Quick question about hydrogen atom perturbation

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Homework Help Overview

The discussion revolves around the perturbation of a hydrogen atom, specifically focusing on the implications of orbital angular momentum in the z-direction being set to zero. The context involves quantum mechanics and the properties of atomic states.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind setting the z-component of orbital angular momentum to zero and its implications for the state of the atom. Questions arise about the quantum numbers associated with angular momentum and the relationship between different atomic states.

Discussion Status

The discussion is active, with participants questioning assumptions and clarifying concepts related to angular momentum. Some guidance has been offered regarding the interpretation of the problem and the significance of the final state in the context of the probability calculation.

Contextual Notes

There is a mention of specific quantum states and their properties, as well as a reference to external resources for further clarification. The original poster expresses uncertainty about the implications of their reasoning.

davon806
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Homework Statement


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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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