3502_07_sept23_LG

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

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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) &amp;lt; 1 | 4 &amp;gt; Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (b) &amp;lt; 2 | 4 &amp;gt; Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (c) &amp;lt; 2 | 3 &amp;gt; 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) &amp;lt; 3 | 4 &amp;gt; Equals zero, because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (e) &amp;lt; 1 | A | 3 &amp;gt; Equals zero. &amp;lt; 1 | A | 3 &amp;gt; = &amp;lt; 1 | b 3 &amp;gt; = b &amp;lt; 1 | 3 &amp;gt; and we know &amp;lt; 1 | 3 &amp;gt; = 0 because non-degenerate eigenfunctions of a Hermitian operator are orthogonal. (f) &amp;lt; 2 | A | 3 &amp;gt; May or may not be zero. &amp;lt; 2 | A | 3 &amp;gt; = &amp;lt; 2 | b 3 &amp;gt; = b &amp;lt; 2 | 3 &amp;gt; and we know from (c) above that this overlap integral need not be zero. (g) &amp;lt; 1 | Z | 4 &amp;gt; 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, &amp;lt; 1 | Z | 4 &amp;gt; = &amp;lt; 1 | d 4 &amp;gt; = d &amp;lt; 1 | 4 &amp;gt; where d is an eigenvalue of Z and we know from (a) above that this overlap integral is zero. (h) &amp;lt; 2 | Z | 3 &amp;gt; May or may not be zero. Continuing from our above logic, &amp;lt; 2 | Z | 3 &amp;gt; = &amp;lt; 2 | e 3 &amp;gt; = e &amp;lt; 2 | 3 &amp;gt; and we know from (c) above that this overlap integral need not be zero. (i) &amp;lt; 2 + 3 | Z | 2 3 &amp;gt; May or may not be zero. This one is a bit tricky. If we expand the integral, we get &amp;lt; 2 | Z | 2 &amp;gt; &amp;lt; 2 | Z | 3 &amp;gt; + &amp;lt; 3 | Z | 2 &amp;gt; &amp;lt; 3 | Z | 3 &amp;gt;. 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, &amp;lt; 2 | Z | 3 &amp;gt; = &amp;lt; Z 2 | 3 &amp;gt; = f &amp;lt; 2 | 3 &amp;gt; where f is the appropriate eigenvalue for Z . But, we already showed in (h) above that &amp;lt; 2 | Z | 3 &amp;gt; = e &amp;lt; 2 | 3 &amp;gt;, 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 &amp;lt; 2 | 2 &amp;gt; e &amp;lt; 2 | 3 &amp;gt; + e &amp;lt; 3 | 2 &amp;gt; e &amp;lt; 3 | 3 &amp;gt;. 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 &amp;lt; 2 | 3 &amp;gt; = c . Then &amp;lt; 3 | 2 &amp;gt; = &amp;lt; 2 | 3 &amp;gt;* =...
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3502_07_sept23_LG - Chem 3502/4502 Physical Chemistry...

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