Chapter 7. The Eigenvalue Problem

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7. The Eigenvalue Problem, December 17, 2009 28 The left-hand side of Eq. 147 can be written as (use p |e(p)e(p)| = Î) N p=1 k=1 = N e(j) | Û −1 | e(p)e(p) | Ĥ | e(k)e(k) | Û | e(i) N p=1 k=1 N U −1 H pk U ki = U −1 HU jp ji where we have denoted the matrix elements by U −1 jp ≡e(j) | Û −1 | e(p) (149) H pk ≡e(p) | Ĥ | e(k) U ki ≡e(k) | Û | e(i) and used the observation that the second double sum in Eq. 149 is the rule for matrix multiplication. This allows us to write the last term (i.e. U −1 HU) as the product of the matrices U −1 , H, andU. Combining Eq. 147 with Eq. 149 proves the theorem. We can go a step further and construct the matrix U from the eigenvectors of Ĥ. Wehavedefined Û through Eq. 145, Acting with e(j)| on this gives |x(i) = Û|e(i) e(j) | x(i) = e(j) | Û | e(i) = U ji (150) Using the completeness relation for the basis set {|e(i)} N i=1, wehave |x(i) = N |e(j)e(j) | x(i) (151) j=1 This means that e(j) | x(i) are the components of the eigenket |x(i) in the basis {|e(i)} N i=1. They are the numbers we obtain when we find the eigenvectors of the matrix H; the eigenvectors are x(i) ={e(1) | x(i), e(2) | x(i), ..., e(N) | x(i) }. The matrix U is (see Eq. 150) ⎛ U = ⎜ ⎝ e(1) | x(1) e(1) | x(2) ··· e(1) | x(N) e(2) | x(1) e(2) | x(2) ··· e(2) | x(N) e(3) | x(1) e(3) | x(2) ··· e(3) | x(N) . . .. . . e(N) | x(1) e(N) | x(2) ··· e(N) | x(N) ⎞ ⎟ ⎠ (152)
7. The Eigenvalue Problem, December 17, 2009 29 Remember that if U ij is a matrix element, the first index (here, i) labelsthe rows and the second index labels the columns. A simple way to construct U accordingtoEq.152istotakethefirst eigenvector of Ĥ and use it as the first column in U, then use the second eigenvector as the second column of U, etc. To construct U, we need to solve the eigenvalue problem for H. Because of this, constructing U is not a shortcut for finding the eigenvectors (from Eq. 145) or the eigenvalues (from Eq. 143). The theorem is however very useful for simplifying some equations and as an intermediate step in some mathematical proofs. In Section 4, Cells 6—7, of the file linear algebra for quantum mechanics.nb, I give an example of the construction of a unitary matrix U that diagonalizes a Hermitian matrix M. In the following displays, the numbers have been rounded to two significant digits. ⎛ M = ⎜ ⎝ −1.1 2.3+0.022i −0.67 + 3.0i −3.5+5.0i 2.3 − 0.022i −2.0 4.2+3.9i −1.1+3.1i −0.67 − 3.0i 4.2 − 3.9i 0.27 −2.5+4.6i −3.5 − 5.0i −1.1 − 3.1i −2.5 − 4.6i 1.2 ⎞ ⎟ ⎠ (153) The eigenvectors are (see Cell 7 of linear algebra for quantum mechanics.nb) x(1) = {−0.33 + 0.26i, −0.26 + 0.34i, −0.024 + 0.53i, 0.60} (154) x(2) = {−0.25 + 0.42i, 0.40 + 0.18i, −0.50 + 0.26i, −0.50} (155) x(3) = {−0.026 − 0.50i, 0.74 + 0.024i, −0.16 + 0.11i, 0.41} (156) x(4) = {−0.18 − 0.54i, −0.12 + 0.26i, 0.36 + 0.49i, −0.47} (157) Construct U by using the eigenvectors as columns: ⎛ ⎞ −0.33 + 0.26i −0.25 + 0.42i −0.026 − 0.50i −0.18 − 0.54i −0.26 + 0.34i 0.40 + 0.18i 0.74 + 0.024i −0.12 + 0.26i U = ⎜ ⎟ ⎝ −0.024 + 0.53i −0.50 + 0.26i −0.16 + 0.11 i 0.36 + 0.49 i ⎠ 0.60 −0.50 0.41 −0.47 (158) We calculate that (see linear algebra for quantum mechanics.nb) ⎛ ⎞ 11.63 0 0 0 U −1 0 −9.87 0 0 MU = ⎜ ⎟ (159) ⎝ 0 0 −4.16 0 ⎠ 0 0 0 0.77
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Chapter 7. The Eigenvalue Problem

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