hw solutions 3 - Physics 519 Homework Set#3 Due in class...

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Physics 519 Homework Set #3 Spring 2013 Due in class 4/24/13 300 pts 1. (100 pts) Sakurai 5.35. Simplified photo-electric effect. You do not need to repeat steps worked out in Sakurai (and also in lectures) to obtain a general formula for the transition rate. The point here is to calculate the resulting matrix elements and obtain a fully explicit expression for the rate and angular distribution. The ground state of a hydrogen atom ( n = 1, l = 0) is subjected to a time-dependent potential as follows: V ( x, t ) = V 0 cos( kz - ωt ) . Using time-dependent perturbation theory, obtain an expression for the transi- tion rate at which the electron is emitted with momentum p . Show, in particular, how you may compute the angular distribution of the ejected electron (in terms of θ and φ defined with respect to the z -axis). Discuss briefly the similarities and the differences between this problem and the (more realistic) photo-electric effect. Note: The initial wave function is given by ψ n =1 ,l =0 ( x ) = 1 π 1 a 0 3 / 2 e - r/a 0 . The final wave function, at least in a very large box of size L , can be approxi- mated to be ψ f = 1 L 3 / 2 e ip · x/ ¯ h . incoming light is ignored by considering the perturbation V ( x, t ) = V 0 2 e i ( kz - ωt ) + e - i ( kz - ωt ) . The e - iωt term leads to absorption of energy by the atom and thus (if the energy exceeds the ionization threshold) to ejection of a pho- toelectron. Following Sakurai’s development or our class notes, one finds, us- ing Fermi’s Golden rule (FGR), that the transition rate into electrons with momentum with solid angle d Ω of p f is dw d Ω = 2 π ¯ h F GR mp f ¯ h 3 L 2 π 3 density of states p f | ( V 0 / 2) e ikz | i 2 matrix element . The required matrix element is p f | e ikz | i = 1 L 3 / 2 d 3 xe - ip f · x/ ¯ h e ikz 1 π 1 a 0 3 / 2 e - r/a 0 = 1 π ( a 0 L ) 3 / 2 d 3 xe - iq · x e - r/a 0 , 1
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where q = p f / ¯ h - k ˆ z . In other words, we need the Fourier transform of the H-atom wave- function. We know the Fourier transform can depend only on q 2 = q · q because the wavefunction is rotationally invariant. Let us pick the z -axis to be along q for the purpose of the integration. Then d 3 xe - iq · x e - r/a 0 = 2 π - 1 1 dc 0 r 2 dre - iqrc e - r/a 0 = 2 π - iq 0 r 2 dr ( e - iqr - e iqr ) e - r/a 0 = 2 π - iq 1 ( iq + 1 /a 0 ) 2 - 1 ( - iq + 1 /a 0 ) 2 = 2 π - iq 4( - iq )1 /a 0 ( q 2 + 1 /a 2 0 ) 2 = 8 π/a 0 ( q 2 + 1 /a 2 0 ) 2 .
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