3502_07_sept23_LG

3502_07_sept23_LG - Chem 3502/4502 Physical Chemistry II...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 7, September 23, 2009 Solved Homework We are given that A is a Hermitian operator such that A 1 = a 1 , A 2 = b 2 , A 3 = b 3 , and A 4 = c 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) < 1 | 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (b) < 2 | 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (c) < 2 | 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) < 3 | 4 > Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (e) < 1 | A | 3 > Equals zero. < 1 | A | 3 > = < 1 | b 3 > = b < 1 | 3 > and we know < 1 | 3 > = 0 because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (f) < 2 | A | 3 > May or may not be zero. < 2 | A | 3 > = < 2 | b 3 > = b < 2 | 3 > and we know from (c) above that this overlap integral need not be zero. (g) < 1 | Z | 4 > Equals zero. If the commutator of two operators is zero, then they share common eigenfunctions and we were told that the functions belong to this set. Thus, < 1 | Z | 4 > = < 1 | d 4 > = d < 1 | 4 > where d is an eigenvalue of Z and we know from (a) above that this overlap integral is zero. (h) < 2 | Z | 3 > May or may not be zero. Continuing from our above logic, < 2 | Z | 3 > = < 2 | e 3 > = e < 2 | 3 > and we know from (c) above that this overlap integral need not be zero. (i) < 2 + 3 | Z | 2 3 > May or may not be zero. This one is a bit tricky. If we expand the integral, we get < 2 | Z | 2 > < 2 | Z | 3 > + < 3 | Z | 2 > < 3 | Z | 3 >. The first question is, just because 2 and 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, < 2 | Z | 3 > = < Z 2 | 3 > = f < 2 | 3 > where f is the appropriate eigenvalue for Z . But, we already showed in (h) above that < 2 | Z | 3 > = e < 2 | 3 >, so e must be equal to f . 7-2 So, if we pull all of the eigenvalues out from our sum of 4 integrals we have e < 2 | 2 > e < 2 | 3 > + e < 3 | 2 > e < 3 | 3 >. Assuming that 2 and 3 are normalized, the first and last terms cancel as e e . However, since 2 and 3 are degenerate, they are not necessarily orthogonal. Let us say that the overlap integral < 2 | 3 > = c . Then < 3 | 2 > = < 2 | 3 >* =...
View Full Document

Page1 / 9

3502_07_sept23_LG - Chem 3502/4502 Physical Chemistry II...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online