Computational Physics

366 Transfer matrix and diagonalisation of spin chains

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n

n


ν

ν

Figure 11.7. Action of the transfer matrix in the six-vertex model. The transfer

matrix connects two adjacent horizontal rows of vertical arrows – the possible

configurations of horizontal arrows are summed over. The indexμindicates the

orientation of the leftmost arrow, andμ′represents the direction of the horizontal

arrows in the iterative multiplication process. In this figure, it labels the second

horizontal link, as in the first step of the multiplication. Only a single configuration

of vertices is shown.

Comparing this with the transfer matrix of the one-dimensional Ising model with

zero magnetic field, show that

tanh(βL)=exp(− 2 βJ).

11.2 In this problem we consider the implementation of the transfer matrix method for
the six-vertex model. The vertices and weights are represented in Figure 11.3. The
transfer matrix connects a row of vertical arrows to the next one as in Figure 11.7.
This implies that for each pair of adjacent rows of vertical arrows a sum must be
performed over all possible configurations of horizontal arrows that are compatible
with the vertical ones, in the sense that the number of ingoing arrows equals the
number of outgoing ones at each site.
The multiplication of the transfer matrix with an arbitrary vectorφis again
carried out site by site, similar to the procedure followed for the Ising model. Let us
first consider the multiplication including the first (leftmost) vertex (j=1). It turns
out to be necessary to introduce a vectorψ(n,μ,μ′)in the multiplication routine.
Here,nrepresents a row configuration of vertical arrows andμandμ′are indices
assuming the values 0 and 1 – they represent the direction of the leftmost horizontal
arrow and the horizontal arrow on the link connecting sitesjandj+1(μ= 0
denotes a right-pointing arrow,μ=1 a left-pointing one), which is the second
horizontal arrow from the left as we are considering the first (leftmost)
vertex.
As the two horizontal arrows of the first site are known, there is a unique relation
between the upper and lower vertical arrows. These arrows correspond to the least
significant bitsνandν′which are given bynmod 2 andn′mod 2 respectively
(ν=0 is an upward-pointing arrow,ν=1 a downward-pointing one). Therefore
we have

ν+μ=ν′+μ′. (E11.1)