# Alternate Muonic Helium

## Main Question or Discussion Point

Hi,

I would like to model an alternate to Muonic helium, and I need some help. I got this idea from a professor of mine who mentioned it off hand in a lecture. The idea is the following:

Model a helium nucleus with one electron, and one muon.

This would have two fermions "orbiting" however, the particles are distinguishable and so none of the indistinguishably ideas around the electrons in the atom would apply.

One of my first question is this: The muon is considerably more massive than the electron, the ratio of mass between the nucleus and the muon is 1 $m_\mu$ is 0.0283466 times smaller than 1 $m_\alpha$. So the muon is $\approx 3\%$ the mass of a helium nucleus. Is this small enough to still approximate a stationary center?

Related Quantum Physics News on Phys.org
If you are referring to gravity as moving the nucleus then you may assume so. Seeing as the relative distance from the nucleus to the the first orbit is incredibly large, gravity would not have an effect on the nucleus.

alxm
Well, your professor probably mentioned it due to the fact there was an http://www.sciencemag.org/content/331/6016/448" [Broken].
The ground-state is a singlet, so as far as that's concerned, the two particles are distinguishable anyway, since they've got opposite spin.

As for your question: Yes, the Born-Oppenheimer approximation was found to hold up well under the circumstances.

Zush: What on earth are you on about? Electrons (or muons) don't have 'orbits', nor stay at any well-defined distance from the nucleus. And the gravitational attraction is over 50 orders of magnitude smaller than the electromagnetic force between the particles.

Last edited by a moderator:
Well, your professor probably mentioned it due to the fact there was an http://www.sciencemag.org/content/331/6016/448" [Broken].
The ground-state is a singlet, so as far as that's concerned, the two particles are distinguishable anyway, since they've got opposite spin.

As for your question: Yes, the Born-Oppenheimer approximation was found to hold up well under the circumstances.

Zush: What on earth are you on about? Electrons (or muons) don't have 'orbits', nor stay at any well-defined distance from the nucleus. And the gravitational attraction is over 50 orders of magnitude smaller than the electromagnetic force between the particles.
Wow, thanks for this article. Do you think anyone will mind If I try and do this without reading the article first, as an exercise. See if I can do physics without seeing it first.

Last edited by a moderator:
I decided to assume that the nucleus was stationary. The Schrodinger equation is then

$$-\frac{\hbar^2}{2m_\mu}\nabla^{2}_{\mu}\Psi(\vec{r}_\mu, \vec{r}_e) -\frac{\hbar^2}{2m_e}\nabla^{2}_{e}\Psi(\vec{r}_\mu, \vec{r}_e)+ke^2\left(-\frac{Z}{r_e}-\frac{Z}{r_\mu}+\frac{1}{r_{e\mu}}\right)\Psi(\vec{r}_\mu, \vec{r}_e)=\hat{E}\Psi(\vec{r}_\mu, \vec{r}_e).$$

I'm going to take

$$H'=\frac{ke^2}{r_e\mu}[/itex] as the perturbation and then divide the wave function into two parts [tex]\Psi(\vec{r}_\mu, \vec{r}_e)=\psi(\vec{r}_\mu)\phi(\vec{r}_e).[/itex] and write $\hat{E}=\hat{E}_\mu+\hat{E}_e$. I'm then going to treat this as two, two-body problems with the Helium nucleus and the Muon, and the Helium nucleus and the electron. Both of these problems have the normal hydrogenic wave function solution [tex]\Phi_{nlm}(r,\theta,\phi)=-\left[\frac{4(n-l-1)!}{(na_0)^3n\left[(n+l)!\right]^3}\right]^{1/2}\rho^l L_{n+l}^{2l+1}(\rho)e^{-\rho/2}Y_{l}^{m}(\theta,\phi)$$

where $\rho=2r/na_0$ and $a_0=\hbar^2/\bar{\mu} e^2$. Here $\bar{\mu}$ is the reduced mass of either one of the systems (I'm running out of symbols...). $L_{n+l}^{2l+1}(\rho)$ and $Y_{l}^{m}(\theta,\phi)$ are the Leguerre polynomials and the Spherical Harmonics respectively. For this Hydrogenic solution, $n,l,m,r,\theta,\phi$ are all dependent on the system, and so are not the same in general for both systems. Then the total unperturbed wave function for the system is

$$\Psi_{\{n \}}(\vec{r}_\mu , \vec{r}_e)=\psi_{\{n_\mu \} }(\vec{r}_\mu)\phi_{\{n_e \}}(\vec{r}_e).$$

Here $\{n\}$ is the set of quantum numbers. Then the ground state of this atom is

$$\Psi_{100,100}(\vec{r}_\mu, \vec{r}_e)=\frac{1}{\pi}\left(a_{0_\mu}a_{0_e}\right)^{-1/2}e^{-(r_\mu/a_{0_\mu}+r_e/a_{0_e})}.$$

Tomorrow, if everything is fine, I will calculate the first correction to energy and other interesting items. Stay tuned.