hw210a - Physics 512 Homework Set #10 Solutions Winter 2003...

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Physics 512 Winter 2003 Homework Set #10 – Solutions 1. This is based on Sakurai, Chapter 5, Problem 28. A hydrogen atom is initially in its ground state (1 s ). At time t = 0 we turn on a spatially uniform electric Feld as follows: ~ E ( t )= n 0 t< 0 E 0 e t/τ ˆ zt 0 a ) Using Frst-order time dependent perturbation theory, compute the probability for the atom to be found in each of the three 2 p states at time t ± τ . You need not evaluate the radial integrals, but perform all other integrations. This electric feld corresponds to a potential V = e E 0 ze t/τ Thus we calculate A k s = i ¯ h Z t 0 e ks t 0 h k ( e ) E 0 t 0 | s i dt 0 = ie E 0 ¯ h Z t 0 e ( ks 1 ) t 0 h k | z | s i dt 0 = ie E 0 ¯ h e ( ks 1 ) t 1 ks 1 h k | z | s i The transition probability is given by P k s = | A k s | 2 = e 2 E 2 0 ¯ h 2 1 2 e t/τ cos ω t + e 2 t/τ ω 2 +1 2 |h k | z | s i| 2 For t ± τ , the decaying exponentials may be dropped, and we fnd P k s ( t ± τ e 2 E 2 0 ¯ h 2 |h k | z | s i| 2 ω 2 2 (1) which is independent o± time t . We now note that this matrix element corresponds to a dipole transition, with corresponding selection rule ` = ² 1 . Furthermore, in terms o± spherical tensors, z = r cos θ = q 4 π 3 rY 0 1 ( θ, φ ) , which is an m =0 component o± a spherical vector. As a result, we ±urthermore have a selection rule δm . So the only possible transition ±rom the 1 s state is to the m =02 p state, ie | 100 i→| 210 i . In this case, the relevant hydrogen wave±unctions are ψ 100 = ± 1 πa 3 0 ² 1 / 2 e r/a 0 ψ 210 = ± 1 2 3 0 ² 1 / 2 r 4 a 0 e r/ 2 a 0 cos θ 1
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Thus h 210 | z | 100 i = 1 2 πa 3 0 Z ± r 4 a 0 e r/ 2 a 0 cos θ ² r cos θ ± e r/a 0 ² r 2 sin θdrdθdφ = 2 π 4 2 4 0 Z 0 r 4 e 3 2 a 0 dr Z π 0 cos 2 θ sin θdθ = 1 3 2 a 4 0 Z 0 r 4 e 3 2 a 0 dr Although the problem does not require it, this radial integral is not too difficult to evaluate. We fnd h 210 | z | 100 i = 128 2 243 a 0 On dimensional grounds, this is a reasonable result, since h z i has to give a length, and the Bohr radius a 0 is what governs the size oF the hydrogen atom. This may now be substituted into (1) to obtain the transition probability P | 210 i←| 100 i ( t ± τ )= ³ 8 9 ´ 5 e 2 E 2 0 a 2 0 ¯ h 2 1 ω 2 ks +1 2 where ¯ = E 210 E 100 = E 0 / 2 2 + E 0 = 3 4 E 0 (with E 0 =13 . 6eV ). b ) What would happen if instead the atom was in the 2 s state to begin with? Since we have the selection rules δ` = ² 1 and δm =0 , transitions to the 1 s state would be Forbidden. However it would still be possible to end up in the 2 p | 210 i state. Note, however, that in this case, ω 0 , so the probability would behave as P ( t ± τ ) e 2 E 2 0 τ 2 ¯ h 2 |h 210 | z | 200 i| 2 where a calculation indicates h 210 | z | 200 i = 3 a 0 . Note, however, that this ex- pression may be large For large τ , and thus must be handled with care. This mixing between the 2 s and 2 p state is similar to that which happens For the linear Stark e±ect, where degenerate time independent perturbation theory was required.
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.

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hw210a - Physics 512 Homework Set #10 Solutions Winter 2003...

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