Prof. Venktesh SinghPh.D.

Central University of South Bihar, Gaya

1

Magnetic moment The property of a magnet that

interacts with an applied field to give a mechanical

moment.

I

A

orb

L

period

eI

;

2

eI

2)(

22 re

rIorb

Lm

e

rmrvmL

e

orb

ee

2

)( 2

2

Spin (I)

Neutron

Magnetic

moment

(N)p

Nm

c

2

T

QI IN

Neutron is charge neutral ! Q = 0 and

therefore NMM should be ZERO ?

Observed Neutron magnetic moment

N)45(91304272.1

3

enneQ

The quark model in its modern form was developed by

Murray Gell-Mann - American physicist who received the

1969 Nobel Prize in physics for his work on the theory of

elementary particles. QM - independently proposed by

George Zweig.3

proton (charge +1) neutron (charge 0)

u dd

u

u

d

Quarks have fractional electric charge ! u electric charge + 2/3

d electric charge -1/3

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1

3

2

3

2

pduu 0

3

1

3

1

3

2nddu

BARYON

4

PROTON

NEUTRON

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Quark Q[e] I and I3 S B

up (u) + 2/3 ½,+½ 0 1/3

down (d) - 1/3 ½, -½ 0 1/3

strange(s) - 1/3 0,0 - 1 1/3

u ddT

QI

321

3

1

3

1

3

2

T

e

T

e

T

e

I

I3= Third component of Isospin

Number of member in the family =2I+1, here I = isospin

For Sigma member 2I+1=3; I=1 and I3=1, 0. -1

For nucleon (p & n) 2I+1=2: I=1/2 and I3= ½ ,(p) and - ½ (n),

Similar to u and d quarks.

Q = e[I3+B/2+S/2] for p, Q=e, I3=1/2, B=1 and S=0 Y=S+B=1

For n, Q=0, I3= -1/2, B=1 and S=0 and Y=S+B=16

For nucleon ( , 0 and +)

For , Q= -e, I3= -1, B=1, -e = e[-1+1/2+S/2] therefore S=-1

and Y=S+B=-1+1=0

For 0, Q=0, I3=0, B=1, 0 = e[0+1/2+S/2] and S=-1 Y=S+B=0

For +, Q=+e, I3= 1, B=1 and S=+1 and Y=S+B=2

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A powerful tool to understand hadron spectrum,

structures and dynamics.

With proper dynamical contents, it is applicable

to multi-quark systems, such as 2 – baryons,

Penta-quarks.

The quark model is a classification scheme for

hadrons in terms of their valence quarks — the

quarks and anti-quarks which give rise to the

quantum numbers of the hadrons.8

Evidence for quarks Or MOTIVATION

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Nucleons

Proton and Neutron have almost equal mass. Strong nuclear

force is charge independent. Vpp ~ Vpn ~ Vnn

Isospin

Proton and Neutron form part of single entity with isospin ½,

analogous to and of spin ½.

Isospin I is conserved in Strong Interactions.

Isospin multiplets

Useful for classification of hadrons.

2I + 1 states in a isospin multiplet |I, I3>. = |0, 0>,

p = |1/2, 1/2>, n = |1/2, -1/2>,+ = |1, 1>, 0 = |1, 0>, - = |1, -1>, ++ = |3/2, 3/2>, + = |3/2, 1/2>,

0 = |3/2, - 1/2> and - = |3/2, - 3/2> 10

1964 – introduced by Gell-Mann & Zweig

Quark Charge Q[e] Isospin|I,I3> Strangeness S

up (u) + 2/3 |1/2, +1/2> 0

down (d) - 1/3 |1/2, -1/2> 0

strange(s) - 1/3 | 0, 0 > - 1

Quark Model

Gives natural explanation for Isospin

I3 = ½ (nu – nd + nd-bar – nu-bar) ni number of i quarks

Isospin works well

Masses of u and d quarks are almost equal

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• Mass increases from 1

generation to the next

• Going down in each

generation, the charges

are:

