A ferromagnet has a nonvanishing total magnetization which results in a finite magnetic field induced outside the magnet. For an Heisenberg antiferromagnet instead, the total magnetization is zero and therefore the "macroscopic" magnetic field induced is also zero in first approximation.

However, since an Heisenberg antiferromagnet can be thought in first approximation as a discrete arrangement of antiparallel magnetic moments (e.g., the Néel state), it is reasonable to imagine that on a very short scale, very close to the system (assume a planar surface), the magnetic field induced is non zero.

Ferromagnet: $\quad\quad\quad\quad\quad\quad\quad\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow$

Antiferromagnet (Néel state): $\quad\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,$

So my questions are:

1) Is this reasoning correct? It is safe to assume a Néel state to calculate the field outside the antiferromagnet?

2) Is it possible to have a rough estimate of how big is this field and how fast it decays with the distance from the surface?

Edit regarding 1) The Neel is an approximate groundstate for an AFM, not the exact (which can be obtained using the Bethe ansatz). Thus, if you calculate the AFM field from the exact groundstate (suppose you could do it), you will get an "exact" field which is probably different from the approximate field calculated using the Néel state. Is the difference between two solutions relevant in condensed matter systems? How good is this approximation?

  • $\begingroup$ Well at this point I cannot award the bounty, because the only answer, although useful, does not take into consideration the quantum nature of the antiferromagnet state. $\endgroup$
    – sintetico
    Dec 6, 2016 at 15:34
  • $\begingroup$ Please note that the bounty was auto-awarded. In fact, the only answer I got does not address the first point of my question (not even try). $\endgroup$
    – sintetico
    Dec 7, 2016 at 14:35
  • $\begingroup$ Hi sintetico. You flagged this post but you did not indicate what went wrong (the bounty system worked as it is intended to work - if you want to propose a modification to that, please take it to Meta Stack Exchange) nor what action you wish to be taken here, so I have declined the flag. Please flag again with a more specific text if there is something here that needs moderator attention. $\endgroup$
    – ACuriousMind
    Dec 7, 2016 at 15:06

1 Answer 1

  1. It is accurate to assume that an antiferromagnetic material has the Néel form as long as it is experimentally verified (e.g. with Magnetic Neutron Scattering). On the theoretical side, whether the Neel state is the ground state depends on the Hamiltonian and dimensionality under consideration. For the Ising model it is an eigenstate, but for the Heisenberg model it is not. If you want an estimate to how well the Neel state works, you can approach it via Mean-field theory and estimate the rms fluctuations. Additionally, you can examine the effects of the excited states perturbatively on the Neel order parameter. Much of this is outlined in this pdf.

Note: However, as is the case for ferromagnets, most antiferromagnets have microscopic domains. Because there is no net dipole moment, getting these domains to line up is much more difficult than for a ferromagnet. Additionally, many antiferromagnets have structures that are more complicated than a simple cubic arrangement of spins such as in the classical Néel state. For example, see the image below from here:

 Ground state of the Heisenberg antiferromagnet on the triangular lattice with long-range antiferromagnetic order.

  1. The overall dipole moment of an antiferromagnetic domain is zero, so the next order term is the quadrupole term. The quadrupole magnetic field goes as ${r^{-4}}$, as compared to the dipole field which goes as ${r^{-3}}$. To get the exact form you would need to calculate the full quadrupole matrix, something that is described in the section on multipole expansion in Griffiths. In any case, the fields are even weaker when taking domain sizes into consideration. Nonetheless, although the fields from antiferromagnets are very weak, you can still measure them. The idea technique for this is scanning SQUID microscopy. Which put simply, is to put a superconducting junction near the material to detect very weak magnetic fields. This was done for example [here] (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.42.257).

Edit: Noticed OP wanted a quantum mechanical explanation. The Néel state is the (approximate) ground state of many quantum mechanical models, but in particular the Ising, Heisenberg, and XY models for the correct sign of the magnetic coupling term $J$ (see below).


However, dimensionality, range of interaction, and finite system sizes are key. In 1D you run into the Mermin-Wagner theorem, and the Néel state is almost never the correct ground state; you can get more details in Ashcroft and Mermin and the Oxford series textbook on Magnetism in Solids.

With regards to how to describe the field coming from a quantum mechanical magnetic moment (either from spin or angular momentum), the dipole formula for the field is exact for spins. That is, a quantum mechanical spin is a pure dipole. With regards to angular momentum, you technically would need to know how to compute the exact current operator $\mathbf{j}=$ together with the Biot-Savart law to get the exact form of the magnetic field. However, because electrons with large angular momenta are typically localized, it is appropriate to use a multipole expansion and keep just the dipole term.

  • $\begingroup$ Thank you for the answer. Thank you for raising the problem of domains which I was not thinking about, and for the reference. I just have some comments on your answer: $\endgroup$
    – sintetico
    Dec 2, 2016 at 14:46
  • $\begingroup$ 1) My greatest concern is that the Neel state, as well as the more complicated state of your picture, are classical approximations of a quantum mechanical ground state, which is a very complicated linear combination of local spin configuration. One can say that the Neel state and the state you draw are semiclassical approximations where magnetic moments are treated as "small magnets" and not like quantum objects. This approximation is valid if one wants only to calculate the magnetic field of the AFM? $\endgroup$
    – sintetico
    Dec 2, 2016 at 14:46
  • $\begingroup$ 2) How severe is the magnetic domain problem? Is there a rough estimate of e.g., the typical or average length scale of this domains? I think this depends in some way on the correlation length of the AF state...can you argue? $\endgroup$
    – sintetico
    Dec 2, 2016 at 14:49
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – KF Gauss
    Dec 2, 2016 at 15:20
  • $\begingroup$ Regarding the edits...I think there is still one missing link. The Neel is an approximate state, not the exact, and we agree on this. And spin can be treated at low energy as classical magnetic moments, and this is also fine. However, if you calculate the AFM field from the exact ground state, you will get a field which is different from the approximate field calculated using the approximate Neel state. Is the difference between two solutions relevant in condensed matter systems? $\endgroup$
    – sintetico
    Dec 2, 2016 at 17:15

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