2003AugQualQUANT2 - Ben Sauerwine Practice for Qualifying...

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Ben Sauerwine Practice for Qualifying Exams Problem Source: CMU August 2003 Qualifying Exam Consider transitions from excited states of a hydrogen atom (2S and 2P) into the ground state (1S). The 2S level is slightly higher in energy than the 2P level ( ) eV E E P S 6 2 2 10 4 × = due to a hyperfine splitting known as the Lamb shift. Denote the Hamiltonian of an isolated atom by . For the sake of clarity, take as the 2S, 2P and 1S states, respectively: 0 H 2 1 , 0 , 0 , 1 2 1 , 0 , 1 , 2 2 1 , 0 , 0 , 2 = = = = = = = = = = = = = = = S S S m m l n m m l n m m l n γ β α (a) An atomic orbital couples to electromagnetic radiation fields via ... + + = DM DE W W W The relevant operators are given by , where i W t Z eE W DE ω cos ˆ = Z ˆ is the position operator, and ( ) t S L B m e W x x DM cos ˆ 2 ˆ 2 + = . For each explain clearly why the matrix element i W αγ i W does or does not vanish. State briefly why the 2S level of hydrogen is long-lived (its lifetime is almost a second) and describe the dominant decay mechanism by which 2S decays to 1S. The dipole transition between and , or the 2S and 1S states, is not allowed due to the parity: These are in the same angular-momentum states, but the Z ˆ operator is odd parity and so must switch parity. The magnetic transition also vanishes: Consider individually and x L ˆ () + + = S S S x ˆ ˆ 2 1 ˆ . Since both wave functions are spherically symmetric, is zero. The spin operator will certainly swap the electron spin, but leave the still-orthogonal angular momentum portion untouched. x L ˆ This is a long-lived state because the dominant means of decay would be to decay to 2P and then decaying to 1S.
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2003AugQualQUANT2 - Ben Sauerwine Practice for Qualifying...

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