+2/3, -1/3, 0, -1

• These are all in multiples

of the elementary

charge H12

Conservation Law

Isospin I is conserved in Strong Interactions

Allows to calculate ratios of cross sections and

Branching Fractions in strong interactions

Isospin addition

+,p |1, 1> |1/2, 1/2> = |3/2, 3/2 > ++

-, p |1, - 1> |1/2, 1/2 > = |3/2, - 1/2 > 0

0, n |1, 0> |1/2, - 1/2 > = |3/2, - 1/2 > 0

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Strange Particles

Discovered in 1947 by Rochester and Butler

V, “fork”, and K, “kink”

Production of V(K0,)

and K via Strong Interaction Weak decay

Associated Production: Strange particles produced in pairs

Strangeness (S)

Additive quantum number Gell-Mann Nishijima

Conserved in Strong and Electromagnetic Interactions

Violated in Weak decays14

Strange Particles

Discovered in 1947 by Rochester and Butler

V, “fork”, and K, “kink”

Production of V(K0,)

and K via Strong Interaction Weak decay Associated Production: Strange particles produced in pairs

Strangeness (S)

Non-zero for Kaons and hyperonsS = 0: pi, p, n, delta, .......S = 1: K+, K0 S = - 1: K-, K0-anti,

Lambda, sigma, ...... S= - 2: Cascade

Naturally explained in Quark Model S = ns-anti

– ns

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Additive quark quantum numbers are related

Q = I3 + ½ (S + B)

Not all independent

Gell-Mann Nishijima predicts Quark Model valid also for

Hadrons

Baryon Number B For quark B = + 1/3

For anti-quark B = - 1/3

Hypercharge Y = S + B is useful quantum number

Quark model gives natural explanation for Isospin and

Strangeness

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Flavour (06) – u (up), d (down), s (strange), c (charm),

t (top), b (bottom) quarks and anti-flavour for anti-

quarks q: u, d, s, c, t, and b

Charge – Q = + 2/3 and – 1/3 (u: +2/3, d: - 1/3, s: - 1/3, c:

+2/3, t: + 2/3, and b: - 1/3

Baryon Number – B = 1/3 – as baryons are made out of

three (03) quarks

Spin – s = ½ - quarks are the Fermions !

Strangeness – Ss = -1 and Ss = 1, and Sq = 0 for q = u,

d, c, t, b and q

Charm – Cc=1 and Cc= -1, and Cq=0 for q= u,d,s,t,b andq

Bottomness –Bb= -1 and Bb=1 and Bq=0 for q=u,d,s,c,t,&q

Topness – Tt = 1 and Tt = - 1, and Tq = 0 for q = u, d, s,

c, b, and q 17

Hypercharge (Y) – Y = B + S + C + B + T

= ( Baryon charge + Strangeness +

Charm + Bottomness + Topness )

I3 (or Iz or T3) – 3’d component of isospin

Charge Q – Gell-Maan-Nishijima formula Q = I3 + Y/2

Quark u d c s t b

Q – electric charge +2/3 -1/3 +2/3 -1/3 +2/3 -1/3

I – isospin 1/2 1/2 0 0 0 0

I3/z – isospin 3’d component +1/2 - 1/2 0 0 0 0

S – strangness 0 0 0 - 1 0 0

C – charm 0 0 + 1 0 0 0

B – bottomness 0 0 0 0 0 -1

T - topness 0 0 0 0 + 1 018

The quark model is the follow-up of the Eightfold Way classificationscheme proposed by Murray Gell-Mann and Yuval Neeman.

The Eightfold Way may be understood as a consequence of flavoursymmetries between various kinds of quarks. Since the strong nuclearforce affects quarks the same way regardless of their flavour,replacing one flavour of a quark with another in a hadron should notchange its mass very much.

