notes08 - 5.73 Lecture #8 Rydberg Klein Rees 8-1 L ast...

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5.73 Lecture #8 Rydberg Klein Rees 8 - 1 Last time: WKB quantization condition for bound eigenstates of almost general V(x) — Connection into bound region from left and right x + (E) p E (x)dx ′ = x h 2 (n + 1/ 2) p E (x) = [ 2m ( E V(x) ) ] 1/2 E n without ψ n ! timing of w.p. as But where do we get V(x)? it moves on V(x)± Certainly not from femtochemistry!± From FREQUENCY DOMAIN SPECTROSCOPY± E v,J V(x) RKR method Next time: Numerical Integration of 1-D Schr. Eq. — see handouts Then begin working toward matrix picture Need background in Ch.2 of CTDL pages 94-121 soon, pages 121-144 by next week Postulates and theorems not to be covered except as needed for solving problems. Today: E v,J spectroscopic notation A(E,J) = E V(x) x e x dV Equilibrium: =→ x e 0 dx A A 1 1 E , J x + x (E) and x + x A A WKB QC applied to E , J Gv ( ),B(v) used to determine x ± (E). Long Range Theory: Ultra Cold Collisions: Atom in Molecule updated 9/18/02 8:57 AM
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5.73 Lecture #8 Rydberg Klein Rees 8 - 2 Someday you will discover that the energy levels of a diatomic molecule are given by E evJ /hc = T e + G(v) + F v (J) cm -1 electronic vibration rotation [ = ν e + Y 00 + ω e (v +1 / 2 ) e x e / 2 ) 2 +… + B e α e / 2 ) [ ] G(v) ] J(J +1 ) DJ 2 (J ) 2 B(v) RKR requires only G(v) and B(v) to get V J (x) where V J (x) = U(x) + h 2 ) J-dependent bare x 2 effective potential centrifugal barrier potential (actually rotational kinetic energy) x R R e mm 12 µ= m 1 + m 2 We are going to derive V 0 (x) directly from G(v), B(v) data. This is the only direct spectrum to potential inversion method! WKB quantization is the basis for this. xE v ) + ( x p E () dx ′ = (h /2)(v + 1/2) v = 0,1, # of nodes v ) v ( In this equation, what we know (E v ) and what we want (V(x) and x at turning points) are hopelessly mixed up. There is a trick! A(E,J) x + (E,J) [ E± V J (x) ] dx x x (E) x + (E) area at E updated 9/18/02 8:57 AM
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5.73 Lecture #8 Rydberg Klein Rees 8 - 3 but, still, we know neither V J (x) nor x ± (E,J)! ! A A integrals (via WKB QC) involving only data input Roadmap: 1 . Show that E and J are numerically evaluable E v,J info here A A [ 2 . independently, E and J determine 2 eqs. in 2 1 1 unknowns give x + (E,J) x (E,J) ] and x + x turning points Do #2 first because it is so easy A = x + E–U( E E x x) h 2 J(J +1 ) dx ′ x 2 = x + 1 dx + x E are zero because integrand is 0 at turning points + contributions from x ± (E, J) 0 0 A E = x + x ! A = x + J J x h 2 ) dx x 2 =− h 2 x + 2 J +1 0 + 0 x x 2 dx ′ + 123 integrand = 0 at x ± A =+ h 2 ( 2 J ) 1 1 J x + x So, if we can evaluate these derivatives from E vJ data, we have V J (x)!
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notes08 - 5.73 Lecture #8 Rydberg Klein Rees 8-1 L ast...

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