A periodic spin chain of sites has

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Computational Physics · High-performance and data-driven computational physics

Scientific Computing for Physicists, Ch Neural networks on physical data: backpropagation, convolution and symmetry

Problem

A periodic spin chain of N=6N=6 sites has s=(s0,…,s5)=(1,−1,1,1,−1,1)\mathbf{s}=(s_0,\dots,s_5)=(1,-1,1,1,-1,1). A convolutional layer with kernel weight +1+1 at offset −1-1 and −1-1 at offset +1+1, z(r)=s(r−1)−s(r+1)z(r) = s(r-1)-s(r+1) (indices mod 6), is applied to it and, separately, to the cyclically shifted chain s′(r)=s(r−1 mod 6)s'(r) = s(r-1 \bmod 6). ((a)) Compute z(r)z(r) for r=0,…,5r=0,\dots,5 from s\mathbf{s}. ((b)) Compute z′(r)z'(r) for r=0,…,5r=0,\dots,5 from s′\mathbf{s}', and check that z′(r)=z(r−1 mod 6)z'(r) = z(r-1 \bmod 6) for every rr, the translation equivariance a circular convolution has by the Fourier shift theorem: shifting a signal multiplies every Fourier coefficient by a phase, leaving convolution with a fixed kernel commuting with the shift. ((c)) On the 13-atom Lennard-Jones cluster (39 inputs) the chapter measured a forward-to-reverse cost ratio of about 5.55.5. If the reverse sweep cost the same fixed multiple of one evaluation at every size, the ratio would grow in proportion to the number of inputs. On that assumption, estimate the ratio for a cluster with 200200 inputs, and say whether the chapter's 3838-atom measurement grew faster or more slowly than this rule predicts. ((d)) The ten square-lattice networks from L=32L=32 are applied, without retraining, to triangular-lattice configurations, and their mean output crosses 12\tfrac12 at S=3.61 J/kBS=3.61\,J/k_B. The ten seeds' crossings have a standard deviation of 0.059 J/kB0.059\,J/k_B; take unumu_{\rm num} as the standard error of their mean, and take an input uncertainty uinput=0.01 J/kBu_{\rm input}=0.01\,J/k_B (from the coupling JJ used to generate the configurations). The comparison value is the triangular lattice's exact D=4/ln⁡3=3.641 J/kBD = 4/\ln 3 = 3.641\,J/k_B (Houtappel 1950), known exactly so uD=0u_D=0. Using E=S−DE=S-D and uval=unum2+uinput2+uD2u_{\rm val}=\sqrt{u_{\rm num}^2+u_{\rm input}^2+u_D^2}, compute EE and uvalu_{\rm val}, state whether the comparison passes at k=2k=2, and say what a pass does and does not show about what the network has learned.

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