I am trying to understand CP-Violation in the Kaon system using feynman diagrams and matrix element. Here is a slide from Mark Thomson corresponding exactly to what I am looking for (http://www.hep.phy.cam.ac.uk/~thomson/lectures/partIIIparticles/Handout12_2009.pdf). As I understand, adding a third quark family induces a complex phase in the CKM matrix that is responsible for cp-violation. However, I find Thomson's slide very confusing (or wrong). First, the rate should be proportional to the matrix element squared $|M_{fi}|^2$ and not $M_{fi}$. If I am correct, I would expect:

$\Gamma(K^0\rightarrow \bar{K^0}) - \Gamma(\bar{K^0}\rightarrow K^0) = |M_{fi}|^2-|M_{fi}^*|^2 =0 $

So no CP violation... Am I understanding this slide wrongly? My idea is that we need to consider the other diagram responsible for kaon oscillations to induce cp violation and obtain

$\Gamma(K^0\rightarrow \bar{K^0}) - \Gamma(\bar{K^0}\rightarrow K^0) \neq 0$

Can someone help me, Thanks!

  • $\begingroup$ In general $|a+b|^2\neq|a|^2+|b|^2$ $\endgroup$
    – fqq
    Commented Jul 22, 2016 at 9:34
  • $\begingroup$ You are right that your notes are glib and schematic on this, and that these "rates" Γ are not actual rates observable directly, in which case you'd need to square amps. Instead, the box diagram theoretical difference, $O(G_F^2)$, is ~$\epsilon \propto \sin \delta$, amplitude contaminations of $K_L$ by $K_S$, which are ultimately proportional to CP-forbidden semileptonic decay differences of physical widths, for example. $\endgroup$ Commented Jun 28, 2018 at 14:11

2 Answers 2


Let us consider what would happen if CP was actually a symmetry. Then, arguing non-perturbatively, the overlap between a $K^0$ state and a $\bar K^0$ after time $t$ would be \begin{equation} \langle \bar K^0| \exp(-i H t) K^0 \rangle. \end{equation} Let $U$ denote the operator implementing CP symmetry, such that $U^{-1}=U$. Then, since $[U,H]$ by assumption, we can write \begin{align} \langle \bar K^0| \exp(-i H t) K^0 \rangle&=\langle U K^0| \exp(-i H t) K^0 \rangle\\ &=\langle K^0| U\exp(-i H t) K^0 \rangle\\ &=\langle K^0| \exp(-i H t)U K^0 \rangle\\ &=\langle K^0| \exp(-i H t) \bar K^0 \rangle. \end{align} This is the overlap between a $\bar K^0$ state and a $K^0$ after time $t$. Then, still assuming CP symmetry, these overlaps would have to be equal order by order in perturbation theory. Specifically, the two diagrams you quote would have to be equal. So at the theory level, CP violation is already demonstrated.

You are correct that the actual oscillation probability in time $t$ is given by \begin{equation} |\langle \bar K^0| \exp(-i H t) K^0 \rangle|^2, \end{equation} so if the two diagrams you quote were the only ones to contribute to the oscillation, the CP violation would not be experimentally detectable in this process, even though see it in the theory. However, when adding complex numbers $|a+b|^2\neq |a|^2+|b|^2$ in general, so by looking at the contribution by other diagrams, as you suggest, the oscillation probalilities might come to differ.

In short, the calculation you show is enough to demonstrate that the theory breaks CP violation. However, you are correct that it does not yet tell you the Kaon oscillation as an observable phenomenon does so. For this you have to consider the additional diagrams that contribute, as you suggest.

  • $\begingroup$ Does the CP violation in the neutral Kaon sector vanish if the CP-violating phase $\delta$ happens to vanish in CKM matrix? In other words, is it the nonzero $\delta$ in CKM matrix that is solely responsible for CP violation in the neutral Kaon sector? @AntonQuelle $\endgroup$
    – SRS
    Commented Jun 25, 2018 at 14:09
  • $\begingroup$ @SRS : In the SM, yes. $\endgroup$ Commented Jun 27, 2018 at 15:01
  • $\begingroup$ @CosmasZachos How can I see that? $\endgroup$
    – SRS
    Commented Jun 27, 2018 at 15:32
  • $\begingroup$ Go to the SM Lagrangian and set δ=0. Now all CKM matrix elements can be made real--this is why K-M got their NP! It is CP invariant. Anything you'd compute with them would not have an Imaginary part. The computation of off-diagonal transition amps is nicely summarized in Li & Cheng, Ch 12.2. $\endgroup$ Commented Jun 27, 2018 at 16:38

To study CP, you want to experiment with eigenstates of CP. The $K^0$ is itself not an eigenstate of CP. But you can combine $K^0$ and $\bar{K^0}$ to form $K^0-\bar{K^0}$ CP even and $K^0+\bar{K^0}$ CP odd eigenstates. When you do that, and look at their possible decays, you find the long-lived CP-odd $K_L$ and the shorter-lived CP-even $K_S$ as physical particles. The $K_S$ CP state will decay to two pions, but the $K_L$ can't, so it decays over a longer time into three pions.

There's an excellent summary of this starting on page 57 of these lectures.

If there's no CP violation, those are good eigenstates, and you'll never see a $K_L$ decay to the two pions indicative of the other state.

Cronin and Fitch (see this summary) found that the $K_L$ did decay to the other state, hence CP violation (that's a much harder experiment than it sounds, because there are lots of other things that could be happening; knowing the two-pion decays really were from $K_L$ involved a lot of work).

So what causes this in terms of your box diagram?

The box diagrams result in mixing of the $K^0$ and $\bar{K^0}$; over time, one state will become the other. If $K^0$ was becoming $\bar{K^0}$ at the exact same rate that $\bar{K^0}$ was become $K^0$, the + and - states wouldn't change. But the phase difference means they're not the same, and the amount of the two K's slowly varies, adding a bit of - to the + state and vice versa. And that means that $K_L$ slowly mixes to a bit of opposite-CP $K_S$: CP violation.

  • $\begingroup$ Does the CP violation in the neutral Kaon sector vanish if the CP-violating phase $\delta$ happens to vanish in CKM matrix? In other words, is it the nonzero $\delta$ in CKM matrix that is solely responsible for CP violation in the neutral Kaon sector? @BobJacobsen $\endgroup$
    – SRS
    Commented Jun 27, 2018 at 15:34
  • $\begingroup$ In the Standard Model, a relatively-real CKM matrix leads to zero CP violation. It's an open question about "solely responsible" though; there are some (speculative?) papers about CP violation in the neutrino sector (i.e. beyond the SM & not yet measured) and CP violation phenomenology. $\endgroup$ Commented Jun 27, 2018 at 15:37

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