Lec7 - Chem 3502/5502 Physical Chemistry II (Quantum...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 7, September 22, 2010 Solved Homework We are given that A is a Hermitian operator such that A φ 1 = a φ 1 , A φ 2 = b φ 2 , A φ 3 = b φ 3 , and A φ 4 = c φ 4 , a b c , and Z is some other operator for which [ A , Z ] = 0. We may then state whether the following integrals are definitely zero, or may be nonzero (a) < φ 1 | φ 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (b) < φ 2 | φ 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (c) < φ 2 | φ 3 > May or may not be zero. Degenerate eigenfunctions are not necessarily orthogonal, even though orthogonal functions having identical eigenvalues can be constructed from linear combinations of them. (d) < φ 3 | φ 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (e) < φ 1 | A | φ 3 > Equals zero. < φ 1 | A | φ 3 > = < φ 1 | b φ 3 > = b < φ 1 | φ 3 > and we know < φ 1 | φ 3 > = 0 because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (f) < φ 2 | A | φ 3 > May or may not be zero. < φ 2 | A | φ 3 > = < φ 2 | b φ 3 > = b < φ 2 | φ 3 > and we know from (c) above that this overlap integral need not be zero. (g) < φ 1 | Z | φ 4 > Equals zero. If the commutator of two operators is zero, then they share common eigenfunctions and we were told that the functions φ belong to this set. Thus, < φ 1 | Z | φ 4 > = < φ 1 | d φ 4 > = d < φ 1 | φ 4 > where d is an eigenvalue of Z and we know from (a) above that this overlap integral is zero. (h) < φ 2 | Z | φ 3 > May or may not be zero. Continuing from our above logic, < φ 2 | Z | φ 3 > = < φ 2 | e φ 3 > = e < φ 2 | φ 3 > and we know from (c) above that this overlap integral need not be zero. (i) < φ 2 + φ 3 | Z | φ 2 φ 3 > May or may not be zero. This one is a bit tricky. If we expand the integral, we get < φ 2 | Z | φ 2 > – < φ 2 | Z | φ 3 > + < φ 3 | Z | φ 2 > – < φ 3 | Z | φ 3 >. The first question is, just because φ 2 and φ 3 are degenerate for the Hamiltonian, are they necessarily degenerate for the operator Z ? The answer is, yes. We can prove this by noting that, by the turnover rule, < φ 2 | Z | φ 3 > = < Z φ 2 | φ 3 > = f < φ 2 | φ 3 > where f is the appropriate eigenvalue for Z . But, we already showed in (h) above that < φ 2 | Z | φ 3 > = e < φ 2 | φ 3 >, so e must be equal to f
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7-2 So, if we pull all of the eigenvalues out from our sum of 4 integrals we have e < φ 2 | φ 2 > – e < φ 2 | φ 3 > + e < φ 3 | φ 2 > – e < φ 3 | φ 3 >. Assuming that φ 2 and φ 3 are normalized, the first and last terms cancel as e e . However, since φ 2 and φ 3 are degenerate, they are not necessarily orthogonal. Let us say that the overlap integral < φ 2 | φ 3 > = c . Then < φ 3 | φ 2 > = < φ 2 | φ 3 >* = c
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This note was uploaded on 04/06/2011 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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Lec7 - Chem 3502/5502 Physical Chemistry II (Quantum...

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