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This seems to be experimentally true, but I don't quite have an intuition as to why. In the Ising model, we usually consider an insulating ferromagnet if $J>0$, where $J$ is the exchange coupling. Does this situation not usually occur in reality?

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2 Answers 2

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First let me make two comments before answering the question.

  1. The difference between metal and insulator rest in the existence of the itinerant electron Fermi surface or not. Ising (or Heisenberg) model is just an effective theory of local moments (localized electrons in the atoms), which contains no information of the itinerant electron, so there is no hope to start from an Ising model and explain the difference between metal and insulator.

  2. The observation that "ferromagnets are mostly metals, while antiferromagnets are mostly insulators" is not quite true. There are ferromagnetic insulators like $\text{Fe}_3\text{O}_4$, which is one of the earliest ferromagnets discovered in human history. There are also examples of antiferromagnetic metals, from the historical ones like $\text{Cr}$ to the most recent ones like the parent compounds of iron-base superconductors (e.g. $\text{Ba}\text{Fe}_2\text{As}_2$).

The different magnets arise from the different magnetic exchange mechanisms in the material. In the following, some most famous exchange mechanisms in the solid are listed, but as the material can be complicated, so the list is far from complete.

  • Ferromagnetic metal: itinerant exchange (RKKY interaction),
  • Ferromagnetic insulator: double exchange,
  • Antiferromagnetic metal: Fermi suface nesting and SDW instability,
  • Antiferromagnetic insulator: superexchange.

In many transition metals (e.g. $\text{Fe}$), the exchange interaction between magnetic ions are mediated by the itinerant (conduction) electrons. The transition metal system contains both the itinerant electrons and the local moments (typically from $d$ orbitals). Local moments just sit on each atom while the itinerant electron travels between the atoms. When the itinerant electron meets the local moment, they mutually polarize each other towards the same orientation. So as the itinerant electrons travel between the atoms, the message of the magnetic orientation is brought from one local moment to another. So eventually all local moments tends to align in the same direction with the itinerant electron, and as the local moment orders, more itinerant electron will be polarized to the ordering orientation to reinforce the ordering. Therefore the ferromagnetism is developed in the metal by this collective behavior. This mechanism is known as the itinerant exchange or the Ruderman-Kittel-Kasuya-Yosida (RKKY) interaction. In the real space, the RKKY interaction between two local moments of the distant $r$ follows the oscillatory behavior $$J_\text{RKKY}(r)\sim-\frac{\cos(k_F r)}{r^3}.$$ As in the dilute limit $k_F r\ll 1$ for many metals, ferromagnetism will dominate.

Antiferromagetic insulators are usually Mott insulators, in which the Fermi surface is gapped out by interaction, and there is no itinerant electrons. In this case, the magnetic correlation must be mediated by another mechanism, which is known as the superexchange. In the simplest Mott insulator (e.g. $\text{Mn}\text{O}$), each magnetic ion $\text{Mn}^{2+}$ would have a single unpaired electron in a $d$ orbital, which can hop between $\text{Mn}$ sites as bridged by the $\text{O}^{2-}$ ion in between. When the electron spins on $\text{Mn}$ are opposite aligned, it can hybridize over the Mn-O-Mn unit and gain kinetic energy.

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However if the electron spins on $\text{Mn}$ are ferromagneitcally aligned, such hybridization will be forbidden by the Pauli exclusion principle. Therefore superexchange favors antiferromagnetism, and the effective exchange interaction is given by $J\sim t^2/U$ where $t$ is the effective hoping integral between $\text{Mn}$ sites, and $U$ is the on-site Coulomb repulsion.

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    $\begingroup$ This is one of the more elaborate and well-written explanations I've come across in this area. Do you have any references for a deeper look? $\endgroup$
    – Xivi76
    Dec 29, 2020 at 9:01
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The main justification for considering the Ising model is that it's exactly solvable in one & two dimensions (and that it shows critical behavior which is universal in some sense). It is not particularly meaningful as an approximation to a real physical system.

The Heisenberg model does a much better job, but it is also a lattice model. If you really want to capture the electronic band structure and the magnetism, the way to go is the Hubbard model. It consists of a (tight-binding) hopping term (with hopping parameter $t$) to describe the kinetics of the electrons, as well as an on-site Coulomb repulsion $U$. Such a setup where the spins are associated with electrons which can move across the lattice is called itinerant magnetism. One can obtain the antiferromagnetic Heisenberg model from the Hubbard model at half-filling in the limit of strong on-site repulsion $U\gg t$. Then, every site is occupied by exactly one electron. Neighboring electrons will be able to swap sites, giving rise to an effective exchange interaction as in the Heisenberg model.

The antiferromagnetic Hubbard model (i.e. half-filling), gives rise to a band gap. That is, the gapless (metallic) band structure one obtains from Bloch theory is modified by the many-body dynamics. This can be understood in mean-field theory and I will outline it below but it requires familiarity with second quantization. An insulator that is insulating because of such an "non-Bloch" band gap is called a Mott insulator.

