Statistical Mechanics - Physics at Oregon State University

9.5. EXACT SOLUTION FOR THE ISING CHAIN. 207
or simply
Z(T, h, N) =
σ1,···,σN
T(σ1, σ2)T(σ2, σ3) · · · T(σN, σ1) (9.77)
Z(T, h, N) = Tr T N
(9.78)
This looks like a very simple expression, and it is. There are many situation
in physics where we can write some complicated expression as the trace or
determinant of a product of simple matrices. Our case is particularly easy,
since all matrices are the same. Making progress in the calculation of matrix
expressions is easiest if we know the eigenvalues and eigenvectors. We need to
solve
T|e >= t|e > (9.79)
In our case the matrix is real and symmetric, and we know that there are
two eigenvectors with real eigenvalues. Finding the eigenvalues of a real and
symmetric two by two matrix can always be done. We need to solve det(T −
tE) = 0, where E is the identity matrix. This gives
or
which has solutions
(e β(J+h) − t)(e β(J−h) − t) − e −2βJ = 0 (9.80)
t 2 − t[e β(J+h) + e β(J−h) ] + e 2βJ − e −2βJ = 0 (9.81)
t± = 1
e
2
β(J+h) + e β(J−h)
± [eβ(J+h) + eβ(J−h) ] 2 − 4[e2βJ − e−2βJ
]
(9.82)
t± = e βJ cosh(βh) ± 1
e
2
2β(J+h) + e2β(J−h) − 2e2βJ + 4e−2βJ (9.83)
t± = e βJ cosh(βh) ± 1
[e
2
β(J+h) − eβ(J−h) ] 2 + 4e−2βJ which leads to the final expression
(9.84)
t± = e βJ
cosh(βh) ± e2βJ sinh 2 (βh) + e−2βJ (9.85)
There are two real solutions as expected for a real and symmetric matrix. Note
that we have t+ > t−. Also, we have t+ + t− = Tr T = e β(J+h) + e β(J−h) > 0
and t+t− = det(T) = e 2βJ − e −2βJ > 0. Therefore both eigenvalues have to be
positive. The corresponding eigenvectors are |e± > and the partition function
in terms of these eigenvectors is