p137bp7sol

p137bp7sol - Physics 137B Fall 2007 Moore Problem Set 7...

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Unformatted text preview: Physics 137B, Fall 2007, Moore Problem Set 7 Solutions 1. (a) The applied perturbation is resonant when the driving frequency ω matches the natural frequency of the system, which in this case is the Bohr frequency for the 1 → 2 transition: ω = ω ba = 3 4 (13 . 6 eV) ~ = 3 4 13 . 6 eV 6 . 582 × 10- 16 eV · s = 1 . 55 × 10 16 s- 1 . (b) Let a denote the 1 s state and b the 2 s state, with wavefunctions ψ a ( r ) = ( πa 3 )- 1 / 2 e- r/a , ψ b ( r ) = (32 πa 3 )- 1 / 2 2- r a e- r/ 2 a , Consider a perturbation H ( t ) =- eEx sin( ωt ). In spherical coordinates x = r sin θ cos φ , so H ba ( t ) = h ψ b | - eEx sin( ωt ) | ψ a i =- eE sin( ωt )( πa 3 )- 1 / 2 (32 πa 3 )- 1 / 2 Z d 3 r r sin θ cos φ 2- r a e- r/ 2 a e- r/a =- eE 4 √ 2 πa 3 sin( ωt ) Z 2 π dφ cos φ Z π dθ sin 2 θ Z r dr r 3 2- r a e- 3 r/ 2 a = 0 , 1 since the φ integral vanishes. The first-order transition amplitude from 1 s to 2 s is proportional to H ba ( t ), so (to first order) this perturbation will not create 1 s → 2 s transitions. (This conclusion is immediate from the electric dipole selection rules, which de- mand that only Δ l = ± 1 transitions are allowed.) 2. For a periodic perturbation H ( t ) = ˆ H sin ωt we define A = 1 2 i ˆ H , so that H ( t ) = ˆ H sin ωt = ˆ H 1 2 i ( e iωt- e- iωt ) = Ae iωt + A † e- iωt . By Eq. (9.10), the first-order transition amplitudes c k ( t ) satisfy i ~ ˙ c k ( t ) = X l H kl ( t ) e iω kl t c l ( t ) = X l A kl e iωt + A † kl e- iωt e iω kl t c l ( t ) ....
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p137bp7sol - Physics 137B Fall 2007 Moore Problem Set 7...

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