The Heisenberg Antiferromagnet and the Lanczos Algorithm Abstract

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BSc Final Year Report - Matthew Holmesˆ⋅ Sˆ0111 ( ↑ ↑ ) = 01 + −2 1 2 1 2=2 000 1 0 011220S (14)And likewise for the opposite polarisation. Similarly, for those staggered pairs in which just one spin isdestroyed, we destroy the pair because of the resultant multiplication by zero.Computing The HamiltonianThe basis { } and ladder operations are conveniently provided for [20] in Fortran with the intrinsic bitprocedures. The present state , is represented using IBITS(I,POS,LEN), with IBCLR(I,POS) andIBSET(I,POS) emulating the ladder operators. In Fig. 2, an example state from an N=3 chaindemonstrates how IBITS is used.01ψ4= ↑2↓1↓0= 100 =IBITS(4,POS,N=3)4Fig. 2 Binary representation in Fortran of a 1D spin chain configuration 4 . The bit positionsPOS are numbered in subscript from the right (from 031 in a 32-bit number). The requiredsegment (chain length, N) of the full number is selected from the right with LEN.The Hamiltonian is computed according to Eq. (8), which can be implemented by following the flowscheme in Fig. 3.DO k1. IBITS=IBITS(i,POS,N)2. Move along IBITS with POS=0N, applying Eq. (13) foreach pair3. If states are flipped, store copy of altered state for use in4.Fig. 3 Flow scheme forconstructing theHamiltonian. Theprocedure followsnaturally from Eq. (8).Where states areflipped, IBCLR sets to 0and IBSET sets to 1.DO k'DO until k'=DO until: k=4. Apply orthonormality relation5. Calculate matrix elementIn this way matrix operations are unnecessary and H can be generated by logic statements. It isinstructive to consider a two spin chain, which is, essentially, the 'unit' on which acts.H ˆ 00 =1 00 + 0 , no ladder term contribution.4ˆ1H 01 = −1 01 + 10 , contribution from 1st ladder term.42ˆ1H 10 = −1 10 + 01 , contribution from 2nd ladder term.42(15)H ˆ 11=1 11 + 04Applying the orthonormality relations to Eqs. (15) we obtain the Hamiltonian for the N=2 chain whichis real symmetric, as expected.6
BSc Final Year Report - Matthew Holmes1 40= −J⋅000−14120012−14000014 H (16)We do not use a closed chain for two spins as this would involve two interactions between a pair,which is unphysical.Ground EigenstatesThe S=1/2 antiferromagnet is one of the few systems for which a non-trivial ground state is exactlyknown [4]. The ferromagnetic, J>0, ground states, obvious from Eq. (16), are always the fullypolarised states. With J. v 1 must not be orthogonal to the desired state, nor can it have equal coefficients. A neworthogonal vector v 2 is produced by subtracting from H|v 1 > the projection along v 1 .ν2ν1−α1ν1= H (18)Thus v 2 is orthogonal to v 1 , < v 1 |v 2 > = 0, whereby,0 = ν H ν −αν ν , and, (19)α11ν Hν1111 11= (20)ν1ν1For the next state, v 3 , we have,ν23ν2−α2ν2− β1ν1= H (21)7
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