I'm reading David Tong's QFT notes, and am having a question about these diagrams: enter image description here When computing the matrix element, the second diagram picks up a minus sign 'due to statistics'. My question is does it matter which diagram has the minus sign, and how can I determine which diagram is the $t$ channel and which is the $u$ channel.

It seems like the $u$ channel is always drawn with crossed legs, but I think we can always treat the $u$ channel as $t$ channel and then assign the minus sign to $t$ instead.

  • $\begingroup$ Link? Which page? $\endgroup$
    – Qmechanic
    Commented Dec 12, 2022 at 19:06

1 Answer 1


If these are diagrams for a probability amplitude, then changing the sign of all diagrams amounts to phase shift, which will not matter when amplitude is squared - e.g., when one calculates the cross-section for the process.

  • $\begingroup$ Thanks, but if we don't square the matrix element, then it doesn't matter which diagram has the minus sign, right? $\endgroup$
    – IGY
    Commented Dec 12, 2022 at 17:26
  • $\begingroup$ And is there a name for the phase? $\endgroup$
    – IGY
    Commented Dec 12, 2022 at 17:27
  • 1
    $\begingroup$ @IGY probability amplitude is not measurable, so you will have to square eventually. In some interpherence experiments the phase may matter (relative to another process) - but I doubt that this is what you are dealing with. Also, the final labels are likely dummy indices, to be summed over in the final cross-section formula - so which diagram has phase is the matter of how you label them and how you draw them (which is Hartree and which is Fock.) In QM $\psi(x)$ and $-\psi(x)$ are the same state. $\endgroup$
    – Roger V.
    Commented Dec 12, 2022 at 17:29

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