15.4 Algorithms for lattice field theories 485
Table 15.2. Values of the renormalised
mass and coupling constant.
Lm g mR mR
8 0.05 0.1 0.456(3) 0.20(4)
12 0.03333 0.04444 0.332(3) 0.18(7)
16 0.025 0.025 0.260(3) 0.13(5)
24 0.016667 0.01111 0.184(2) 0.12(6)
32 0.0125 0.00625 0.1466(7) 0.10(4)
Values obtained from(15.53)and(15.54), for dif-
ferent lattice sizes. Various methods (see later
sections) are used.
of freedom to the field at each site (the Car–Parrinello method is based on a similar
trick; see Chapter 9). As we have seen in Chapters 7 and 8 , for a dynamical system
the probability distribution of the coordinate part can be obtained by integrating
out the momentum degrees of freedom, and this should be the desired distribution
e−S[φ]. Therefore, we simply add a kinetic energy to the action in order to obtain a
classical Hamiltonian (which should not be confused with the field theory’s quantum
Hamiltonian):
Hclass=
∑
n
p^2 n
2
+S[φ]. (15.60)
Integrating out the momentum degrees of freedom of the classical partition function,
we obtain the Boltzmann factor of the action back again (up to a constant):
∫
[Dpn]e−Hclass[pn,φn]=Const·e−S[φ]. (15.61)
The Andersen method
The classical Hamiltonian gives rise to classical equations of motion which can be
solved numerically. These equations yield trajectories with constant energy (up to
numerical errors). But we want trajectories representing the canonical ensemble,
and inChapter 8we studied various methods for obtaining these. In the Andersen
refreshed molecular dynamics method, the momenta are refreshed every now and
then by replacing them with a new value drawn from a random generator with a
Maxwell distribution. In field theories, one often replaces all momenta at the same
time with regular intervals between these updates (the method is usually denoted
as thehybrid method). That is, first the equations of motion are solved for a number
of time steps, and thenallmomenta are replaced by new values from the Maxwell
random generator. Then the equations of motion are solved again for a number of