Now you are in the subtree of Special Topics in Many-Body Theory, Spring 2016 project.

# Physical origin of antiferromagnetism; basic FM/AF differences

Although humans have used magnetism for navigation and other purposes since antiquity, understanding how it emerges from the underlying physics of a solid or insulator is quite complex. In fact, only in the past few decades can we really claim to have understood magnetism. Magnetism is also a reminder that some of the seemingly abstract phenomena we discuss in this course can have important real applications. For instance, for a few hundred dollars you can purchase a 100 GB hard disk that is rewritable many times, stable for years with no power, and contains more information than an entire ancient library. Future read/write heads for such hard disks may be based on CMR (colossal magnetoresistance) and GMR (giant magnetoresistance) effects, which were discovered quite recently.

$$$$E = -\sum_{\langle i j \rangle} J s_i s_j$$$$
where the classical spin variables take values $\pm 1$ and the sum is over nearest neighbors. We will say a bit more about this model in a moment, but it is first important to understand how such a model can emerge from the underlying quantum mechanics of the underlying mobile, interacting electrons.

A standard model for understanding the origin of magnetism is the Hubbard model Hamiltonian
$$$$H = -t \sum_{\langle i j \rangle,\sigma} (c^\dagger_{i\sigma} c_{j\sigma} + h.c.) + \sum_i (\epsilon-\mu) (n_{i\uparrow} + n_{i \downarrow})+ \sum_i U n_{i\uparrow} n_{i \downarrow}.$$$$
(Actually several different variants are called the Hubbard model.) This model is relatively simple if $U$ is a small perturbation, as then we have nearly free electrons from the hopping part $t$ with a weak attractive or repulsive interaction. We will be more interested in the case where a repulsive $U \gg t$, as we will show that in this case the system is effectively antiferromagnetic at half filling (one electron per site on average). If $t=0$ the model is trivially solvable (the ground state is any assignment of one electron per site).

Similar to the above Hubbard model is the Anderson model of a single impurity coupled to noninteracting conduction electrons: $H = H_0 + H_1$,
$$\begin{eqnarray} H_0 &=& \sum_{k,\sigma} \epsilon_k (c^\dagger_{k\sigma} c_{k\sigma} + h.c.) + (\epsilon-\mu) (n_{d\uparrow} + n_{d \downarrow})+ U n_{d \uparrow} n_{d \downarrow},\cr H_1 &=& \sum_{k\sigma} V_{kd} (c^\dagger_{k\sigma} c_{d\sigma} + h.c.) \end{eqnarray}$$
Even this model that only has interactions at one point can show surprising correlation effects like the Kondo effect when the values of $\mu$ and $U$ are chosen so that

First, note that the above quantum Hubbard model has an $SU(2)$ spin symmetry and hence should give rise, in the localized limit, to a spin model with Heisenberg rather than Ising symmetry. Suppose we start with the above degenerate ground states at $t=0$ and ask how they are split once $t$ is turned on: we are essentially obtaining an effective Hamiltonian for the basis of states with one electron per site, valid on energy scales much less than the $t=0$ gap (either $\mu - \epsilon$ or $U +\epsilon - \mu$). Let us assume the symmetric repulsive case with $\epsilon=-U/2$ and $\mu=0$.

Consider a single nearest-neighbor pair of sites. First-order perturbation theory in the degenerate basis doesn't do anything, since $t$ moves the system to have two electrons on one site and one empty site, which takes the system out of the degenerate ground-state manifold. In second-order perturbation theory, there is no coupling induced if the spins on both neighboring sites are both up or both down. If site 1 is up and site 2 is down, there is both a spin-flip coupling and a spin-preserving coupling:
$$$$H_{12} = {2 t^2 \over U} ((S^z_1 S^z_2 -\frac{1}{4}) + 2 S^+_2 S^-_1).$$$$
Here the $S_i$ are spin-half operators. Adding the case with site 1 down and site 2 up, and ignoring an overall constant energy, we finally obtain the antiferromagnetic Hamiltonian
$$$$H = J \sum_{\langle i j \rangle} {{\bf S}_i \cdot {\bf S}_j}.$$$$
Here $J = \frac{4 t^2}{U}$ and the $S$ operators are spin-half quantum operators.
It is possible to obtain ferromagnetic interactions instead from slightly different microscopic models, and some real ferromagnets are itinerant (the electrons whose spins produce the ferromagnetism are mobile), corresponding to the Stoner instability we discussed in the context of Fermi liquids.

Note that, as usual, obtaining a significant magnetic interaction depends on underlying electric rather than magnetic interactions. Aside from the symmetry difference between the above antiferromagnet and an Ising model, another important difference is that specifying the (static) energy in a classical model does not specify the dynamics, while specifying the Hamiltonian of a quantum problem determines both the energy levels on the dynamics. We will see some practical consequences of this below when discussing spin waves.

