. Perhaps it looks effectively like a nearest-neighbor square lattice Ising Hamiltonian with an effective $J$, but a real material will not have a completely uniform lattice, it will have disorder, and it will have longer-range and more complicated (including non-magnetic) interactions. Furthermore, as you say, lattice non-uniformities will change as one tunes the temperature, an effect completely outside of the purview of the Ising model on a fixed lattice.

The reason why the Ising model is useful is that it manages to capture certain universal properties close to the phase transition, which are independent of all the model-dependent properties I mentioned above*. These effects will change where the critical temperature is, but they will not change the critical exponents or the amplitude ratios near the critical temperature.

* This is assuming the short-range Ising model at least remains a good approximation. For example, if a system has long-range interactions, or strong enough disorder, it will change the universal properties. But one can define long-range and disordered Ising models to treat these.

$\endgroup$
1
  • 2
    $\begingroup$ Thank you, but I already know why the Ising and Heisenberg models etc. are useful. I am not challenging the assumptions of these models. I just want to how much the effective $J$ typically changes in real materials across the phase transition due to the ordering itself (not quenched disorder or static inhomogeneity). $\endgroup$ – KF Gauss Oct 9 '19 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.