3502_13_octo08

3502_13_octo08 - Chem 3502/5502 Physical Chemistry II...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 13, October 8, 2010 Solved Homework (Homework for grading is also due today) We are asked to demonstrate the orthogonality of the functions Φ ( φ ) that are the φ -dependent components of the spherical harmonics. We know that they are eigenfunctions of L z , and thus, since they are non-degenerate eigenfunctions of a Hermitian operator, that they must be orthogonal, but demonstrating this point explicitly serves as a check on our accuracy, if you like. So, the question is, given two eigenfunctions Φ characterized by different eigenvalues m l and m l ´, is it true that m l φ ( ) Φ m l φ ( ) = 0? The variable φ ranges from 0 to 2 π . So, we evaluate the above integral as Φ m l φ ( ) Φ m l φ ( ) = e im l φ ( ) * e i m l φ ( ) 0 2 π d φ = e im l φ ( ) e i m l φ ( ) 0 2 π d φ = e i m l m l ( ) φ 0 2 π d φ = 1 i m l m l ( ) e i m l m l ( ) φ 0 2 π = e 2 π i m l m l ( ) i m l m l ( ) 1 i m l m l ( ) We may use the relationship e 2 π i m l m l ( ) = cos 2 π m l m l ( ) [ ] + i sin 2 π m l m l ( ) [ ] to evaluate the first part of the solved integral. Since m l and m l ´ are integers, their difference is also an integer. The cosine of an integral multiple of 2 π is 1 and the sine of an integral multiple of 2 π is 0. Substituting this simplification into the eqs. above provides
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13-2 Φ m l φ ( ) Φ m l φ ( ) = e 2 π i m l m l ( ) i m l m l ( ) 1 i m l m l ( ) = 1 i m l m l ( ) 1 i m l m l ( ) = 0 Thus proving the orthogonality of the eigenfunctions. Note that if m l and m l ´ had been equal to one another, the original integrand would have simplified to e 0 d φ = d φ , the integral for which would be 2 π , and as we've already seen, this gives rise to the normalization constant (2 π ) 1/2 included in Φ . The Rigid Rotator We’ve previously considered two masses connected by a spring in solving the vibrational Schrödinger equation. The solutions were the quantum mechanical harmonic oscillator wave functions. Now, if we replace the spring with a solid rod (no vibration) and permit the system to rotate about an axis perpendicular to the rod, it will rotate about its center of mass. As we discussed in lecture 3, the kinetic energy for a rotating system is T = l 2 2 I (13-1) where l is the angular momentum and I is the moment of inertia. Although our original discussion considered only a single particle orbiting a fixed position, the moment of inertia generalizes to multiple particles as I = m i r i 2 i = 1 N (13-2) where there are N total particles each having a distinct mass m and distance from the center of mass r . When there are only two particles, one can show with some straightforward algebra that I = μ R 2 (13-3) where μ is the reduced mass defined by μ = m 1 m 2 m 1 + m 2 (13-4)
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13-3 and R is the length of the rigid rod connecting them. It would behoove you to memorize eqs. 13-3 and 13-4.
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3502_13_octo08 - Chem 3502/5502 Physical Chemistry II...

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