Calculating Electron Energy Shift with Perturbation Theory

[MathematicalPhysicist]

Homework Statement

Regard the nucleus of charge Ze as a sphere of radius R0 with uniform density.
Assume that R0<<a0 where a0 is Boher radius/

1. Derive an expression for the electrostatic potential V(r) between the nucleus and the electrons in the atom. If V0(r)=-Ze^2/r is the potential from a point charge, find the difference dV=V(r)-V0(r) due to the size of the nucleus.

2. Assume one electron is bound to the nucleus in the lowest bound state. What is its wave function when calculated using the potential V0(r) from a point nucleus?

3. Use first-order perturbation theory to derive an expression for the change in the ground state energy of the electron due to the finite size of the nucleus.

Homework Equations

e- is the electron's charge.

The Attempt at a Solution

1. I believe it's should be [tex]V(r)=\frac{-Ze^2}{r-R0}[/tex] and the difference: [tex]\delta V(r)=-\frac{R_0}{r}\frac{Ze^2}{r-R0}[/tex].

2. I believe the wave function is that of the solution of the hydrogen potential, i.e:
[tex]\psi_{1,0,0}=\sqrt{\frac{1}{\pi a^3_0}} exp(-r/a_0)[/tex], because for n=1,l=0,m=0 it's ground state of the electron.

3. I am not sure, but I think I need to expand dV by r>>R0, but after that I don't know how to procceed.

Any hints?

 

[jdwood983]

3. I am not sure, but I think I need to expand dV by r>>R0, but after that I don't know how to procceed.

Any hints?

What is the equation to find the energy of the first-order perturbation?

 

[MathematicalPhysicist]

[tex]E^{(1)}_n=<n|H1|n>[/tex], where H1 is the perturbation of the hamiltonian H, but I am not sure what H1 is in this question, is it dV, and if it is so, did I get it right in 1?
If it's so, then [tex]E^(1)_n=\int_{R0}^{a0} \psi_{n}(r)\delta V(r) \psi_{n}(r)dr[/tex]

 

[jdwood983]

Part 1 does not look correct to me. You might want to consider both Gauss's law and the boundary that at [itex]R_0=r[/itex], the potential must be [itex]V_0=-Ze^2/r[/itex]. In your current form, your potential will go to infinity at [itex]R_0=r[/itex], which we know is not true at all!

 

[MathematicalPhysicist]

Do you mean that the potential should satisfy:
[tex]1/r^2 d/dr (d/dr r^2V(r))=4\pi e/(4\pi r^3/3)=3e/r^3[/tex].

 

[jdwood983]

Do you mean that the potential should satisfy:
[tex]1/r^2 d/dr (d/dr r^2V(r))=4\pi e/(4\pi r^3/3)=3e/r^3[/tex].

There should be another factor or two in there (say [itex]Z[/itex], another [itex]e[/itex] and [itex]R_0[/itex]?), but that is along the lines.

My suggestion is that you first find the electric field inside the sphere of radius [itex]r[/itex] that does not include the electron (nucleus only). As a small helper/hint, it will be proportional to [itex]1/R_0^3[/itex].

 

[MathematicalPhysicist]

I am not sure I get to the answer.
[tex]1/r^2 d/dr (d/dr (r^2 \phi(r)))=-3Ze/R^3_0[/tex] I get to two constants of integration for which I have only one boundary condition, the one which you gave in your previous post.

 

[jdwood983]

Through Gauss's Law, you should find the electric field to be (for [itex]r<R_0[/itex])

[tex]
E=-\frac{Zer}{R_0^3}
[/tex]

So that the potential is

[tex]
-\frac{dV}{dr}=eE(r)=-\frac{Zer}{R_0^3}
[/tex]

 

[MathematicalPhysicist]

E should be without the minus sign, but thanks now I understand this question.

P.s
what of question 2 did I get it right, or not?

 

[jdwood983]

E should be without the minus sign, but thanks now I understand this question.

Right, and my second equation there should have [itex]e^2[/itex] in it.

There should be an accounting of [itex]Z[/itex] in your wavefunction; it should look like

[tex]
\Psi_{100}=\sqrt{\frac{Z^3}{\pi a_0^3}}\exp\left[-\frac{Zr}{a_0}\right]
[/tex]

And as for the third part, you should be using [itex]\delta V[/itex] for your perturbed potential:

[tex]
\delta E=\int d^3r\Psi^2\delta V
[/tex]

 

[MathematicalPhysicist]

The limits of the integral is between R0 to infinity or something else.

btw, I get an integral of the form:
[tex]\int_{R_0}^{\infty} exp(-r/a_0)/r dr[/tex]
obviously this integral can be evaluated by power series, my question is how do I know that the upper limit is finite or zero?

Thanks in advance.

 

[jdwood983]

The limits of the integral is between R0 to infinity or something else.

btw, I get an integral of the form:
[tex]\int_{R_0}^{\infty} exp(-r/a_0)/r dr[/tex]
obviously this integral can be evaluated by power series, my question is how do I know that the upper limit is finite or zero?

Thanks in advance.

The nucleus exists between [itex]0\leq r\leq R_0[/itex], so those two should be your limits, not [itex]R_0[/itex] and [itex]\infty[/itex].

 

[MathematicalPhysicist]

stiil as I mentiond earlier, I get a singularity in r=0 for the integrand:
exp(-2r/a0)/r well I expand with [tex]exp(-r/a0)=\sum (-2r/a0)^n/n![/tex], how to reconcile this?

 

[jdwood983]

I think you are messing up your potential. Given what I told you for [itex]-dV/dr[/itex], you should find

[tex]
V(r<R_0)=-\frac{Ze^2}{2R_0}\left[3-\left(\frac{r}{R_0}\right)^2\right]
[/tex]

(the 3 comes from matching the boundary so that the potential is continuous at [itex]R_0=r[/itex] ) So that you get

[tex]
\delta V=-\frac{Ze^2}{2R_0}\left[3-\left(\frac{r}{R_0}\right)^2-\frac{2R_0}{r}\right]
[/tex]

so when you multiply this by the square of the wavefunction, you get

[tex]
\delta E=\int d^3r\Psi^2\delta V=\int_0^{R_0}r^2dr\int_{-1}^1d(\cos\theta)\int_0^{2\pi}d\phi\Psi^2\delta V=\frac{2Z^3}{a_0^3}\int_0^{R_0}\exp\left[-\frac{2Zr}{a_0}\right]\delta V\,r^2dr
[/tex]

 

[MathematicalPhysicist]

I think my mistake was I integrated wrt to dr and not to d^3r.

Thanks, I get it now.