The mean-field theory of the classical Ising Hamiltonian can be derived heuristically by imagining that each individual spin moves in the field of its $z$ nearest neighbors and possibly an external field $H$. This gives the self-consistency equation
$$$$\langle s \rangle = {e^{\beta (J z \langle s \rangle+H \langle s \rangle)} - e^{-\beta (J z \langle s \rangle+H \langle s \rangle)} \over e^{\beta (J z \langle s \rangle+H \langle s \rangle)} + e^{-\beta (J z \langle s \rangle + H \langle s \rangle)}} = \tanh(\beta (J z +H)\langle s \rangle)$$$$
This has a phase transition at $H=0$ between ordered and disordered states at the temperature $k T = \beta^{-1} = J z$. At zero temperature the system is perfectly ordered for both the ferromagnet and antiferromagnet. The Ising model can be understood as an anisotropic and classical limit of the Heisenberg model which we have derived from the underlying electrons, for cases when the electrons move in an anisotropic crystal, say.

Now we can state another interesting difference between the quantum and classical models. On a bipartite lattice, a lattice that can be divided into two symmetric sublattices A and B so that every bond of the model joins one site from sublattice A to one from sublattice B (like the square lattice, for example), there is no thermodynamic difference in zero magnetic field between the classical ferromagnet and antiferromagnet. The idea is that flipping the sign of the coupling constant is the same as flipping all the spins on one sublattice, so that any quantity which sums over all spin configurations should be independent between the ferromagnet and antiferromagnet. Hence, for example, the critical temperature in the above mean-field theory is the same for the ferromagnet and antiferromagnet. Under the symmetry transformation, a magnetic field would go into a "staggered field" that alternates from one sublattice to another.

Is there such a symmetry in the quantum Heisenberg model? One can see that there is not by considering just two quantum Heisenberg spins: then with a spin-symmetric coupling there is a singlet and a triplet, but there is not a symmetry between the singlet and triplet since they contain different numbers of states. With two classical Ising spins, there are two opposite-spin states and two same-spin states, so there is indeed a symmetry.
This has some surprising consequences, especially in low dimensions.

Let us return to the quantum Heisenberg model and suppose that we are at zero temperature, so that all we need is the ground state of the model
$$$$H = J \sum_{\langle i j \rangle} {{\bf S}_i \cdot {\bf S}_j}.$$$$
For negative $J$ (the ferromagnetic case), the ground state is very simple: any state with all spins pointing in the same direction is an eigenstate of the model, and is shown by a little thought to be the lowest-energy eigenstate, since it minimizes the energy of each individual bond. For positive $J$ these states (all spins up, for example) are still eigenstates but now have high energy. What is surprising is that the ground state is now very complicated, rather than simply being the Ne\'el state of alternating spins. To understand why the ground state is complicated, note that it is not possible to minimize every bond independently in the antiferromagnetic case by putting every nearest-neighbor pair of spins into a singlet configuration. Antiferromagnetic spin-liquid states where spins form singlet pairs in a disordered way ("resonating valence bond" states) have attracted a great deal of attention as models for the environment in which holes move in the cuprates.

Let us return to the ferromagnet and construct one type of excitation above the ground state with all spins up, say. Consider first lowering one spin at site $i$: this state can be written $S_i^- |0 \rangle$. This is not an eigenstate, but the action of $H$ on this state is simple: the only off-diagonal part transfers the down spin to one of the nearest-neighbors. This is just like the hopping term in a tight-binding model for electrons, so we can take over the well-known form for the spectrum: for a square/cubic lattice of coordination number $z$ and lattice spacing $a$, the excitation energy at momentum ${\bf k}$ is
$$$$\epsilon(k) = E_k - E_0 = 2 S J \sum_{i=1}^z \sin^2(\frac{1}{2} {\bf k} \cdot {\bf r}_i).$$$$

In 2 and 3 dimensions, the antiferromagnet still orders on the square/cubic lattice, but the ground state has reduced staggered moment compared to the classical ground state (in two dimensions, this reduction is about 40 percent). In one dimension, Bethe guessed the exact ground state (the first example of the so-called Bethe ansatz for 1D interacting problems), which turns out to have only power-law correlations. If you are curious how the solution is accomplished, it actually consists of starting from the exact high-energy eigenstate of all spins aligned, then constructing spin-wave excitations as above, then finding a way to describe an interacting gas of many spin-wave excitations. So the spin waves become like particles ("magnons") moving in 1D. An elementary discussion is in the textbook of Fradkin.