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(15.96)
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(A + zB + z-lB*}u
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zl-nB + z-lB*}u zn =
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These equations are solved by putting
(15.97)
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The three equations (15.96) then become the same one:
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(A + zB + z-lB*}u
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(15.98)
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where the sign =+= in (15.97) is associated with w . Thus, for w+ and for w-, there are n values of z:
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(k = 0,1, ... ,2n - I)
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where
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(15.99)
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1,3,5,
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,2n-l
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(15.100)
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For each k, two eigenvalues Ak are determined by the equation
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(A + zkB + z;;lB*)u
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(15.101)
and Ak is to be associated with w according to (15.100). This determines 2n eigenvalues each for w . To find Ak , note that according to (15.91) and (15.92) det
1, 1
det IBI
det IB*I
det IA + zkB + z;;lB*1
SPECIAL TOPICS IN STATISTICAL MECHANICS
Therefore the two values of Ak must have the forms
Ak=e Yk
(k=0,1, ... ,2n-1)
(15.102)
The value of Yk may be found from the equation 1Tr(A + ZkB + z;;lB*) = He Yk + e- Yk ) = coshYk
(15.103)
Evaluating the trace with the help of (15.91), (15.92), and (15.99) we obtain 7Tk cosh Yk = cosh 2<1> cosh 2(J - cos - sinh 2<1> sinh 2(J n (k = 0, 1, ... , 2n - 1) (15 .104) if Yk is a solution to (15.104) then -Yk is also a solution. But this possibility has already been taken into account in (15.102). Therefore we define Yk to be the positive solution of (15.104). It is easily verified that Yk = Y2n-k (15.105 ) o < Yo < Yl < .,. < Yn The first is obvious. The second can be seen by noting that Ykl k = (7Tln) sin (7Tkln)/sinYk' which is positive for k s n. A plot of Yk as a function of <I> is shown in Fig. 15.3. As n ~ 00 these curves merge into a continuum. The eigenvalues of n are the same as those of w , respectively. Therefore n are products of commuting plane rotations, although this fact is not obvious from (15.84). The 2 n eigenvalues of V can now be written down immediately with the help of lemma 3 of the last section: eigenvalues of V - are e 1/ 2 Yo YI Y4 ... Y2n - 2) ( (15.106) 1 / 2 y, Y3 Ys .. ' YIn-I) eigenvalues of V + are e ( (15.107) where all possible choices of the signs are to be made independently.
f-------~~--------+cP
//, /
....... ....... / / /' /' / /, ....... -1'0 /',/,/ / __ --------.................... ": 1'1
-I'n
----/(J .......
= k'T
.......
....... .......
.......
Fig. 15.3 The solutions of (15.104).
THE ONSAGER SOLUTION
EIGENVALUES OF V
As we have explained, the set of eigenvalues of V consists of one-half the set of eigenvalues of V+ and one-half that of V-. The eigenvalues of V are all positive and of order en. Therefore all eigenvalues of V are positive and of order en. This justifies the formula (15.61). To find explicitly the set of eigenvalues of V, it would be necessary to decide which half of the set of eigenvalues of V + and Vshould be discarded. We are only interested, however, in the largest eigenvalue of V. For such a purpose it is not necessary to carry out this task. Suppose the 2 n X 2 n matrix F transforms (15.80) into diagonal form and the n 2 X 2 n matrix G transforms (15.81) into diagonal form:
F[ HI + U)Y+ 1F- 1 G[HI - U)Y-1G- 1
(15.108) (15.109)
where Vl are diagonal matrices with half the eigenvalues of (15.106) and (15.107), respectively, appearing along the diagonal. It is possible to choose F and G in such a way that FUF- 1 and GUG- 1 remain diagonal matrices. Then F and G merely permute the eigenvalues of U along the diagonal. But the convention has been adopted that U has the form (15.78). Hence F and G either leave U unchanged or simply interchange the two submatrices 1 and -1 in (15.78). That is, F and G either commute or anitcommute with U. This means that
HI O)FV+F- 1
(15.110) (15.111)
Vi) = HI O)GV-G-
where the signs can be definitely determined by an explicit calculation. * For our purpose this determination is not necessary. We may write
(15.112)
FV+F- 1 GY-G -1