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This is the result in 4.3 I 4.4 The mean-field equations for the three-state Potts model H=-1 8070; 0; = 1, 2, 3 (ij) (4.57)
This is the result in 4.3
I
4.4 The mean-field equations for the three-state Potts model H=-1 8070; 0; = 1, 2, 3 (ij) (4.57) can be derived as follows using the result obtained in problem 4.3: (i) Show that eqn 4.57 is equivalent to Problems 65 H=-1 51.5; (ij) where 5 = ( ) ( 5512).(-32) () show that the mean-field equations (ii) Putting H. become Jz 38H0/2 - 1 e38H0/2 +2 NH Fmf = -NkT In(eHo + 2e -BH0/2) + 2Jz (It is easiest to focus on an ordered state where ( 6 ) predom- inates so that H = 0 by symmetry.) (iii) Expand Fmf in (s), and show that it contains a cubic term. By sketching the mean-field free energy for suitable values of the coefficients in the expansion show that the transition is first- order. (iv) Verify that the transition is at kTc = 3J2/8 In 2 and that the jump in the magnetization is J2/2. 4.3 Show that for a Hamiltonian H=-151.5; (ij) and a trial Hamiltonian Ho = -.. the mean-field equations may be written H. = Jz(5), Imf = -kT In Zo + JzN(578/2 where the subscript 0 denotes an average in the ensemble defined by Ho, N is the number of spins, and z is the coordination number of each spin. 4.4 The mean-field equations for the three-state Potts model H=-1 8070; 0; = 1, 2, 3 (ij) (4.57) can be derived as follows using the result obtained in problem 4.3: (i) Show that eqn 4.57 is equivalent to Problems 65 H=-1 51.5; (ij) where 5 = ( ) ( 5512).(-32) () show that the mean-field equations (ii) Putting H. become Jz 38H0/2 - 1 e38H0/2 +2 NH Fmf = -NkT In(eHo + 2e -BH0/2) + 2Jz (It is easiest to focus on an ordered state where ( 6 ) predom- inates so that H = 0 by symmetry.) (iii) Expand Fmf in (s), and show that it contains a cubic term. By sketching the mean-field free energy for suitable values of the coefficients in the expansion show that the transition is first- order. (iv) Verify that the transition is at kTc = 3J2/8 In 2 and that the jump in the magnetization is J2/2. 4.3 Show that for a Hamiltonian H=-151.5; (ij) and a trial Hamiltonian Ho = -.. the mean-field equations may be written H. = Jz(5), Imf = -kT In Zo + JzN(578/2 where the subscript 0 denotes an average in the ensemble defined by Ho, N is the number of spins, and z is the coordination number of each spinStep by Step Solution
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