16 II. The variational pr<strong>in</strong>cipleS<strong>in</strong>ce the matrix-elements h ij are fixed, acf( ˆψ, ˆψ; τ) can be seen as a differentiable functionof the coefficients b i .Wenowneedtomaximizethisfunctionwithrespecttoaconstra<strong>in</strong>t,namely, that the solution has unit norm:1= ˆψ mm2 = ˆψ | ˆψ = b i b j χ i | χ j µ= b 2 i ,(II.43)µi,j=1as the basis functions χ i were assumed to be orthonormal. In summary, we need to maximizethe functionmmF (b 1 ,...,b m )= b i b j h ij − ξ( b 2 i − 1),(II.44)i,j=1where ξ is a Lagrange multiplier, with respect to the coefficients b i . S<strong>in</strong>ce the H-matrix issymmetric, this results <strong>in</strong> the equations:0= ∂F m=2 h ij b j − 2ξb i , i =1,...,m, (II.45)∂b i⇒ ξb i =j=1i=1mh ij b j , i =1,...,m. (II.46)j=1We have thus arrived at the eigenvalue equation H b = ξb. S<strong>in</strong>ceH is a symmetric matrix,we can f<strong>in</strong>d m solutions b i correspond<strong>in</strong>g to real eigenvalues ξ i . If we now compute theRayleigh coefficient of the function ˆψ i = mj=1 b i,jχ j generated from such a solution, wef<strong>in</strong>d:ˆψi |T(τ) | ˆψ mmi = b i,j b i,k χ j |T(τ) | χ k = b i,j b i,k h jk (II.47)=j,k=1mξ i b 2 i,j = ξ i ,j=1i=1j,k=1(II.48)where we used the eigenvalue relation mk=1 b i,kh jk = ξ i b i,j .Thus,thesolutioncorrespond<strong>in</strong>gto the largest eigenvalue ξ 1 maximizes the Rayleigh coefficient among the basis functions χ i ,as we claimed.Lemma II.9: The second largest eigenvalue ξ 2 of Equation II.40 satisfies ξ 2 ≤ λ 2 .Proof. Let ˆψ 1 and ˆψ 2 be the functions generated from the eigenvectors b 1 , b 2 and theireigenvalues ξ 1 , ξ 2 ,respectively. Consideral<strong>in</strong>earcomb<strong>in</strong>ation ˆψ := x ˆψ 1 + y ˆψ 2 ,forx, y ∈ R.If ˆψ is normalized, we f<strong>in</strong>d, by orthonormality: 1=ˆψ | ˆψ = x 2 ˆψ1 | ˆψ 1 + y 2 ˆψ2 | ˆψ 2 +2xy ˆψ1 | ˆψ2 = x 2 + y 2 . (II.49)µµµµ
II.4. The Ritz method and the Roothan-Hall method 17Moreover, as ˆψ 1 and ˆψ 2 diagonalize the H-matrix:ˆψ1 |T(τ) | ˆψ mm2 = b 1,i b 2,j χ i |T(τ) | χ j = b 1,i b 2,j h ij=i,j=1mξ 2 b 1,i b 2,i =0.i=1i,j=1(II.50)(II.51)Therefore, comput<strong>in</strong>g the Rayleigh coefficient for ˆψ yields: ˆψ |T(τ) | ˆψ= x 2 ˆψ1 |T(τ) | ˆψ 1 + y 2 ˆψ2 |T(τ) | ˆψ 2 +2xy ˆψ1 |T(τ) | ˆψ2(II.52)= x 2 ξ 1 + y 2 ξ 2 =(1− y 2 )ξ 1 + y 2 ξ 2 = ξ 1 − y 2 (ξ 1 − ξ 2 ). (II.53)Depend<strong>in</strong>g on the choice of y, thisvalueisbetweenξ 2 and ξ 1 .Ifwenowhadλ 2