1549380323-Statistical Mechanics Theory and Molecular Simulation

Renormalization Group 639

tanhK′= tanh^3 K

K′= tanh−^1

[

tanh^3 K

]

. (16.9.15)

Eqn. (16.9.15) defines the RG transformation as

R(K) = tanh−^1

[

tanh^3 (K)

]

. (16.9.16)

We must remember, however, that eqn. (16.9.16) is particular to the block spin trans-
formation in eqn. (16.9.10). From eqns. (16.9.14) and (16.9.15), we obtain the new
Hamiltonian as

Θ′ 0 ({σ′}) =N′g(K)−K′

∑N′

i=1

σ′iσ′i+1, (16.9.17)

where the spin-independent functiong(K) is given by

g(K) =−

1

3

ln

[

cosh^3 K

coshK′

]

−

2

3

ln 2. (16.9.18)

andN′is a constant. Thus, apart from theN′g(K) term in eqn. (16.9.17), the new
Hamiltonian has the same functional form as the original Hamiltonian,but it is a
function of the new spin variables and coupling constantK′.
The block spin transformation of eqn. (16.9.10) could be applied againto the
Hamiltonian in eqn. (16.9.17), leading to a new Hamiltonian Θ′′ 0 in terms of new spin
variablesσ′′ 1 ,....,σ′′N′′and a new coupling constantK′′. It is a straightforward exer-
cise to show that the coupling constantK′′would be related toK′by eqn. (16.9.15).
Repeated application of the block spin transformation is, therefore, equivalent to it-
eration of the RG equation. Since the coupling constantKdepends on temperature
throughK=J/kT, this iterative procedure can determine if, for some temperature,
an ordered phase exists. Recall that an ordered phase corresponds to the fixed point
condition in eqn. (16.9.7). From eqn. (16.9.16), this condition for thepresent example
becomes
K= tanh−^1

[

tanh^3 K

]

. (16.9.19)

In terms ofx= tanhK, the fixed point condition is simplyx=x^3. SinceK≥0, the
only possible solutions to this fixed point equation arex= 0 andx= 1.
To understand the physical content of these solutions, considerthe RG equation
away from the fixed point:x′=x^3. SinceK=J/kT, at highT,K→0 andx=
tanhK→ 0 +. At low temperature,K→∞andx→ 1 −. If we view the RG equation
as an iteration or recursion of the form

xn+1=x^3 n, (16.9.20)

and we start the recursion atx 0 = 1, then each successive iteration will yield 1. How-
ever, for any valuex= 1−ǫless than 1 (here,ǫ >0), eqn. (16.9.20) eventually iterates
to 0. These two scenarios are illustrated in Fig. 16.15. The iteration of eqn. (16.9.20)