Transcript Slide 1

16.451 Lecture 13:
Isospin and symmetries
16/10/2003
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• Nucleon: T = ½, mt =  ½. For a nucleus, by extension: mt = ½ (Z - N).
• If neutrons and protons are really “identical” as far as the strong interaction is
concerned, then nuclei with the same mass number but (Z,N) interchanged ought to
be very similar. These are called “mirror nuclei”, e.g. 11B (5,6) and 11C (6,5)
• Energy spectra line up after correction for Coulomb energy difference in the
ground state. 
J
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Two nucleon (NN) system and isospin:
Pauli exclusion principle: wave function for identical fermions must be antisymmetric
if the particle labels are exchanged
How do we tell what symmetry the isospin configurations have? T = 0 or 1 for NN.
Use symbolic representation:  = ½ and  = - ½
The 4 configurations (m1, m2) are:
(), (), () , ()
() and () are symmetric – exchanging the symbols (1,2) has no effect.
These correspond to total isospin (T, mT) = (1, 1) and (1, -1)
(), () states correspond to mT = 0, but they have mixed symmetry. 
Solution: make symmetric and antisymmetric combinations of the mixed states:
symmetric:
() + ()  () + ()
anti –
() - ()  () - () = - {() - () }
Bottom line:
:
(T=1, mT = 0)
T = 1 states are symmetric, T = 0 antisymmetric.
(T=0, mT=0
(Same for spin, S)
implications of isospin symmetry:
(Krane, ch. 11)
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• pp and nn systems are always T = 1
• np system is () , ie it can be partly T = 1 and partly T = 0
• for a nucleus, mT = ½ (Z-N) and T = |mT|, ie lowest energy has smallest T
• Example: “isobaric triplet”
14C, 14N, 14O:
14O:
Z = 8, N= 6
mT = + 1, T = 1
14C:
Z = 6, N = 8
mT = -1, T = 1
14N
mT = 0, T = 0 (g.s) and T = 1 (8 MeV)
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Final example, evidence for isospin:
Consider the deuteron,
2H
= (np) bound state (d)
Quantum numbers: mT = 0, T = 0
J = 1 +
(S = 1, L = 0,  = (-1)L )
How do we know it has T = 0 ?
“Isospin selection rules”:
The reaction:
1)
isospin analysis:
2)
isospin analysis:
d + d 
 
 
00  00
d + d 
 
 
00  10
 +
4He
occurs, but
(T = 1 deuteron also works)
° +
4He
does not
(only T = 0 prevents this!)
Bottom line: T is conserved by the strong interaction. Energy depends on T but not on mT
Isospin and Quarks:
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There are a total of 6 quarks in the Standard Model (u,d,s,c,t,b – more later!) but
only two play a significant role in nuclear physics: u and d.
Not surprisingly, isospin carries over into the quark description: the “up” quark has
isospin T = ½ “up” and similarly for the “down” quark:
Quark “flavor”
Spin, s
Charge, q/e
Isospin projection, mt
u (“up”)
1/2
+ 2/3
1/2
d (“down”)
1/2
- 1/3
-1/2
Isospin addition for the proton: p = (uud),
neutron: n = (udd),
mt = ½ + ½ - ½ = ½ 
mt = ½ - ½ - ½ = - ½ 
What about the delta? Addition of 3 x isospin- ½ vectors: T = 1/2 or 3/2;
T = 3/2 is the :
++ = (uuu), + = (uud), ° = (udd), - = (ddd) 
What about antiquarks? same isospin but opposite mt
 e.g. pion: (+, °, - )
1
1

  u d,
mt 
2
 2   1, etc... 
Back to the neutron:
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
Lifetime: 885.7  0.8 sec (world average)
n  p  e   e

Neutron Beta Decay: n  p  e   e
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• a fundamental Weak Interaction process
• lifetime  is relatively long:
 ~
1

