16.451 Lecture 16: Beta Decay Spectrum 29/10/2003 e e p n (and related processes...) Goals: • understand the shape of the energy spectrum • total decay rate sheds light on the underlying weak interaction mechanism f if if M 2 2 r d r V M i f if 3 * ) ( f f dE dn/ (transitions / sec) Starting point: “Fermi’s Golden Rule” again! (lecture 6) matrix element density of states 1
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16.451 Lecture 16: Beta Decay Spectrum 29/10/2003 (and related processes...) Goals: understand the shape of the energy spectrum total decay rate sheds.
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16.451 Lecture 16: Beta Decay Spectrum 29/10/2003
eepn
(and related processes...)
Goals:
• understand the shape of the energy spectrum• total decay rate sheds light on the underlying weak interaction mechanism
fifif M 22
rdrVM ifif3* )(
ff dEdn /(transitions / sec)
Starting point: “Fermi’s Golden Rule” again! (lecture 6)
matrix element
density of states
1
Simplest model is to take a pointlike interaction with an overall energy scale “G”: (Fermi, 1934 – almost right!)
Matrix element Mif?
n
p
e
e
rdrrrrGM inefpif3
,***
, )()()()(
(the interaction is proportional to the wavefunctionoverlap of initial and final state particles at the samepoint in space)
W-
e
e
n
p
Standard Model description: an extended interaction, but the range is only about 0.002 fmwhich is just about zero!
The Standard Model can ‘predict’ the value of Gin terms of model parameters, whereas in Fermi’stheory, it remains to be determined from experiment.
2
Spin considerations: electron and neutrino
There are two possibilities for the angular momentum coupling of the two leptons!
10, orSSss tote
For neutron decay: eepn
Angular momentum: both can contribute to neutron decay!S
2
1
2
1
Subtle point: because the leptons are emitted with a definite helicity (lecture 15),we can deduce a correlation between their directions of motion in the two cases:
Gamow-Teller decay, S = 1:
ep
p
e- and travel in opposite directions
)(LHes
)(RHs
Fermi Decay, S = 0:
)(LHes
)(RHs
ep
p
e- and travel in the same direction
3
More on spin correlations:
Fermi Decay, S = 0: Leptons travel in the same direction::
ep
p
eepn
pp
S = 0
Recoiling proton spin is in the same directionas the initial neutron spin.
Gamow – Teller Decay, S = 1:
Leptons travel in the opposite direction:
ep
p
pp
S = 1
Recoiling proton spin is in the opposite directionas the initial neutron spin, i.e. a “spin flip”
4
How to proceed?
As before, assume a pointlike interaction, but allow for different coupling constantsfor the Fermi (F) and Gamow-Teller (GT) cases.
rdrrrrGMM inefpVFif3
,***
, )()()()(
Fermi case, S = 0: (coupling constant: “GV” because the operator transforms like a space vector.)
Gamow–Teller, S = 1: (coupling constant: “GA” because the operator transforms like an axial vector, i.e. like angular momentum.)
rdrrrrGMM inefpAGTif3
,***
, )()()()(
Experimentally, the coupling constants are very similar:
25.1/ VA GG
(These are evaluated by comparing different nuclear beta decay transitions, whereangular momentum conservation restricts the total lepton spin states that contribute)
5
)3(~2 222
AVfifif GGM
Transition rate: eepn
for the neutron:
since there are 3 times as many ways for the leptons to be emitted with S = 1(ms= 1, 0, -1) as with S = 0.
For now, let us work out a generic matrix element, since the expressions are thesame for both apart from the coupling constants:
rdrrrrGM inefpif3
,***
, )()()()(
Rp p
q
electron
neutrino
recoil
Electron and neutrino are represented byplane wave functions of definite momentum:
.,/
)( etcV
rpiere
6
Rp p
q
electron
neutrino
recoil)( qppR
//)( 11** rpierqpie Rve VV
The integral for Mif extends over all space regions for which the nucleon wave functions (n,p) are non-zero: Rmax ~ 1 fm (in nuclei, use R ~ 1.2 A1/3 fm ) ...
But, the recoil momentum pR is no larger than the Q-value for the reaction, ~ MeV ...
1197
1~/
MeV.fm
MeV.fmmax c
RQrpR
Vve
1**
rdrrrrGM inefpif3
,***
, )()()()(
Matrix element:
This is a great simplification: the lepton wave functions are just a constant over theregion of space that matters to calculate the matrix element
7
Summary so far for the transition rate calculation:
fnipffifif rdV
GM
23
,*,2
22 22
The remaining integral is referred to as the nuclear matrix element:
rdrrM nipfnuclear3
,*, )()(
When beta decay occurs in a nucleus, the initial and final wave functions of the protonand neutron need not be exactly the same, so in general:
1nuclearM
However, in the case of the free neutron, there are no complicated nuclear structureeffects, and so the matrix element is identically 1:
fifV
G 2
22
When this occurs in a nucleus, the beta decay rate is the fastest possible, and the transition is classified as “superallowed”
8
Density of states factor:
Just like the calculation we did for electron scattering, but now there are two lightparticles in the final state!
Rp p
q
electron
neutrino
recoil ff dE
dn
We want to work outthe number of equivalentfinal states within energyinterval dEf of Ef.
Final state momenta are quantized in volume V (lecture 6)
3322 44
h
V
h
Vdqqdppdndndn e
Energy conservation, massless neutrino: QcqKK eR
But the nucleon is much heavier than the other particles: 02/)( 2 MpK RR
dqcdEconstKf e
9
Put this all together for the density of states factor:
dpch
qpV
dqc
dndn
dE
dn e
ff 6
2222)4(
And finally, for the transition rate:
dpqph
Mc
G nuclearif22
6
222 )4(2
free neutron: Mnuclear = 1
mixed transition: G2 = GV2 + 3 GA
2
Exercise: plug in all the units and check that the transition rate is in sec-1
Notice: This is actually a partial decay rate, because the electron momentum p isspecified explicitly. if here gives the rate at which the decay occurs for a givenelectron momentum falling within dp of p this predicts the momentum spectrum!
10
Electron momentum spectrum:
2222
22
)()(.)()(
.)()(
e
ifo
KQpconstqpconstpN
dpqpconstNdppN
Rp p
q
electron
neutrino
recoil
approx: KR = 0
N(p)
N(Ke)
Predicted spectral shapes,Krane, figure 9.2:
(Note: max. K(e-) = Q )
11
Comparison to Reality: (Krane, fig. 9.3: e+ and e- decays of 64Cu)