Qualitative question about the Stark effect

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The discussion centers on the Stark effect in Rydberg atoms, specifically addressing the number of sublevels for quantum states n=7 and n=8. Participants clarify that for these states, there are n² eigenfunctions, but only four sublevels remain due to selection rules (Δl=±1 and Δm=0). The conversation highlights the significance of angular momentum states L=7 and L=8, which are approximately degenerate and can mix under perturbation. The complexity of the mixing matrix elements and their implications on degeneracy are also emphasized.

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Clara Chung
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Homework Statement
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Relevant Equations
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242121

For e ii) The answer is
242122

Why are there only 4 sublevels?
We haven't learned about degenerate perturbation theory, the only thing mention in lecture is
242123

which I don't understand so I only memorize the good eigenfunctions for n=2. Could you explain why there are still only 4 sublevels for n=7 and 8?
 
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Clara Chung said:
which I don't understand so I only memorize the good eigenfunctions for n=2. Could you explain why there are still only 4 sublevels for n=7 and 8?
You do understand the Hydrogen atom Stark effect and not the Rydberg atom? (I'm not sure what you need here)...
 
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hutchphd said:
You do understand the Hydrogen atom Stark effect and not the Rydberg atom? (I'm not sure what you need here)...
For n=7 and n=8 there will be n^2 eigenfunction... if the states that will be mix are the states that satisfy the selection rules... i.e. delta l=+-1 and delta m=0, there will be so many... not only the two eigenstates in the hydrogen atom...?
 
Clara Chung said:
For n=7 and n=8 there will be n^2 eigenfunction... if the states that will be mix are the states that satisfy the selection rules... i.e. delta l=+-1 and delta m=0, there will be so many... not only the two eigenstates in the hydrogen atom...?
He is talking about the n=9 Rydberg states only. Within this are the allowed states L=0,1,...,8 each with multiplicity 2L+1. Only the highest angular momentum states are "far enough away" from the "nucleus/inner electron complex" to be approximately degenerate. So we are down to L=7,8. So how will this subset be affected by perturbation?
 
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hutchphd said:
He is talking about the n=9 Rydberg states only. Within this are the allowed states L=0,1,...,8 each with multiplicity 2L+1. Only the highest angular momentum states are "far enough away" from the "nucleus/inner electron complex" to be approximately degenerate. So we are down to L=7,8. So how will this subset be affected by perturbation?
If only L=7,8 are approximately degenerate... There are still many levels that can be mixed... Like l=7, ml=0,1,2,3,4,5,6,7 with l=8, ml=1,2,3,4,5,6,7...??
 
Clara Chung said:
If only L=7,8 are approximately degenerate... There are still many levels that can be mixed... Like l=7, ml=0,1,2,3,4,5,6,7 with l=8, ml=1,2,3,4,5,6,7...??
Yes all those states (let's be precise ml=0, ±1,..., ±L for L=7,8). But what about the (mixing) matrix elements. Are some zero? Are some the same size? This is not an easy exercise...but it is good. Clearly some levels will still be degenerate.
 
hutchphd said:
Yes all those states (let's be precise ml=0, ±1,..., ±L for L=7,8). But what about the (mixing) matrix elements. Are some zero? Are some the same size? This is not an easy exercise...but it is good. Clearly some levels will still be degenerate.
How to know whether they have the same size? :(
 
There are two states you know don't connect (create any off diagonal elements). What are they?

You say you have not had degenerate perturbation theory...why are you (we) looking at this problem? It is not obvious to me exactly how it simplifies completely as advertised.
 
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hutchphd said:
There are two states you know don't connect (create any off diagonal elements). What are they?

You say you have not had degenerate perturbation theory...why are you (we) looking at this problem? It is not obvious to me exactly how it simplifies completely as advertised.
Thank you for your help.. This appears in an exam of an introductory course on atomic physics... I guess I will leave it...
 
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Perhaps someone else on the Forum sees the route to the requested solution. Also you can keep this problem in your head and, as you learn more, re-examine the possible solutions. To have such mental touchstones is not a bad thing.
 

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