Free neutron vs bound neutron decay

[anorlunda]

I was curious as to the difference in free neutron decay and the lack of decay when bound in a nucleus. I found the following statement here http://en.m.wikipedia.org/wiki/Neutron#Free_neutron_decay

When bound inside of a nucleus, the energetic instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to appear by beta decay, and thus participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons in a nucleus are not necessarily so.

I don't understand that. Is the wave function of the neutron different when free or bound? The phrase "is balanced" sounds wrong; balanced by who or what?

Perhaps they mean that instead of a Feynman diagram with a neutron in and proton/electron/anti neutrino out, we should use a diagram with a nucleus in and a nucleus(with higher atomic number)/electron/anti neutrino out. That would suggest to me the wave function of the nucleus supersedes the wave functions of the constituent nucleons.

Would someone please explain those two Wikipedia sentences, or point me to a more understandable source?

Follow up question: Since the binding energy per nucleon is different in different elements, is the stability of a bound neutron a function of which element/isotope it is bound in?

 

[bcrowell]

For example, let's say we have a 16O nucleus. It's not possible for one of its neutrons to undergo beta decay because the resulting system would be 16F (plus an electron and a neutrino). But 16F has a higher mass-energy than 16O, so this would violate conservation of mass-energy.

Similarly, a free proton can't beta decay because of conservation of mass-energy.

 

[mfb]

I don't understand that. Is the wave function of the neutron different when free or bound?

Sure, it is bound, so it has a different energy. Even more important, a produced proton would have a different energy as well - too high to be accessible in the decay.

Perhaps they mean that instead of a Feynman diagram with a neutron in and proton/electron/anti neutrino out, we should use a diagram with a nucleus in and a nucleus(with higher atomic number)/electron/anti neutrino out. That would suggest to me the wave function of the nucleus supersedes the wave functions of the constituent nucleons.

That's another (more general) way to see it, yes.

Follow up question: Since the binding energy per nucleon is different in different elements, is the stability of a bound neutron a function of which element/isotope it is bound in?

Sure.

 

[anorlunda]

I wish Wikipedia would explain as simply as you guys did. Thanks.

 

[anorlunda]

I've been thinking more on this and I would like a follow up question please.

I've taught that when two electrons bind in a singlet pair, the identity and individual quantum states of the constituents disappear. Only the quantum states of the singlet may be observed.

Should I think of a multi nucleon nucleus in a way analogous to the singlet pair?

 

[mfb]

Yes.