# Perturbation first order hydrogen - Coulomb model

1. May 1, 2016

### JaneHall89

1. The problem statement, all variables and given/known data
Assume that there is a deviation from Coulomb’s law at very small distances, the Coulomb potential energy between an electron and proton is given by

$V_{mod}(r)=\begin{cases} -\frac{q^{2}}{4\pi\varepsilon_{0}}\frac{b}{r^{2}} & 0<r\leq b\\ -\frac{q^{2}}{4\pi\varepsilon_{0}}\frac{1}{r} & r>b \end{cases}$

(a) Find the first order correction for the ground state using the standard integrals

2. Relevant equations

Ground State - $\psi_0=\frac{2}{a_0^{3/2}}e^{-r/a_0}$

To find the first order correction - $E_{1}^{(1)}$=<ψ0| $\delta\hat{\textrm{H}}$|ψ0>

Perturbation - $\delta\hat{\textrm{H}}=-\dfrac{q^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)$ for 0<r≤b
Standard integrals
$\int_{0}^{x}$ e-u du = 1 - e-x
-$\int_{0}^{x}$ u e-u du = 1 - e-x - xe-x

3. The attempt at a solution
Using $E_{1}^{(1)}$=$\int_{-∞}^{∞}$ψ(0) * $\delta\hat{\textrm{H}}$ ψ(0)
$E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})$ dr

So I've taken the constants out and applied a 'sandwich integral' with the perturbation Hamiltonian... This problem is I have no idea where to start making this easier for me, ive been staring at it for aleast and hour. Can someone help please?

Last edited: May 1, 2016
2. May 1, 2016

### blue_leaf77

Where is the volume element for the integration $d^3\mathbf{r}$?

3. May 1, 2016

### JaneHall89

r is radius not a vector in this question. So integration is only on radius. I added dr

4. May 1, 2016

### blue_leaf77

That's not the point of my question. Any integral must take the form of $\int f(x) \, dx$, where is your $dx$ in your calculation of the integral?

5. May 1, 2016

6. May 1, 2016

### blue_leaf77

That's still not completely correct, you are integrating over the entire volume which means the integration element must be that of a volume element. Find the correct form of $d^3\mathbf{r}$ (sometimes also written as $dV$) in spherical coordinate, for example take a look at https://en.wikipedia.org/wiki/Spherical_coordinate_system.

7. May 1, 2016

### JaneHall89

But the perturbation hamilton that I have is independent of Φ,θ

8. May 1, 2016

### blue_leaf77

The form of the volume element is entirely determined by the coordinate system being used, not by the integrand.

9. May 1, 2016

### JaneHall89

$E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})$ dr dθ dΦ

10. May 1, 2016

### blue_leaf77

Still something missing, look for $dV=?$ in the wikipedia link above.

11. May 1, 2016

### JaneHall89

$E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})$ r2sin θ dr dθ dΦ =
$E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (b-r)$ sin θ dr dθ dΦ

12. May 1, 2016

### blue_leaf77

Yes, that's right. Now all $r$'s will be in the numerator and the integral should be analytically solvable.

13. May 1, 2016

### JaneHall89

hmm i'm still not seeing how to make use of the standard integrals given

14. May 1, 2016

### blue_leaf77

I mean this equation:
$$E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}} \int_{0}^{2pi} \int_{0}^{pi}\int_{0}^{b}e^{\frac{-2r}{a_0}} (b-r) dr d\theta d\phi$$
Do the integrals over the angles first if the appearance makes obscure your mind. You will then be left with the integral over $r$, which takes the form of the formula you posted under the "Relevant equations" in your starting post.

15. May 1, 2016

### JaneHall89

$\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (b-r) dr = \int_{0}^{b}-r(e^{\dfrac{-2r}{a_{0}}})+b(e^{\dfrac{-2r}{a_{0}}})$

This is hurting me, I cant see how these work with standard integrals?? Everything im trying on my paper is not working :/ Have I gone wrong somewhere else and thats why I cant see it

Last edited: May 1, 2016
16. May 1, 2016

### klw289

Integrate via a substitution using u=2r/a

17. May 1, 2016

### blue_leaf77

One more time, don't leave the integration element behind whenever you write an integral.
As to the method to solve, do you know integration by substitution? It's the same as the one someone above suggested.

18. May 2, 2016

### JaneHall89

I have done the subs after a very long day of staring at this and I think its correct but its incredibly messing and hard to simplify.
For the first integral...
= ((a0b) /2) x (1-e-2b/a0)
Second
= (a02 / 4) x ( 1 - e - 2b/a0 - 2b/a0 e-2b/a0)
So I'll add them to complete.
I have Just to double check that the limits are from 0 to 2b/a0?

19. May 2, 2016