I did some reading on the GIM mechanism today, and simply fail to understand how it works. I understand how the CKM-matrix can be used to do the basic calculation of the probability of, say, observing an up quark after a strange quark's decay over the weak interaction. However, I don't understand how the CKM matrix is applied to these Feynman diagrams:

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I see that the vertices for the weak interactions of the quarks are labeled with the corresponding matrix elements, but how do I take these two Feynman diagrams and infer that they cancel each other out?

  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Commented Aug 20, 2019 at 8:40

1 Answer 1


Well, the two interfering amplitude diagrams do not quite cancel each other out: they almost cancel each other out. That is to say in the notional limit that the mass of the u and the mass of the c were identical, the two diagrams would be identical except for the minus sign of the Cabbibo (CKM for 3 generations) matrix influence on the vertices, which you reassure us you are comfortable with.

To the extent the masses of the two quarks in the internal lines differ, the effect of them on the respective propagators differ, and so the respective results of the loops differ. In fact, good SM books compute the nonvanishing, but vastly suppressed amplitude. It is a function of $m_c/m_u$ which goes to 0 as that ratio goes to 1. Something like $\propto g^4 \frac{m_c^2}{M_W^2} ( 1-m_u^2/m_c^2)$.

So you might sensibly object that the term in the parenthesis is much closer to 1 than it is to 0. But, look at the factor multiplying it, $\alpha^2 m_c^2/M_W^2$, and how small it is: do this. (Still, if that were a part of your puzzlement, perversely, the introduction of c actually increases the $\Delta S=1$ rate instead of suppressing it! Historically, the rate was used to bound the mass of the then hypothetical c from above!)

When the third generation is introduced, the 3x3 unitary analog matrix (CKM) performs the same function. The quark mass ratios are bigger, but the couplings are suppressed at the vertices, so, if I recall, the effect of neglecting the 3rd generation is not dramatic.

  • Extra credit hypothetical: In an imaginary world where the masses of u and c differ by one part per million, but are huge, say half the mass of the W, would you have this strangeness-changing neutral current amplitude be suppressed, or not?
  • 1
    $\begingroup$ Based on what you wrote about the amplitude, the FCNC would not be suppressed if we had huge quark masses. Thank you for that detailed explanation. But I'm asking myself how to get to this amplitude. I've only had an introductory lecture to particle physics, so quantum field theory is a bit beyond me, but is there an equation I can understand that uses the matrix and the quark masses to finally obtain the amplitude (let's say ignoring third-generation quarks for simplicity)? $\endgroup$
    – Keno
    Commented Aug 20, 2019 at 5:56
  • $\begingroup$ Ah, sorry... I have no shortcut to Feynman diagram calculation, so, a bypass of QFT..... $\endgroup$ Commented Aug 20, 2019 at 12:27

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