Statistical Mechanics - Physics at Oregon State University

186 CHAPTER 8. MEAN FIELD THEORY: CRITICAL TEMPERATURE.
−3
−2
x
−1
y
5
4
3
2
1
0
0
Figure 8.4: β = 2 , J = 1 , q = 3
If the slope of the left hand side of 8.129 at h ′ = 0 is larger than the slope
of the exponent, there has to be at least one intersection of the two curves for
some positive value of h ′ . Hence if
or
β sinh(2βJ)
cosh 2 2β
>
(βJ) q − 1
tanh(βJ) > 1
q − 1
1
2
3
(8.131)
(8.132)
there is a solution of 8.129 with a value of h ′ > 0. What happens if the slope
of the left hand side is smaller? One has to look at the curvatures. It turns out
that there are no solutions in that case. The condition for more solutions will
always happen if q > 2, since for small values of the temperature the hyperbolic
tangent approaches the value one. For q = 2, however, there is no solution.
Hence the one dimensional Ising chain in the Bethe approximation does not
show a phase transition. But the point T = 0 is still special, since in the limit
the equation is obeyed.
The critical temperature in the Bethe approximation is therefore given by
J
kBTc =
coth −1 (q − 1)
(8.133)
which can be compared to the mean field result of kBTc = qJ. For a twodimensional
square lattice q = 4, and we find kBTc = 2.885J. This is much
closer to the exact answer 2.269J than the mean field result 4J. The Bethe
approximation greatly improves the estimates of the critical temperature.
We can solve coth(βJ) = q − 1 using
eβJ + e−βJ eβJ = q − 1 (8.134)
− e−βJ