Antiferromagnetic/ferromagnetic coupling

[qubits]

What is antiferromagnetic/ferromagnetic coupling, and what is the difference between the two? I'm referring to coupling between two two-level half-flux quantum systems for example.

 

[ZapperZ]

What is antiferromagnetic/ferromagnetic coupling, and what is the difference between the two? I'm referring to coupling between two two-level half-flux quantum systems for example.

I'm assuming what you meant by "half-flux quantum systems" is spin 1/2 systems.

First of all, this is a VERY difficult subject that falls into quantum magnetism. So I'm guessing you're asking this because you're studying either quantum magnetism, or solid state physics. If that is the case, then the "coupling" is the Heisenberg spin-exchange interaction, defined within the Heisenberg spin Hamiltonian as

$$H = \sum_{ij}J_{ij}\mathbold{S_i} \cdot \mathbold{S_j}$$

where $$J_{ij}$$ is the Heisenberg coupling of nearest-neighbor spins. In a "simple" system (pay attention to that word), a ferromagnetic coupling implies that J < 0, whereas antiferromagnetic coupling means that J > 0. The exact determination of what J is for any given system is usually impossible to find exactly. This is where one has to make approximations within the many-body picture.

Zz.

 

[santoshroy]

In any physical system we are concerned with energy lowering that leads to stability of the system. So the hamiltonian that is presented above H=sum (ij) J(ij) S(i).S(j) is a simplestic one illustrating the fact.

1.For a antiferromagnetic system neighbouring spins are opposite (+1/2) and (-1/2) so J is +ve that leads to an overall -ve sign>>> means H= -ve ( ... ) this hamiltonian will have -ve eigen energy leading to a bound state.

2.For ferromagnetic spins are (+1/2) (+1/2) so by similar argument J should be -ve...

This hamiltonian is simple but it hides the actual physics. The J cointains Coulomb and spin exchange terms within it...