Does antineutrino capture preferentially form neutrons?

In summary, the process of antineutrino capture can change quark flavor, with the most preferred reaction being p+ν=n+e+. However, at high energies, the conversion of proton into n is still preferred over conversion into any specified flavor of Λ. This is due to the different CKM matrix elements involved.
  • #1
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Besides the energetic preference (lower threshold, and more phase space above)?
Antineutrino capture is a weak process, so it can and does change quark flavour.
p+ν=n+e+
is actually
uud+ν=udd+e+
that is
u+ν=d+e+
But given enough energy (like cosmic ray neutrinos), do antineutrinos also get captured:
p+ν=Λ+e+?
Because this is just
udu+ν=uds+e+
that is
u+ν=s+e+
As you see, even though baryon charges match, a process
p+ν=Ξ+e+
would be obstructed, because Ξ has 2 s quarks. But process
p+ν=Λb+e+
should be just
u+ν=b+e+

Obviously these processes are impossible below energy threshold, and above they have a phase space factor. But at high energies, does proton conversion into n vs conversion into any specified flavour of Λ approach ratio of unity, or will any difference remain?
 
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  • #2
You would have different CKM matrix elements in there. And ##|V_{ud}| > |V_{us}| > |V_{ub}|##. So I would say yes, ##p+\bar{\nu}\to n+e^+## is preferred over ##p+\bar{\nu}\to \Lambda+e^+## even setting aside phase space constraints.

Edit: changed ##\nu\to\bar{\nu}##, see below
 
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  • #3
But it should be ##p+\bar{\nu} \rightarrow n+ e^+##.
 
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  • #4
Indeed, I just copied the reactions as noted in the OP (all ##\nu## there should also be ##\bar{\nu}## but the text correctly says antineutrino).
 
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