Taylor expansion fine structure

  • #1
darkshadow28
I have to do a Taylor expansion of the energy levels of Dirac's equation with a coulombian potential in orders of (αZ/n)^2 , but the derivatives I get are just too large, I guess there is another approach maybe?
This is the expression of the energy levels
upload_2017-10-13_19-28-6.png

And i know it has to end like this:
upload_2017-10-13_19-28-57.png
 

Answers and Replies

  • #2
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You only need two derivatives. The second one will be ugly but you evaluate it at ##\alpha Z=0##, which means you can skip many terms as soon as you know that appears as factor.
 
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  • #3
Charles Link
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A simpler way is to first expand ## \sqrt{(j+1/2)^2-(\alpha Z)^2} ## as a Taylor series. It greatly simplifies the denominator. Upon squaring the term in brackets ##[ \, ] ##, I get ## n^2(1-\frac{(\alpha Z)^2}{n(j+1/2)} +...) ##. This makes ##1/ [1+(\frac{\alpha Z}{n})^2(1-\frac{(\alpha Z)^2}{n(j+1/2)} )]^{1/2}-1 ##. With this very last Taylor expansion with a ## f(x)=\sqrt{1+x} ##, it is important to also keep the second order term of the ## (\frac{\alpha Z}{n})^2 ## factor. I almost got complete agreement with the result you posted, but I got a 1/4 instead of 3/4 in the fourth order term. ## \\ ## Editing: One additional item, and I think I got their result: In taking the final result in the form ## f(x)=\frac{1}{1+x}=1-x+x^2-x^3+ ... ##, it is important to retain the ## x^2 ## term of the ## \frac{1}{2} (\frac{ \alpha Z}{n})^2 ## term. (Originally, I tried to use ## \frac{1}{1+x} \approx 1-x ##. That did not supply sufficient accuracy.) This will then give the result you provided with the 3/4.
 
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  • #4
darkshadow28
A simpler way is to first expand ## \sqrt{(j+1/2)^2-(\alpha Z)^2} ## as a Taylor series. It greatly simplifies the denominator. Upon squaring the term in brackets ##[ \, ] ##, I get ## n^2(1-\frac{(\alpha Z)^2}{n(j+1/2)} +...) ##. This makes ##1/ [1+(\frac{\alpha Z}{n})^2(1-\frac{(\alpha Z)^2}{n(j+1/2)} )]^{1/2}-1 ##. With this very last Taylor expansion with a ## f(x)=\sqrt{1+x} ##, it is important to also keep the second order term of the ## (\frac{\alpha Z}{n})^2 ## factor. I almost got complete agreement with the result you posted, but I got a 1/4 instead of 3/4 in the fourth order term. ## \\ ## Editing: One additional item, and I think I got their result: In taking the final result in the form ## f(x)=\frac{1}{1+x}=1-x+x^2-x^3+ ... ##, it is important to retain the ## x^2 ## term of the ## \frac{1}{2} (\frac{ \alpha Z}{n})^2 ## term. This will then give the result you provided with the 3/4.
Thanks, if I try doing from ## n^2(1-\frac{(\alpha Z)^2}{n(j+1/2)} +...) ## i get the correct terms, but I'm getting an extra n(j+1/2) in the Taylor series of the square root.
 
  • #5
Charles Link
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In your first Taylor series, you need to write ## (j+1/2) \sqrt{1-(\frac{\alpha Z}{j+1/2})^2}=(j+1/2)(1-\frac{1}{2} \frac{(\alpha Z)^2}{(j+1/2)^2}+...) ##. The ## j+1/2 ## out front gets canceled by the ## -(j+1/2) ## in the denominator. And next, factor out an ## n^2 ## before you square the result.
 
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