I'm currently going over some undergraduate exams on particle physics and I'm having problems with a specific reaction, namely

$$ \pi^- + p \to K^- + \Sigma^+ $$

which, in my opinion, is not allowed due to strangeness violation ($0 \rightarrow -2$), so it could only be a weak interaction. Since it's purely hadronic, though, it should be a strong interaction. Can anyone explain to me where I'm wrong? Or is the professor's answer wrong? He marked the "not allowed" answer with "wrong" (no further explanation).

  • $\begingroup$ That's a good reason why it's not allowed. Of course, there could be multiple reasons, and it's not like nature checks them in a specific order. $\endgroup$
    – MSalters
    Oct 20, 2014 at 11:31
  • $\begingroup$ wrong signs ? pi+p can go to K+ sigma+ :inspirehep.net/record/6000 $\endgroup$
    – anna v
    Oct 20, 2014 at 17:36

2 Answers 2


This reaction is not possible, but for a nontrivial reason.

As you note, the non-conservation of strangeness forces this to be a weak interaction, but there is nothing stopping baryons from partaking in weak interactions (as in e.g. this reaction). However, no charge is exchanged in the process, so it would be due to a weak neutral current, and these do conserve strangeness.

The proton has an ($\mathrm{uud}$) configuration and the pion contains ($\mathrm{d \overline u}$); the Σ is ($\mathrm{uus}$) and the K is ($\mathrm{s \overline u}$). This means that you would need to turn two down quarks into strange quarks, i.e. something like this:

Such a flavour-changing neutral current is as yet hypothetical, and has not been observed. (See for instance, the Z boson Feynman rules.)


This reaction can proceed through multiple charged weak interactions (i.e. instances of a virtual $W$ boson being emitted or absorbed, the flavor-changing weak processes of the standard model); provided sufficient center-of-mass energy, of course. Schematically:

$$ \small{ \begin{array}{cc} \pi^- \!\! \equiv \begin{Bmatrix} \overline u \cr d \end{Bmatrix} \cr p ~ \equiv \begin{Bmatrix} u \cr u \cr d \end{Bmatrix} \end{array} \rightarrow \begin{bmatrix} \overline u \cr \! u + W^- \!\! \cr u \cr u \cr d \end{bmatrix} \equiv \begin{bmatrix} \overline u \cr u \cr \! W^- + u \cr u \cr d \end{bmatrix} \rightarrow \begin{bmatrix} \overline u \cr u \cr s \cr u \cr d \end{bmatrix} \rightarrow \begin{bmatrix} \overline u \cr u \cr s \cr u \cr u + W^- \!\! \end{bmatrix} \equiv \begin{bmatrix} \overline u \cr u \cr s \cr \! W^- + u \! \cr u \end{bmatrix} \rightarrow \begin{bmatrix} \overline u \cr u \cr s \cr s \cr u \end{bmatrix} \equiv \begin{array}{cc} \begin{Bmatrix} \overline u \cr s \end{Bmatrix} \equiv K^- \cr \begin{Bmatrix} u \cr u \cr s \end{Bmatrix} \equiv \Sigma^+ \end{array} } $$

But surely the cross section of this reaction, $\sigma[~\pi^- + p \rightarrow K^- + \Sigma^+~]$,
is expected to be quite small compared to $\sigma[~\pi^- + p \rightarrow K^0 + \Sigma^0~]$, for instance.


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.