16.451 Lecture 16:
Beta Decay Spectrum
29/10/2003
n  p  e   e
(and related processes...)
Goals:
• understand the shape of the energy spectrum
• total decay rate sheds light on the underlying
weak interaction mechanism
Starting point: “Fermi’s Golden Rule” again! (lecture 6)
matrix element
2
if 
M if

2
f
(transitions / sec)
M if 

 *f

V (r )  i d 3 r
 f  dn/ dE f
density of states
1
2
Matrix element Mif?
Simplest model is to take a pointlike interaction with an overall energy scale “G”:
(Fermi, 1934 – almost right!)




M if  G  *p, f (r ) e* (r ) * (r )  n,i (r ) d 3r

e
p
e
(the interaction is proportional to the wavefunction
overlap of initial and final state particles at the same
point in space)
n
e
p
n
W-
e
Standard Model description: an extended
interaction, but the range is only about 0.002 fm
which is just about zero!
The Standard Model can ‘predict’ the value of G
in terms of model parameters, whereas in Fermi’s
theory, it remains to be determined from experiment.
3
Spin considerations: electron and neutrino
There are two possibilities for the angular momentum coupling of the two leptons!



se  s  S tot,

S  0 or 1
For neutron decay:
n  p  e   e
Angular momentum:

1
2


1
2


S
both can contribute to neutron decay!
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:
Fermi Decay, S = 0:

se

s
(LH )
(RH )
Gamow-Teller decay, S = 1:

pe

se
(LH )

p

s
(RH )
e- and  travel in the same direction

pe

p
e- and  travel in opposite directions
4
More on spin correlations:
n  p  e   e
Fermi Decay, S = 0:
Leptons travel in the same direction::

pe

pp

p
Recoiling proton spin is in the same direction
as the initial neutron spin.
S = 0
Gamow – Teller Decay, S = 1:
Leptons travel in the opposite direction:

pe

p
S = 1

pp
Recoiling proton spin is in the opposite direction
as the initial neutron spin, i.e. a “spin flip”
How to proceed?
5
As before, assume a pointlike interaction, but allow for different coupling constants
for the Fermi (F) and Gamow-Teller (GT) cases.
Fermi case, S = 0:
(coupling constant: “GV” because the operator transforms
like a space vector.)




M if  M F  GV  *p, f (r ) e* (r ) * (r )  n,i (r ) d 3r

Gamow–Teller, S = 1:
(coupling constant: “GA” because the operator transforms
like an axial vector, i.e. like angular momentum.)




M if  M GT  GA  *p, f (r ) e* (r ) * (r )  n,i (r ) d 3r

Experimentally, the coupling constants are very similar:
GA / GV   1.25
(These are evaluated by comparing different nuclear beta decay transitions, where
angular momentum conservation restricts the total lepton spin states that contribute)
n  p  e   e
Transition rate:
2
if 
M if

for the neutron:
6
2
f
~ (GV2  3 G A2 )
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 the
same for both apart from the coupling constants:
M if  G
 *  * 

(r ) e (r )  (r )  n,i (r ) d 3r

p

pR
recoil

 *p, f
electron

q
Electron and neutrino are represented by
plane wave functions of definite momentum:

neutrino
e (r ) 
e
 
ip  r / 
, etc.
V
7
Matrix element:

p

pR

electron
recoil
e*
M if  G

q
v*

 *p, f
 *  * 

(r ) e (r )  (r )  n,i (r ) d 3r



pR   ( p  q )
neutrino
 e
1
V
  
i ( p  q)  r / 

 e
1
V
 
i pR  r / 
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 ...
 
pR  r / 

Q Rmax
c
~
1 MeV.fm
 1
197 MeV.fm
e* v* 
1
V
This is a great simplification: the lepton wave functions are just a constant over the
region of space that matters to calculate the matrix element 
8
Summary so far for the transition rate calculation:
2
if 
M if

2
f
2 G 2

 V2

 *f , p
 i ,n d r
3
2
f
The remaining integral is referred to as the nuclear matrix element:
M nuclear 

* 
f , p (r )
 3
 i , n (r ) d r
When beta decay occurs in a nucleus, the initial and final wave functions of the proton
and neutron need not be exactly the same, so in general:
M nuclear  1
However, in the case of the free neutron, there are no complicated nuclear structure
effects, and so the matrix element is identically 1:
2 G 2
if 
f
2
 V
When this occurs in a nucleus, the beta decay
rate is the fastest possible, and the
transition is classified as “superallowed”
9
Density of states factor:
Just like the calculation we did for electron scattering, but now there are two light
particles in the final state!

p

pR
recoil
electron

q
neutrino
We want to work out
the number of equivalent
final states within energy
interval dEf of Ef.
Final state momenta are quantized in volume V

dn  dne  dn   4 p 2 dp

Energy conservation, massless neutrino:
V
3
h
K e const
 c dq

dn
dE f
(lecture 6)


   4 q 2 dq


V
3
h



K R  K e  cq  Q
But the nucleon is much heavier than the other particles:
dE f
f
K R  ( pR ) 2 / 2M  0
Put this all together for the density of states factor:
f
dn

dE f
dne  dn

c dq
 (4 )
2
10
V 2 p 2q 2
6
h c
dp
And finally, for the transition rate:
if
2
2
2 (4 )
2 2
 G
M nuclear
p
q dp
6
c
h
2
free neutron: Mnuclear = 1
mixed transition: G2 = GV2 + 3 GA2
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 is
specified explicitly. if here gives the rate at which the decay occurs for a given
electron momentum falling within dp of p  this predicts the momentum spectrum!
11
Electron momentum spectrum:
approx: KR = 0
N ( p) dp  N o if  (const.)  p q dp
2 2

N ( p)  (const.)  p 2 q 2
recoil
N(p)

p

pR
 (const)  p 2 (Q  K e ) 2
electron

q
neutrino
N(Ke)
Predicted spectral shapes,
Krane, figure 9.2:
(Note: max. K(e-) = Q )
Comparison to Reality:
(Krane, fig. 9.3: e+ and e- decays of 64Cu)
12
Too many low energy eN(p)
N(K)
???
N(p)
Coulomb
effects –
solution next
lecture!
N(K)
Too few low energy e+