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The Heisenberg Antiferromagnet and the Lanczos Algorithm Abstract

The Heisenberg Antiferromagnet and the Lanczos Algorithm Abstract

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BSc Final Year Report - Mat<strong>the</strong>w HolmesIf only <strong>the</strong> ground or lower levels are required <strong>the</strong> <strong>Lanczos</strong> method enables much larger systems to betreated than would o<strong>the</strong>rwise be <strong>the</strong> case. That said, <strong>the</strong> treatment of 1D chains is still limited to N~20spins [19] even on powerful systems [20]. At first glance this may seem small but it presents a~10 6 x10 6 matrix, which is large!Initiating <strong>the</strong> Procedure<strong>The</strong> initial vector v 1 , must have some projection along <strong>the</strong> desired eigenstate v. This is usually satisfiedby r<strong>and</strong>omising <strong>the</strong> coefficients [20,19] of v 1 ,CALL RANDOM_NUMBER(v1)v1= v1/SQRT(DOT_PRODUCT(v1,v1))Remember, v 1 is a vector in <strong>the</strong> original basis { i } <strong>and</strong> has coefficients. Generally <strong>the</strong>n,v 1 =c 1 1 +c 2 2 +…c ,. It is tempting to choose equal coefficients (as <strong>the</strong> author did) in <strong>the</strong> hope someprojection along <strong>the</strong> desired state. <strong>The</strong> <strong>Lanczos</strong> method will not work if this is done [20] as v 2 will bezero <strong>and</strong> <strong>the</strong> iteration fails. A 'proof' is proposed below.Consider <strong>the</strong> 'symmetric vector' (of equal projection, c, in all basis states) v, = c{ ψ i}; i = 1. (29)Normalising, we obtain <strong>the</strong> initial vector v 1 ,c= { ψ } = { ψ } { ψ }1 i2 i=cηc1 , (30)ηias <strong>the</strong> i are <strong>the</strong>mselves normalised such that |v|=[c 2 |{ i }| 2 ] 1/2 =c 1/2 . Substituting Eq. (30) into Eq. (20),we have for 0 ,α1η{ } H { }ψ iψ i0 = (31i)εεα0=ii=ηη2{ ψ } { ψ } = { ψ } εi(31ii)Where ε is <strong>the</strong> sum along any row or column of H. Substituting (Eq. 31ii) into Eq. (18), we obtain,ε1{ ψ } −ε{ ψ } 02=ii= . (32)ηηFrom Eq. (32) a new vector v 2 cannot be produced <strong>and</strong> we conclude that an initial symmetric vector isunsuitable when representing a symmetric matrix. Such a vector is orthogonal to all eigenstates. Thisresult is physically intuitive, as a symmetric vector represents a state of zero magnetisation.Application & AnalysisOrthogonalityWhen using <strong>the</strong> <strong>Lanczos</strong> method it is vital that <strong>the</strong> basis is orthogonal. Fig. 4 shows <strong>the</strong> increasingoverlap between successive states in <strong>the</strong> absence of any remedial action. Without such action, rogueeigenvalues are introduced to <strong>the</strong> spectra [14]. This makes it difficult to check for accuracy with anexact routine when setting up <strong>the</strong> <strong>Lanczos</strong> method.9

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