Glauber Dynamics for Ising Model I AMS Short Course

Transcription

1 Glauber Dynamics for Ising Model I AMS Short Course January 2010

2 Ising model Let G n = (V n, E n ) be a graph with N = V n < vertices. The nearest-neighbor Ising model on G n is the probability distribution on { 1, 1} V n given by µ(σ) = Z(β) 1 exp β σ(u)σ(v), (u,v) E n where σ { 1, 1} V n. The interaction strength β is a parameter which has physical interpretation as 1/temperature.

3 Three regimes High temperature (β < β c ):

4 Three regimes low temperature (β > β c ),

5 Three regimes critical temperature (β = β c ),

6 Glauber dynamics The (single-site) Glauber dynamics for µ is a Markov chain (X t ) having µ as its stationary distribution. Transitions are made from state σ as follows: 1 a vertex v is chosen uniformly at random from V n. 2 The new state σ agrees with σ everywhere except possibly at v, where σ (v) = 1 with probability where e βs(σ,v) e βs(σ,v) + e βs(σ,v) S(σ, v) := w:w v σ(w). Note the probability above equals the µ-conditional probability of a positive spin at v, given that that all spins agree with σ at vertices different from v.

7 Glauber dynamics The (single-site) Glauber dynamics for µ is a Markov chain (X t ) having µ as its stationary distribution. Transitions are made from state σ as follows: 1 a vertex v is chosen uniformly at random from V n. 2 The new state σ agrees with σ everywhere except possibly at v, where σ (v) = 1 with probability where e βs(σ,v) e βs(σ,v) + e βs(σ,v) S(σ, v) := w:w v σ(w). Note the probability above equals the µ-conditional probability of a positive spin at v, given that that all spins agree with σ at vertices different from v.

8 Glauber dynamics The (single-site) Glauber dynamics for µ is a Markov chain (X t ) having µ as its stationary distribution. Transitions are made from state σ as follows: 1 a vertex v is chosen uniformly at random from V n. 2 The new state σ agrees with σ everywhere except possibly at v, where σ (v) = 1 with probability where e βs(σ,v) e βs(σ,v) + e βs(σ,v) S(σ, v) := w:w v σ(w). Note the probability above equals the µ-conditional probability of a positive spin at v, given that that all spins agree with σ at vertices different from v.

9 Cut-off Consider a sequence of Markov chains, (Xt n), and write d n(t) and tmix n (ɛ) for the distance to stationary and mixing time, respectively, of the nth chain. A sequence of Markov chains has a cutoff if tmix n (ɛ) tmix n 1 as n. (1 ɛ) A cutoff has window size {w n } if w n = o(t n mix ) and lim α lim α lim inf d n (t n n mix + αw n) = 1 d n (tmix n + αw n) = 0. lim sup n

10 A conjecture For the Glauber dynamics on graph sequences with bounded degree, tmix n = Ω(n log n). (T. Hayes and A. Sinclair) Conjecture (Y.P.): If the Glauber dynamics for a sequence of transitive graphs satisfies tmix n = O(n log n), then there is a cut-off.

11 Mean field case (Complete Graph) Take G n = K n, the complete graph on the n vertices: V n = {1,..., n}, and E n contains all ( n 2) possible edges. The total interaction strength should be O(1), so replace β by β/n. The probability of updating to a +1 is then where S is the total magnetization e β(s σ(v))/n e β(s σ(v))/n + e β(s σ(v))/n S = n σ(i). i=1 The statistic S is almost sufficient for determining the updating probability. The chain (S t ) is the key to analysis of the dynamics.

12 Mean field case (Complete Graph) Take G n = K n, the complete graph on the n vertices: V n = {1,..., n}, and E n contains all ( n 2) possible edges. The total interaction strength should be O(1), so replace β by β/n. The probability of updating to a +1 is then where S is the total magnetization e β(s σ(v))/n e β(s σ(v))/n + e β(s σ(v))/n S = n σ(i). i=1 The statistic S is almost sufficient for determining the updating probability. The chain (S t ) is the key to analysis of the dynamics.

13 Mean field case (Complete Graph) Take G n = K n, the complete graph on the n vertices: V n = {1,..., n}, and E n contains all ( n 2) possible edges. The total interaction strength should be O(1), so replace β by β/n. The probability of updating to a +1 is then where S is the total magnetization e β(s σ(v))/n e β(s σ(v))/n + e β(s σ(v))/n S = n σ(i). i=1 The statistic S is almost sufficient for determining the updating probability. The chain (S t ) is the key to analysis of the dynamics.

14 Mean field has t mix = O(n log n) If β < 1, a path-coupling argument can be used to show that t mix = O(n log n).

15 High temperature mean-field Theorem (L.-Luczak-Peres) Let (Xt n) be the Glauber dynamics for the Ising model on K n. If β < 1, then t mix (ɛ) = (1 + o(1)) n log n 2(1 β) and there is a cut-off. In fact, we show that there is window of size O(n) centered about t n = 1 n log n. 2(1 β) That is, and lim sup d n (t n + γn) 0 as γ. n lim inf d n (t n + γn) 1 as γ. n

16 Critical temperature mean-field Theorem (L.-Luczak-Peres) Let (X n t ) be the Glauber dynamics for the Ising model on K n. If β = 1, then there are constants c 1 and c 2 so that c 1 n 3/2 t mix c 2 n 3/2.

