1549380323-Statistical Mechanics Theory and Molecular Simulation

640 Critical phenomena

..

Unstable Stable

x =1 x = 0

K =^8 K = 0

Fig. 16.15RG flow for the one-dimensional Ising model.

generates arenormalization group flowthrough the one-dimensional coupling-constant
space. The fixed point atx= 1 is called anunstable fixed pointbecause any value of
x 0 other than 1, when iterated through the RG equation, flows away from this point
toward thestable fixed pointatx= 0. As the stable fixed point is approached, the
coupling constant decreases until it reachesK= 0, corresponding to infinite temper-
ature! At the unstable fixed point (x= 1),K=∞andT= 0. The absence of a fixed
point for any finite, nonzero value of temperature tells us that there can be no ordered
phase (and hence no critical point) in one dimension. Note, however, that in one di-
mension, perfect ordering exists atT= 0. Although this is not a physically meaningful
ordered phase (T= 0 can never be achieved), this result suggests that ordered phases
and critical points are associated with the unstable fixed points of the RG equations.
Recall that we also obtained ordering atT= 0 from the exact analytical solution of
the one–dimensional Ising model in Section 16.6.
Let us make one additional observation about theT= 0 unstable fixed point by
analyzing the behavior of the correlation lengthξatT= 0.ξhas units of length, but if
we choose to measure it in units of the lattice spacing, then it can onlydepend on the
coupling constantKorx= tanhK, i.e.,ξ=ξ(x). Under the block spin transformation
of eqn. (16.9.10), as Fig. 16.14 indicates, the lattice spacing increases by a factor of 3
as a result of coarse graining. Thus, in units of the lattice spacing,ξmust decrease by
a factor of 3 in order to maintain the same physical distance. Thus,

ξ(x′) =

1

3

ξ(x). (16.9.21)

More generally, if we had taken our blocks to havebspins, eqn. (16.9.21) suggests that
ξshould transform as

ξ(x′) =

1

b

ξ(x). (16.9.22)

In addition, the RG equation would becomex′=xb. We now seek a functional form for
ξ(x) that satisfies eqn. (16.9.22). In fact, only one functional form ispossible, namely

ξ(x)∼

1

lnx

. (16.9.23)

This can be shown straightforwardly as follows:

ξ(x′) =ξ(xb)∼

1

lnxb

=

1

blnx

=

1

b

ξ(x). (16.9.24)