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Inspired by this: Electrical neutrality of atoms

If I have a wavefunction of the 'reduced mass coordinate' for a hydrogen like atom made from an electron and a positron, what is the spatial charge distribution?

When we solve the hydrogen atom, we change into coordinates of the center of mass, and the separation distance with the reduced mass. Here, the masses of the constituent particles are the same. So the center of mass is equidistant from the positron and electron, and so discussing r and -r is just swapping the particles. Since the probability distribution for all the energy levels of the hydrogen atom are symmetric to inversion (images can be seen here http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html ), this seems to say no matter what energy level positronium is in, the charge distribution is neutral? Since the energy level basis is complete, this seems to say we can't polarize a positronium atom without dissociating it!? This doesn't make sense to me, so I'm probably making a big mistake here.

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Positronium is rather neutral indeed but can be polarized as any atom. An external electric field acts differently on $e^-$ and $e^+$. –  Vladimir Kalitvianski Apr 20 '11 at 21:44
    
""Positronium is rather neutral .." ALL atoms are rather neutral indeed. The sepecial thing with positronium is that any volume element is neutral! –  Georg Apr 21 '11 at 10:10
    
@Georg The notion of neutrality always implies some sort of interaction. If the interaction is weak and short-lasting (a fast charged projectile), the positronium may look as a neutral system. Otherwise it is not. –  Vladimir Kalitvianski Apr 21 '11 at 13:08

2 Answers 2

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Okay we have the center of mass coordinate $r_{cm} = (r_e + r_p)/2$, and the reduced mass coordinate $r = r_e - r_p$. So given the wavefunction $\psi(r_{cm},r)$ what you are asking is just a change of basis from $|r_{cm},r\rangle$ to $|r_e,r_p\rangle$. So you just need to consider

$$\langle r_e,r_p|r_{cm},r\rangle = \delta\left(r_{cm}-(r_e+r_p)/2\right) \ \delta\left(r-(r_e-r_p)\right)$$


To get more at what appears to be confusing you, let's focus on how positronium can be polarized. The spherical harmonics will have a defined even or odd parity to inversion, and so yes the square of the wavefunction (probability density) is invariant to inversion. But this does not mean all superpositions of these harmonics will have this property.

Consider for instance just s + p_z orbital. On one side the wavefunction amplitude will add constructively, while on the other it will add deconstructively. The density is no longer symmetric to inversion. This is the standard chemistry description of orbitals, so probably a good way to get follow up reading is searching for hybrid orbitals.

Here's a link I found with some images for you:
http://www.uwosh.edu/faculty_staff/gutow/Orbitals/N/What_are_hybrid_orbitals.shtml

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Two remarks: 1) The reduces mass coordinate is just a difference $r_e - r_p$. 2) I like your explanation of polarization - any asymmetric function can be represented as a spectral sum. –  Vladimir Kalitvianski Apr 21 '11 at 8:40
    
@Vladimir Oops, I didn't even realize I did that. Thanks for catching that mistake. –  Edward Apr 21 '11 at 9:35
    
Yes, the most important thing to see is that the $r$ in the wave function for positronium is the "separation" of the two particles. So, the probability distribution $\psi^{*}\psi$ isn't the probability density of finding a particle in position $x$, it is the probability density of finding the particles separated by a distance $r$. (or if you prefer, the probability of finding the absolute value of the dipole moment of the system to be $qr$ –  Jerry Schirmer Apr 21 '11 at 12:59

Normally we speak of charge distribution in terms of atomic form-factors. It means Born approximation which means in turn fast charged projectiles or photons.

Para- and orto-positroniums have different magnetic fields due to different spin orientation.

For slow charged projectiles its field strongly polarizes the atom and the charge distribution is not the same as in a "free" atom.

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I'm not asking about scattering. Given $\psi(r)$ where $r$ is the reduced mass coordinate, and thus contains information about both the electron and positron, how do we go from this to a probability density for just the electron, and the probability density for just the positron? –  John Apr 20 '11 at 23:05
    
Also, any bound state should be able to be written in terms of the complete basis given by the no-external potential energy levels. So applying an external electric field for example, how does this "polarized" positronium look since the basis doesn't seem to allow charge separation to give polarization. –  John Apr 20 '11 at 23:06
    
Because of the non-mainstream propaganda on your blog, please don't link to your blog in answers. –  Edward Apr 20 '11 at 23:56
    
In order to find something about one particle in a compound system, you just express its position via center of inertia and relative variables: $\psi (r_1,r_2) = \Psi (R)\psi (r)$. You can easily see that in a compound system any particle is in a mixed state. And quasi-particles can be in pure states. –  Vladimir Kalitvianski Apr 21 '11 at 8:44
    
@Edward If you see an evident error in my results, tell me in my blog or here, whatever. The article about form-factors I refer to is a classical subject with classical results rather than a "non mainstream propaganda", shame on you. –  Vladimir Kalitvianski Apr 21 '11 at 9:25

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