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Ortho- and Para-Positronium Decays

by Mantella
Tags: decays, ortho, parapositronium
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Mantella
#1
May15-14, 02:18 PM
P: 6
One of my homework problems asks what number of photons are released in the decay of ortho- and para-positronium, and I'm somewhat confused. From the point of view of 4 momentum conservation I understand that at least two photons must be emitted in the decay process. In the case of ortho-positronium, the overall angular momentum of 1 must be conserved as well. The simplest answer is that 3,5,7,... photons are created with alternating spins, but wouldn't it also be possible for two photons or, for that matter, any even number of photons to be created with opposite spins and an orbital momentum of one? What about even more esoteric photon spin and orbital angular momentum combinations that satisfy the angular momentum conservation? I don't have much experience with orbital angular momentum, and my teacher hasn't described it in any detail.

Also, for para-positronium the decay products are an even number of photons with alternating spin, but to the best of my understanding parity must be conserved as well. Electron parity of +1, positron parity of -1, (I think) while photons have a parity of -1. (+1)(-1) ≠ (-1)(-1). Again my understanding of particles and their interactions is pretty limited.

Parity for a ortho-positronium does seem to be conserved however.

On a side note: does orbital angular momentum have directions like spin or is it always greater than or equal to 0?
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andrien
#2
May16-14, 04:20 PM
P: 1,020
What you are looking for is not angular momentum conservation, you need to know about C-parity. You need to concentrate on ground state of Positronium here, this is where you prefer singlet and triplet state as para-positronium and ortho-positronium.

Transformation of triplet state under charge conjugation goes as (replace ##b^+## by ##d^+##),
##b_1^+d_1^+|0> → d_1^+b_1^+|0> → -b_1^+d_1^+|0>## ( ##b^+ ##and ##d^+## anticommute) , hence triplet state has odd parity and it can decay to odd number of photons since photon's C-parity is -1.

Transformation of singlet state goes as ( apart from a factor of ##\frac{1}{√2}##),
## [b_1^+d_2^+-b_2^+d_1^+]|0> → [d_1^+b_2^+-d_2^+b_1^+]|0> → [b_1^+d_2^+-b_2^+d_1^+]|0>##, hence singlet state has even parity and it can decay to even number of photons.
( 1 refers to spin up and 2 refers spin down).


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