Consider u, d, s quarks:

Then the quarks lie in thefundamental representation, 3(called the triplet) of the flavourgroup SU(3) : [3]

Mathematically, this replacementmay be described by elements ofthe SU(3) group.

The anti-quarks lie in the complex conjugate representation 3 : [3]19

Triplet in SU(3)flavour group : [3] Anti-triplet in SU(3)flavour group : [3]

Y = 2 (Q – I3) e.g. u-quark: Q = +2/3, I3 = +1/2 and Y = 1/3

I3 I3

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Colour Three (03) – Red. Green and Blue

Triplet in SU(3)colour group [3]

Anti-colour Three (03) – anti-Red, anti-Green and anti-Blue

anti-Triplet in SU(3)colour group [3]

The quark Colours (Red. Green and Blue) combine to be

Colourless or neutral

The quark Anti-colours (anti-Red, anti-Green and anti-Blue)

also combine to be Colourless

All hadrons Colour neutral = Colour Singlet in the

SU(3)colour group21

mu =1.7 – 3.3 MeV/c2,

md = 4.1 – 5.8 MeV/c2,

ms = 70 – 130 MeV/c2,

mc = 1.1 – 1.5 GeV/c2,

mb = 4.1 – 4.4 GeV/c2 and

mt ~ 180 GeV/c2

The quantum number, “Colour” has been introduced by

Greenberg in 1964 to describe the state ++(uuu) (Q = +2,

J=3/2), discovered by Fermi in 1951 as +p resonance:

++(uuu) p(uud)+ +(du).

The state ++(uuu) with all parallel spins to achieve J=3/2

is forbidden according to the Fermi statistics (without

colour).

Masses of the Quarks – The current quark mass is also

known as the mass of the

“Naked” (Bare) quark.

The constituent quark mass is

the mass of a “Dressed” current

quark, i.e., for quarks

surrounded by a cloud of virtual

quarks and gluons – mu(d) ~ 350

MeV/c2

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Fermions Fermions

Leptons QuarksElectric charge

Fermions

Generation

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Gell-Mann (1964) – Hadrons are NOT fundamental but they

are build from “valence quarks”.

Baryon charge BB = 1 Bm = 0

Constraints to build hadrons from quarks –

Strong colour interaction (Red, Green and Blue)

Confinement

Quarks must from colour-neutral hadrons

State function for baryons – anti-symmetric under

interchange of two quarks

Since all baryons are colour neutral, the colour part of A

must be anti-symmetric i.e., a SU(3)colour singlet

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Possible states A –

Or a linear combination of (i) and (ii) –

where a2 + b2 = 1

(i)

(ii)

(iii)

a

b

Consider flavour space (u, d, s quarks) SU(3)flavour group

Possible states - |flavour> - (i) – anti-symmetry

For baryons (ii) – symmetry

(iii) – mixed symmetry25

|Meson> = |qq>

Quark Anti-quarkTriplet in SU(3)flavour group – [3] Anti-triplet in SU(3)flavour

group – [3]

From group theory : the nine state (Nonet) made out of a

pair can be decomposed into the trivial representation, 1 called the

singlet, and the adjoin representation, 8 called the octet.

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Bound states of qq

Zero net colour charge

Zero net baryon number B = (+ 1/3) + (- 1/3) = 0

Ground state (L=0)

Meson “spin” (total angular momentum) is given by the

qq spin state.

