19
$\begingroup$

Both the Ising and the Heisenberg Models describe spin lattices with interaction on first neighbors. The Hamiltonian in each case is quite similar, despite the fact of treating de spins as Ising variables (1 or -1) or as quantum operators. In the Ising case it looks like

$$H_\textrm{Ising} = -~J \sum_{\langle i\ j\rangle} s^z_{i}\ s^z_{j}$$

where J is the coupling constant ($J>0$ for ferromagnet and $J<0$ for anti-ferromagnet), $\langle i\ j\rangle$ represents sum over first neighbors and $s^z$ is the spin in z direction. On the other hand, the Heisenberg model is

$$H_\textrm{Heisenberg} = -~J \sum_{\langle i\ j\rangle} \hat{S}_{i} \cdot \hat{S}_{j}$$

where the only difference lies in the spins being operators. (In both cases I took away the interaction with an external field for simplicity)

My Question is: What new phenomena does treating the spins as operators brings? I can see that $\hat{S}_{i}\ .\ \hat{S}_{j}$ takes account of the spin in every direction and not just z, but I can't see the physical implication of that.

Improve this question
$\endgroup$
8
  • $\begingroup$ The question feels a bit broad (and moreover, the question in the title is different from the question in the question). Basically, you are asking for a list, which does not fit well the format of the site. You could try to narrow it down: What kind of properties do you care (or not care) about, and why? $\endgroup$
    – Norbert Schuch
    Sep 15, 2016 at 17:31
  • 1
    $\begingroup$ You are right, it was kind of vague... sorry! Im specifically interested in two things. Understanding why the heisenberg model has spin wave solutions while the ising model does not and how are the phase transitions in both models. $\endgroup$
    – P. C. Spaniel
    Sep 15, 2016 at 17:47
  • $\begingroup$ Ah, then why don't you either (i) narrow down your question to ONE of these or (ii) post a new question? Remember that the idea of this site is to build some kind of Q&A knowledgebase. The more specific the questions are, the better they will serve as a reference for future users. $\endgroup$
    – Norbert Schuch
    Sep 15, 2016 at 17:48
  • 1
    $\begingroup$ Regarding " Understanding why the heisenberg model has spin wave solutions while the ising model does not" -- spin waves require a continuous symmetry, so they only make sense in the Heisenberg model in the first place! $\endgroup$
    – Norbert Schuch
    Sep 15, 2016 at 17:50
  • $\begingroup$ thank you very much for your answers. What does "Spin waves require continuous symmetry" means? $\endgroup$
    – P. C. Spaniel
    Sep 17, 2016 at 18:21

2 Answers 2

Reset to default
12
$\begingroup$

As this is a list-like question, let me list a few things (without much discussion -- feel free to ask specific questions about individual points). Each item mentions what the Heisenberg model (HM) has as opposed to the Ising model (IM).

  • continuous symmetry vs. discrete symmetry

  • as a consequence: gapless excitations whenever the symmetry is broken (i.e. in all cases except the 1D antiferromagnet -- in that case, however, there are gapless modes due to the Lieb-Schultz-Mattis theorem)

  • as a consequence thereof: no spontaneous symmetry breaking in one and two dimensions at finite temperature (as opposed to the 2D HAFM), this is the Mermin-Wagner theorem

  • non-commuting terms: The eigenstates will typically not have a simple form (as opposed to the IM, which has commuting terms)

  • the IM Hamiltonian has integer eigenvalues (times $J$), while we cannot easily characterize the eigenvalues of the HM

Some caveats, however:

  • some of these properties hold because of continuous vs. discrete symmetry, rather than classical vs. quantum

  • some of the properties hold because of commuting vs. non-commuting rather than discrete vs. continuous symmetry

  • some of these properties only hold for (infinite) lattices, others already on the level of just a few spins

Improve this answer
$\endgroup$
8
  • $\begingroup$ Great. Two more come to mind: (1) the HM (at least when $J< 0$ following the OP convention) can never be smoothly connected to a classical ground state (whilst preserving the continuous symmetry of the Hamiltonian), even when the ground state spontaneously breaks the symmetry, and (2) due to its larger quantum effects, the HM can have phases other than spontaneously breaking its spin symmetry (or the trivial phase), such as valence bond order (spontaneously breaking translation invariance) or even spin liquid physics (which does not break any symmetry and has emergent fractional excitations). $\endgroup$
    – Ruben Verresen
    Sep 15, 2016 at 18:04
  • $\begingroup$ @RubenVerresen I'm not sure I fully follow the reasoning behind the first statement (though I kind of agree with it) ... Do you have a specific lattice in mind, or is this a general statement? E.g., what would happen in infinite dimensions? Regarding (2), note that also the Ising AFM has more rich physics on frustrated lattices! $\endgroup$
    – Norbert Schuch
    Sep 15, 2016 at 18:19
  • $\begingroup$ Thanks for setting me straight on (2). I wonder to what extent one could argue that HM physics is strictly richer than IM physics in terms of possible phases. As for (1) it was a case of thinking out loud, but in the case of spontaneous symmetry breaking I would argue the following: if we start with a Neel ground state and only allow smooth changes that preserve the $SO(3)$ symmetry, then the low energy theory has to stay linear. So it is sufficient to argue the perhaps more natural statement that a classical ground state can never have low energy modes with a linear dispersion. $\endgroup$
    – Ruben Verresen
    Sep 15, 2016 at 20:28
  • $\begingroup$ I don't know the most efficient way to argue the last statement, but I can imagine using that since the low energy theory is basically described by a massless relativistic field, we know its ground state entanglement must obey a certain scaling law (and in particular be non-zero). It would be interesting to prove the more general statement: "The ground state of the Heisenberg AFM --on any lattice-- can never be smoothly connected to a classical ground state whilst preserving $SO(3)$ symmetry." At least I can't imagine a counter-example, but I might just be lacking creativity :) $\endgroup$
    – Ruben Verresen
    Sep 15, 2016 at 20:33
  • $\begingroup$ @RubenVerresen "if we start with a Neel ground state" -- does it mean that you assume a square lattice? In that case, since the model is gapless, I'm not sure what exactly you mean by "smoothly connected" (since I guess "classical ground state" means gapped). But what if we have a model on a lattice where it, say, dimerizes? $\endgroup$
    – Norbert Schuch
    Sep 15, 2016 at 20:38
5
$\begingroup$

One of the main differences is that the Ising model lies on a discrete symmetry (the $Z_2$ symmetry) while the Heisenberg model lies on a continuous one (rotational symmetry). It will affect the phase transitions that these models undergo.

In particular, because of the Mermin-Wagner theorem, there can be no finite-temperature phase transition of the Heisenberg model in $d=2$ (leaving apart the very special case of the BKT transition).

This is not the case of the Ising model that undergoes a phase transition at finite temperature from a high-$T$ disordered state to a low-$T$ ordered state in $d=2$ (the exact solution has even been calculated by Onsager and later by others).

There's probably much more to it than this particular case, feel free to edit my answer if you feel like adding something.

Improve this answer
$\endgroup$
2
  • $\begingroup$ If I'm not mistaken, the Heisenberg model also does not exhibit the Kramers-Wannier duality in d= 2 unlike the Ising model. $\endgroup$
    – user106422
    Sep 15, 2016 at 17:57
  • $\begingroup$ I do not know about this duality, feel free to edit my answer or to post another one about it :) $\endgroup$
    – Dimitri
    Sep 16, 2016 at 12:20

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