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