# Antiferromagnetic ground state due to superexchange

## Summary:

How does a virtual two-step tunneling process explain antiferromagnetic ordering?

For context, I'm generally new to condensed matter and many-body QM and am working through Altland and Simons' Condensed Matter Field Theory. I'm thinking in general about magnetic ordering.

I've seen a Heisenberg-like spin Hamiltonian derived by starting with a Hubbard Hamiltonian

$$H = -t\sum_{<ij>}a^{\dagger}_{i\sigma}a_{j\sigma} + U\sum_{j}n_{j+}n_{j-}$$

where the first sum is over nearest neighbors. As long as ##U/t >> 1## we can treat the tunneling terms as a perturbation. This gives us something like

$$H = -2\frac{t^{2}}{U}(1 + a^{\dagger}_{i\sigma}a^{\dagger}_{j\sigma'}a_{i\sigma'}a_{j\sigma}) = 4\frac{t^2}{U}(\vec{S}_{i}\cdot\vec{S}_{j}-\frac{1}{4})$$

From the first equality, it makes sense to me that this Hamiltonian is accounting for a process in which we essentially flip two spins at sites ##I## and ##j##. Furthermore, because our model only accounts for tunneling which a spin does not change, I think this process has to be one in which one of the spins hops to the neighboring site, and the other hops to the site that was left unoccupied (please correct me if I'm wrong!). The second equality makes it clear to me that the ordering of the ground state will be antiferromagnetic, due to the positive coupling constant.

This is all fine mathematically. I think what I am confused by is the physical connection between these two equalities. I've seen the antiferromagnetic ordering explained as originating from a virtual process in which spins reduce their energy by briefly hopping to a neighboring site. How exactly does the possibility of this process lower the energy of the ground state? Maybe asked another way, why would a ferromagnetic state, in which this virtual process is unavailable due to Pauli exclusion, necessarily raise the energy of the system?

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dRic2
Gold Member
Disclaimer: I'm very far from being able to understand Atland's book, so I didn't really follow the post either. I just wanted to share my thoughts, hoping they make sense.

How exactly does the possibility of this process lower the energy of the ground state?
Well, at equilibrium the energy of the system has to be minimum only if you fix the volume and the entropy. But here you are asking which configuration of spins represents the equilibrium condition, so the entropy is not fixed. If you flip some spins the entropy will generally change. So, in order to find the equilibrium state, it's better to minimize the Helmholtz's free energy ##F##, which requires only to fix the volume and the temperature of the system (but not the entropy). So we should look at the minimum of ##F = E - TS##. So if you flip some spins, ##S## (the entropy) increases and so ##F## decreases. On the other hand ##E## might increase. It's a tug of war between ##E## and ##S## in order to minimize ##F##.

PS: How many times did I just use the word "so"? Well, at equilibrium the energy of the system has to be minimum only if you fix the volume and the entropy. But here you are asking which configuration of spins represents the equilibrium condition, so the entropy is not fixed. If you flip some spins the entropy will generally change. So, in order to find the equilibrium state, it's better to minimize the Helmholtz's free energy ##F##, which requires only to fix the volume and the temperature of the system (but not the entropy). So we should look at the minimum of ##F = E - TS##. So if you flip some spins, ##S## (the entropy) increases and so ##F## decreases. On the other hand ##E## might increase. It's a tug of war between ##E## and ##S## in order to minimize ##F##.
Thank you for the response! However, my question pertains more to quantum mechanics than to which Legendre transform of the energy is being minimized.

dRic2
Gold Member

Sorry to post again, but does anyone know whether it would be better to ask this question in a different forum?Maybe it's awkward because it's sort of a QFT question about a condensed matter system?

I recommend you read my reply to a related question in the following post:

To quote the most relevant part:

1) Local moment picture/Heisenberg model, (Insulators): Electrons are almost entirely bound to a particular atom, however there exists enough orbital overlap such that some hopping between neighbors can occur. Recall that for a 1D quantum box of length ##L##, the energy is proportional ##L^{−2}##, thus hopping allows electrons to effectively wander through a larger box, thereby reducing the kinetic energy.

• physlosopher
Recall that for a 1D quantum box of length ##L##, the energy is proportional ##L^{−2}##, thus hopping allows electrons to effectively wander through a larger box, thereby reducing the kinetic energy.