Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 8 This is a Hermitian matrix with real elements. Let ψ = {ψ 1 , ψ 2 , ψ 3 } be a vector. Then Aψ =0means(use(Aψ) i = 3 j=1 A ij ψ j ) Aψ = ⎛ ⎜ ⎝ ⎞ A 11 ψ 1 + A 12 ψ 2 + A 13 ψ 3 ⎟ A 21 ψ 1 + A 22 ψ 2 + A 23 ψ 3 A 31 ψ 1 + A 32 ψ 2 + A 33 ψ 3 ⎠ = ⎛ ⎜ ⎝ 3ψ 1 +2ψ 2 +4ψ 3 2ψ 1 +1.2ψ 2 +3.1ψ 3 4ψ 1 +3.1ψ 2 +4ψ 3 ⎞ ⎟ ⎠ = 0 (36) If a vector is equal to zero then all its components are zero, and Eq. 36 means 3ψ 1 +2ψ 2 +4ψ 3 =0 (37) 2ψ 1 +1.2ψ 2 +3.1ψ 3 =0 (38) 4ψ 1 +3.1ψ 2 +4ψ 3 =0 (39) The notation Aψ = 0 is a shorthand for these three equations. This is a homogeneous system of three linear equations. Similarly if φ = {2.1, 6.2, 4.3} then is shorthand for Aψ = φ (40) 3ψ 1 +2ψ 2 +4ψ 3 =2.1 (41) 2ψ 1 +1.2ψ 2 +3.1ψ 3 =6.2 (42) 4ψ 1 +3.1ψ 2 +4ψ 3 =4.3 (43) This is an inhomogeneous system of three linear equations. If the matrix A has an inverse A −1 , we can solve Eq. 40 by using A −1 Aψ = A −1 φ (44) and which together lead to A −1 Aψ = Iψ = ψ (45) ψ = A −1 φ (46) You have learned in your mathematics courses about Cramer’s rule, which allows you to solve Eq. 40. This is tedious to implement and I prefer to
7. The Eigenvalue Problem, December 17, 2009 9 use Mathematica to give me the inverse matrix (see the file WorkBook7 The eigenvalue problem.nb). Mathematica tells me that A −1 = ⎛ ⎜ ⎝ 160.333 −146.667 −46.6667 −146.667 133.333 43.3333 −46.6667 43.3333 13.3333 Acting with A −1 on φ gives (see WorkBook7.nb) End of Example ⎞ ⎟ ⎠ (47) ψ = A −1 φ = {−773.3, 705.0, 228.0} (48) Exercise 1 Prove that if a matrix H is Hermitian then so is its inverse. Note a very important fact. If A −1 exists then Aψ = φ has a solution. This also means that the solution of Aψ = 0 is ψ = 0 (because A −1 Aψ = A −1 0 = 0 and A −1 Aψ = Iψ = ψ). The solution ψ = 0 is of no interest, and we say that if A −1 exists then Aψ = 0 does not have a solution. How does this relate to the eigenvalue problem The eigenvalue equation writtenintheform (A − λI)ψ = 0 (49) is a homogenous system of equations. If the operator (Â − λÎ)−1 (50) exists then (A − λI)ψ = 0 does not have a solution. Therefore (A − λI)ψ = 0 has a solution if and only if (A − λI) −1 does not exist, which happens exactly when det(A − λI) =0 (51) Note that this condition does not involve the unknown vector ψ; it depends only on λ (which we don’t know) and on the matrix elements A ij (which we do know). This is an equation for λ. One can prove, based on the properties of determinants, that the left-hand side of Eq. 51 is a polynomial of order N in λ, which is called the characteristic polynomial of the matrix A. The fundamental theorem of algebra tells us that Eq. 51 must therefore have
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Chapter 7. The Eigenvalue Problem

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