Statistical Physics

7.4 The One-Dimensional Ising Model 107calculation of the free energy of an interacting system is impossible in general.However, in one dimension, the free energy of certain systems can be calculatedexactly. The Ising model is one such system, and so here we calculate its freeenergy exactly.We investigate a system of unit length consisting of N spins aligned linearly.The ith spin is denoted by σ i , which is equal to either +1 (up) or −1(down). We impose a periodic boundary condition so that σ N+1 = σ 1 .Thespins interact with their nearest-neighbor spins. A microscopic state given bya spin configuration (σ 1 ,σ 2 , ···,σ N ) has the following energy E:N∑N∑E({σ i })=−J σ i σ i+1 − µB σ i , σ i = ±1 , (7.46)i=1where the magnitude of the magnetic moment µ has been included in the couplingconstant J, andB is the external magnetic field. The partition functionis the following sum over the microscopic states:Z = ∑ {σ i}e −βE({σi})i=1= ∑ ∑... ∑ ∏N A (σ i ,σ i+1 ) , (7.47)σ 1 σ 2 σ Ni=1where[A (σ i ,σ i+1 )=exp βJσ i σ i+1 + 1 2 βµBσ i + 1 ]2 βµBσ i+1. (7.48)Let us consider the sum over the ith spin σ i :∑σ iA (σ i−1 ,σ i ) A (σ i ,σ i+1 ) . (7.49)We notice that this can be considered as a matrix product of a two-by-twomatrix A with itself, where( )A(+, +) A(+, −)A =. (7.50)A(−, +) A(−, −)To simplify the treatment, we introduce the notation K = βJ and b = βµB.The matrix elements of A areA (+, +) = exp(βJ + 1 2 βµB + 1 )2 βµB =e K+b , (7.51)A (+, −) =A (−, +) = exp (−βJ) =e −K , (7.52)A (−, −) =exp(βJ − 1 2 βµB − 1 )2 βµB =e K−b . (7.53)