# Why do antiferromagnets occur at lower temperature than ferromagnets?

The minimal model for describing magnets is the Heisenberg Hamiltonian

$$H = -\frac{1}{2}J\sum_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$$

Where $$i,j$$ are nearest neighbors and the factor of $$1/2$$ is for double counting.

If $$J$$ is positive, spins will want to align to save energy (ferromagnets), and if it is negative they will anti-align (antiferromagnets). Ultimately $$J$$ comes about from Pauli exclusion and electrons not wanting to sit in the same orbital (Coulomb repulsion).

But if I look at a table of ferromagnets here, I see transition temperatures up to 1400 K. On the other hand, the highest transition temperature for antiferromagnets is a measly 525 K, with most being below room temperature.

Why do antiferromagnets generally occur at significantly lower temperatures than ferromagnets?

One can argue that maybe $$\vert J\vert$$ is larger in ferromagnets than antiferromagnets (as one of the current answers does), but this just begs the question. Why should that be the case (assuming it is true)? I don't see an experimentally-verified theoretical basis for asserting $$\vert J_{\mathrm{AFM}}\vert < \vert J_{\mathrm{FM}}\vert$$.

This question came up in a class I am teaching to talented senior undergraduates.

For example, Iron has $$J$$ of roughly 0.3 eV and La$$_2$$CuO$$_4$$ of 0.13 eV. Iron has a 1000K transition temperature and La$$_2$$CuO$$_4$$ of about 325K.
• I guess I must have found an incorrect value for La214 then. Let's say the cause is smaller $J$, why should it be the case that antiferromagnets have smaller values in the first place? In the limit of very strong Coulomb interactions (Hubbard model), we still have antiferromagnetism winning over ferromagnetism – KF Gauss Mar 26 at 11:52