5.6: Example of Golden rule - beta decay
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A nucleus decays via the reaction n→pe−ˉν. to form a electron and antineutrino, releasing energy E0.
The simplest form for the matrix element describing nuclear β-decay is given by the so-called Fermi ansatz Vmk=GFM/Ω where Ω is the normalisation volume for the wavefunctions, |M|2≈1 is the wavefunction overlap between initial and final nuclear states and GF is a constant.
We can work in the COM reference frame, so the kinetic energy of the nucleus is zero. Momentum is conserved, so the final state has nuclear, electron and neutrino momentum P+p+q=0 while the energy released goes into the electron and neutrino, which for simplicity we treat as massless: E0=Ee+qc The proton and neutron are heavy compared with the electron and neutrino. Given that momentum must be conserved, the kinetic energy must be concentrated in the lighter particles.
The density of final states for the electron is given by the phase space volume
dn=d3pd3r(2πℏ)3
with a similar expression for the neutrino. Number of states in a volume of phase space is given by the number of electron states, times the number of neutrino states, provided energy is conserved:
dn=d3pd3r(2πℏ)3d3qd3r(2πℏ)3δ(Ee+qc−E0)
Using the relativistic relation E2=p2c2+m2c4 implies dpdE=Epc2
the normalisation volume is just ∫d3r=Ω, and rotational invariance gives d3p=4πp2dp.
All of which which simplifies the integral to
dn=Ω24π4ℏ6c6Ee√E2e−m2ec4E2νδ(Ee+Eν−E0)dEedEν
where Ee is the electron energy and Eν is the neutrino energy. What can actually be measured is the electron energy, so we integrate over the neutrino energies,
dndEe=Ω24π4ℏ6c6Ee√E2e−m2ec4(E0−Ee)2
This is the distribution of electron energies from beta decay: the rate fo emission of electrons at a particular energy is given by the Golden Rule
R=2πℏG2FM24π4ℏ6c6Ee√E2e−m2ec4(E0−Ee)2
Figure 5.6.1 shows the simplest case of beta-emission: neutron decay. Conservation laws tell us that the electron energy must lie between its rest mass (0.51MeV) and the total energy available (0.7823MeV). But the entire shape can be deduced from geometry.
Figure 5.6.1