Two possible qq total spin states: S = 0, 1

S = 0: pseudoscalar mesons

S = 1: vector mesons

Angular Momentum L

For lightest mesons

Ground state L=0 between quarks

bbggrr|3

1|

Consider ground state mesons consisting of light quarks

(u, d, s), mu ~ 0.3 GeV, md ~0.3 GeV, ms ~ 0.5 GeV

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Parity P

Intrinsic quantum number of quarks and leptons

P = + 1 for fermions and P = - 1 for anti-fermions

Composite system of two particles with orbital angular

momentum L:

P(qq)= Pq Pq (-1)L = (+1)(-1)(-1)L = - 1 for L=0

Quantum Field Theory tells us that

Fermions and anti-fermions: Opposite parity

Bosons and anti-bosons : Same parity

Choose: Quarks and Leptons : Pq,L = +1

Anti-quarks and anti-leptons : Pq,L = - 129

Total Angular Momentum J

J = L + S include quark spins

S = 0 qq spins anti-aligned or

JP = 0– Pseudo – scalar mesons

S = 1 qq spins aligned or

JP = 1– Vector mesons

Quark Flavours

ud, us, du, ds, su, sd non-zero flavour states

uu, dd, ss Zero net flavour states

have identical additive quantum numbers

Physical states are mixtures30

Classification of mesons –

Quantum numbers –

Spin s

Orbital angular momentum L

Total angular momentum J = L + s

Properties with respect to Poincare transformation –

(i) Continues transformation Lorentz boost

(3 parameters – b) UB~eiba

Casimir operator (invariant under transformation)

M2 = pp

(ii) Rotations (3 parameters – Euler angle - UR ~ eiJ

Casimir operator: J2

(iii) Space Time shifts (4 parameters – a) Ust ~exp(iax)

x’ x + a

10 parameters of Poincare Group31

Discrete operators –

(iv) Parity transformation – Flip in sign of the spacial

coordinate r = - r and Eigenvalue P = 1,

here P = (- 1 )L+1

(v) Time reversal – t - t and eigenvalue T = 1

(vi) Charge conjugation – C = - C, here C = (-1)L+s

C = parity operator and eigenvalue C = 1

General CPT Theorem – P. C. T = 1

Due to the fact that discrete transformations correspond

to the U(1) group they are multiplicative.

Properties of the distinguishable, not continuum, particles

are defined by M2 or M, J2 or J, P, C

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33

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Hadrons are not fundamental, but they are built from

valance quarks, i.e. Quarks and anti-Quarks, which give

the quantum numbers of the hadrons

|Baryon>=|qqq> and |Meson>=|qq>

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Baryon spin wave functions (spin)

Combine Three spin ½ quarks

Consider J = 3/2 The spin wave function for the |3/2, 3/2> state is

|3/2, 3/2> = 36

Generate other states using the ladder operator J

Giving the J = 3/2 states :

All symmetric under

interchange of any two spins

Consider J = 1/2

First consider the case where the first 2 quarks are in a |0, 0> state is

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Anti-symmetric under exchange 1 2.

Third-quark J = ½ states can also be formed from the state with

the first two quarks in a symmetric spin wave function.

Can construct a three – particle state | ½, ½ >(123) from

38

Taking the linear combination

with a2 + b2 = 1. Act upon both sides with J+

which with a2 + b2 = 1 implies

Giving 39

Symmetric under interchange 1 2.

Three – quark spin wave functions

Symmetric under

interchange of any

2 quarks

J = 3/2

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Anti-symmetric under

interchange of 1 2.J = 1/2

J = 1/2 Symmetric under

interchange of 1 2.

spin flavour must be symmetric under interchange

of any two quarks.

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Hadron charge Quark content

B Q s S Y=S+B

+ e ud 1/3 –1/3 = 0 +2/3+1/3

=1

=0 0+0=0 0+0=0

K+ e us 1/3-1/3 = 0 +2/3+1/3

=1

=0 0+1=1 1+0=1

p+ e uud 1/3+1/3

+1/3= 1

2/3+2/3-

1/3=1

=1/2 0+0+0

= 0

0+1=1

n0 0 ddu 1/3+1/3

+1/3 = 1

-1/3-1/3

+2/3=0

=1/2 0+0+0

= 0

0+1=1

– - e sss 1/3 +1/3

+1/3 = 1

-1/3-1/3

-1/3= -1

=3/2 -1-1-1

= -3

-3+1=-2

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