Statistical Mechanics - Physics at Oregon State University

8.3. MEAN FIELD RESULTS. 165
σ=±1
In a similar way we find
σk=±1
σk=±1
e β(h+mqJ)σ
σ1=±1
N
= 2 N cosh N (β(h + mqJ)) (8.37)
e 1
2 βNqJm2
Tr Ske −β(Hmf −hM) =
σ1=±1
· · ·
σN =±1
· · ·
σN =±1
N
β(h+mqJ)
σke i=1 σi =
σk
σke β(h+mqJ)σk
σke β(h+mqJ)
i=k
e β(h+mqJ)σi =
i
σ=±1
σi=±1
e β(h+mqJ)σi
e β(h+mqJ)σ
N−1
2 N cosh N−1 (β(h + mqJ)) sinh(β(h + mqJ)) (8.38)
This expression is independent of k indeed, as required. Combining the last
two formulas we find that m(T, h), the average of the spin variable, has to obey
m = tanh (β(mqJ + h)) = tanh βqJ (m + h
qJ )
(8.39)
The next question we have to answer is what are the solutions for m? We
see that there are two ways to modify the nature of the solutions. We can either
change βqJ, which changes the slope of the hyperbolic tangent at the origin,
of change h
qJ , which shifts the whole curve. The magnitude of the coupling
constant J or the number of neighbors q together do only set the scales for the
temperature and the magnetic field. If we express the temperature in unit of
qJ
and the magnetic field in units of qJ then the solutions for m do not depend
kB
directly on q and J anymore. Hence we define h ∗ = h
qJ and β∗ = βqJ. The
resulting equation we have to solve is
m = tanh (β ∗ (m + h ∗ )) (8.40)
This has the form m = f(m) and a standard technique to solve such an
equation is to draw a graph. We look at the graph of y = f(m) and check where
it intersects the straight line y = m. In figure 8.1 we show the results for β ∗ = 1
2
and h ∗ = 2.
In this case we see that there is one solution. In general, there will always be
at least on solution, since the hyperbolic tangent stays between minus and plus
one, and the line y = m has to intersect it. If one draws a picture of the left
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