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I'm trying to determine if the reaction

$$n\rightarrow p + \pi^-$$

is allowed. First of, this doesn't list this as one of the decay modes of the neutron, so I suspect that it should not be allowed. However, I would like to be able to argue this fact without reference to the link.

I have found that this reaction violates parity, because $P(n)=P(p)=+1$, but $P(\pi^-)=-1$, leading to the inequality $P(n) \neq P(p)P(\pi^-)$. This excludes the electromagnetic and strong force, and leaves me only with the possibility that the reaction is due to the weak force.

The reaction conserves angular momentum $\vec J$, baryonic number $B$ and electric charge $Q$, so no luck there.

I know that this reaction should respect $CP$, or at least $CPT$ invariance, but I'm having a hard time determining the effect of charge conjugation ($C$) on this reaction, because for each of the particles, applying $C$ changes the particle, and so I can't assign an eigenvalue $\pm 1$. For instance, $C\,|n\rangle = |\bar n\rangle$ is not an eigenvalue equation.

How can I determine if this reaction respects $CP$, or is there something else I should look at that I have failed to see so far? And if this reaction is indeed allowed, is the only way to confirm this to check every conservation law?

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1 Answer 1

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You missed a rather important conservation law: $$m_nc^2 < m_pc^2 + m_\pi c^2$$

But in general, yes, the only way to really confirm that a reaction is allowed is to check all the conservation laws. This is why we have tables of decay modes. Other people have checked the conservation laws (and done experiments to back it up) so you don't have to. I mean, it's still an important educational exercise, but in the "real world" when you just want to know if a particle decays in a certain way, you can simply check the tables.

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  • $\begingroup$ But could we not assign enough kinetic energy to the neutron, so that the total energy of the neutron is at least equal to the sum of the mass-energy of the produced particles? $\endgroup$
    – Bendik
    Commented Mar 13, 2016 at 9:31
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    $\begingroup$ That doesn't work; consider the reaction in the neutron's center-of-mass frame to see why. $\endgroup$
    – David Z
    Commented Mar 13, 2016 at 9:39
  • $\begingroup$ Of course... When you have stared at something for to long I guess this is what happens :p Thanks! $\endgroup$
    – Bendik
    Commented Mar 13, 2016 at 9:40
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    $\begingroup$ Note that $\gamma + \rm n \to p + \pi^-$ is allowed --- it's called "photoproduction" of pions. $\endgroup$
    – rob
    Commented Mar 14, 2016 at 18:19
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    $\begingroup$ I don't think its good advice to suggest relying on decay tables at all times. I think its useful to be able to figure this sort of problems on your own. $\endgroup$
    – JeffDror
    Commented Mar 27, 2016 at 13:10

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