# How does the process $\bar{K}^0+p \to \Lambda^0+\pi^+$ work?

I was asked to find an energy for the $$\bar{K}^0+p \to \Lambda^0+\pi^+$$ interaction for an exercise and I got curious about how the Feynman diagram for this process would look like.

In terms of quarks, we have $$s\bar{d}+uud\to u\bar{d}+uds$$, so in the end the $$s$$ quark of the kaon is "swapped" with one of the up quarks of the proton, but I'm not sure about how this is done. My best guess is that the up quark undergoes a flavour change into a strange quark via weak interaction and the strange changes to an up by a similar process. Then the two W bosons produced on the flavour change would have to decay into quark-antiquark pairs that then annihilate, but that doesn't seem right to me. How does this interaction really work?

• Why would you consider rare Weak interactions when you can just swap quarks in a strong process? Commented Jan 4, 2021 at 18:39
• I didn't know that was possible, I don't know a lot about strong processes. Could you elaborate? Commented Jan 4, 2021 at 18:48
• There is an example strong force Feynman diagram for Delta decay on hyperphysics that may be useful to look at: hyperphysics.phy-astr.gsu.edu/hbase/Forces/feyns.html Commented Jan 4, 2021 at 22:26

In a effective theory, you are exchanging a virtual $$K^{\pm}$$ in the $$t$$-channel. The $$t$$-channel means there is no annihilation/creation of $$q\bar q$$ pairs.

That is, the proton's up quark is sent up to the neutral kaon, while the kaon's strange quark is sent down to the proton. In doing so, the proton becomes a $$\Lambda^0$$, and the neutral kaon becomes a $$\pi^+$$.

The exchange kaon has the quarks flowing in opposite directions (it's space like in the $$t$$-channel anyway), so you can consider it as a superposition of:

(1) The proton emits a $$K^+$$: $$p\rightarrow \Lambda^0 + K^+$$, followed by the $$K^0$$ absorbing it: $$K^0 + K^+ \rightarrow \pi^+$$

(2) The $$K^0$$ emits a $$K^-$$: $$K^0\rightarrow \pi^+ + K^-$$, followed by the proton absorbing it: $$p + K^- \rightarrow \Lambda^0$$

You should be able to draw that without any flavor changing weak currents, nor any $$q\bar q$$ annihilation.

In the $$s$$-channel, the $$d\bar d$$ annihilate, leaving a virtual $$\Sigma^0$$, which then decays into a lambda and a pion via the creation of $$d\bar d$$ pair:

$$K^0 + p \rightarrow \Sigma^+ \rightarrow \pi^+ + \Lambda^0$$