First order perturbation theory

  • Thread starter Joe_UK
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The potential of an electron in the field of a nucleus is:

-Ze^2/(4 Pi Epsilon0 r) r > r0
-Ze^2/(4 Pi Epsilon0 r0) r <= r0

where r0 is the fixed radius of the nucleus.

What is the pertubation on the normal hydrogenic Hamiltonian?

Calculate the change in energy of the 1s state to the first order.

Hint: If r0 is small e^(-r0/a0) ~ 1-r0/a0

-----

So I know the normal hydrogen Hamiltonian potential part is -Ze^2/(4 Pi Epsilon0 r) and I have been fiddling around with trying different terms added onto this such as:

-Ze^2/(4 Pi Epsilon0 (r0-r)) or -Ze^2/(4 Pi Epsilon0 r0/r)

However I cant find a pertubation that really seems to make sense or fit the original conditions. I think that the pertubation should also possibly be a constant not dependant on r, but I cant seem to find any at all that fit. I tried

-Ze^2/(4 Pi Epsilon0 r0)

but this just gave a potential 2x too big at r0. I know how to find the change in energy by <u*|H'|u> once I have the pertubation, but its just eluding me.

Cheers,
Joe
 

Answers and Replies

  • #2
diazona
Homework Helper
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Well, the perturbation is what you add on to the original Hamiltonian to give the perturbed Hamiltonian:
[tex]H = H_0 + H'[/tex]
You can easily rearrange that to
[tex]H' = H - H_0[/tex]
But you know [itex]H[/itex] and [itex]H_0[/itex], so it should be a simple matter to find [itex]H'[/itex] by subtracting one from the other. It's slightly "unsimple" (I hesitate to say complicated) because [itex]H[/itex] is a piecewise function, with different definitions in two regions. All that really means, though, is that you need to calculate [itex]H'[/itex] separately in each of the two regions. Do one subtraction for [itex]r > r_0[/itex] and one for [itex]r \le r_0[/itex].
 

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