Quick question about hydrogen atom perturbation

davon806
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Homework Statement


a.jpg

I have already solved the problem, but I don't really understand why the orbital angular momentum in the z-direction has to be taken to 0 ?

Homework Equations

The Attempt at a Solution


Suppose the component of orbital angular momentum in the z-direction is non-zero, that means there are extra angular momentum apart from l = 1 in the 2p state ? So when excitation occurs,it is no longer a "2p" state? Just a guess :nb)
 
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What is the quantum number related to the projection of angular momentum?
 
DrClaude said:
What is the quantum number related to the projection of angular momentum?
Oh,do you mean that the problem asks you to compute the probability from 1s -> |210> because m = 0 in the z-direction? So basically the remaining two n = 2 states have nothing to do with this question?
 
davon806 said:
I have already solved the problem, but I don't really understand why the orbital angular momentum in the z-direction has to be taken to 0?

Suppose the component of orbital angular momentum in the z-direction is non-zero, that means there are extra angular momentum apart from l = 1 in the 2p state? So when excitation occurs,it is no longer a "2p" state? Just a guess :nb)
Check out http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/vecmod.html to see why your guess isn't correct.

The problem asked you to calculate the probability for a given final state. It could have specified a different final state and given you a different expression for the probability.
 
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