17 Low temperature mean-field If β > 1, then t n mix > c 1e c 2n. This can be established using Cheeger constant there is a bottleneck going between states with positive magnetization and states with negative magnetization. Arguments for exponentially slow mixing in the low temperature regime go back at least to Griffiths, Weng and Langer (1966) Our results show that once this barrier to mixing is removed, the mixing time is reduced to n log n.

18 Low temperature mean-field If β > 1, then t n mix > c 1e c 2n. This can be established using Cheeger constant there is a bottleneck going between states with positive magnetization and states with negative magnetization. Arguments for exponentially slow mixing in the low temperature regime go back at least to Griffiths, Weng and Langer (1966) Our results show that once this barrier to mixing is removed, the mixing time is reduced to n log n.

19 Low temperature mean-field If β > 1, then t n mix > c 1e c 2n. This can be established using Cheeger constant there is a bottleneck going between states with positive magnetization and states with negative magnetization. Arguments for exponentially slow mixing in the low temperature regime go back at least to Griffiths, Weng and Langer (1966) Our results show that once this barrier to mixing is removed, the mixing time is reduced to n log n.

20 Low temperature mean-field If β > 1, then t n mix > c 1e c 2n. This can be established using Cheeger constant there is a bottleneck going between states with positive magnetization and states with negative magnetization. Arguments for exponentially slow mixing in the low temperature regime go back at least to Griffiths, Weng and Langer (1966) Our results show that once this barrier to mixing is removed, the mixing time is reduced to n log n.

21 Magnetization chain: key equation If S t = n i=1 X t (i), then for S t 0, [ ] St E[S t+1 S t F t ] n tanh(βs t/n).

22 β < 1 When β < 1, using the inequality tanh(x) x for x 0 shows that for S t 0, ( E[S t+1 F t ] S t 1 1 β ) n Need [2(1 β)] 1 n log n steps to drive E[S t ] to n. Additional O(n) steps needed for magnetization to hit zero. (Compare with simple random walk.) Can couple two versions of the chain so that the magnetizations agree by the time the magnetization of the top chain hits zero. Once magnetizations agree, use a two-dimensional process to bound time until full configurations agree. Takes an additional O(n) steps.

23 β < 1 When β < 1, using the inequality tanh(x) x for x 0 shows that for S t 0, ( E[S t+1 F t ] S t 1 1 β ) n Need [2(1 β)] 1 n log n steps to drive E[S t ] to n. Additional O(n) steps needed for magnetization to hit zero. (Compare with simple random walk.) Can couple two versions of the chain so that the magnetizations agree by the time the magnetization of the top chain hits zero. Once magnetizations agree, use a two-dimensional process to bound time until full configurations agree. Takes an additional O(n) steps.

24 β < 1 When β < 1, using the inequality tanh(x) x for x 0 shows that for S t 0, ( E[S t+1 F t ] S t 1 1 β ) n Need [2(1 β)] 1 n log n steps to drive E[S t ] to n. Additional O(n) steps needed for magnetization to hit zero. (Compare with simple random walk.) Can couple two versions of the chain so that the magnetizations agree by the time the magnetization of the top chain hits zero. Once magnetizations agree, use a two-dimensional process to bound time until full configurations agree. Takes an additional O(n) steps.

25 β < 1 When β < 1, using the inequality tanh(x) x for x 0 shows that for S t 0, ( E[S t+1 F t ] S t 1 1 β ) n Need [2(1 β)] 1 n log n steps to drive E[S t ] to n. Additional O(n) steps needed for magnetization to hit zero. (Compare with simple random walk.) Can couple two versions of the chain so that the magnetizations agree by the time the magnetization of the top chain hits zero. Once magnetizations agree, use a two-dimensional process to bound time until full configurations agree. Takes an additional O(n) steps.

26 β < 1 When β < 1, using the inequality tanh(x) x for x 0 shows that for S t 0, ( E[S t+1 F t ] S t 1 1 β ) n Need [2(1 β)] 1 n log n steps to drive E[S t ] to n. Additional O(n) steps needed for magnetization to hit zero. (Compare with simple random walk.) Can couple two versions of the chain so that the magnetizations agree by the time the magnetization of the top chain hits zero. Once magnetizations agree, use a two-dimensional process to bound time until full configurations agree. Takes an additional O(n) steps.

27 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.

28 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.

29 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.

30 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.

31 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.

32 β = 1. Why n 3/2? Expanding tanh(x) = x x 3 /3 + in the key equation yields E[S t+1 S t F t ] 1 3 Need t = Θ(n 3 2α ) steps for E[S t ] = n α. ( ) 3 St. n By comparison with nearest-neighbor random walk, need additional n 2α steps to hit zero. Total time to hit zero is O(n 3 2α ) + O(n 2α ) The choice α = 3/4 makes the powers equal, and gives hitting time with expectation n 3/2. Once the magnetizations agree, need additional O(n log n) to make the configurations agree.