# hw6solutions - HOMEWORK ASSIGNMENT 6 PHYS852 Quantum...

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HOMEWORK ASSIGNMENT 6 PHYS852 Quantum Mechanics II, Spring 2008 Topics covered: Applications of time-dependent perturbation theory, Fermi-Golden rule Please feel free to keep only the term closest to resonance when computing amplitudes or probabilities. 1. AC Stark Effect : Consider a two level atom with bare Hamiltonian H 0 = E e | e )( e | + E g | g )( g | . We want to drive this atom with an oscillating electric field, vector E ( vector r,t ) = E 0 cos( ωt ) vectore z . The interaction between the atom and the field is the usual V = vector d · vector E ( t ). Here the dipole moment operator is of course vector d = e vector R . a.) Find the expression for the perturbation operator as a two-by-two matrix in the basis {| e ) , | g )} . Compute the matrix elements explicitly for the hydrogen atom transition 1S to 2S, and also for the 1S to 2P transitions. What can you say about the ability to drive the 1S to 2S transition via the electric-dipole interaction? Answer : V = eE 0 Z cos( ωt ) eE 0 cos( ωt ) parenleftbigg ( e | Z | e ) ( e | Z | g ) ( g | Z | e ) ( g | Z | g ) parenrightbigg For all transitions we have ( e | Z | e ) = ( g | Z | g ) = 0 due to parity. We also have ( g | Z | e ) = ( e | Z | g ) . So only ( e | Z | g ) needs to be computed for each transition. For the 1S to 2S transitions we have | e ) = | 200 ) and | g ) = | 100 ) , thus we find ( 200 | Z | 100 ) = 1 2 a 3 0 integraldisplay 0 drr 3 parenleftbigg 1 r 2 a 0 parenrightbigg e r/a 0 integraldisplay π 0 sin θ cos θ = 0 So we see that the 1S to 2S electric dipole transition if forbidden. For the 1S to 2P we have | e ) = | 210 ) and | g ) = | 100 ) . We have chosen m = 0 for the 2P level, because V has azimuthal symmetry so that m is conserved. ( 210 | Z | 100 ) = 1 2 2 a 4 0 integraldisplay 0 drr 4 e r/a 0 integraldisplay 0 sin θ cos 2 θ = a 0 2 2 integraldisplay 0 dρρ 4 e ρ integraldisplay 1 1 duu 2 = 4 2 a 0 This gives V = 4 2 ea 0 E 0 cos( ωt ) parenleftbigg 0 1 1 0 parenrightbigg . 1

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b.) Consider the 1S and 2P levels as | g ) and | e ) , respectively. If the atom starts in state | g ) , use perturbation theory to compute is the probability as a function of time to make the transition to state | e ) . Your answer should be valid up to second-order in the probabilities. Answer : Now P g e ( t ) = | c (1) e ( t ) | 2 So we need c (1) e = i planckover2pi1 integraldisplay t 0 dt 1 V eg ( t 1 ) e a t 1 , where ω a = ( E e E g ) / planckover2pi1 = 1 . 64 × 10 16 s 1 . Inserting the matrix element V eg ( t 1 ) gives c (1) e ( t 1 ) = i 4 2 ea 0 E 0 planckover2pi1 integraldisplay t 0 dt 1 cos( ωt 1 ) e a t 1 . Keeping only the near-resonant term gives c (1) e ( t 1 ) = i 2 2 ea 0 E 0 planckover2pi1 integraldisplay t 0 dt 1 e i ( ω ω a ) t 1 = 2 2 ea 0 E 0 planckover2pi1 bracketleftbig e i ( ω ω a ) t 1 bracketrightbig ( ω ω a ) = 4 i 2 ea 0 E 0 planckover2pi1 e i ( ω ω a ) t/ 2 sin (( ω ω a ) t ) ω ω a Thus we find P g e ( t ) = | c (1) e ( t ) | 2 = d 2 | E 0 | 2 planckover2pi1 2 sin 2 (( ω ω a ) t ) ( ω ω a ) 2 , where d = 4 2 ea 0 is the 1S to 2P dipole moment.
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