Statistical Mechanics - Physics at Oregon State University

220 CHAPTER 9. GENERAL METHODS: CRITICAL EXPONENTS.
Z(N, ˜ J, ˜ h) =
σ2,σ4,··· i odd
e ˜ h( 1
2 (σi−1+σi+1)) 2 cosh( ˜ J(σi−1 + σi+1) + ˜ h) (9.150)
This looks like something for a partition function where only even spins play
a role. On the right hand side we have exponential like things. Can we write
this in a form
Z(N, ˜ J, ˜ h) =
e E(σ2,σ4,···)
(9.151)
σ2,σ4,···
where the form of the energy is similar to the original energy? Clearly,
the value of the coupling constant will be different, and also the zero of energy
can be shifted. Therefore we try
E(σ2, σ4, · · ·) =
i odd
2g + ˜ J ′ σi−1σi+1 + 1
2 ˜ h ′
(σi−1 + σi+1)
(9.152)
The question is, is this possible, and if yes, what are the functions g( ˜ J, ˜ h) ,
˜J ′ ( ˜ J, ˜ h) , and ˜ h ′ ( ˜ J, ˜ h).
The answer whether it is possible can be given easily. Because we wrote
the forms symmetric for interchanges of left and right, we need to consider only
three combinations of neighboring even spins, (+, +) , (+, −) = (−, +) , and
(−, −). That gives us three equations, from which we can find three independent
functions!
The equations to solve are
(+, +) : e ˜ h 2 cosh(2 ˜ J + ˜ h) = e 2g+ ˜ J ′ + ˜ h ′
(+, −) : 2 cosh( ˜ h) = e 2g− ˜ J ′
(−, −) : e −˜ h 2 cosh(−2 ˜ J + ˜ h) = e 2g+ ˜ J ′ − ˜ h ′
(9.153)
(9.154)
(9.155)
The road to solving these is a bit tricky, but here is one approach. Multiply the
first and the third, which gives
4 cosh(2 ˜ J + ˜ h) cosh(−2 ˜ J + ˜ h) = e 4g+2 ˜ J ′
Multiply by the square of the second equation and we get
16 cosh(2 ˜ J + ˜ h) cosh(−2 ˜ J + ˜ h) cosh 2 ( ˜ h) = e 8g
(9.156)
(9.157)
which gives us g. We can also divide the product of the first and third by the
square of the second, and now we find
cosh(2 ˜ J + ˜ h) cosh(−2 ˜ J + ˜ h)
cosh 2 ( ˜ = e
h)
4 ˜ J ′
(9.158)