First order perturbation energy correction to H-like atom

You'll have some terms involving ##R## and ##a## that need to be simplified using the definition of the Bohr radius.
  • #1

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
 
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  • #2
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##
 
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  • #3
Ed Sheeran said:

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.
 

1. What is the first order perturbation energy correction to H-like atom?

The first order perturbation energy correction to H-like atom is a mathematical approach used in quantum mechanics to account for the effects of external perturbations on the energy levels of a hydrogen-like atom. It is based on the first-order approximation method, where the perturbation is considered to be small compared to the unperturbed energy.

2. How is the first order perturbation energy correction calculated?

The first order perturbation energy correction is calculated by using the first-order perturbation formula, where the energy correction is given by the product of the first-order perturbation term and the corresponding matrix element. The matrix element is calculated using the wavefunctions of the unperturbed and perturbed states of the atom.

3. What are the applications of the first order perturbation energy correction?

The first order perturbation energy correction is used in various fields of physics, including atomic and molecular physics, quantum chemistry, and solid-state physics. It is also used to study the effects of external perturbations on the energy levels of atoms and molecules, such as the Stark effect and the Zeeman effect. Additionally, it is used to calculate transition probabilities and spectral lines in spectroscopy.

4. How does the first order perturbation energy correction affect the energy levels of a hydrogen-like atom?

The first order perturbation energy correction causes the energy levels of a hydrogen-like atom to shift from their unperturbed values. This shift depends on the strength of the perturbation and the energy level being considered. The correction can either increase or decrease the energy level, and it can also split degenerate energy levels into different levels.

5. Are there higher order perturbation corrections available for H-like atoms?

Yes, there are higher order perturbation corrections available for H-like atoms, such as the second and third order perturbation corrections. These higher order corrections provide a more accurate calculation of the energy levels, especially for strong perturbations. However, the first order perturbation correction is often sufficient for weak perturbations and is easier to calculate.

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