Amplitudes in QFT are typically real. I'd like to understand the physical meaning of an amplitude having a phase. I know of three ways that amplitudes can get a phase:

  • If the couplings have an imaginary component
  • If there is a trace over the spin matrices, $ \gamma _\mu $ producing a $ i \epsilon _{ \alpha \beta \gamma \delta } $.
  • If a particle has a significant decay width we allow its propagator to have an imaginary contribution, $$ \frac{i}{p^2-m^2} \rightarrow \frac{i}{p^2-m^2+i m \Gamma } $$

I have heard many times that phases have to do with CP violation, but I'm not able to make the connection. In particular I'd like to know

  1. What is the meaning behind the different sources of phases in the amplitude mentioned above?
  2. Are there any other sources of phases that I'm missing?
  3. Is it true that if the above sources were gone then all amplitudes to all orders would be real (or is it possible for extra $i$'s to sneak in due to things like Wick rotation)?
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    $\begingroup$ Why do you say that "Amplitudes in QFT are typically real" ? Probability amplitudes are complex numbers. $\endgroup$
    – Trimok
    Commented Aug 27, 2014 at 13:59
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    $\begingroup$ @Trimok: True, that's why they don't need to be real. But if you calculate the amplitudes contributing to S matrix elements, ${\cal M}$, in QED for example, you will find that they are going to be real (even though they didn't apriori "have to be"). This is also mentioned in pg 232 on Peskin and Schroeder. $\endgroup$
    – JeffDror
    Commented Aug 27, 2014 at 14:05
  • $\begingroup$ I found a paper, where modifications of QED are analysed in the process $e^+e^- \to \gamma\gamma$. Some modifications, like implementing a scalar boson (vertex $Se^+e^-$) seems to lead ((more precisely the pseudo-scalar part) to complex amplitudes (see end page $509$, page $510$) . See also the general structure of the cross section pages $503 \to 506$ (the array page $506$ is interesting). $\endgroup$
    – Trimok
    Commented Aug 29, 2014 at 10:50
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    $\begingroup$ Interesting, I have yet to go through it in detail, but my most naive guess is then that having a pseudo scalar boson may break CP symmetry? $\endgroup$
    – JeffDror
    Commented Aug 29, 2014 at 18:46

1 Answer 1


My apologies if this answer is more simplistic that you were looking for, but you did ask why phases are related to CP-violations.

Most QFTs are CPT invariant, so a CP-violation is also a T-violation.

If you consider an S-matrix element to be the amplitude for a transition from $a$ to $b$, i.e. $\langle a|b\rangle$, then the T-version should be $\langle b|a\rangle$, so if no T-violation, the two amplitudes should be the same, so the amplitude is real.

All the sources of phase you mention seem to be about the introduction of phases into a perturbative approximation to the true amplitude, so the same argument should apply if a perturbative approximation up to a particular order is designed in a way where it (the perturbative approximation) is also CPT-invariant.

  • $\begingroup$ Thanks for your response. I didn't think of going back to basics and bra-ket notation. What you write is interesting but I don't think its the full story for a a few reasons: 1) The amplitudes in QFT aren't exactly brakets. 2) I don't think this incorporates the third case for when particles go on shell. 3) It seems to me that there must be a deep conspiracy in order for all the imaginary components to always cancel if we don't have CP violation and I'd like to understand better how this works $\endgroup$
    – JeffDror
    Commented Sep 6, 2014 at 13:01

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