03s_anoscvibrot

# 03s_anoscvibrot - MIT OpenCourseWare http/ocw.mit.edu 5.80...

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MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Lecture # 3 Supplement Contents 1. Anharmonic Oscillator, Vibration-Rotation Interaction . . . . . . . . . . . . . . . . . . 1 2. Energy Levels of a Vibrating Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1. Anharmonic Oscillator, Vibration-Rotation Interaction The Hamiltonian is P 2 k J 2 H = R + x 2 + a x 3 + (1.1) 2 µ 2 2 µR 2 �� � �� � harmonic oscillator non rigid rotor First we must re-express R 1 2 in terms of a quantity whose matrix elements we know such as the displacement x . x = R R e x R = R e + x = R e 1 + (1.2) R e 2 1 1 x x R 2 R 2 e 1 2 R e + 3 R e power series expansion (1.3) where we truncate expansion after ( x/R e ) 2 . The Hamiltonian operator becomes P 2 k J 2 2 x 3 x 2 H = 2 µ R + 2 x 2 + a x 3 + 2 µR 2 e 1 R e + R e 2 (1.4) Let us choose a basis {| v � | J �} where | v is the harmonic oscillator basis and | J is the rigid rotor basis. The rotational matrix elements we need are J | J 2 | J = J ( J + 1) δ JJ (1.5) J | constant | J = δ JJ constant (1.6) J | vibr . coord . | J = δ JJ vibr . coord . (1.7) 2 and B e (in energy units) = J. (1.8) 2 µR e 2 1
5.76 Lecture # 3 Supplement Page 2 Thus the rotational expectation values of the Eq.(1) Hamiltonian become H P 2 + k x 2 + a x 3 + B e J ( J + 1) 2 x + 6 µB e x 2 (1.9) J | | J = 2 µ 2 1 R e 2 where 1 was replaced by 2 µB e . R 2 2 e Now we need some x matrix elements. Let k γ = = ω = (1.10) γ 1 / 2 v + 1 x v,v +1 = (1.11) 2 γ 1 x v,v 2 +2 = [( v + 1)( v + 2)] 1 / 2 (1.12) 2 γ v + 1 x 2 = 2 (1.13) v,v γ x 3 = ( v + 1)( v + 2)( v + 3) 1 / 2 (1.14) v,v +3 8 γ 3 3 / 2 v + 1 x v,v 3 +1 = 3 (1.15) 2 γ Remember that the harmonic oscillator part of our Hamiltonian is diagonal in the harmonic oscillator basis we have chosen. 1 3 2 6 µB e 2 vJ | H | v J = v + 2 δ JJ δ vv + ax vv δ jj + B e J ( J + 1) δ vv δ JJ R e x vv δ JJ + 2 x vv δ JJ (1.16) Note that the Hamiltonian matrix is completely diagonal in J. The remaining problem is how to arrange the H matrix now that we have two indices J and v . The Hamiltonian matrix is a super-matrix consisting of a v , v matrix of J , J matrices. However, since there are no matrix elements off-diagonal in J , it is convenient to alter our perspective and think of a set of single v , v matrices, one for each value of J . Thus the Hamiltonian matrix is given by v | H | v = v + 1 2 + B e J ( J + 1) + B e 2 γ 6 µ 2 v + 1 2 J ( J + 1) (1.17) 3 / 2 1 / 2 v + 1 2 B e v + 1 v | H | v + 1 = 3 a 2 γ R e 2 γ J ( J + 1) (1.18) 3 / 2 1 / 2 v | H | v 1 = 3 a 2 v γ 2 R B e e 2 v γ J ( J + 1) (18a) v | H | v + 2 = 3 γ B e 2 2 µ J ( J + 1)[( v + 1)( v + 2)] 1 / 2 (1.19) v | H | v 2 = 3 γ B e 2 2 µ J ( J + 1)[( v 1) v ] 1 / 2 (19a) 1 / 2 ( v + 1)( v + 2)( v + 3) v | H | v + 3 = a 8 γ 3 (1.20) v ( v 1)( v 2) 1 / 2 v | H | v 3 = a 8 γ 3 (20a)

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5.76 Lecture # 3
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