Computational Physics

486 Computational methods for lattice field theories

steps and so on [ 13 – 16 ]. The exact dynamical trajectory plus the momentum update
can be considered as one step in a Markov chain whose invariant distribution is the
canonical one. We do not obtain the exact dynamical trajectory but a numerical
approximation to it, and the errors made can be corrected for in a procedure which
will be discussed in the next section. In Chapter 8 we mentioned that the Andersen
method does indeed lead to the canonical distribution of the coordinate part. We
shall prove this statement now.
First, it is useful to consider ‘symmetric’ Markov steps: these consist of an integ-
ration of the equations of motion over a timet/2, then a momentum refreshing,
and then again an integration over a timet/2. Such a step can schematically be
represented as follows:

i,Pi

t/ 2

−→m,Pm Refresh :m,Prm

t/ 2

−→f,Pf.

Energy conservation during the microcanonical trajectories implies

H(i,Pi)=H(m,Pm); (15.62a)

H(m,Prm)=H(f,Pf). (15.62b)

The steps occur with a probability

T(i,Pi→f,Pf)=δ(f−microcanonical)exp(−P^2 rm/ 2 ), (15.63)

where the delta-function indicates thatfis uniquely determined by the microca-
nonical trajectory, which depends of course on the initial configurationi,Pi,on
the refreshed momentumPrm, and on the integration time (which is fixed).
The trial steps are ergodic, and the master equation of the Markov chain
∑

′,P′

ρ(,P)T(,P→′,P′)=

∑

′,P′

ρ(′,P′)T(′,P′→,P) (15.64)

will have an invariant solution. However, the detailed balance condition for this
chain is slightly modified. The reason is that we need to use the time-reversibility
of the microcanonical trajectories, but this reversibility can only be used when we
reverse the momenta. Therefore we have

ρ(′,P′)

ρ(,P)

=

T(,P→′,P′)

T(′,−P′→,−P)

(15.65)

(note thatρ(,P)=exp[−P^2 / 2 −S()]is symmetric with respect toP↔−P).
The transition probability in the denominator of the right hand side corresponds
to the step in the numerator traversed backward in time (see the above diagram
of a symmetric trial step). The fraction on the right hand side is clearly equal to
exp[(P^2 mr−Pm^2 )/ 2 ]. Using Eq. (15.62), it then follows that the invariant distribution
is given asρ(,P)=exp[−P^2 / 2 −S()].