,
~
 
*
f
V (r )  i d 3 r

2
(lecture 6!)
large  implies small transition rate , therefore ‘weak’ interaction V(r)
compare to  resonance decay: +  p + °, a strong interaction process,
with  = 5.7 x 10-24 seconds!!!
• precision studies of neutron decay are a very important testing ground for the
“Standard Model” of fundamental interactions, as we shall see....
• interaction is almost pointlike, that is, the neutron disappears and the decay
products appear almost instantaneously at the same place. (Fermi theory)
• modern picture (Assignment 3):
W- boson
Closer look: electron energy spectrum
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(m  0, K  p )
e
p
n
e
“before”
“after”
1) mn  m p  me  K p  K e  K



2) p p  pe  p  0
Define the “Q – value”:
(energy cons.)
(momentum)
(in general, Q > 0 for a reaction to proceed)
Q  mn  m p  me  K p  K e  K
From Particle Data Group entries:
Q = 0.78233  0.00006 MeV
( 60 eV!)
Electron Energy Spectrum from “PERKEO” expt. at ILL reactor, France:
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Bopp et al., Phys. Rev. Lett. 56, 919 (1986)
Fit to expected spectrum
shape including detector model.
Note: presence of neutrino
affects this shape dramatically –
otherwise it would be a sharp
peak at a value determined by
momentum /energy conservation!
Endpoint smeared with detector resolution
Case study: “state of the art” neutron lifetime measurement
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Outline of the method:
decay rate:
n  p  e   e
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dN
N
 
dt

measure rate by counting decay protons in a given time interval (dN/dt) and
normalizing to the neutron beam flux (N)
decay proton, to detector
transmitted beam
neutron
detector
incoming n beam
decay volume, length L
Ideally done with “cold neutrons”, e.g. from a reactor, moderated in liquid hydrogen...
Issues: 1. precise decay volume ? 2. proton detection ?
3. beam normalization ? ...
Neutron beam distribution – definitely not monoenergetic:
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• ~ MeV neutrons from a reactor are “moderated” by scattering in a large tank of
water (“thermal”) or liquid hydrogen (“cold”)
• after many scatterings, they come to thermal equilibrium with the moderator and
are extracted down a beamline to the experiment
• velocity distribution is “Maxwellian”: energies in the meV range (kT = 26 meV @ 293K)
Krane, Fig 12.4
Neutron detection at low energy: (Krane, ch. 12)
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• several light nuclei have enormous neutron capture cross sections at low energy:
(recall, cross sectional area of a nucleus, e.g. 10B is about 0.2 barns, lecture 4)
(barns)
• key feature: cross sections scale as 1/velocity at low energy
10B
+ n   + 7Li + 2.79 MeV
 = ovo/v with o = 3838 b
kinetic energy of ionized
fragments can be converted
into an electrical signal  detector
Put this together for detector counting rates as shown:
N beam
N


dN / dt
N decay
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N decay 
dNbeam
dt
decay proton, to detector
transmitted beam
incoming n beam
N beam
10B
neutron
detector
N det

decay volume, length L
L  vT
N det v

~ N beam
( o v o )
Decays and transmitted detector
counts accumulated for time T
neutrons must be captured to
be detected!
N beam  (const)  N det
(DC beam, thermal
energy spread)
L
T

  (const) 
N det
L
N decay
Experimental details (all in vacuum, at ILL reactor, France):
• use Penning trap to confine decay protons
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 ~
• let them out of the trap after accumulation interval T
N det
L
N decay
• measure the ratio Ndet/Ndecay as a function of trap length L  slope gives 
N decay
proton
detector
N det
L
 particle
detectors
for capture
products
thin 10B foil
to capture
beam n’s
variable length
Penning trap
(16 electrodes)
protons spiral
around field lines
when let out of
the trap
Amazing results:
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Rate versus trap length L
proton energy
from timing:
about 34 keV
(after kick –
raw energy is
only 0.75 keV)
Result: 893.6  5.3 seconds (1990)
PDG average: 885.7  0.8 (2003)