1
$\begingroup$

enter image description here The image attached above describes the 4 non-zero helicity combinations for electron-positron annihilation giving muon - anti muon in the final state out of total 16 helicity combinations of this process. Am I right to say that in any of these diagrams, whether I talk of initial state or final state, we see total spin angular momentum $S = 1$ because photon is of spin 1? i.e Am I right to say that intermediate photons too obey spin conservations?

Or, does the total Spin=1 of final state or initial state in any of these diagrams merely a consequence of helicity conservation and thus it has nothing to do with the spin of photon?

Other way of asking the same question:

  1. Suppose, I take first diagram. $S_{total,initial}=1,S_{z,total,initial}=1$. Now can I say that $S_{total,initial}=1$ because photon has spin 1(Of course, we can add spin of electron and positron to get spin 1, but I am thinking to get the answer by considering propagator photon's spin)? If the answer is yes, then:

  2. Now suppose. I have LH electron and LH positron. Now such a combination has $S_{total,initial}=1,S_{z,total,initial}=0$. Now, again I have total spin 1 which is same as photon. Then why does the photon not couple to LH electron and LH positron simultaneously?

$\endgroup$
  • $\begingroup$ @CosmasZachos I edited the question! My question is: yes, the initial spin state is 1. But, my query is- Does the total initial state has spin 1 because photon has spin 1? Or because mathematically, other hellicity combinations cancel out and you are left with total of spin1 state. i.e. Are the other helicity combinations zero because they do not give Spin 1 total state?(Like LH electron and LH positron which have S=1, S_z=0 which goes against what I think that photon being spin 1 dhould allow this hellicity combination) $\endgroup$ – kbg Nov 14 '17 at 23:48
  • 1
    $\begingroup$ @CosmasZachos Just a small question: In case I have LH electron and LH positron, am I in a total spin state 1 or 0? (Of course total $S_{z}=0$) $\endgroup$ – kbg Nov 15 '17 at 1:50
  • $\begingroup$ @CosmasZachos Sir, could you explain the reason again. I did not get this point. $\endgroup$ – kbg Nov 15 '17 at 8:33
2
$\begingroup$

I think it might be a good idea to identify Thomson's 2009 lectures you are referring to. He actually explains the suppression of the remaining 12 combinations you appear to be worrying about, at high energies, $E\gg m_\mu\gg m_e$.

The QED action couples the electron to chiral combinations similar to the kinetic term, $$eA_\mu \bar{\psi} \gamma^\mu \psi= eA_\mu (\bar{\psi} L \gamma^\mu R\psi + \bar{\psi} R \gamma^\mu L\psi),$$ where L and R are the left and right projectors, respectively. That is, a photon couples L electrons to R positrons and R electrons to L positrons, only.

This is in sharp contrast to the mass terms for both electrons and muons, which do the very opposite, $$ m\bar{\psi} \psi = m(\bar{\psi}LL \psi +\bar{\psi}RR \psi), $$ coupling L electrons to L positrons and R electrons to R positrons. So, the fermion propagators involving the masses introduce effective couplings upon radiative corrections (such as the vertex correction, magnetic moments, etc) which would couple all 4 initial with all 4 final state combinations. (You are already aware of this for spin triplet, J=1 orthopositronium, but also spin singlet, J=0 parapositronium.)

You also know that at low energies, an effective (nonrenormalizable) QED vertex is the Pauli moment, $\propto F_{\mu \nu} \bar{\psi}\sigma^{\mu\nu} \psi$, coupling R with R and L with L like the mass term. So the photon couples to these combinations too, at low energy.

However, at high energies, one may imagine the masses dropping out, in which case there is no way to have these rogue RR and LL couplings--the systematic formal way is to see them suppressed in the amplitudes by powers of m/E. Also in this limit, chirality and helicity are the same, and they are good quantum numbers, preserved in the process. So the 4 amps you list survive, and the 12 rogue ones are suppressed, vanishing in such an idealized massless theory. You need no knowledge of the intermediate state, one or many photons, for that matter, once you know what the initial and final states can be.

It is easy to use the rotation matrices in your helicity amplitudes, $d^1_{1~1}(\theta)$=$(1+\cos \theta)/2$, etc., to rotate from the initial to the final state, square, and get the cross-sections provided--but I understand this is not your problem. Again, thinking about the spin of the photon is counterproductive (it is off shell, and 2 photons would introduce more mootness, etc.), and the moment the amps are restricted in the high-energy limit, you can get the angular distributions by inspection, which is the point here. A full quantum calculation would include everything, including the rogue stuff, with full energy suppression behavior... This is just a quick-and-dirty shortcut, but, at high energies, a good one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.