Recently, I was calculating this observable:

$\langle p s|\bar\psi(0)\gamma^{\mu}\psi(0)|ps\rangle$

Where we only consider the QED case. $\psi$ corresponds to massless Dirac fermion field, p is the fermion momentum, and s is the helicity.

There are three Feynman diagrams associated with this observable in one loop level.

1) enter image description here 2) enter image description here 3) enter image description here

Using dimensional regularization to regularize both UV and IR divergences,

for 1), I have the answer

$\frac{\alpha}{2\pi}(\frac{1}{\epsilon_{UV}}-\frac{1}{\epsilon_{IR}})\bar u(p,s)\gamma^\mu u(p,s)$

for 2), I have the answer

$-\frac{\alpha}{2\pi}(\frac{1}{\epsilon_{UV}}-\frac{1}{\epsilon_{IR}})\bar u(p,s)\gamma^\mu u(p,s)$

and 3) has the same result as 2)

Since it's 4 vector, the sum of these one-loop diagrams should be Lorentz-invariant, which means it should be zero.(It's related to the vector current conservation)

However if I sum 1)+2)+3), it's non vanishing. Someone told me I need to put a 1/2 factor in both 2) and 3), the self-energy diagram of which is in the external legs. Now here is my question.

Why is there a 1/2 factor in both diagram 2) and 3)? How does it come from?

Someone told me it's from the reduction of Greens function, but I can't figure out how LSZ reduction can give such a 1/2 factor. If it indeed comes from reduction, can anyone show me how 1/2 appears from it explicitly?

  • 1
    $\begingroup$ Write out the wick contractions. You'll find that the vertex contributes twice for combinatorial reasons. $\endgroup$ – Aaron Jun 19 '19 at 10:35
  • $\begingroup$ I know what you mean. However actually there are only two possible contractions, cuz $\psi$ only annihilate particles and create antiparticles. And the two would be cancelled by the 1/2 factor which comes from perturbation expansion up to $e^2$ order $\endgroup$ – James Liu Jun 19 '19 at 13:37
  • $\begingroup$ I upload a picture which I drew just now. imgur.com/TFMMB5U Hope it can help our discussion more. In one loop level, there are six possible connected contractions corresponded to three Feynman diagrams right? Thus, each two contractions correspond to one diagram. I still don't figure out where the double comes from. If the diagram I drew is wrong just let me know, thanks. $\endgroup$ – James Liu Jun 19 '19 at 14:17
  • $\begingroup$ I mean for each Feynman diagram, the symmetry factor is the same here. $\endgroup$ – James Liu Jun 19 '19 at 14:18
  • $\begingroup$ Sorry, I realized my mistake and had tried to remedy it, but you caught me haha. You're absolutely right in your diagrammatics. Can you explain why you expect this diagram to be zero? As far as I understand, the Ward identity says $k_\mu M^\mu = 0$ for on shell particles; why should $M^\mu = 0$ generically for 4-vectors? $\endgroup$ – Aaron Jun 19 '19 at 15:12

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