Understanding Neutron Decay in Elementary Particle Physics

[elas]

In ‘Introduction to Elementary Particles’, David Griffiths makes the following two statements:

a neutron decays into a proton, an electron, and an antineutrino (1.8)

And later:
but the following decay is not observed
an antineutrino plus a neutron decay into a proton and an electron (1.13)

But why was it expected? Surely the expected decay for (1.13) would be to a proton, electron and two antineutrinos

Clearly I am missing something, but I cannot see what; any clarification would be appreciated.

 

[Norman]

First of all (1.13) is not a decay- it is a scattering event. Second, when you bring the antineutrino to the left hand side of (1.8) it must be changed into a neutrino by crossing symmetry. Read pages 20-21 for information on crossing symmetry. The idea here is the concept of conservation of lepton number. In 1.12 lepton number is conserved, while in 1.13 it is not. While there may be reactions where lepton number is not conserved, the assumption at the time was that lepton number was conserved and the expectation in the scattering experiment was that if the neutrino and its antiparticle were distinct particles they would carry different lepton numbers (by the definition of antiparticle) and therefore reaction (1.12) would occur but (1.13) would not.

 

[ZapperZ]

In ‘Introduction to Elementary Particles’, David Griffiths makes the following two statements:

a neutron decays into a proton, an electron, and an antineutrino (1.8)

And later:
but the following decay is not observed
an antineutrino plus a neutron decay into a proton and an electron (1.13)

But why was it expected? Surely the expected decay for (1.13) would be to a proton, electron and two antineutrinos

Clearly I am missing something, but I cannot see what; any clarification would be appreciated.

There are many reasons, but one most obvious is conservation of spin angular momentum. The latter was "expected" because the initial situation before decays has the possibility of spin of either zero or 1 (1/2 from antineutrino, 1/2 from neutron). The situation after the decay has the same possibility (1/2 from proton, 1/2 from electron). So just going by this conservation along, such decay channel can't be ruled out. So there are other criteria why this decay isn't possible.

Zz.

 

[elas]

Thanks for the replies, but I am still not clear for the following reasons-

First of all (1.13) is not a decay- it is a scattering event

Griffiths does not use the term "scattering" but at this point is also unclear about decay, so I checked with 'Wikipedia' and they clearly state that Cowan and Reines experiment used decay events.

conservation of spin angular momentum.

On the left of 1.13, antineutrino (1/2 spin); neutron (0 spin).
On the right of 1.13 (proton (1/2 spin); electron (1/2 spin)

I assume that the antiparticle 1/2 spin of the antineutrino cancels the particle 1/2 of the proton within the (0 spin) neutron. But where is the antiparticle half spin, needed on the right of the equation, in order to reduce the total particle spin to 1/2.
Which brings me back to my original question - why did Cowan and Reines expect to find something predicted by an unbalanced equation?

 

[Janitor]

...
On the left of 1.13, antineutrino (1/2 spin); neutron (0 spin).
On the right of 1.13 (proton (1/2 spin); electron (1/2 spin)...

Why do you believe a neutron is spinless?

 

[zefram_c]

Thanks for the replies, but I am still not clear for the following reasons-

First of all (1.13) is not a decay- it is a scattering event

Griffiths does not use the term "scattering" but at this point is also unclear about decay, so I checked with 'Wikipedia' and they clearly state that Cowan and Reines experiment used decay events.

A "decay" in particle physics typically means a process where one initial particle is converted into other stuff. The term may be used liberally in some contexts; I don't know.

On the left of 1.13, antineutrino (1/2 spin); neutron (0 spin).
On the right of 1.13 (proton (1/2 spin); electron (1/2 spin)

Neutron has spin 1/2.

There simply exists no mechanism within the SM to run (1.13). Of course, this means very little, since the SM is not final. Even considering extensions of the SM, it may be the case that (1.13) would still not proceed, since it violates lepton number conservation. This conservation law is largely empirical - it does not lie on equal footing with eg. conservation of energy or charge. If neutrinos were Majorana particles, (1.13) could proceed. There is simply no evidence that it does.

 

[elas]

Neutron has spin 1/2.

My mistake, thanks for correction and for explanation, can now proceed with studies

 

[MarieMuch]

I have two question, hope someone can help me...

1) Is the neutron decay isotropic? Or does it somehow depends on the spin direction?
2) Does anybody know a book, or even a code dealing with the calculation of the
energy distribution of the electron/neutrino after the decay?

 

[quantumdude]

1) Is the neutron decay isotropic? Or does it somehow depends on the spin direction?

Neutron decay is anisotropic. That's because the neutron decays weakly, and the weak interaction violates parity conservation. This was predicted by Lee and Yang, and confirmed by Madame Wu.

2) Does anybody know a book, or even a code dealing with the calculation of the
energy distribution of the electron/neutrino after the decay?

It's been a while since I've looked it up, but I would think that any book that treats fermi statistics would provide you with the machinery for this. Someone else will probably have a more specific answer.