On several texts it is stated that the semileptonic decays are different for the $K^0$ and $\overline{K^0}$, and each only gives leptons of particular charges (see www.hep.ph.ic.ac.uk/~dauncey/will/lecture16.pdf).

The allowed decays are $K^0\rightarrow l^+\nu\pi^-$ and $\overline{K^0}\rightarrow l^-\bar{\nu}\pi^+$ while $\overline{K^0}\not\rightarrow l^+\nu\pi^-$ and $K^0\not\rightarrow l^-\bar{\nu}\pi^+$.

Are $\overline{K^0}\rightarrow l^+\nu\pi^-$ and $K^0\rightarrow l^-\bar{\nu}\pi^+$ not allowed or only very suppressed with respect to the other semyleptonic decays? Why is the following diagram wrong/not allowed?

enter image description here

  • $\begingroup$ The leptons carry the charge of the virtual W emitted by the s or $\bar s$ involved. Can you write down the quark content of your states that you learned in your particle physics course and monitor this charge flow? $\endgroup$ – Cosmas Zachos Oct 21 '17 at 21:49
  • $\begingroup$ In fact, I see your reference does just that and rightly calls the rule obvious. Can you explain in your question what you do and do not understand about that? This is the unique way to conserve charge in every W vertex, no? $\endgroup$ – Cosmas Zachos Oct 21 '17 at 21:55
  • 1
    $\begingroup$ Yes, quite clear. Your now proposed doubly weak decay is allowed, of course, but suppressed prodigiously w.r.t. the singly weak one your text considers. Can you estimate by how much? $\endgroup$ – Cosmas Zachos Oct 21 '17 at 23:50

To lowest order in the electroweak coupling these processes are given by a sequence of subprocesses like: $$ q_1\to W^* q_2 \to (l \nu_l)q_2 $$ $W^*$ is an off shell W boson. Take a look at the flavour content of the hadrons involved and the quarks that are being held responsible for the decays. What values can the charges of the W bosons have in each case?

| cite | improve this answer | |
  • $\begingroup$ Thanks for your answer. I understand why the main decay of $K^0$ does not involve a negative lepton. I just wanted to know if the diagram I wrote is allowed (even if being very suppressed with respect to the other one). $\endgroup$ – Mil Oct 28 '17 at 14:08
  • $\begingroup$ Oh I see. I'm sorry for the inconvenience. As far as I can notice, rather than not being allowed, your process is highly suppressed, as Cosmas already pointed out. Because it is tree level it's straightforward to establish how much, check its vertices and compare with the one being taken into account. $\endgroup$ – Johann Liebert Oct 30 '17 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.