Is there an analytical expression for the probability of finding two consecutive parallel spins in the one-dimensional antiferromagnetic Ising model (with constant J
I am thinking it should be a function p(x) of x=-J/kT, which satisfies the conditions:
p(0)= ½ (fully disordered system)
p(infinity)=0 (fully ordered system)
For example: p = 1/(1+exp(-J/kT)) satisfies the conditions above, but of course is not unique in that property.
Maybe someone can point me to a solution and save me the time playing with the equations (or Mathematica)....
A closely related question is the x-dependence of the nearest neighbor spin-spin correlation function. In a linear Ising chain it is given by =Tanh(x) meaning at T=0 neighboring spins are fully correlated/point in the same direction and at x->infinity neighboring spins are fully uncorrelated meaning =0. If this is not exactly what you are looking for, you may ultimately find that this is a very useful quantity and also, in the context where this is discussed in the literature, you may find the answer to your original question.
Thank you, Christian, that's certainly useful and I will explore that route.
I think I have found now the solution to my original question, doing it in the old fashion way of writing probabilities and partition functions. I obtained that the probability of finding two neighbouring parallel spins is indeed: p = 1/(1+exp(-J/kT)), as I had guessed in my previous comment!
Yes, using the spin-spin correlation function I obtain the same result. There is a subtlety in a way the J is described (it depends on how one writes the Hamiltonian), but taking
p =
leads to the same result after some algebra. Thanks, Christian!
This is a nice problem for application of the transfer matrix method.
Assume the chain is periodic, with N+1 spins, N=1, 2, ..., and define
X = 2*cosh(βJ), Y = -2*sinh(βJ) (eigenvalues of the transfer matrix, with J defined positive for anti-ferromagnetic coupling).
Then the probability P of parallel spins, and probability Q of anti-parallel spins are respectively
P = exp(-βJ)*(XN + YN)/(XN+1 + YN+1), Q = exp(βJ)*(XN - YN)/(XN+1 + YN+1).
The N = 1 case assumes double interactions between the two spins. The expressions can be simplified when N becomes large. Note that there is an interesting frustration effect when N+1 is odd.
@Ricardo. Your definition of p does not define a probability, but a mean value.
Thank you, Kåre. I will think slowly about your post on the application of the transfer matrix method (I am out of my comfort zone in this topic :-).
About my definition of p, I am not sure I understand your comment. I think that the probability that a given NN pair has two parallel spins is the mean value of
1/2 (1 + SiSi+1)
over the whole chain, given that the quantity above is 0 for antiparallel pairs and 1 for parallel pairs. Right? (I corrected the expression in my post above, as typed it wrong).
Then p=1/2 (1 + ) = 1/2 (1+tanh(J/kT)) = 1/(1+exp(-2J/kT))
If you do not have a magnetic field, this can be neatly mapped on a free particle problem: indeed, any spuin configuration
s1, s2, s3,
can be mapped onto
s1, w1, w2, w3
where s1 is the same, and w1 is 1 if there is a ``wall'' , that is, if s2 is different from s1,and zero otherwise. The total energy of an ising antiferromagnet is then
-J sum over all i of w_i
and the probability that a w_i is zero, viz that two spins are parallel, is now easily computed.
For a magnetic field, you do need the transfer matrix formalism, which also gives you interesting effects at finite N, such as the corrections induced by periodic boundary conditions when the total number of spins is odd. But the free particle picture, with its limitations, I find very instructive.
Ricardo> I am not sure I understand your comment.
For good reasons. It was completely decoupled from my mind that probabilities are uniquely determined from average values in this case, since there are only two possibilities involved. Sorry for that! I agree with your result, for an open chain with free (summed-over) boundary spins. For which the result becomes much easier to derive, by transformation to disorder variables wi = si si+1, as commented by Francois.
My closed-chain result is more complicated, because a periodicity property must be built into the spin-spin correlation function. I think it is pretty hard to solve that case without a transfer matrix, even without a magnetic field, because the "last-to-first" interaction ends up depending on all independent disorder variables. For a chain with an odd number of spins, there is frustration effect which becomes visible if one takes the limit βJ to infinity at fixed N. There must always be at least one frustrated bond in the system, the probability that this is the pair under investigation is 1/N (instead of 0, as for a chain with even N).
I took another look at my solution. With t = tanh(βJ) it can be written
P = 1/[1+exp(2βJ)] * (1 - tN)/(1+tN+1) when N+1 is even (unfrustrated chain)
P = 1/[1+exp(2βJ)] * (1 + tN)/(1-tN+1) when N+1 is odd (frustrated chain)
For the last case we may write t = 1-ε ≈ exp(-ε) when βJ is large. Then one finds
P = ε/[1-exp(-(N+1)ε)] -> 1/(N+1) when ε -> 0.
My definition of J have the opposite sign than Ricardo.
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