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Unformatted text preview: PH 216, Problem set 5, Due 5/4 1. In class we derived σ abs = 4 π 2 αω ni h n  ~x  i i · ˆ  2 δ ( ω ω ni ) , for a plane EM wave of frequency ω with linear polarization direction ˆ incident upon a bound electron in state  i i . (a) Define the oscillator strength f ni ≡ 2 m e ω ni ~ h n  ~x  i i · ˆ  2 . Show that X n f ni = 1 . (b) Show that Z dωσ abs ( ω ) = 2 π 2 c e 2 m e c 2 . Under what conditions is this valid? Solution: (a) We start with the definition of f ni : f ni ≡ 2 m e ω ni ~ h n  ~x  i i · ˆ  2 . Using ω ni ≡ E n E i ~ , and letting ~x · ˆ = x , f ni simplies to f ni = 2 m e ( E n E i ) ~ 2 h n  x  i i 2 . Summing over n gives X n f ni = 2 m e ~ 2 X n ( E n E i ) h n  x  i i 2 = 2 m e ~ 2 X n ( E n E i ) h i  x  n ih n  x  i i = m e ~ 2 X n ( E n E i ) ( E i E n ) h i  x  n ih n  x  i i . 1 Using h n  [ H,x ]  i i = ( E n E i ) h n  x  i i and h i  [ H,x ]  n i = ( E n E i ) h i  x  n i , we get X n f ni = 1 i ~ X n h i  x  n ih n  im ~ [ H,x ]  i i  h i  im ~ [ H,x ]  n ih n  x  i i , but we know that p x = im ~ [ H,x ] , so = 1 i ~ X n h i  x  n ih n  p x  i i  h i  p x  n ih n  x  i i , = 1 i ~ h i  [ x,p x ]  i i = h i  i i = 1 . Kaching! (b) The scattering cross section can be rewritten in terms of f ni as σ abs = 4 π 2 α ~ 2 m e f ni δ ( ω ω ni ) = 2 π 2 α ~ m e f ni δ ( ω ω ni ) . Performing the integration over ω gives Z dωσ abs ( ω ) = 2 π 2 α ~ m e Z dωf ni ( ω ni ) δ ( ω ω ni ) . The integral is simply Z dωf ni δ ( ω ω ni ) = X n f ni = 1 , and thus Z dωσ abs ( ω ) = 2 π 2 α ~ m e = 2 π 2 c e 2 m e c 2 . 2. This problem provides a crude model for the photoelectric effect. Consider the hydro gen atom in its ground state (you may neglect the spins of the electron and proton). At time t = 0, the atom is placed in a high frequency uniform electric field that points in the zdirection, ~ E ( t ) = E ˆ z sin ωt. We wish to compute the transition probability per unit time that an electron is ejected into a solid angle lying between Ω and Ω + d Ω. (a) Determine the minimum frequency, ω , of the field necessary to ionize the atom. 2 (b) Using Fermi’s golden rule for the transition rate at firstorder in timedependent perturbation theory, obtain an expression for the transition rate per unit solid angle as a function of the polar angle θ of the ejected electron (measured with respect to the direction of the electric field). HINT: The matrix element that appears in Fermi’s golden rule describes a transi tion of the negativeenergy bound electron in its ground state to a positiveenergy “free” electron. The wave function of the latter is actually quite complicated, since one cannot really neglect the effects of the longrange Coulomb potential....
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This document was uploaded on 10/31/2011.
 Spring '09
 Polarization

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