C566lecture2_000

# C566lecture2_000 - 1 Lecture 2 C566 Spring 2011 Rotational...

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Unformatted text preview: 1 Lecture 2 C566 Spring 2011 Rotational spectra of molecules- See Chpt. 4 of Hollas We will start with diatomic molecules, then work our way up to polyatomics (Hollas dives right in to treat all molecules in the first section. Additional reference: G. Herzberg Molecular Spectra and Molecular Structure Vol. I. Diatomic Molecules Approximate treatment of molecules H = H el + H vib + H rot + H nuclear spin + … ψ = ψ el ⋅ψ vib ⋅ψ rot ⋅ψ nuclear spin ⋅ … = E el + E vib + E rot + E nuc spin + … ψ rot is approximated as a rigid rotor. Again, we shift from the 6 degrees of freedom to separate out translation of the diatomic (in center-of-mass coordinates, X = M-1 ⋅∑ m i x i , etc. for Y,Z, M = m 1 + m 2 ) and rotational motion about the center of mass. R is fixed, reduced mass is µ = m 1 m 2 /M. m 1 r 1 = m 2 r 2 shows where the center of mass is along the internuclear bond. For heavy- light, the COM is close to the heavy center, and the reduced mass is close to the light mass. For a homonuclear, the COM is in the center, and the reduced mass = m/2. The Hamiltonian operator now must include a full three dimensional treatment, but with no variation in R. V(r, θ , φ ) = 0, so the only operator is the T operator. m 1 m 2 R r 1 r 2 QC 451 .H53 z y x µ R φ θ 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ − = θ θ θ θ φ θ µ µ sin sin 1 sin 1 1 1 2 2 ˆ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 R R R R z y x H h h R= constant, so all derivatives with respect to R are zero. µ R 2 = I, the moment of inertia. First, we assume that our wavefunction is separable into theta and phi variables. The next step is to divide through by ΘΦ and − h 2 /2I. ) ( ) ( ) , ( ) , ( sin sin 1 sin 1 2 ) , ( ˆ 2 2 2 2 φ θ φ θ ψ φ θ ψ θ θ θ θ φ θ φ θ ψ Φ Θ = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ + ∂ ∂ − = E E I H h 2 2 2 2 2 2 2 ) sin (cos sin 1 sin 1 h IE − = ∂ Θ ∂ + ∂ Θ ∂ Θ + ∂ Φ ∂ Φ θ θ θ θ θ φ θ Multiplying by sin 2 θ , then rearranging (bring over E term, set whole expression equal to 0 = M J 2 − M J 2 ) you prove separability, and show that the phi-dependent portion of the...
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C566lecture2_000 - 1 Lecture 2 C566 Spring 2011 Rotational...

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