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Chirality, helicity, and electron-photon vertices, and polarised electron beams

  1. Dec 20, 2011 #1
    I am having some conceptual difficulties here. Lets start with the electron-photon vertex piece of the QED Lagrangian:

    [itex]-e\overline{\psi}\gamma^\mu\psi A_\mu[/itex]

    Now we can write this in the chiral representation as

    [itex]-e\overline{\psi}_L\sigma^\mu\psi_R A_\mu - e\overline{\psi}_R\overline{\sigma}^\mu\psi_L A_\mu[/itex]

    So this implies that every time an electron (in a chirality eigenstate) interacts with a photon then its chirality gets flipped. Ok I'll roll with that for now...

    Next, there is helicity. Imagine our electron propagating along, then it softly emits a photon and continues with almost unchanged momentum. It's helicity must be flipped because a unit of angular momentum gets carried off by the photon. If it was a really hard event then ok helicity might not flip, but at least spin projected along some constant direction must flip.


    Now what if I am at the SLC and I have made myself a polarised electron beam? This means we have a beam of electrons in a known helicity eigenstate right? But how can this beam stay polarised given my above comments? Shouldn't the spins be flipping all over the place each time an interaction with a photon occurs? Which should be all the time?

    Since this does not happen I guess I am wrong that every interaction with a photon flips the spin of an electron. So what is wrong with my reasoning?

    Also I am not very clear on the benefits of colliding a polarised electron beam with a positron beam, as opposed to an unpolarised beam. I initially thought it must be because different beam polarisations would behave differently under weak interactions, but that is related to chirality, not helicity.

    I'm so confused.
  2. jcsd
  3. Dec 20, 2011 #2
    But ok chirality and helicity become basically the same thing in the high energy limit, so does this mean that if I create two left-handed helicity electron beams and try to collide them then no interactions will take place? This would be a little bit mind blowing for me if it is true, after all they both have electric charge so naively I would expect them to scatter. Does it mean that repulsion within a bunch is similarly suppressed? Similarly is it bad for the SLC polarized e- beam, i.e. do they only scatter off half of the positrons in the unpolarised e+ beam?
  4. Dec 23, 2011 #3


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    NO! You made a mistake here. Photons do *NOT* flip chirality! The interaction should have been

    [tex]-e\overline{\psi}_L\sigma^\mu\psi_L A_\mu - e\overline{\psi}_R\overline{\sigma}^\mu\psi_R A_\mu[/tex]

    Just like the Kinetic term. It is the MASS term that flips chirality.
  5. Dec 23, 2011 #4
    Oh, well that certainly would make more sense. I must have gotten my gamma matrix in the wrong representation or something. Must go check.
  6. Dec 23, 2011 #5
    Ahh, bugger I see what I did. The gamma matrix in the Weyl basis indeed flips the left and right chirality pieces of the spinor fields, but of course there is a [itex]\gamma^0[/itex] hidden in the [itex]\overline{\psi}[/itex] which flips it back again. Duh.
  7. Dec 23, 2011 #6


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    quite so! you do these calculations a million times you don't even think about them anymore... :wink:
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