1. Monte Carlo Simulation of Canonical Distribution The idea is to generate states i,j,… by a stochastic process such that the probability  (i) of state i is given by the appropriate distribution –> canonical, grand canonical etc.  states are generated and the desired quantity A i (energy, magnetization,…) is calculated for each state = lim 1/   A i  i For the canonical distribution,  (i) = exp(-  E i )/ Z where Z =  exp(-  E i )

2. How do we do this using a computer? Consider a system of N classical spins which can be up or down. The total number of microstates is M = 2 N H = -J  s i s j i<j We could generate configurations randomly and calculate E(i) and weight its contribution by exp(-  E(i)) =  E(i) exp(-  E(i)) /  exp(-  E(i)) Very inefficient since M = 2 N is exponentially large. We can never generate all states if they have equal probability and many configurations make a small contribution We want to use importance sampling!

3. Importance sampling =  [A(i)/  (i)] exp(-  E(i))  (i)  [1/  (i)] exp(-  E(i))  (i) If we generate the microstates with probability  (i) = exp(-  E(i))/  exp(-  E(i)) then = (1/n)  A(i) How do we obtain  (i) ?

4. Markov process Suppose the system is in state i. The next state is selected with a transition probability P(j  i) that does not depend on the previous history of the system. This process produces states with a unique steady-state probability distribution (after a transient) The steady state probability  (j) is an eigenvector with eigenvalue 1 of the transition matrix  (j) =  P(j  i)  (i) i

5. Eg. if he is in room 2, then P(3  2) = P(1  2) = 1/2 Similarly, P(1  3) = P(2  3) = P(4  3) = 1/3 What are the transition probabilities? What fraction of the time will the student spend in each room in a steady state? Consider the following example A student changes rooms at regular intervals and uses any of the doors leaving the room with equal probability

6. Hence P(j  i) = 0 1/2 1/3 0 1/2 0 1/3 0 1/2 1/2 0 1 0 0 1/3 0 Eigenvalues are 1, -1/2, -1/4  1/2 (11/12) 1/2 Eigenvector of largest eigenvalue is ( 1/4, 1/4, 3/8, 1/8) Hence after a long time we reach a steady state with  (1)= 1/4  (2)= 1/4  (3)=3/8  (4)=1/8 Note:   (i) = 1 (normalization) P(j  i)  (i) = P(i  j)  (j) (detailed balance)

7. Ising Model Suppose system is in state i. Pick a site  at random and consider flipping it s  = - s . The final state can be the same (i) or different (j). After n steps  (f) = lim P(f  i) =  P(f  i n-1 ) P(i n-1  i n-2 ) … P(i 1  i) n  approaches a limiting distribution independent of the initial state i. We require  (f) to be normalized and satisfy  (m)/  (j) = exp[-  (E(m)-E(j)] for all pairs m,j Normalization means  P(j  m) = 1 j and P(j  m)  (m) = P(m  j)  (j) “detailed balance” Hence  (m) =  P(j  m)  (m) =  P(m  j)  (j) j j  (m) is a stationary probability distribution

8. Metropolis Algorithm 0) establish an initial microstate 1) pick site  randomly 2) compute the energy change if the spin is flipped  E = E(new) – E(old) 3) determine the value of A(i) 4) if  E  0, then flip it and proceed to 7 5) if  E>0, then compute w=e -  E 6) generate a random number r 7) if r  w accept the new state otherwise remain in the old 8) repeat steps 1) to 7) 9) Calculate and - 2

9. Periodic boundary conditions

10. Specific heat and magnetic susceptibility C v = - 2 kT 2  = = - 2 kT e.g. Ising Model S i =  1 on a square lattice of N=L 2 sites In the limit L , the exact results are known

12. In the limit as L  the system undergoes a phase transition The exact T c = 2.269185 The specific heat diverges logarithmically C v ~ ln|T-T c | The susceptibility diverges as  ~ |T-T c | -  with  =7/4

13. Monte Carlo Simulation of the Ising Model

14. This is an example of an order- disorder transition F = E - TS energy(order) versus entropy(disorder) In d=1, the ground state at T=0 has all spins aligned parallel  Low energy excitations correspond to domain walls   E = 2J  S = k ln(N)  F =2J- kT ln(N) < 0 The ordered phase is unstable at finite T>0 towards the formation of defects (domain walls)

15. d=2 On the square lattice, the ground state has all spins aligned parallel  Low energy excitations consist of compact clusters(domains) of overturned spins   r=8   E = 2J r,  S = k ln(3 r ) r is the perimeter of the cluster Hence  F  [2J- kT ln(3)] r r>>1 At low T,  F is positive but vanishes at a finite T