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Combining Fine Structure Corrections

  • #1

Homework Statement



We are to combine the Relativistic Kinetic Energy, Spin-Orbit Interaction, and Darwin fine structure correction terms into a single formula for the energy shift in the Hydrogen atom. The formula must depend only on j = l +/- 1/2, but not l, and must be valid for all l, including l = 0.

Homework Equations



The above corrections are given as:
https://mywebspace.wisc.edu/dpfahey/web/PF01.bmp [Broken]

The Attempt at a Solution



Well, [itex] \Delta {E}_{n,total} = \Delta {E}_{n,kin} + \Delta {E}_{n,so} + \Delta {E}_{n,D}[/itex]

Where [itex]<S \cdot L> = \left[j(j + 1) - l(l + 1) - s(s +1) \right ] [/itex]
So, presumably, we just add the given corrections, and collect/eliminate like terms. I began doing this until I became confused by stipulation of dependence on j only, and not l.

So my (simple) question is: If the formula will depend on j, and j depends on l, then how will the formula not depend on l?

Also, how will the resultant formula be good for all l, as one of the correction terms does not allow for l = 0?
 
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Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
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Homework Statement



We are to combine the Relativistic Kinetic Energy, Spin-Orbit Interaction, and Darwin fine structure correction terms into a single formula for the energy shift in the Hydrogen atom. The formula must depend only on j = l +/- 1/2, but not l, and must be valid for all l, including l = 0.

Homework Equations



The above corrections are given as:
https://mywebspace.wisc.edu/dpfahey/web/PF01.bmp [Broken]

The Attempt at a Solution



Well, [itex] \Delta {E}_{n,total} = \Delta {E}_{n,kin} + \Delta {E}_{n,so} + \Delta {E}_{n,D}[/itex]

Where [itex]<S \cdot L> = \left[j(j + 1) - l(l + 1) - s(s +1) \right ] [/itex]
So, presumably, we just add the given corrections, and collect/eliminate like terms. I began doing this until I became confused by stipulation of dependence on j only, and not l.

So my (simple) question is: If the formula will depend on j, and j depends on l, then how will the formula not depend on l?

Also, how will the resultant formula be good for all l, as one of the correction terms does not allow for l = 0?

You will have to break it down into three cases.

First consider l=0 (in which case, j is obviously l+1/2 =1/2). Add the kinetic and darwin corrections

Now consider l is not zero. Break this up into two subcases. First consider j=l-1/2. So replace all the "l"s by j+1/2 and add the kinetic and spin-orbit.

Now do j=l+1/2, repeat as above.

If I recall correctly, something quite miraculous happens. I think that all three results end up identical. But don't quote me on that.

Patrick
 
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  • #3
dextercioby
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Of course all 3 turn equal, else the formula would be much more complicated.
 
  • #4
Ah, thanks to both of you for the advice. It's great to see it turn out!
 

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