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)

N(p)

N(p)N(K)

N(K)

Too many low energy e-

Too few low energy e+

???

Coulombeffects –solution nextlecture!

12