Would it be possible for a neutron to lose a positron and become an antiproton?
Or would would it need to be the decay of an antineutron to antiproton instead?
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4 Answers
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It would violate the law of conservation of baryons. Baryons (half-integer-spin particles, i.e. s=1/2, 3/2, 5/2,... interacting through the strong force) cannot be created at will, but must conserve the total baryonic number: protons and neutrons both have $+1$ baryon number, while their antiparticles, antiproton and antineutron, have baryon number $-1$ each.
Thus, if the neutron were to decay into an antiproton and a positron (plus, I presume, a neutrino to conserve the lepton number), it would conserve the total charge and the lepton number ($-1$ for the positron and $+1$ for the neutrino), but it would violate the conservation of total baryon number.
This statement is nothing more than a condensed result of many years of searching for these kinds of events, without ever finding one. Why nature should conserve baryon number is not known, at present. In fact, many theories predict that conservation of bayon number is not an absolute law, but just a by-product of doing physics in the low-energy limit. In fact, many experiments are searching 8so far without any success) for violation of the conservation of baryon number, in particular for the free decay of protons. Since protons are the lightest baryons, their decay could occur (if at all) into non-baryonic particles, thus violating the conservation of baryon number.
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$\begingroup$ Note that the experimental lower bound on the time between $\Delta B=2$ oscillations between neutrons and antineutrons is surprisingly short, roughly $10^8\,\text{s}$. By far the most probable pathway for a $n\to\bar p e^+\nu$ would be a $n\to\bar n$ transition followed by an ordinary weak decay. See the particle data group for references. $\endgroup$– rob ♦Commented Sep 5, 2014 at 1:57
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1$\begingroup$ The decay $n\rightarrow\bar{p}+e^{+}+\bar{\nu}$ is actually probably the preferred one, since in the standard model, $B$ and $L$ are separately nonconserved in sphaleron processes. Howver, $B-L$ is still absolutely conserved in the standard model. $\endgroup$– Buzz ♦Commented May 8, 2016 at 18:28
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A neutron contains (on average) 1 up quark and 2 down quarks. The decay to a proton occurs when a down quark emits a W$^-$ particle and changes to an up quark. This gives a proton with two up quarks and 1 down quark. The W$^-$ particle decays to an electron and anti-neutrino.
However an antiproton contains 2 up antiquarks and 1 down antiquark, which is completely different to the 1 up quark and 2 down quarks in a neutron. Even if it were possible for antiquarks to turn into quarks (it isn't) the decay would require all three quarks to change at the same moment.
So no, the decay you describe cannot occur.
answered Sep 4, 2014 at 14:08
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$\begingroup$ This is not really an explanation. Quarks can turn into each other, like for instance in the standard beta decay of the neutron where a down quarks decays into an up quark (plus stuff). Thus one might naively ask: why can't a quark decay into an anti-quark, plus stuff? The reason is: baryon number conservation. Only interactions violating baryon number conservation can do that, and none exists within the standard model (except for an unobservable exception noted above). $\endgroup$ Commented Sep 5, 2014 at 7:31
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I'm not a particle physicist, but my understanding is that baryon number is not strictly conserved in the standard model. See, e.g., http://en.wikipedia.org/wiki/Baryon_number#Conservation . And there are also theories beyond the standard model in which baryon number is not conserved; e.g., many people do seem to think that proton decay is a reasonable thing to expect and look for. B-L (baryon number minus lepton number) is, AFAIK, considered more fundamentally likely to be conserved. So although the decay process $n\rightarrow \bar{p}+e^+$ violates conservation of B, perhaps a more fundamental reason not to expect it to happen is that it would go from $B-L=1$ in the initial state to $B-L=-2$ in the final state.
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1$\begingroup$ Chiral anomaly in the standard model does indeed lead to the violation of baryon number conservation, but the expected rate is so small to be considered unobservable, G. 't Hooft, "Symmetry breaking through Bell-Jackiw anomalies," Phys. Rev. Lett. 37, 8 (1976). Proton decay outside the standard model instead leads to potentially observable effects, hence the numerous experiments already carried out, or still being carried out (albeit to no avail, so far). $\endgroup$ Commented Sep 5, 2014 at 7:28
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According to Wikipedia, the antineutron is one anti-up and two anti-downs; it does indeed decay exactly as the question suggests something should, to a positron and an antiproton (and a neutrino)
The only definite source for this that I can find (http://www.in2p3.fr/physique_pour_tous/questions/reponses/antimatiere.htm) is in French; essentially, it states in pertinent part that if A decays to X, Y, and anti-Z (for any given A), then an anti-A decays with the same half-life to anti-X, anti-Y, and anti-anti-Z (which is to say Z). It is harder to observe, though, because antimatter tends to bump into matter and annihilate itself before decaying, especially when the decay has as long a half-life as a neutron does.
