Statistical Mechanics - Physics at Oregon State University

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212 CHAPTER 9. GENERAL METHODS: CRITICAL EXPONENTS. Γj = 1 Z(T, h = 0, N) σ0,···,σN−1 N−1 βJ σ0σje i=0 σiσi+1 (9.102) where we use periodic boundary conditions and have σ0 = σN . In order to evaluate this expression we use a simple trick. First we make the problem more complicated by assuming that all bonds can be different. This means that the coupling constant between spins i and i+1 is Ji. We treat these coupling constants as independent parameters. In the end we need, of course, Ji = J. We define a function S by S(J0, .., JN−1) = σ0,···,σN−1 e β Jiσiσi+1 (9.103) This is similar to a partition function, and we have that Z = S(J, J, J, · · ·). Next we take partial derivatives with respect to J0 to Jj−1. This gives us ∂ ∂ · · · ∂J0 ∂J1 Since σ 2 i This leads to ∂ ∂Jj−1 S = σ0,···,σN−1 = 1 this simplifies to ∂ ∂ · · · ∂J0 ∂J1 ∂ ∂Jj−1 S = β j gj = 1 β j S e β Jiσiσi+1 [βσ0σ1][βσ1σ2] · · · [βσj−1σj] σ0,···,σN−1 ∂ ∂ · · · ∂J0 ∂J1 (9.104) e β Jiσiσi+1 σ0σj (9.105) ∂ ∂Jj−1 S Ji=J (9.106) Therefore we would like to evaluate S. This is done using transfer matrices like we did in the previous section. We define T (i) βJi e e = −βJi (9.107) e −βJi e βJi or T (i) ′ σ,σ ′ = eβJiσσ . The definition of S has a sum in the exponent and we write this exponent of a sum as the product of exponents in the form S = e βJ0σ0σ1 βJ1σ1σ2 βσN−1σN−1σ0 e · · · e (9.108) From this we see that σ0,···,σN−1 S = Tr T (0) T (1) · · · T (N−1) (9.109) If we need to calculate the trace or determinant of a matrix, it is always very useful to know the eigenvalues and eigenvectors. The eigenvectors of the matrices T (i) are easy, they are e± = 1 √ (1, ±1) independent of the index i. 2
9.6. SPIN-CORRELATION FUNCTION FOR THE ISING CHAIN. 213 The eigenvalues do depend on the index i, but they are easy to find,they are λ (i) ± = e βJi ± e −βJi . We can now evaluate S(J, J, · · ·) and find S(J, J, · · ·) = < e+|T (0) T (1) · · · T (N−1) |e+ > +< e−|T Ji=J (0) T (1) · · · T (N−1) |e− > = λ Ji=J N + +λ N − (9.110) where we have defined λ± = eβJ ± e−βJ . Next we discuss the derivative with respect to Ji. Because the factor Ji occurs in one place only, we see ∂ S = Tr T ∂Ji (0) T (1) · · · ∂T(i) · · · T ∂Ji (N−1) The derivative of the matrix is easy, and we define and find U (i) = ∂T(i) ∂Ji U (i) βJi e −e = β −βJi −e−βJi eβJi (9.111) (9.112) (9.113) The eigenvectors of this matrix are again e± independent of the index i, but the eigenvalues are now interchanged, e+ goes with βλ (i) − and vice versa. We are now in the position to calculate the spin correlation function. We have ∂ ∂ · · · ∂J0 ∂J1 which is equal to ∂ ∂Jj−1 S = Tr U (0) · · · U (j−1) T (j) · · · T (N−1) (9.114) < e+|U (0) · · · U (j−1) T (j) · · · T (N−1) |e+ > + < e−|U (0) · · · U (j−1) T (j) · · · T (N−1) |e− > (9.115) If we now calculate this expression at Ji = J we find β j λ j −λ N−j + + β j λ j +λ N−j − (9.116) Since the average magnetization is zero, the spin correlation functions is equal to the pair distribution function, we we get Γj = gj = λj−λ N−j + + λ j +λ N−j − λN + + λN − (9.117)
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Statistical Mechanics Henri J.F. Ja
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IV CONTENTS 4.3 Gas of poly-atomic
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VI CONTENTS
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VIII INTRODUCTION Introduction. The
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X INTRODUCTION In chapter nine we d
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2 CHAPTER 1. FOUNDATION OF STATISTI
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158 CHAPTER 8. MEAN FIELD THEORY: C
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