First-order perturbation

In summary: The integral, therefore, is:∫ 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
  • #1
rovert
7
0

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?
 
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  • #2
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
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
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
rovert said:
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
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
rovert said:
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
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
rovert said:
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

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
 
Last edited by a moderator:

What is first-order perturbation?

First-order perturbation is a mathematical tool used to approximate the behavior of a system when a small perturbation or disturbance is applied to it. It is based on the assumption that the system's response to the perturbation can be described by a linear function.

How is first-order perturbation used in scientific research?

First-order perturbation is commonly used in many fields of science, including physics, chemistry, and biology. It is used to study the effects of small changes or disturbances on a system and to make predictions about the system's behavior under different conditions.

What are the limitations of first-order perturbation?

First-order perturbation is only accurate for small perturbations, and it assumes that the system's response is linear. It also does not take into account higher-order effects, which may become significant for larger perturbations. Therefore, it is important to use caution when applying first-order perturbation and to consider other methods for more accurate predictions.

What are some real-world applications of first-order perturbation?

First-order perturbation has many applications in various fields. In physics, it is used to study the behavior of particles in quantum mechanics and to calculate energy levels in atoms. In chemistry, it is used to analyze chemical reactions and to predict the stability of molecules. In biology, it is used to study the effects of mutations on genetic systems and to understand the stability of biological networks.

How does first-order perturbation differ from higher-order perturbation?

First-order perturbation only considers the effects of a small perturbation, while higher-order perturbation takes into account larger perturbations and their nonlinear effects. Higher-order perturbation can provide more accurate predictions, but it also requires more complex calculations. Therefore, first-order perturbation is often used as a simpler approximation for systems with small perturbations.

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