Perturbation theory energy shift for hydrogen atom

philip041
Messages
104
Reaction score
0
I'm trying to follow some working by lecturer;

Treating delK (previously found in first bit of question), show that the energy En of the usual hydrogenic state [nlm> is shifted by some expression given.

basically we start with

<br /> <br /> \[<br /> \frac{1}{2m_{0}c^{2}} \left\langle nlm\right|\left(\hat{H_{0}} - V\left(r\right)\right)^{2}\left|nlm\right\rangle<br /> \]<br /> <br />

and it goes to

<br /> <br /> <br /> \[<br /> \frac{1}{2m_{0}c^{2}} \left\langle nlm\right|\left(E_{n}+\frac{\alpha\hbar c}{r}\right)^{2}\left|nlm\right\rangle<br /> \]<br /> <br />

I don't understand what he has done top replace the H - V?

Cheers
 
Physics news on Phys.org
Hello Philip041 :smile:

Could you describe what delK is? From what I can see, this looks like the first order relativistic correction to the hydrogen atom, commonly found in the fine structure correction (minus the spin orbit coupling).
 
philip041 said:
I don't understand what he has done top replace the H - V?

Hi philip041! :smile:

Well, H0|nlm> = En|nlm> by definition.

What is V(r)?

(it's presumably defined so that V(r)|nlm> is approximately (ahc/r)|nlm>)
 
In that step Professor Heath has taken the potential given in the question (4th line down from the question number) and worked on it using the definition of alpha.

V\left(r\right) = -\frac{e^{2}}{4\pi\epsilon_{0}r}

\alpha = \frac{e^{2}}{4\pi\epsilon_{0}\hbar c}

\alpha\hbar c = \frac{e^{2}}{4\pi\epsilon_{0}}

So, V\left(r\right) = -\frac{\alpha\hbar c}{r}

\left(H_{0}-V\left(r\right)\right) = \left(H_{0} + \frac{\alpha\hbar c}{r}\right)


Ta Da!
 
Last edited:
I looked at your profile yesterday(!) by chance and thought 'only Jew Dave would call himself that', I'm assuming it is Jew Dave? Or Gayer than Gay Dave.. Cheers for the heads up. Hope revision's going well. I think I will be getting a 2:2 at this rate, I'm holding out for Bristol to flood so we can't take exams...
 
I hope my previous post doesn't constitute bad banter, they are legitimate nicknames of people on my course...
 
The question is: which phil are you??

Dave
 
How many Daves and Phils are there in your place? :smile:

and isn't there a tiny-dave or a tiny-phil? :-p
 
I can think of at least 3 Phil's in this lecture course, and there's probably a similar number of Daves. I suppose if you want I can be tiny-dave too, I'm only 6'2''...

Dave
 
  • #10
I'm the cool one
 
  • #11
demonstrated by association with this forum
 
Back
Top