Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 12 This is the system of equations (3 − λ 1 )ψ 1 (1) + 2ψ 2 (1) + 4ψ 3 (1) = 0 (59) 2ψ 1 (1) + (1.2 − λ 1 )ψ 2 (1) + 3.1ψ 3 (1) = 0 (60) 4ψ 1 (1) + 3.1ψ 2 (1) + (4 − λ 1 )ψ 3 (1) = 0 (61) We seem to have a difficulty here. If {ψ 1 (1), ψ 2 (1), ψ 3 (1)} is a solution then {βψ 1 (1), βψ 2 (1), βψ 3 (1)} is also a solution. This is not a real problem, since, as noted earlier, we can fix β by requiring that the eigenvector be normalized. This is the only meaningful eigenvector in quantum mechanics. I can look at Eqs. 59—61 and see right away that ψ = {0, 0, 0} is a solution. This is not physically acceptable, because it cannot represent a state of the system. Are there other solutions besides {0, 0, 0} Since this is a homogeneous system of equations, the answer is yes: we have chosen λ = λ 1 , which means that det[A − λI] = 0 and the system has a solution. If we divide Eqs. 59—61 by ψ 3 (1), we obtain (3 − λ 1 ) ψ 1(1) ψ 3 (1) +2ψ 2(1) = −4 ψ 3 (1) 2 ψ 1(1) ψ 3 (1) +(1.2 − λ 1) ψ 2(1) = −3.1 ψ 3 (1) 4 ψ 1(1) ψ 3 (1) +3.1ψ 2(1) = λ 1 − 4 ψ 3 (1) The homogeneous system of three equations and three unknowns has become an inhomogeneous system of three equations and two unknowns (ψ 1 (1)/ψ 3 (1) and ψ 2 (1)/ψ 3 (1)). Mathematica gives (see WorkBook7, Section 2, Cell 2) so the eigenvector ψ(1) is ψ 1 (1) = −0.542 ψ 3 (1) (62) ψ 2 (1) = −0.914 ψ 3 (1) (63) ψ(1) = {−0.542 ψ 3 (1), −0.914 ψ 3 (1), ψ 3 (1)} = ψ 3 (1){−0.542, −0.914, 1} (64) Of course, ψ 3 (1) is not yet determined. The unknown ψ 3 (1) is a multiplicative constant, which I determine by imposing the normalization requirement 1=ψ(1) | ψ(1) =2.13 |ψ 3 (1)| 2 (65)
7. The Eigenvalue Problem, December 17, 2009 13 I solved Eq. 65 for ψ 3 (1) and I got two solutions, 0.685 and −0.685. They are equivalent, since they differ by a factor whose absolute value is 1; such “phase factors” make no difference in any calculation of an observable quantity. I use the positive root: ψ 3 (1) = 0.685 (66) and with this, I have ψ(1) = 0.685{−0.542, −0.914, 1} = {−0.371, −0.626, +0.685} (67) If you are the kind of person who worries about the mysteries of mathematics, you might wonder what happens if you use Eqs. 60 and 61, or perhaps Eqs. 59 and 61. I did this (see WorkBook7, Section 2, Cell 2) and no matter which pair of equations I pick, I get the same result for ψ 1 (1) and ψ 2 (1)! This must be so, if the theory is consistent, but it seems rather miraculous that it happens. However, no miracle is involved. Let us assume that we did not calculate the eigenvalue λ and left it undetermined in Eqs. 59—61. We can take Eqs. 59 and 60 and solve for the ratios ψ 1 (1)/ψ 3 (1) and ψ 2 (1)/ψ 3 (1). We obtain two solutions that depend on λ. Next we take Eqs. 59 and 61 and solve for ψ 1 (1)/ψ 3 (1) and ψ 2 (1)/ψ 3 (1), obtaining two new solutions for those ratios. But the ratios obtained in the first calculation must be equal to the ratios obtained from the second calculation. This equality gives me an equation for λ. Icalculateλ from it and then I solve Eqs. 60 and 61, with this value of λ, forψ 1 (1)/ψ 3 (1) and ψ 2 (1)/ψ 3 (1). The result I get must be equal to the one obtained previously. This demand for consistency of the system of homogeneous equations, Eqs. 59—61, is fulfilled only for special values of λ. These are the eigenvalues. Exercise 3 Implement the recipe outlined above for the homogeneous system (3.2 − λ)ψ 1 +2.4ψ 2 = 0 (68) 2.4ψ 1 +(3.6 − λ)ψ 2 = 0 (69) Solve each equation for ψ 1 /ψ 2 and demand that the ratio (ψ 1 /ψ 2 ) obtained from Eq. 68 be equal to the ratio obtained from Eq. 69. This gives you an equation for λ; call it EQ. Then calculate the characteristic polynomial of the matrix and test whether it is the same as EQ. After that, go ahead and find the eigenvalues and the eigenvectors.
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Chapter 7. The Eigenvalue Problem

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