Consider the reaction $p+p\rightarrow \pi^++d$ (where $d$ is deuteron) which occurs via the strong interaction. From what I have read (in e.g. Williams 1992 (p326)) it would seem there is nothing preventing this from happening via the weak interaction (or for that matter the electromagnetic).

Is this the case? i.e. can $p+p\rightarrow \pi^++d$ occur solely through the weak interaction? And if not how do we know it must be via the strong (without looking at e.g. cross-sections)?


Just for completeness the only quark changes that occur are the creation of $d$ and $\bar d$ ($d$ here is the down quark).

  • $\begingroup$ Have you tried to draw the corresponding Feynman diagram? If yes, what's the problem; if no, why not? $\endgroup$
    – ACuriousMind
    Commented Feb 5, 2017 at 14:26
  • $\begingroup$ @ACuriousMind Yes I have - but I am unsure how to tell from the Feynman diagram which processes it is (I can only find information on when an interaction is definitely weak or EM but not strong). $\endgroup$ Commented Feb 5, 2017 at 14:30
  • $\begingroup$ Hint: write the elementary particles making up every particles and look for what would turn to what for the theory part, for the real world part, look it up in the pdg booklet. $\endgroup$
    – Ismasou
    Commented Feb 5, 2017 at 14:30

3 Answers 3


Obviously, $p+p\to d + \pi^+$ can occur in a world in which the weak interaction is turned off. In particular, it respects all the conservation laws of the strong interaction. (If you want to think in terms of Feynman diagrams: One of the $u$ quarks in the proton can pair produce, via gluons, a $d\bar{d}$ pair $u\to d +\bar{d}+u$, and the $u\bar{d}$ forms a charged pion. The proton is now a neutron, and can capture on the other proton to form a deuteron.)

What is it that we might mean by $p+p\to d+\pi^+$ occuring via the weak interaction? Clearly, the $d$ and $\pi^+$ are strongly interacting particles, and if we turn off the strong interaction these states cannot be formed. What we mean by $n\to p+e+\bar{\nu}$ being caused by the weak interaction is that it needs to involve at least one weak interaction (and many strong interactions). This is not the case here.

Having said this, it is of course not forbidden for one weak interaction to occur during the process $p+p\to d+\pi^+$. This could be a $u\to d+W^+$ transition followed by $W^+\to\pi^+$, but it could be other things as well. What this really describes is a weak correction to a strong interaction rate.

How would we ever know that this happens, given that the weak interaction is weak? The trick is to look for effects that are forbidden in the strong interaction, for example a parity violating spin asymmetry in $p+p\to d+\pi^+$. These effects are small, but they have been seen in experiment.

  • $\begingroup$ Are we basically saying that it can occur by both the strong and the weak but since the strong dominates, it usually occurs via the strong route. $\endgroup$ Commented Feb 5, 2017 at 15:22
  • 1
    $\begingroup$ Yes, but remember that this is QM, so there is no sense in which a given decay is caused by a specific Feynman diagram. Indeed the leading weak correction is usually an interference term between a strong and a weak amplitude. $\endgroup$
    – Thomas
    Commented Feb 5, 2017 at 15:33
  • $\begingroup$ "The trick is to look for effects that are forbidden in the strong interaction, for example a parity violating spin asymmetry" Exactly. Of course, it's not easy. $G^0$ needed a custom designed and built 10 million dollar detector and months of dedicated Hall C beam to pick out the parity violating scattering amplitude for $p(e,e'p')$ from the merely electromagnetic background. Go figure how much harder it's going to be when the dominate term is strong. $\endgroup$ Commented Feb 5, 2017 at 17:43
  • $\begingroup$ @dmckee : The effect is small, but it has been observed in several experiments, see arxiv.org/abs/1303.4132 $\endgroup$
    – Thomas
    Commented Feb 5, 2017 at 18:15

So the other answers have said what you need to know but I thought I would draw the Feynman diagram to help.

Bit of explanation as to how I decided on what Feynman diagram to draw

The $p + p \rightarrow \pi^{+} + d$ is the same as $ p+p \rightarrow \pi^{+} + n +p $ so we can see one of our protons survives. All we need now is to see how the $p \rightarrow \pi^{+} + n$ might occur.

Looking at the difference in quark content we know we have $uud \rightarrow u\bar{d} + udd$, so a $ud$ on both sides, lets try to see if we can see a process which only involves the others, ie $u \rightarrow d + u\bar{d}$. The charges agree but the solitary $u$ could lose +1 charge to get to be the solitary $d$. I knew the $W^{+}$ (from the weak interaction) could carry this charge so we plug that in to see if it works.

enter image description here Hey presto.

When comparing a decay process through different forces the decay time is key, decays through the weak interaction take much longer than those through the strong. Therefore if you wished to determine whether or not something decayed through the strong interaction you could measure its decay time. I won't go into that too much here since it doesn't seem to be the focus of your question.

  • $\begingroup$ I think there is a problem with angular momentum for a W to go to a pion directly. The W is a spin one the pi is zero. $\endgroup$
    – anna v
    Commented Feb 5, 2017 at 16:44
  • $\begingroup$ @annav I don't think you're right about that; the standard decay chain for the $\pi^+$ is via the $W^+$ into $\mu^+ +\nu_\mu.$ The pion has two distinguishable spin-1/2 quarks and they can either spin together (+1, -1) or opposite (0) so it should be spin-1. $\endgroup$
    – CR Drost
    Commented Feb 5, 2017 at 16:50
  • $\begingroup$ With that said this answer still has the problem that $u\mapsto d+\pi^+$ is in fact allowed by the strong force because $\pi^+ = u\bar d,$ so if it happens it happens more via the strong force than the weak one. $\endgroup$
    – CR Drost
    Commented Feb 5, 2017 at 16:52
  • $\begingroup$ a second particle is necessary , imo, the two quarks have to be constrained to spin 0, whereas the mu and nu_mu are free . can a vector turn into a scalar ? $\endgroup$
    – anna v
    Commented Feb 5, 2017 at 17:13
  • $\begingroup$ if you have access to a library look at inspirehep.net/record/358110?ln=en . there is always another particle $\endgroup$
    – anna v
    Commented Feb 5, 2017 at 17:20

A proton is an $uud$, a neutron is an $udd$, and a $\pi^+$ is $u\bar{d}$

One of the protons to turn into a neutron an up quark must turn into a down quark. This can happen with a gluon radiating a $d\bar{d}$ pair, the down going with the proton turning in into a neutron and the $u$ joining the $\bar{d}$ making a $\pi^+$. All the vertices are strong.

An up can turn into a down quark, given energy from the reaction, to make a neutron through the weak interaction in a proton proton scatter (it is the main fusion in stars) but in this case the rest of the interaction goes to an electron and an electron neutrino. To bind up in a deuteron the energies must be of the order of MeV.

To get a $\pi^+$ two extra vertices have to enter, the $W^+$ interacting with a quark from a radiated gluon, making the probability infinitesimally small.

Here is the minimum Feynman diagram of such a reaction,


It conserves off mass shell the angular momentum, and has two weak vertices. In comparison with the strong production this is not a detectable reaction due to the weak coupling constant. The phase space for the end product of a proton and neutron to bind into a deuteron is very small unless the scattering energies are of nuclear size, order of MeV

  • $\begingroup$ You can use this website to draw Feynman diagrams. Hope it helps. $\endgroup$ Commented Feb 5, 2017 at 19:28

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