In the reaction $\nu_e + p \rightarrow e^-+ \Delta^{++}$, a Feynman diagram would be:

enter image description here

So my question is, could a neutrino emit that $W^+$ boson (knowing that the boson's mass is much higher than the neutrino's)?

Thanks in advance.

  • 2
    $\begingroup$ Do you know what "off-shell" means? $\endgroup$
    – JEB
    Dec 15, 2022 at 16:34
  • $\begingroup$ No, I haven't studied QFT yet. $\endgroup$
    – user9867
    Dec 15, 2022 at 16:36
  • 2
    $\begingroup$ The W "emitted" is a virtual one, and you never observe it directly: it is "an intermediate state", and its "mass" doesn't stop you.... The magic of QM! $\endgroup$ Dec 15, 2022 at 16:39
  • 1
    $\begingroup$ @conradDell wrong. It is the invariant mass, the length of the four vector that determines what interactions and decays a particle can have without violating conservation laws. hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Dec 15, 2022 at 18:40
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    $\begingroup$ @conradDell If a lone neutrino moving fast enough could decay to $W+e$, then you find the threshold speed in the $x$ direction, $v_x$, and the threshold in the $-x$ direction, $u_x$, repeat for $y, z$ and $\vec v + \vec u$ is the absolute rest frame of space. Can't have that. $\endgroup$
    – JEB
    Dec 15, 2022 at 22:55

2 Answers 2


If you are asking about the vacuum reaction

$$ \nu\to We $$

then the answer is no, that can’t happen. In the rest frame of the neutrino, the final state would be more massive than the initial state, which can’t conserve energy.

If you allow for a massless neutrino, the reaction is still forbidden, because the massless initial state would have no rest frame, but the massive final state would.

In an interaction picture, like you’ve drawn, it’s not really correct to say that the $W$ boson is “emitted.” It’s better to say that the neutrino-proton scattering is mediated by the field associated with the weak charged current. The Feynman diagram, which shows this field interaction as a “virtual particle,” is really a shorthand for an integral over a continuum of momentum exchanges which the $W$-associated field can mediate.


Look at this process in the center-of-momentum frame, at high energy so we can ignore the lepton masses, then the initial state 4-momenta are:

$$ p^{\mu}_p = (E_p, \vec p) $$ $$ p^{\mu}_{\nu_e} = (p, -\vec p) $$

and the final state's are:

$$ p^{\mu}_{\Delta} = (E_{\Delta}, \vec p') $$ $$ p^{\mu}_e = (p', -\vec p') $$

Since energy and momentum are conserved at vertices:

$$p^{\mu}_W = p^{\mu}_{\nu_e} - p^{\mu}_e = (p-p', -\vec p -\vec p') $$

so the mass of the $W$ is:

$$M^2_W=p^{\mu}_Wp_{\mu\,W} $$ $$M^2_W = (-p-p')^2 - (\vec p-\vec p')\cdot(\vec p-\vec p')$$ $$M^2_W = p^2+p'^2 -2pp' - (p^2+p'^2 -2\vec p \cdot \vec p')$$ $$M^2_W = -2pp'(1+cos{\theta_{CM}}) < 0$$

So the mass-squared is negative, and $p^{\mu}_W$ is space-like (hence @rob's answer saying you can't time-order the vertices).

That's the nature of $t$-channel virtual particles.


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