Computational Physics

A8 Linear algebra problems 597

sparseness. In most cases we need only the lowest part of the eigenvalue spectrum
of the discretised operator, as it is only for these eigenvalues that the discrete
eigenfunctions represent the continuum solutions adequately. We can then use the
Lanczosorrecursionmethod. This method yields the lowest few eigenvalues and
eigenvectors inO(N)steps, thus gaining two orders of efficiency with respect to
Householder’smethod[ 20 – 22 ].
The Lanczos method works for Hermitian matrices, of which the Hamiltonian
matrix for a quantum system is an example. We denote the matrix to be diagonalised
byA. We start with an arbitrary vectorψ 0 , normalised to one and construct a second
vector,ψ 1 , in the following way:

Aψ 0 =a 0 ψ 0 +b 0 ψ 1. (A.125)

We requireψ 1 to be normalised and perpendicular toψ 0 ;ψ 1 ,a 0 andb 0 are then
defined uniquely:

a 0 =〈ψ 0 |A|ψ 0 〉

b 0 ψ 1 =Aψ 0 −a 0 ψ 0

||ψ 1 || =1. (A.126)

Next we construct a normalised vectorψ 2 by lettingAact onψ 1 :

Aψ 1 =c 1 ψ 0 +a 1 ψ 1 +b 1 ψ 2. (A.127)

The vectorψ 2 is taken perpendicular toψ 0 andψ 1 and normalised, soψ 2 ,a 1 ,b 1
andc 1 are determined uniquely too. We proceed in the same way, and at steppwe
have

Aψp=

∑p−^1

q= 0

c(pq)ψq+apψp+bpψp+ 1 (A.128)

whereψp+ 1 is taken orthogonal to all its predecessorsψq,q≤p. Using the fact
thatAis Hermitian we find forq<p−1:

c(pq)=〈ψq|Aψp〉=〈Aψq|ψp〉=0. (A.129)

The last inequality follows from the fact that we have encounteredqalready in the
iteration procedure, soψpis perpendicular toAψq. In the same way we find for

c(pp−^1 ):
c(pp−^1 )=〈ψp− 1 |Aψp〉=〈Aψp− 1 |ψp〉=bp− 1. (A.130)

So afterpsteps we obtain

Aψp=bp− 1 ψp− 1 +apψp+bpψp+ 1 , (A.131)

whereψp+ 1 is normalised and orthogonal toψpandψp− 1. By the foregoing analysis,
it is then orthogonal to all the previousψq(q<p−1). We see thatA, expressed