First order perturbation for hydrogen

[bobred]

Homework Statement

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

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

(a) Specify the perturbation
(b) Find the first order correction for the ground state
(c) Show that the answer in (b) can be approximated by
$E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R$ where $E_R=\frac{e^{2}}{8\pi\varepsilon_{0}a_0}$ is the Rydberg energy.

Homework Equations

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

$E_{1}^{(1)}=\int_{0}^{b}\left| \psi_0 \right|^2 \delta\hat{\textrm{H}} r^2\,\textrm{d}r$

$b \ll r$

The Attempt at a Solution

(a) $\delta\hat{\textrm{H}}=-\dfrac{e^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)$

so with $b \ll r$ we have $e^{-2b/a_0}\approx 1$
(b)
$E_{1}^{(1)}=\dfrac{e^2 b}{2\pi \varepsilon_{0}a_0^2}$

I think the above is correct, I just can't see how to get part (c).

 

[bobred]

Sorry, I was stupid I have it now.

 

[herculed]

Sorry, I was stupid I have it now.

Hi, what had you done wrong initially?

 

[bobred]

I found the first order perturbation which included the exponential terms. To get part c what I should have done was to set the exponential to unity then perform the integration. So part b should have included the exponential terms.
A better way I feel is to note that $b/a_0 <<1$ then take a 2nd order Taylor series and insert this into the integral in b.

 

[paranormal]

(c) To approximate the answer in (b), we can use the fact that b \ll r and the exponential term e^{-2b/a_0}\approx 1 to simplify the expression. Additionally, we can use the definition of the Rydberg energy, E_R=\frac{e^{2}}{8\pi\varepsilon_{0}a_0}, to rewrite the expression as:

E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R\left(\int_{0}^{b}r^2\left| \psi_0 \right|^2 \textrm{d}r\right)

We can further simplify this using the expression for \psi_0 given in the Homework Equations section:

E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R\left(\frac{2}{a_0^{3/2}}\int_{0}^{b}r^2e^{-2r/a_0}\textrm{d}r\right)

Using integration by parts, we can evaluate this integral to get:

E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R\left(\frac{2}{a_0^{3/2}}\left[-\frac{a_0^3}{2}e^{-2r/a_0}\right]_{0}^{b}\right)

Finally, we can substitute in the values for b and a_0 and simplify to get the desired result:

E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R\left(\frac{2}{a_0^{3/2}}\left[-\frac{a_0^3}{2}e^{-2b/a_0}\right]\right)\approx\dfrac{e^2 b}{2\pi \varepsilon_{0}a_0^2}=E_{1}^{(1)}