At half-filling, the tight-binding dispersion exhibits a property called perfect nesting: $\varepsilon_{k+Q}=-\varepsilon_k$ where $Q=\frac{\pi}{a}(1,1,1)$, i.e. $Q$ is the reciprocal vector from the $\Gamma$-point to the edge of the Brillouin zone. One can show that the susceptibility for external fields with $q=Q$ is divergent, i.e. an instability towards a spin density wave with $q=Q$ occurs. The fact that $Q$ reaches to the edge of the Brillouin zone means that the spin density wave corresponds to an antiferromagnetic ground state. In contrast, for the Hubbard model at low filling, an instability occurs for $q=0$, leading to a ferromagnetic ground state.

One can include the effects of the spin density wave in mean-field theory where $\Delta\propto \langle S_Q^z\rangle$ (the "gap") attains a finite expectation value. After a few lines of calculation, one can arrive at the following mean-field Hamiltonian:

$$H = \sum_{\sigma}\sum_{k}'\psi_{k\sigma}^\dagger \boldsymbol{\varepsilon}_k^\sigma\psi_{k\sigma}+K_0,\qquad \psi_{k\sigma}=\begin{pmatrix} c_{k\sigma}\\ c_{k+Q,\sigma} \end{pmatrix},\quad\boldsymbol{\varepsilon}_k^\sigma=\begin{pmatrix} \varepsilon_k & -\sigma\Delta\\ -\sigma\Delta & -\varepsilon_k \end{pmatrix}.$$

Here, the sum over $k$ extends only over half of the Brillouin zone, called the "magnetic zone". The $\psi$ vector is actually a vector of annihilation operators. This is merely a mathematical trick and nothing deep. But it allows us to write the Hamiltonian in this compact quadratic form.

One can now diagonalize the matrix $\boldsymbol{\varepsilon}_k^\sigma$:

$$\psi_{k\sigma}^\dagger \boldsymbol{\varepsilon}_k^\sigma\psi_{k\sigma} = \begin{pmatrix}\gamma_{k\sigma}^{c\dagger} & \gamma_{k\sigma}^{v\dagger}\end{pmatrix}\begin{pmatrix} E_k & 0\\ 0 & -E_k \end{pmatrix}\begin{pmatrix}\gamma_{k\sigma}^c\\\gamma_{k\sigma}^v\end{pmatrix},\qquad E_k=\sqrt{\varepsilon_k^2+\Delta^2}$$

This is called a Bogoliubov transformation and is very similar to what one does in the BCS theory of superconductivity. The superscript $c$ and $v$ denote "conduction" and "valance", respectively. Indeed, $\pm E_k$ is now the band dispersion of the conduction ($+$) and valance ($-$) electrons which are themselves superpositions of electrons with wave vectors $k$ & $k+Q$. You can see that for $\Delta\to 0$, the band dispersion reduces to the tight-binding one ($E_k\to\varepsilon_k$). For finite $\Delta$, however, a gap will open, centered at the Fermi surface, where the dispersion has a jump of $2\Delta$. This turns the metal into a (Mott) insulator.

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  • $\begingroup$ Thank you for your answer! I understand most aspects of this explanation. However, I was looking for a more sort of model-independeny and intuitive answer as to why localized electrons are most likely to result in an antiferromagnetic ground state and itinerant electrons prefer a spin-aligned configuration. Appeals to the Hubbard model are indeed very useful, but does it apply in most situations? $\endgroup$
    – Xcheckr
    Sep 27, 2014 at 15:48
  • $\begingroup$ Hmm, sorry if the answer is a bit on the theoretical side; I just happen to be a condensed matter theorist. I'm having trouble offering a conclusive intuitive answer but I'd say you're asking the wrong way: it's not a question of why valence electrons prefer AFM, whereas conduction electrons prefer FM. I'd say the question is why AFM gives rise to a band gap while FM does not. $\endgroup$ Sep 27, 2014 at 16:40
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    $\begingroup$ After all, it's not like a conductor consists only of conduction electrons. Most of its electrons are in the valence states. The difference is just that the conduction states are "easily accessible" (in that there's no gap) while in an insulator there is a gap, so the conduction states are not thermally excited. $\endgroup$ Sep 27, 2014 at 16:43
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    $\begingroup$ thanks for that comment. It is a different way of looking at things than I had considered before. I really appreciate that. Thanks! As you can tell I'm a condensed matter experimentalist-- which is why I was looking for an "intuitive" explanation. $\endgroup$
    – Xcheckr
    Sep 27, 2014 at 17:30
  • $\begingroup$ @Xcheckr: Physical Intuition, in the end, is nothing but a summary of mathematical investigations - so from this point a theoretical study is already best you can do... $\endgroup$ Sep 27, 2014 at 18:12

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