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ps6b - From these equations we can easily see that the...

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Unformatted text preview: From these equations we can easily see that the Lz—eigenstates are ¢i§(|1,0,0) :i: i|0, 1, 0)) and l0, 0, 1). Considering similar expressions for L1 and Ly it is easily checked that the three states are eigenstates of L2 with eigenvalue 2&2, which confirms that these are states with l = 1. Comparing with the derivation for the hydrogen atom, the only differences are the radial wave functions and the frequency of the emitted photon. Lets write the relevant wave functions explicitly. On the cartesian basis mom : <%)3/4e‘%(“2+y2+22) . (10) =WGG%YMV%fiamw an $00193 = (%)3/4 @— WWWAZZ) 27;“? (12) 2 571 :3; ($l5/4re*ifi”"21’lo<97¢> (13) It is straightforward to check that the same radial wave function appears in the other two I = 1 eigenstates. The radial integral appearing in the calculation of the decay rate is Ems [ms-l l—J (_)m-—] “‘0 =Li % (WE)2‘/0°° dr r4e”mhgrz (15) = same a)“ w = 2% (17) With this result and taking into account that in this case the frequency of the emitted photon is the frequency of the oscillator, we get for the decay rate £_2W2 d9 _- 27r 2172c2 1 (5m,o€3 + 55ml + 5m,—1)(€,i + 53)) (18) ...
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