15.5 Reducing critical slowing down 493
In the SW method, any configuration can be reached from any other configuration,
because there is a finite probability that the lattice is partitioned intoLdsingle-
spin clusters which are then given values+and−at random. Furthermore it is
clear that the method does not generate periodicities in time and it remains to be
shown that the SW method satisfies detailed balance. We do this by induction. We
show that the freezing/deleting process for some arbitrary bond does not destroy
detailed balance, so carrying out this process for every bond in succession does not
do so either.
Every time we delete or freeze a particular bondijwe change the Hamiltonian
of the system:
H→H 0 +Vij. (15.78)
His the Hamiltonian in which the bond is purely Ising-like.H 0 is the Hamiltonian
without the interaction of the bondij, andVijrepresents an interaction between the
spins atiandjwhich is either∞(in the case of freezing) or 0 (if the bond has
been deleted); the remaining bonds do not change. We write the detailed balance
condition for two arbitrary configurationsSandS′for the system with Hamiltonian
Has follows:
T(S→S′)
T(S′→S)=e−β[H(S
′)−H(S)]
=T 0 (S→S′)
T 0 (S′→S)
e−βJ(s′is′j−sisj)
, (15.79)whereT 0 is the transition probability for the HamiltonianH 0 and we have explicitly
split off the contribution from the bondij. In the last equality we have used the
detailed balance condition for the system with HamiltonianH 0.
In the SW algorithm, we must decide for a bondijwhether we delete or freeze
this bond. The transition probability of the system after this step can be written as
T(S→S′)=Tf(S→S′)Pf(S)+Td(S→S′)Pd(S). (15.80)Here,Pd,f(S)is the probability that we delete (d) or freeze (f) the bondijin spin
configurationS.Td(S→S′)is the transition probability with a deleted bond, and
thereforeTd=T 0 , andTfis the transition probability when the bond is frozen. The
latter is equal toT 0 in the case that the spinssi,sjare equal in bothSandS′and it
is zero in the case that they are unequal inS′(they must be equal inS, otherwise
they could not have been frozen).
Let us consider the detailed balance condition for the transition probability in
(15.80):
T(S→S′)
T(S′→S)=
Tf(S→S′)Pf(S)+Td(S→S′)Pd(S)
Tf(S′→S)Pf(S′)+Td(S′→S)Pd(S′)
=e−β[H(S′)−H(S)]
.
(15.81)
We show that this condition is indeed satisfied, using(15.79). Let us assume that
siandsjare equal in bothSandS′. In that casePf(S)= 1 −exp(− 2 βJ)and