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GR8677 #48 |
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Problem
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Atomic }Transition
Transitions go from state one to another state that's completely unique relative to the first state. If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. This is forbidden.
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Alternate Solutions |
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Comments |
thebigshow500
2008-10-13 22:12:48 | Can anyone explain if other choices are something to do with wave function? (e.g. parity, orthogonality, etc.) Or are these choices just simply nonsense? :( |  | lattes
2008-10-13 09:18:54 | it might be useful to make an analogy with electromagnetism. It's known that a pulsating sphere does not emit radiation because of its spherical symmetry. An analogy can be made with wave-function. (D). For further reading check out Beiser's book on modern physics pag. 165. | | curie
2008-10-08 05:55:58 | Why does spherically-symmetric imply same angular momentum?
rkraman
2008-10-15 14:35:02 |
This has to do with the symmetry properties in mechanics. If the Lagrangian (and hence the Hamiltonian) describing a system is symmetrical with respect to any coordinate, then the canonical momentum corresponding to that coordinate is conserved. So if there is translation symmetry in  , the the canonical momentum which is the linear momentum  is conserved. Similarly if the Hamiltonian is rotationally invariant, it is symmetrical with respect to  , then the corresponding angular momentum  , which is nothing but the angular momentum  , is conserved. This is a specific instance of a fundamental theorem in Physics called the Noether's theorem.
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|  | evanb
2008-06-20 11:56:30 | "completely unique relative..."
I'm not sure exactly what that means, but if it means that m must also change that's not quite right.
The selection rules are that NO TRANSITION OCCURS UNLESS:
 m = {-1, 0, +1} and
l = {-1 , +1}.
 has ore options. | |
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