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# hw212a - Physics 512 Homework Set#12 Solutions Winter 2003...

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Physics 512 Winter 2003 Homework Set #12 – Solutions 1. The differential cross section for the ejection of an electron with momentum k f by an incident photon of momentum k ( ω = c | k | ) and polarization ˆ (the photoelectric effect) may be written as d = α | k f | 2 πm ¯ L 3 | f | e ik · r p · ˆ | i | 2 where the matrix element refers to the initial (bound) and final (free plane wave) electron states. Derive this expression using the quantum theory of radiation (instead of the classical treatment shown in class). We begin by computing the transition rate for this process to occur. For the rate, Fermi’s golden rule for harmonic perturbations gives w = 2 π ¯ h | f ; 0 | V | i ; { n ( α ) k = 1 } | 2 ρ f ( E f ) Here we have used a notation where i and f denote the initial and final electron state, while the remaining pieces are a shorthand label for the photons in the Fock space. For absorption of a photon, we have V = e mc A (+) · p = e mc 2 π ¯ hc 2 1 L 3 / 2 k,α 1 ω a α ( k ) p · ˆ ( α ) e ik · r Evaluating the matrix element in the photon Fock space, we see that the lowering operator annihilates the initial photon, leaving the Fock vacuum (which is normalized so that 0 | 0 = 1 ). This selects out only one particular term in the sum (corresponding to the single initial state photon). As a result f ; 0 | V | i ; { n ( α ) k = 1 } = e mc 2 π ¯ hc 2 ωL 3 f | e ik · r p · ˆ | i Squaring this and using Fermi’s golden rule yields w = (2 π ) 2 e 2 m 2 ωL 3 | f | e ik · r p · ˆ | i | 2 = (2 π ) 2 e 2 m 2 ωL 3 m | k f | ¯ h 2 L 2 π 3 | f | e ik · r p · ˆ | i | 2 ρ f ( E f ) = αc | k f | 2 πm ¯ | f | e ik · r p · ˆ | i | 2 d where, for the final state, we have used the free electron density of states ρ ( E f ) = m | k f | ¯ h 2 L 2 π 3 d 1

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Finally, this may be converted to a differential cross section by dividing by the incident (single photon) ﬂux c/L 3 . This gives the desired result d = α | k f | 2 πm ¯ L 3 | f | e ik · r p · ˆ | i | 2 Note that this matrix element is only for the electron states. Moreover, the final ejected electron wavefunction must be normalized as ψ 1 /L 3 / 2 . Taking this into account,
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hw212a - Physics 512 Homework Set#12 Solutions Winter 2003...

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