Information Theory, Inference, and Learning ... - Inference Group

Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.31.2: Direct computation of partition function of Ising models 409is to note that if the system is translation-invariant along its length then weneed to do only one iteration in order to find the properties of a system of anylength.The computational task becomes the evaluation of an S × S matrix, whereS is the number of microstates that need to be considered at the boundary,and the computation of its eigenvalues. The eigenvalue of largest magnitudegives the partition function for an infinite-length thin strip.Here is a more detailed explanation. Label the states of the C columns ofthe thin strip s 1 , s 2 , . . . , s C , with each s an integer from 0 to 2 W −1. The rthbit of s c indicates whether the spin in row r, column c is up or down. Thepartition function isZ = ∑ exp(−βE(x)) (31.26)x= ∑ ∑· · · ∑ ()C∑exp −β E(s c , s c+1 ) , (31.27)s 1s 2s Cwhere E(s c , s c+1 ) is an appropriately defined energy, and, if we want periodicboundary conditions, s C+1 is defined to be s 1 . One definition for E is:E(s c , s c+1 ) =∑∑∑J x m x n + 1 4J x m x n + 1 4J x m x n . (31.28)(m,n)∈N :m∈c,n∈c+1(m,n)∈N :m∈c,n∈cc=1(m,n)∈N :m∈c+1,n∈c+1This definition of the energy has the nice property that (for the rectangularIsing model) it defines a matrix that is symmetric in its two indices s c , s c+1 .The factors of 1/4 are needed because vertical links are counted four times.Let us defineM ss ′ = exp ( −βE(s, s ′ ) ) . (31.29)Then continuing from equation (31.27),Z = ∑ ∑· · · ∑ [ C]∏M sc,sc+1 s 1s 2s Cc=1(31.30)= Trace [ M C] (31.31)= ∑ aµ C a , (31.32)+−++−−++−s 2 s 3+−−+++Figure 31.13. Illustration to helpexplain the definition (31.28).E(s 2 , s 3 ) counts all thecontributions to the energy in therectangle. The total energy isgiven by stepping the rectanglealong. Each horizontal bondinside the rectangle is countedonce; each vertical bond ishalf-inside the rectangle (and willbe half-inside an adjacentrectangle) so half its energy isincluded in E(s 2 , s 3 ); the factor of1/4 appears in the second termbecause m and n both run over allnodes in column c, so each bond isvisited twice.For the state shown here,s 2 = (100) 2 , s 3 = (110) 2 , thehorizontal bonds contribute +J toE(s 2 , s 3 ), and the vertical bondscontribute −J/2 on the left and−J/2 on the right, assumingperiodic boundary conditionsbetween top and bottom. SoE(s 2 , s 3 ) = 0.where {µ a } 2Wa=1 are the eigenvalues of M. As the length of the strip C increases,Z becomes dominated by the largest eigenvalue µ max :Z → µ C max . (31.33)So the free energy per spin in the limit of an infinite thin strip is given by:f = −kT ln Z/(W C) = −kT C ln µ max /(W C) = −kT ln µ max /W. (31.34)It’s really neat that all the thermodynamic properties of a long thin strip canbe obtained from just the largest eigenvalue of this matrix M!ComputationsI computed the partition functions of long-thin-strip Ising models with thegeometries shown in figure 31.14.As in the last section, I set the applied field H to zero and considered thetwo cases J = ±1 which are a ferromagnet and antiferromagnet respectively. Icomputed the free energy per spin, f(β, J, H) = F/N for widths from W = 2to 8 as a function of β for H = 0.