Proof of allowed and forbidden electron state transition.

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Homework Statement


One way to establish which transitions are forbidden is to compute the expectation value of the electron’s position vector r using wave functions for both the initial and final states in the transition. That is, compute ∫ΨfrΨidτ where τ represents an integral over all space, and Ψf and Ψi are the final and initial states. If the value of the integral is zero, then the transition is forbidden.

Use this procedure to show that a transition from a L=1, mL=0 to a L=0 state is allowed.

Homework Equations


∫ΨfrΨidτ
R21(r)=Are^(-r/2a), A=1/(a^(5/2)2√6)
Y10(θ,φ)=1/2√(3/π)cosθ

The Attempt at a Solution


Just plug in values and solve. Easy!

But wait, I don't know what ψf is. The first state is the 2p state so I can find it's wave equation but the L=0 state has no other given quantum numbers.

I know that n>0, L<n and |mL|≤L so from what is given, the final state is n>0, L=0 and mL≤0.

So what do I do about the value of n and mL? How do I find the wave equation for the final state?
 
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If L = 0, then there is only one possible value for mL.
 
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Oops. |mL|≤L, L=0, mL=0

I believe this would be the 3s state. Time to solve the integral.
 
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Send BoBs said:
Oops. |mL|≤L, L=0, mL=0
Yes

I believe this would be the 3s state. Time to solve the integral.
Why would n = 3? The problem statement in the first post doesn't give any information about the initial or final value of n. I believe you only need to worry about the angular integrals over ##\theta## and ##\varphi##.
 
TSny said:
Yes

Why would n = 3? The problem statement in the first post doesn't give any information about the initial or final value of n. I believe you only need to worry about the angular integrals over ##\theta## and ##\varphi##.

I believe the integral over r is also important here since the integral is over 3D space. So r ##\theta## and ##\varphi## are all important. I was going into this with the assumption that ψ(r,##\theta##,##\varphi##)=Rnl(r)Ylml(##\theta##,##\varphi##) would be how I describe each states wave equation. Of course this requires me take the quantum number n from a given chart of number configurations and results in a rather lengthy integral that already makes me think my method is wrong.

But why would the radial wave function not play a part here?
 
Send BoBs said:
But why would the radial wave function not play a part here?
There is no particular reason why the integral over r would be zero, so you can take it to be non-zero. In contrast, the integral over θ, φ can be shown to be zero except for special cases.
 
I think I have a better understanding now.

If I can show that each r,##\theta##,##\varphi## integral ≠ 0 then the transition should be valid. I know that the radial wave integral will be some non zero value and the ##\varphi## integral would just be 2π so I should be able to compute the ##\theta## integral and have my answer.
 
Send BoBs said:
If I can show that each r,##\theta##,##\varphi## integral ≠ 0 then the transition should be valid. I know that the radial wave integral will be some non zero value and the ##\varphi## integral would just be 2π so I should be able to compute the ##\theta## integral and have my answer.
That sounds good.
 
Send BoBs said:
I think I have a better understanding now.

If I can show that each r,##\theta##,##\varphi## integral ≠ 0 then the transition should be valid. I know that the radial wave integral will be some non zero value and the ##\varphi## integral would just be 2π so I should be able to compute the ##\theta## integral and have my answer.
And don't forget that the spherical harmonics are orthonormal ...
 
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kuruman said:
And don't forget that the spherical harmonics are orthonormal ...
The integral includes ##\mathbf{r}##.
 
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  • #11
DrClaude said:
The integral includes ##\mathbf{r}##.
Yes, of course. Thanks for the reminder.
 
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