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Suppose the QED with massless electron and assume the process $$ \gamma \to e\bar{e}, $$ driven by the triangle loop.

Is this process possible; precisely, is the quantity $$ \Gamma \simeq (2\pi)^{4}\int d\rho_{2}|M_{\gamma \to e\bar{e}}|^{2}\delta(p_{e} + p_{\bar{e}} - p_{\gamma}), $$ with $d\rho_{2}$ being 2-particle phase space, non-zero?

Name YYY
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    I'm voting to close this as a homework-like question; essentially the question boils down to computing it in massless QED and there has been no attempt made whatsoever by the OP. – JamalS Jan 06 '17 at 20:56
  • @JamalS : It seems that the question may not require the computation of above quantity. I just write down the expression for the decay width, which may be zero because some simple reason. – Name YYY Jan 06 '17 at 21:04
  • For the record, to tree level and to one loop-order, the amplitude vanishes. Off the top of my head, I cannot think of any symmetry argument as to why the amplitude should vanish to all orders in perturbation theory. EDIT: oh, wait, what about Furry's theorem? – AccidentalFourierTransform Jan 06 '17 at 21:06
  • Possible duplicate of Decay of massless particles – AccidentalFourierTransform Jan 06 '17 at 21:20

1 Answers1

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Mostly in response to @Name YYY 's sound remark: Kinematics and helicity (= chirality for massless fermions assumed here) ensure the process vanishes.

The massless products need to share the photon's mass and momentum, so they have to be collinear and moving in the same direction, from energy-momentum conservation.

So they have to share the photon's helicity. Take a right handed photon. It should decay to a right-handed electron and a right-handed positron, helicities 1=1/2+1/2; but these are not coupled by the vector coupling γ : the left-handed electron couples to the right-handed positron; while the right-handed electron couples to the left-handed positron.
The decay does not go.

Edit to address chiral extension in comment: The no-go argument, of course, also applies to chiral gauge fields, and gets corrected with masses increased as a perturbation. So, a putative massless W- would also not decay to a masses electron and antineutrino, since the electron would be left-handed and the antineutrino right-handed, so they'd add up to helicity zero.
In fact, as a pedagogical lark, let's stand the conventional helicity argument for the stability of a massive charged pion against decay to putative massless leptons on its head. In the rest frame of the π-, the electron and antineutrino would come out back-to-back. The -1/2 helicity of the electron would subtract from the helicity of the right handed antineutrino to spin -1 on that axis, unavailable to the parent pion; so chirality relaxes this through the mass of the heaviest lepton, and the decay rate goes as the fifth power of the electron mass. (This is wildly subdominant to the fifth power of the muon mass, so this is why the pion decays to muons instead, in practice, which would be counter-intuitive at first blush!) But now appreciate this suppression would be there even if the weak coupling were not chiral, since a vector like coupling would allow the positive electron helicity mode, and negative antineutrino helicity, also subtracting to +1, also unavailable. The culprit, as indicated to start with, is always the γ-matrix coupling. Jacob&Wick ever rule!

Cosmas Zachos
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  • Thank You! It seems that this argument isn't applied when the gauge field is chiral (interacts with both left fermions and antifermions), right? – Name YYY Jan 08 '17 at 12:06
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    It applies for everybody massless, e.g., a massless W, massless electron, and massless anti neutrino.... Recall the helictity argument of the charged pion decay? – Cosmas Zachos Jan 08 '17 at 13:34