# Is positronium really an orbiting pair of electron/positron ?

1. Jan 30, 2014

### xortdsc

Hi,

it states on the one hand that it is just an electron and a positron orbiting around their common center, held together by electro-magnetic forces due to their charge. On the other hand it states it behaves very similar to a hydrogen atom just that the proton is replaced by the positron and due to its much lower mass the positron's movement will not be negligible (as it is for the proton). But still the energy of the system seems to be quantized (so the spinning particles are not continuously spiraling into each other until annihilation) and the electron/positron seems to move in orbitals just as electrons in a hydrogen-atom (s-/p-orbitals and so on) which would be somewhat in conflict of stating that the electron is actually orbiting the nucleus/positron because in hydrogen the electron would not orbit the nucleus but instead being "spread out" into an orbital cloud and not being in a single spot at any time.
So now what is the valid picture here ? The spread out electron-cloud orbital thing or actually (point) particles orbiting each other ?

2. Jan 30, 2014

### Staff: Mentor

Its a positron orbiting an electron or vice-versa.

http://en.wikipedia.org/wiki/Positronium

Since its a QM system, it follows the electron-cloud orbital thing but its an unstable configuration as they will quickly annihilate each other and go out in a blinding flash of light as an even number of gamma ray photons.

3. Jan 30, 2014

### ChrisVer

Well, something orbiting the other doesn't really make sense in the Quantum Mechanic picture. What happens is that it's a bound state (unstable), in the same way the electron is a bound state to the hydrogen atom. If I would be asked how could that happen in the case of QM (forgetting the relativistic quantum mechanics allowing annihilation) it would mean that the potential could have a local minimum. Of course things are a bit different, because in the case of H you had the induced mass equal to the electron mass (because you assumed the mass of the proton>>> the mass of electron). In the positronium case the induced mass is not such...
In classical mechanics view, that would mean that both particles would rotate around a point that's not identified as one of the particles (in Hydrogen atom you had the electron rotating around a point that coincided with where the proton was).

4. Jan 31, 2014

### xortdsc

So it should be possible to calculate the electron density distribution for the 1s orbital relative to the systems common center of mass, right ? Since it should be radially symmetric it should only be a function of r (where r=0 is the common center of mass of both particles). The positron will always need to be at the opposite side for any hypothetical electron-position chosen from the distribution.
If this is possible can someone give me the corresponding function ?

5. Jan 31, 2014

### Staff: Mentor

The wave functions are actually unchanged from that of a hydrogen atom, provided you use the reduced mass for positronium. In the ground state, the system is in a spherically symmetric orbital centered on the position of the center-of-mass.

Since the interparticle vector $\mathbf{r}$ is colinear with the position vector of the electron $\mathbf{r}_\mathrm{e}$ when the center of mass is taken as the origin, the angular distribution of the electron follows the angular distribution of $\mathbf{r}$, so the electron is in a spherically symmetric orbital centered on the position of the center-of-mass, as you expected.

6. Jan 31, 2014

### xortdsc

Okay, so it means that the probability of both particles at the common center of mass (r=0) is >0 which corresponds to the probability of annihilation of the entire structure ?

7. Jan 31, 2014

### ChrisVer

I don't think that you can speak about annihilation in the QM view point (which comes as a result of antiparticle existence which are not coming from QM but from QFT). The annihilation is an interaction procedure and it occurs by interactions. It can happen if the electron and positron are not in the same position as well...
For the probability of that to happen you have to calculate the cross section of the interaction
$e^{+}e^{-} \rightarrow γ γ$

In the view of QM, you would have to calculate the probability for your particles to leave that "local minimum" of the potential as I said, which corresponds to their bound state....

8. Feb 1, 2014

### xortdsc

okay, thank you very much :)

9. Feb 1, 2014

### Staff: Mentor

As ChrisVer mentioned, you have to take into account the cross-section for electron-positron annihilation. But the wave function still plays a role. The fact that in an s state the probability of finding the two particle at the same place is not zero, annihilation is quite fact, and the lifetime of the 1s state is 1.3x10-10 s against annihilation. In contrast, the Lyman-α line has been observed: the 2p state has a lifetime of 3.19x10-9 s against decay to the 1s state. In other words, the 2p state has time to decay by emitting a photon before annihilation takes place.

A nice source of information on positronium is S Berko and H N Pendleton, Ann. Rev. Nuc. Part. Sci. 30, 543 (1980). Unfortunately, it's behind a paywall, so you have to have an institutional access.

Last edited: Feb 1, 2014
10. Feb 1, 2014

### snorkack

Is the probability of annihilation, for all s states of positronium, strictly proportional to the probability of electron and positron occurring at the same spot?
(That probability is zero for all states of positron orbiting electron, but all those states can emit photons to decrease their angular momentum by 1)

Also, is the ratio between the annihilation probability of para and ortho positronium the same for all s states?

All s states 3s or above can decay to any lower lying p state, as well as annihilate, but 2s obviously cannot. Thus the stablest positronium ought to be 2s ortho positronium.

11. Feb 1, 2014

### Staff: Mentor

For s states, to first order, yes, the decay rate is $\Gamma \propto | \psi(0) |^2 \propto 1/n^3$.

I don't get what you're saying here. There is no "positron orbiting electron", only positron and electron orbiting their common center of mass.

There is a difference in the decay rate between para and ortho positronium. For the 1s state of para-Ps, $\Gamma = 8 \times 10^{9}\ \mathrm{s}^{-1}$, while for ortho-Ps it is $\Gamma = 7 \times 10^{6}\ \mathrm{s}^{-1}$.

The lifetime of the 2s ortho state is 1.1x10-6 s, indeed quite high.

12. Feb 1, 2014

### snorkack

Electron that "orbits around" a positron, or proton or any other point charge, possessing nonzero angular momentum, has zero probability of being anywhere on the axis of orbit, including the nucleus. This applies to all p states and even more to any d or higher states.

It is only the s states, that oscillate through the nucleus rather than orbit around it, which have a nonzero probability of interacting with the nucleus by annihilation or electron capture.

So:
For p, d etc. states of positronium, annihilation is impossible, but radiation to a lower angular momentum state is always possible
For 1s state, radiation is impossible (it is the ground state) but annihilation is possible and fastest of all s states
For 2s state, radiation is forbidden (like for all hydrogen like atoms), annihilation is possible but 8 times slower than for 1s state
For 3s and all higher s states, radiation to all lower p states up to 2p is possible, but so is annihilation direct from the high s state... so these should be competing processes of comparable speed.

13. Feb 1, 2014

### Staff: Mentor

Annihilation is possible: you have to consider the annihilation cross-section. The electron and positron need not be exactly at the same place to annihilate, only be within the cross-section.

It is forbidden for electric dipole transitions. My guess is that other decay processes are slower than annihilation.

14. Feb 1, 2014

### Bill_K

Each QFT interaction takes place at a point, but what you're forgetting is that electron-positron annihilation is a second-order process, with two photons emitted. The Feynman diagram has two vertices, meaning that the external legs for the electron and positron do not exactly meet.

15. Feb 1, 2014

### ChrisVer

Do you mean like that one photon could have spin + and the other - so that the 0 angular momentum change would still be valid? just asking to make sure I understood what you mean by a 2nd order process..

16. Feb 1, 2014

### Bill_K

Singlet states of even L and triplet states of odd L decay into an even number of photons. Triplet states of even L and singlet states of odd L decay into an odd number of photons. Of course annihilation from higher states is greatly suppressed in favor of decay to the ground state followed by annihilation from there.

Serious efforts are underway to create positronium hydride, di-positronium molecules, and even positronium BE condensates.