Computational Physics

11.2 The one-dimensional Ising model and the transfer matrix 341

necessary, but for matrix sizes up to 10 000×10 000, diagonalisation of the transfer
matrix is a trivial numerical exercise (Appendix A8.2).
Let us now calculate the expectation value of the spin at sitem. As a result of
translation invariance, the result is independent ofm. This expectation value is
given by

〈sm〉=

∑

{si}sm

∏L

i= 1 exp[βJsisi+^1 +βHsi]

∑

{si}

∏L

i= 1 exp[βJsisi+^1 +βHsi]

. (11.11)

Introducing again the transfer matrix and using the same argument as above to
retain the largest eigenvalueλ 0 with eigenvalueφ 0 only, we obtain

〈sm〉=

∑

s 1 〈s^1 |T

m− (^1) sˆmTL−m+ (^1) |s 1 〉
∑
s 1 〈s^1 |T
L|s 1 〉 =
〈φ 0 |ˆsm|φ 0 〉λL 0
λL 0
=〈φ 0 |ˆs|φ 0 〉. (11.12)
In a spinor representation, in which the spin-up and -down states are(1, 0)and
(0, 1)respectively, the operatorsˆis the Pauli matrixσz. There is no subscriptmin
the rightmost expression as the ground state eigenvectorφ 0 is independent ofm.
We can also calculate the correlation function. This is defined as
g(m−n)=〈snsm〉−〈sm〉〈sn〉. (11.13)
Denoting the eigenvalues ofTbyλα(and higher indices) and the corresponding
eigenvectors byφα(for the one-dimensional Ising model,αassumes only the two
values 0 and 1), we have
g(m−n)=

1

Z

∑

φα···φ

〈φα|Tm−^1 |φβ〉〈φβ|ˆsm|φγ〉〈φγ|Tm−n|φδ〉

×〈φδ|ˆsn|φ〉〈φ|TL−n+^1 |φα〉−〈φ 0 |ˆs|φ 0 〉^2. (11.14)

The last eigenvectorφαin the first term is the same as the first eigenvector because
of the periodic boundary conditions. One is usually interested in the long-range
behaviour of the correlation function:

1 |m−n|L. (11.15)

This suggests that the major contribution to the correlation function comes from
replacing all eigenvalues in (11.14) by the largest one,λ 0. However, inspection
shows that the two terms in (11.14) cancel exactly for this choice. The major con-
tribution to the correlation function is therefore obtained by replacing the transfer
matrices acting between positionsnandmby the largest-but-one eigenvalue,λ 1.
In the limit (11.15), all the other choices are much smaller. This results in

g(m−n)=|〈φ 0 |s|φ 1 〉|^2

(

λ 1

λ 0

)|m−n|

. (11.16)