First order perturbation energy correction to H-like atom

[Ed Sheeran Fan]

Homework Statement

Real atomic nuclei are not point charges, but can be approximated as a spherical distribution with radius $R$, giving the potential

$$ \phi(r) = \begin{cases}
\frac{Ze}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) &\quad r<R\\
\frac{Ze}{r} &\quad r>R \\
\end{cases}$$

We can view the modified Hamiltonian as having a perturbation

$$H^\prime = \begin{cases}
-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r} &\quad r<R\\
0 &\quad r>R\\
\end{cases} $$

Calculate the first order correction to the energy of the $1s$ state.

Given answer: $\frac{2}{5}\frac{Z^4e^2R^2}{a^3}$, where $a$ is the Bohr radius.

Homework Equations

Nondegenerate case, first order energy correction to the $n$'th unperturbed energy $$E_n^1 = H^\prime_{nn}= \langle\psi_n^0 | H^\prime| \psi_n^0 \rangle$$

Radial wavefunction of lowest state of hydrogen/hydrogen like ions
$$\psi(r) = \sqrt{\frac{Z^3}{\pi a^3}}e^{-Zr}$$

*think there may be a typo in the radial wavefunction, as the exponential should have a factor of $1/a$, making it be $e^{-Zr/a}$

The Attempt at a Solution

Tried to evaluate

$$E^1_{nlm=100}=\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \frac{Z^3}{\pi a^3}e^{-2Zr/a}\times (-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r})
\times r^2sin\theta \,dr \,d\theta \,d\phi$$

but the integral seemed to fall apart. Wondering 1) if this gives the right answer and 2) is there a better way, e.g. some way to evaluate this without using integration by parts 4 times.***EDIT***
Found the/a solution in another book
They use the approximation $e^{-2Zr/a} = 1$ since $a \gg R$ after which the answer falls out
Feel a bit cheated after having done 3 pages of algebra

 

[Gene Naden]

I can appreciate your frustration, having done much unnecessary algebra myself. However, the nucleus is of very small radius compared to the electron orbital. This was established by Rutherford scattering.

You didn't need to integrate over $\theta$ and $\phi$. Just $\int \psi^* \psi H^\prime 4 \pi r^2 dr$

 

[vela]

The Attempt at a Solution

Tried to evaluate

$$E^1_{nlm=100}=\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \frac{Z^3}{\pi a^3}e^{-2Zr/a}\times (-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r})
\times r^2sin\theta \,dr \,d\theta \,d\phi,$$ but the integral seemed to fall apart. Wondering 1) if this gives the right answer and 2) is there a better way, e.g. some way to evaluate this without using integration by parts 4 times.

This integral isn't really that bad. One of the things you do want to learn is how to do calculations like this efficiently.

Do the angular integrals first to get:
$$E^1_{nlm=100}=4\pi \int_{0}^{R} \frac{Z^3}{\pi a^3}e^{-2Zr/a}\left[-\frac{Ze^2}{R}\left(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}\right) + \frac{Ze^2}{r}\right] r^2 \,dr $$
Let $\rho = r/R$. In terms of this variable and pulling some constants out front, you get
$$E^1_{nlm=100}=4 Z^3 \left(\frac{Ze^2}{R}\right) \frac{R^3}{a^3} \int_0^1 e^{-2Z (R/a)\rho}\left[-\left(\frac{3}{2}-\frac{1}{2}\rho^2\right) + \frac{1}{\rho}\right]
\rho^2 \,d\rho.$$ If you want to make it look even simpler, let $\alpha = 2ZR/a$. Then you have
$$E^1_{nlm=100}=\frac 12 \left(\frac{Ze^2}{R}\right) \alpha^3 \int_0^1 e^{-\alpha\rho}\left(\rho - \frac{3}{2}\rho^2+\frac{1}{2}\rho^4\right)\,d\rho.$$ At this point, integration by parts is probably the best way forward. Start with the $\rho^4$ term. Integrate by parts twice, and you'll get an integral you can combine with the $\rho^2$ term, and so on.

Another technique in evaluating stuff like this uses the fact that
$$\rho^n e^{-\alpha \rho} = \left(-\frac{\partial}{\partial \alpha}\right)^n e^{-\alpha \rho},$$ so, for example, you have
$$\int_0^1 \rho e^{-\alpha \rho}\,d\rho = \int_0^1 \left(-\frac{\partial}{\partial \alpha}\right)e^{-\alpha \rho}\,d\rho = -\frac{\partial}{\partial \alpha} \int_0^1 e^{-\alpha \rho}\,d\rho.$$ The integral is simple to evaluate, and then you just differentiate the result.

After you integrate, however, you still have more work to do to get the answer you were given.