= Solution
The charged weak current changes one <neutron> <down quark> into an <up quark>:
$$
d\to u+W^{-*},\qquad W^{-*}\to e^-+\bar\nu_e,
\qquad\boxed{n(udd)\to p(uud)+e^-+\bar\nu_e.}
$$
The other two valence <quarks> are spectators in this tree-level <Feynman diagram>. The intermediate <W boson> is highly virtual; the available <neutron> decay energy is far below its on-shell rest energy. Charge is conserved at each vertex, and the <electron>'s lepton number is balanced by the <electron antineutrino>.
The reversed quark-level charged current is allowed as an interaction, but a \b[free] <proton> cannot decay as $p\to n+e^++\nu_e$ because $m_p<m_n+m_e$. <Four-momentum conservation> leaves no accessible final-state <Lorentz-invariant phase space>. The absence of this free-proton beta-plus decay is therefore kinematic, not an absence of a $u\to d$ weak vertex. Binding-energy differences in a nucleus, or incoming energy in a collision, can change whether the corresponding conversion is possible. Exotic baryon-number-violating <proton> decay is a separate question and is not generated by this ordinary charged-current diagram.
\Image[/past-exam-of-the-sub-atomic-physics-course-of-the-university-of-oxford/2016-neutron-beta-quark-diagram.png]
{title=Neutron beta decay with two spectator quarks and a virtual W boson}
{description=Original tree-level quark diagram. Fermion-flow arrows distinguish the emitted electron from the electron antineutrino; the W is off shell.}
Back to article page
