Nucleon: T = ½, m t = ½. For a nucleus , by extension: m t = ½ (Z - N).

• 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

16.451 Lecture 13: Isospin and symmetries 16/10/2003 1

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: () + () () + () (T=1, mT = 0)

anti – : () - () () - () = - {() - () } (T=0, mT=0

Bottom line: T = 1 states are symmetric, T = 0 antisymmetric. (Same for spin, S)

2

implications of isospin symmetry: (Krane, ch. 11)

• 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:

14C: Z = 6, N = 8mT = -1, T = 1

14O: Z = 8, N= 6mT = + 1, T = 1

14N mT = 0, T = 0 (g.s) and T = 1 (8 MeV)

3

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) d + d + 4He occurs, but

isospin analysis: (T = 1 deuteron also works)

2) d + d ° + 4He does not

isospin analysis: (only T = 0 prevents this!)

0000

0100

Bottom line: T is conserved by the strong interaction. Energy depends on T but not on mT

4

Isospin and Quarks:

There are a total of 6 quarks in the Standard Model (u,d,s,c,t,b – more later!) butonly two play a significant role in nuclear physics: u and d.

Not surprisingly, isospin carries over into the quark description: the “up” quark hasisospin 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), mt = ½ + ½ - ½ = ½

neutron: n = (udd), 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: (+, °, - )

...,1,2

1

2

1 etcmdu t

5

Back to the neutron:

Lifetime: 885.7 0.8 sec (world average)

eepn

6

Neutron Beta Decay:

• a fundamental Weak Interaction process

• lifetime is relatively long: (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):

23* )(~,1

~ rdrV if

eepn

W- boson

7

Closer look: electron energy spectrum

n p

e

e“before”

“after”

0)2

)1

ppp

KKKmmm

ep

epepn

),0( pKm

Define the “Q – value”: (in general, Q > 0 for a reaction to proceed)

(energy cons.)

KKKmmmQ epepn

(momentum)

From Particle Data Group entries: Q = 0.78233 0.00006 MeV ( 60 eV!)

8

Electron Energy Spectrum from “PERKEO” expt. at ILL reactor, France:

Bopp et al., Phys. Rev. Lett. 56, 919 (1986)

Endpoint smeared with detector resolution

Fit to expected spectrumshape including detector model.

Note: presence of neutrinoaffects this shape dramatically –otherwise it would be a sharppeak at a value determined bymomentum /energy conservation!

9

Case study: “state of the art” neutron lifetime measurement 10

Outline of the method:

N

dt

dNdecay rate:

eepn

measure rate by counting decay protons in a given time interval (dN/dt) and normalizing to the neutron beam flux (N)

incoming n beam

decay volume, length L

decay proton, to detector

transmitted beam

neutron detector

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 ? ...

11

Neutron beam distribution – definitely not monoenergetic:

• ~ 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

12

Neutron detection at low energy: (Krane, ch. 12)

• 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)

• key feature: cross sections scale as 1/velocity at low energy

(barn

s) 10B + n + 7Li + 2.79 MeV

= ovo/v with o = 3838 b

kinetic energy of ionizedfragments can be convertedinto an electrical signal detector

13

Put this together for detector counting rates as shown:

incoming n beam

decay volume, length L

decay proton, to detector

transmitted beam

10B neutron detector

beamdetdet ~

)v(

vN

NN

oo

dt

dNN beam

decay decay

beam

/ N

N

dtdN

N

LN

Nconst

T

LNconstN

decay

detdetbeam )()(

TL v

Decays and transmitted detector counts accumulated for time T

beamN

(DC beam, thermalenergy spread)

neutrons must be captured tobe detected!

14

Experimental details (all in vacuum, at ILL reactor, France):

• use Penning trap to confine decay protons

• let them out of the trap after accumulation interval T

• measure the ratio Ndet/Ndecay as a function of trap length L slope gives

protons spiralaround field lineswhen let out of the trap

protondetectordecayN

thin 10B foilto capture beam n’s

particledetectorsfor captureproducts

detN

variable lengthPenning trap(16 electrodes)

L

LN

N

decay

det~

15

Amazing results:

proton energyfrom timing:about 34 keV(after kick –raw energy isonly 0.75 keV)

Rate versus trap length L

Result: 893.6 5.3 seconds (1990)

PDG average: 885.7 0.8 (2003)

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