I apologize for gaps in my knowledge that may make this obvious, but shouldn't the emission of a W boson from a neutron violate conservation of energy? The mass of a neutron is 939.565 MeV/$c^2$, so it would seem that this is also the maximum mass energy of the products of the decay. Otherwise energy would have to come from somewhere else. The W- boson has a mass of 80.379 GeV/$c^2$ and the proton has a mass of 938.272 MeV/$c^2$, adding up to 81.317 GeV. Doesn't this mean that the decay products have 80.377 GeV more energy than the original neutron? It also seems that the neutron doesn't have to gain this energy from somewhere else because free neutrons decay this way. Where does this energy come from?


3 Answers 3


The $W$-Boson has a mass of $M_W\approx 80\,$GeV. It also has a decay width $\Gamma \approx 2\,$GeV. That means the mass of a $W$ made in the lab is randomly distributed according to the Breit-Wigner distribution:

$$ f(M) = \frac{2\sqrt 2 M_W\Gamma\sqrt{M_W^2(M_W^2+\Gamma^2)}}{\pi\sqrt{M_W^2+\sqrt{M_W^2(M_W^2+\Gamma^2)}}} \times \frac 1{(M^2-M_W^2)^2+M_W^2\Gamma^2}$$

which is non-zero at beta-decay energies, but it is highly suppressed.

Note that if you consider the old-fashion perturbation theory picture of a down quark emitting a $W^-$ boson that then decays into an electron/anti neutrino pair:

$$ d\rightarrow u + W^-$$ $$ W^-\rightarrow e^- +\bar\nu_e $$

in which the $W^-$ has "borrowed" energy from the vacuum according to the Heisenberg Uncertainty Principle, you are only telling half the story.

You must also consider the spontaneous creation of a $W^+$ and the leptons, followed by absorption of the $W^+$ by a down quark:

$$|0\rangle \rightarrow W^++e^- +\bar\nu_e$$ $$W^++d \rightarrow u$$

In the fully covariant formulation, the $W$-vertices aren't time-ordered. The intermediate boson is neither emitted nor absorbed; rather, it is exchanged.

  • $\begingroup$ @nielsnielsen wikipedia "Breit-Wigner". $\endgroup$
    – JEB
    Jun 15, 2021 at 5:04

The W-boson is virtual, in the sense that it pops out of the vacuum as needed on borrowed funds, makes its brief cameo appearance to fulfill its weak-interaction job description, and then pays back the debt and exits stage left (metaphorically speaking). In this sense its own mass-energy doesn't show up in the reaction products.

Per Roger J. Barlow, it may help if you think of the intermediate (proton plus W boson) state in n→pW→peν as like when an electron tunnels through a potential barrier. With E<V, conservation of energy forbids the electron to be inside the barrier, yet it passes through.

Remember that Fermi's original model for beta decay agreed well with the facts for the momenta of the reaction products (at low energies) yet did not contain the W at all.

  • 1
    $\begingroup$ It may (?) help if you think of the intermediate $pW$ state in $n \to pW \to p e \nu$ as like when an electron tunnels through a potential barrier. With $E<V$ conservation of energy forbids the electron to be inside the barrier, yet it passes through. $\endgroup$ Jun 15, 2021 at 8:18
  • $\begingroup$ @RogerJBarlow, good point, should I edit? $\endgroup$ Jun 15, 2021 at 16:36
  • $\begingroup$ If you like.... $\endgroup$ Jun 16, 2021 at 13:07

Particle physics is modeled by quantum field theory and is based on the kinematics of special relativity. Both special relativity and quantum theory violate intuitions coming from just knowing classical mechanics conservation laws. Neutron decay is a quantum mechanical phenomenon and a result of the mass energy relations introduce by special relativity.

All particles are described by an energy and momentum four vector, whose "length" is the invariant mass of the particle when free, the neutron before decay at its center of mass, in this instance. The decay can be calculated using quantum field theory, with specific integrals described by Feynman diagrams. In a Feynman diagram there are input real particles, in this case the neutron, and output real particles , in this case the proton, the electron and the electron_antineutrino. Real means that the "length" of the four vector is equal to the mass of each particle as given in PDG..

In the Feynman diagram you refer in the question:


there is a W, it is true. It is called a W because it has all the quantum number and charges attributes of a real, on mass shell W, but it is off mass shell. Of mass shell means that the four vector in the calculation of the Feynman diagram is not equal to the mass of the W but varies between the limits of integration, which integration certainly obeys energy and momentum conservation. It is way off mass shell as you observe, but it is not a real W, it is a virtual W.

Please see this answer of mine which discusses virtual particles .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.