Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 10 N solutions. Because they are eigenvalues of A and A is Hermitian, these solutions mustallberealnumbers. I will assume here that you have heard of determinants. If not, take heart: computers can calculate them for you. Mathematica has the function Det[A] that returns the determinant of the matrix A. § 8 Back to physics. IfAisanobservablethentheeigenvaluesofthe matrix A are the values that we observe when we measure the observable A. The “quantization” of the values that the observable can take occurs because (A − λI)ψ = 0 has solutions only for those special values of λ for which the matrix A − λI does not have an inverse. These are the values of λ for which det[A − λI] = 0. Note that the spectrum of A depends only on the matrix A, as it should. If we use N functions in our orthonormal, complete basis set, the matrix A can have only N eigenvalues. The are not necessarily distinct, because some roots of the characteristic polynomial might be equal to each other. As we increase the number of functions in the orthonormal basis set, we increase the number of eigenvalues. If we use an infinite basis set, we will have an infinite number of eigenvalues. These eigenvalues will be discrete, however. We cannot possible use an infinite basis set and therefore we never get the exact eigenvalues and eigenfunctions. By luck, or perhaps because of some mathematical theorem unknown to me, if we are intelligent in our choice of basis set, we get a good approximation to the smallest eigenvalues even with a relatively small basis set. As you’ll see in an example given later, using roughly 40 basis functions gives accurate values for about the 20 lowest eigenvalues. § 9 An example of an eigenvalue problem: nondegenerate eigenvalues. Let us examine the eigenvalue problem for the matrix A given by Eq. 35. The characteristic equation is (see Eq. 51) ⎛ ⎜ det ⎝ 3 − λ 2 4 2 1.2 − λ 3.1 4 3.1 4− λ ⎞ ⎟ ⎠ =0 (52) If you find tedious algebra soothing, you can calculate this determinant. IprefertouseMathematica, and the result is (see Section 2 of the file WorkBook7 The eigenvalue problem.nb) −λ 3 +8.2λ 2 +9.21λ − 0.03 = 0 (53)
7. The Eigenvalue Problem, December 17, 2009 11 This is the characteristic polynomial of the matrix A. It is a third-order polynomial because we used a three-dimensional basis set and therefore A is a3× 3 matrix. The roots of this equation are the eigenvalues of A. They are (see Section 2ofWorkBook7) ⎫ λ 1 = −1.00391 ⎪⎬ λ 2 =0.0032479 (54) ⎪⎭ λ 3 =9.20007 The eigenvalues are real numbers because A is Hermitian. Exercise 2 Use Mathematica to check that det A = λ 1 λ 2 λ 3 . How do we find the eigenvectors For the eigenvector ψ(1) corresponding to the eigenvalue λ 1 ,insertλ 1 in the equation Aψ = λψ to obtain Aψ(1) − λ 1 ψ(1) = 0, (55) We have modified the notation so that ψ(1) = {ψ 1 (1), ψ 2 (1), ψ 3 (1)} is the eigenvector corresponding to the eigenvalue λ 1 .IfIinsertλ 2 in Aψ = λψ, I get a different equation Aψ(2) − λ 2 ψ(2) = 0 (56) for a different eigenvector, ψ(2), corresponding to the eigenvalue λ 2 . Similarly, Aψ(3) − λ 3 ψ(3) = 0 (57) gives the third eigenvector ψ(3). Because I have three values for λ, Ihave three distinct equations for ψ, and their solutions are ψ(1), ψ(2), and ψ(3). Each symbol ψ(1), ψ(2), ψ(3) represents a vector in R 3 .Thecomponents of ψ(α), α =1, 2, 3, are denoted by With this notation, ⎛ ⎜ (A − λ 1 I)ψ(1) = ⎝ ψ(α) ={ψ 1 (α), ψ 2 (α), ψ 3 (α)} (58) ⎞ 3 − λ 1 2 4 ⎟ 2 1.2 − λ 1 3.1 ⎠ 4 3.1 4− λ 1 ⎛ ⎜ ⎝ ψ 1 (1) ψ 2 (1) ψ 3 (1) ⎞ ⎟ ⎠ = 0
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Chapter 7. The Eigenvalue Problem

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