First-order perturbation

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Homework Statement


Consider a perturbed hydrogen atom whose Hamiltonian, in atomic units, is:

H= -1/2(∆^2) – ½ + b/(r^2) (∆ should be upside down), where b is a

positive constant. The Schrodinger equ. for this hamiltonian can be solved

exactly for the energy eigenvalues. The results for the ground state is:

E = -1/[2(B^2)], where B = [1 + (1 + 8b)^(1/2)]/2.

Use first-order perturbation theory in which the perturbation is b/(r^2) to

compute the ground-state energy of the purturbed system for b=0.01,

b=0.10, b=0.50, b=1.0, and b=10.0.


Homework Equations





The Attempt at a Solution



I am not sure how to start this problem. Any hints to get me started?
 

Answers and Replies

  • #2
OlderDan
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I'm not getting the connection between your Hamiltonian expression and the Hydrogen atom, but the first couple of pages of this link gives a nice introduction to perturbation theory. I assume you have a solution to the Hamiltonian without the b/(r^2) term. (Of course you do if it's the hydrogen atom.) That is the H_o in the perturbation derivation.

http://web.uconn.edu/~ch351vc/pdfs/pert0.pdf
 
  • #3
dextercioby
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If you consider the perturbation to the ground state of the H-atom, then we know that the level is nondegenerate and therefore what perturbation theory to use. In you case, the shift is energy to first order is

[tex] \Delta E \sim \left\langle \Psi_{100}(r)\left |\frac{1}{r^2}\right |\Psi_{100}(r)\right\rangle [/tex]

Daniel.
 
  • #4
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OK, now I have:
∫ π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) (b/r2) π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) dr

which is equal to:
b∫ π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) (1/r2) π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) dr
Now I am stuck again because I need help with this integral. Anyone?
Thanks
 
  • #5
OlderDan
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OK, now I have:
∫ π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) (b/r2) π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) dr

which is equal to:
b∫ π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) (1/r2) π^(1/2)(Z/a0)^(3/2)e^(-Zr/a0) dr
Now I am stuck again because I need help with this integral. Anyone?
Thanks

Does it help if you think in three dimensions and do a volume integral in spherical coordinates?
 
  • #6
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OdlerDan, it actually makes it worse for me to do it in spherical coordinates.
The major problem here is I am not up to parr on the math it takes to do quantum mechanics (I have a mainly chemistry background).
I have pulled all the constants out and am now left with:
∫ e^(-Zr/a0) (1/r2) e^(-Zr/a0) dr .
Anyone know any tricks for this integral?
 
  • #7
OlderDan
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OdlerDan, it actually makes it worse for me to do it in spherical coordinates.
The major problem here is I am not up to parr on the math it takes to do quantum mechanics (I have a mainly chemistry background).
I have pulled all the constants out and am now left with:
∫ e^(-Zr/a0) (1/r2) e^(-Zr/a0) dr .
Anyone know any tricks for this integral?

The point is that the volume element in spherical coordinates has an r² factor as well as the sinθdθdφ. Since your function has no angular dependence, the angle integrals are simple. Doesn't the r² factor make life easier for you when doing the r-integral? In any case, the hydrogen wave function is 3-dimensional and normalized that way. That's the integral you need to be doing.
 
  • #8
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I now see what you are saying. See how bad I am at this stuff. I didn't really even know what you meant at first with the spherical coordinates. By using this r^2 sin(other stuff), I will be able to eliminate the 1/(r^2) term in my integral, correct? Is the other stuff in the volume element just as you have in your post, or in other words, can you show me how it will look with my problem? Thanks again, you have helped a lot already.
 
  • #9
OlderDan
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I now see what you are saying. See how bad I am at this stuff. I didn't really even know what you meant at first with the spherical coordinates. By using this r^2 sin(other stuff), I will be able to eliminate the 1/(r^2) term in my integral, correct? Is the other stuff in the volume element just as you have in your post, or in other words, can you show me how it will look with my problem? Thanks again, you have helped a lot already.

I tried to find a good drawing for you. This is the best one I could find.

http://www.mas.ncl.ac.uk/~nas13/mas251/Section511.pdf [Broken]

The volume element in spherical coordinates is a "cube" in the infinitesimal limit of a volume bounded by sides dr, rdθ and rsinθdφ, so dV = r²sinθdφdθdr
 
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