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Coupling in the Ising model

  1. Dec 8, 2007 #1
    I was trying to understand why for every spin configuration of a ferromagnetic system, there exists a corresponding isoenergetic state of an antiferromagnetic system.
    Can I treat an antiferromagnetically coupled 1-D ising model as a combination of two interpenetrating sublattices which are ferromagnetically coupled. If so, then how do I visualise this in 2 and 3 dimensions and extend the argument to d-dimensions?

    Basnically, how are these two systems (ferromagetically and antiferromagnetically correlated)
    thermodynamically equivalent?

    P.S: obviously, in the absence of a magnetic field.
  2. jcsd
  3. Dec 9, 2007 #2
    I assume that in 2 and 3 dimensions you want a square or cubic lattice (or some other bipartite lattice).

    Your Hamiltonian is
    H = -J \sum_i s_i s_{i+1}
    So, if J > 0 then we have ferromagnetic (FM) coupling, and if J < 0 then we have antiferromagnetic (AFM) coupling.

    In 1 dimension, if you label every other site as (a) or (b), ie:
    a b a b a b a .. etc

    then to map AFM onto FM introduce a variable which is a flipped spin on the b sublattice (taking the first site as i=0):
    \sigma_i = (-1)^i s_i

    Then in the AFM case the Hamiltonian becomes
    H = -|J| \sum_i \sigma_i \sigma_{i+1}
    which is the same Hamiltonian as for the FM case
  4. Dec 9, 2007 #3
    would it work the same way for a d-dimensional lattices as well?
    wht kind of variable would i choose for nearest and next nearest neighbor interactions?
    I am pretty confused! This does seem simple but I cant understand this physically at all.
    Last edited: Dec 9, 2007
  5. Dec 9, 2007 #4
    Yes, it works for higher dimensional lattices, as long as they are bipartite. In general, you have to figure out how it will work for any particular lattice. You have to come up with some sort of transformation so that A lattice sites are surrounded only by B lattice sites. For instance in the 2D square lattice, your transformation is [tex]\sigma_i = (-1)^{i+j}s_i[/tex]

    This won't work for all lattices. A triangular lattice in 2-dimensions for instance can't be transformed in this way. In fact, antiferromagnetism is frustrated on a triangular lattice.
  6. Dec 9, 2007 #5
    I just searched for frustration. So you basically mean that the ground state for a triangular lattice does not have a unique ground state. Hence non zero entropy. Interesting.
    Is that because the hamiltonian for the ground state can be written in more than one ways?
    Last edited: Dec 9, 2007
  7. Dec 10, 2007 #6
    No, when you rework the Hamiltonian you are just writing the same thing in a different way. It has nothing to do with frustration.

    The frustration comes from the geometry. If you just look at a single equilateral triangle and try to put spins on the corners with AFM ordering, you have a problem. Pick one of them to be up, another to be down, and then no matter whether you pick the third to be up or down it will still have a bond with a like spin, which AFM doesn't like. The same principle extends to the triangular lattice.
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