Statistical Physics 3 December 1st, 2010
Solution 9
a) The partition function looks like:
Z= ∑
σ1,σ3,...
∑σ2exp Kσ1σ2+12H(σ1+σ2)
exp Kσ2σ3+12H(σ2+σ3)
∑σ4exp Kσ3σ4+12H(σ3+σ4)
exp Kσ4σ5+12H(σ4+σ5) . . .
We pose:
Cexp
K0σ1σ3+1
2H0(σ1+σ3)
=
= ∑
σ2
exp
Kσ1σ2+1
2H(σ1+σ2)
exp
Kσ2σ3+1
2H(σ2+σ3)
=
= exp 1
2H(σ1+σ3)
∑
σ2
exp(Kσ2(σ1+σ3) +Hσ2) =
= exp 1
2H(σ1+σ3)
2 cosh(K(σ1+σ3) +H) (1)
b) We evaluate the equation (1) atσ1=σ3=1,σ1=−σ3=1 andσ1=σ3=−1.
eH2 cosh(2K+H) =CeK0+H0 2 cosh(H) =Ce−K0
e−H2 cosh(2K−H) =CeK0−H0 After some manipulation we obtain:
H0=H+12lncosh(2K+H)
cosh(2K−H)
K0=14lncosh(2K+H)cosh(2K−H) cosh2H
C2=4 cosh(H) (cosh(2K+H)cosh(2K−H))12 c) IfK∗=0, it is clear thatK∗0 =0 andH0=H ∀H.
IfH∗=0 the second equation gives
e2K= e2K+e−2K 2
which is satisfied ifK=0, which is a particular case of the previous solution.
The solutionK=∞corresponds to the solution atβ =∞that isT =0, which is the ferromag- netic point of the Ising model in one dimension, so we expect to find it as a fixed point. In order to see this it is worth to make the substitution: x=e−K andy=e−H (and of coursex0=e−K0
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Statistical Physics 3 December 1st, 2010
andy0=e−H0) so the infinite valueK=∞ corresponds more properly tox=0. In particular, substituting in the first equation of point b) we find the renormalization relations:
x04= (1+x(1+y4 2)2x4
y2)(y2+x4)
y02= y2(y2+x4)
1+x4y2
C= 1+yy2(1+x4y2)(y2+x4) (1+y2)2x4
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The caseK∗=0 corresponds to x∗ =1 and the equations are satisfied ∀y. The caseH∗=0 corresponds toy∗=1 and the equations are satisfied ifx=0 (ferromagnetic) orx=1 (param- agnetic).
d) For smallH:
H0 = H+1 2ln
cosh(2K+H) cosh(2K−H)
= H+1 2ln
cosh(2K) +sinh(2K)H cosh(2K)−sinh(2K)H
= H+1 2ln
1+tanh(2K)H 1−tanh(2K)H
= H+Htanh(2K) (2)
and:
K0 = 1 4ln
cosh(2K+H)cosh(2K−H) cosh2H
= 1
4ln((cosh(2K) +sinh(2K)H)(cosh(2K)−sinh(2K)H))
= 1
4ln (cosh2(2K)
= 1
2ln((cosh(2K))
⇔tanh(K0) = cosh12(2K)−cosh−12(2K) cosh12(2K) +cosh−12(2K)
= cosh(2K)−1 cosh(2K) +1
= tanh2(K) (3)
e) Substitutingv=tanh(K)the ferromagnetic caseK∗=∞corresponds tov∗=1 and the obtained equations (for smallH) are:
v0=v2 H0=H
1+ 2v
1+v2
The matrix of first derivatives is
2v 0
2H 1−v2
(1+v2)2 1+ 2v
1+v2
!
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Statistical Physics 3 December 1st, 2010
To linearize the transformation around(v∗=1,H∗=0)means to approximate the transforma- tion at the first order in the Taylor expansion: we just need to evaluate the first derivative matrix at(v∗=1,H∗=0)whhich gives:
2 0 0 2
An equilibrium point is stable if all the eigenvalues of the linearized application have absolute value less than 1. This matrix has two eigenvalues λ1 =λ2 =2. So the point is not stable.
Physically this means that as soon as the temperature is different than zero the system is no more ferromagnetic (so no transition phase).
Aroundv∗=0 and smallH, the linearized transformation is:
0 0 2H 1
We conclude that the variety is stable along thevaxes because the eigenvalue is 0. So as soon asT 6=0 the system tends to the point v=0. But we can say nothing about H, because the eigenvalue is 1: this means that it behaves as a power law and not as an exponential low.
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