Exercises 4174.4184.424.4224.4244.4264.4284.434.4320.01 0.014 0.018 0.022 0.026lnλ^01 /L^2
Figure 12.9. The logarithm of the largest eigenvalue of the transfer matrix versus
the inverse of the square of the strip widthL. The straight line has a slopeπ/6 and
is adjusted in height to fit the data.whereE 0 is a guess of the trial energy (which should be equal to−lnλ 0 ,λ 0 is the
largest eigenvalue),Nis the actual number of walkers andN 0 is the target number
of walkers. This term aims at stabilising the population size to the target numberN 0.
The simplest information we obtain is the largest eigenvalue, which is given as
exp(Etrial), where the average value ofEtrialduring the simulation is to be used
(with the usual omission of equilibration steps). This can be used to determine
central charges. In Table 12.3 we compare the values of this quantity for the Ising
model with those obtained by a Lanczos diagonalisation of the transfer matrix.
The agreement is seen to be excellent. For theXYmodel, the eigenvalues cannot
be found using direct diagonalisation and we can check the MCTM method only
by comparing the central charge obtained with the known value: 1 in the low-
temperature phase and 0 at high temperatures. InFigure 12.9we show the results
forβJ=1.25. The points in a graph of the form lnλ 0 vs. 1/L^2 lie on a straight
curve with a slope ofπ/6(c=1).
Exercises
12.1 In this problem we consider the virial expression for the energy[20].
In a path-integral QMC simulation for a particle in one dimension in a potential
V(x)we want to find the energyEas a function of temperatureT= 1 /(kBβ).Wedo
this by using the thermodynamic relation
E=−
∂lnZ
∂β