AIMS-pp-07.pdf

From Nuclei to Quarks

*

 

1910: Rutherford discovers nucleus

*

 

1932: Chadwick discovers neutron

*

 

1940: Pions discovered

*

 

1950s: Huge zoo of strongly interacting particles discovered

*

 

1963: Quarks hypothesized (Gell-Mann/Zweig)

Isospin

set of unitary 2x2 matrices of determinant 1

•  After the discovery of the neutron, further experiments showed

that neutrons and protons looked pretty much the same except for electric charge (Q=0 for neutron, Q=+1 for proton)

•  Heisenberg proposed that they actually were the same: that they

were 2 different states of one particle: the nucleon!

The new symmetry was called Isospin

Spin-1/2 particle

in ordinary space in Isospin spaceNucleon in

up proton

isospin up (proton)

isospin down (neutron)

Isospin Invariance: Include

QED:

Charge conservation:

Baryon number conserved

Phenomenological

formula for charge +1 for proton

0 for neutron

•  Isospin symmetry is an

approximate symmetry

•  But it is a good approximation,

since electromagnetism is much weaker than the force holding the nucleus together

p = i = 1

2 ,i3 = + 1 2

n = i = 1

2,i3 = − 1 2

New isospin states from old:

the deuteron

(like orthopositronium)

i =1,i₃ =1 = 1 2, 1 2 1 2 , 1

2 = p p

i =1,i₃ = 0 = 1 2 1 2, 1 2 1 2,−

1

2 +

1 2,−

1 2 1 2, 1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1

2

(

p n + n p

)

i =1,i₃ = −1 = 1 2,−

1 2

1 2,−

1

2 = n n

⎫ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎪ triplet

i = 0,i₃ = 0 = 1 2 1 2, 1 2 1 2,−

1

2 −

1 2,−

1 2 1 2, 1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1

2

(

p n − n p

)

1940s: 3 states of nearly equal mass observed in nuclear scattering expts:

Agrees with observation! the pion!

50% 50%

Isospin conservation

Q = ℑ₃ + B

2 =

+1 π+

0 π0

−1 π−

⎧ ⎨ ⎪⎪

⎩ ⎪ ⎪

Set B= 0 for pions ⇒pions are 3 states of an isospin triplet

π+ = i =1,i₃ = +1

π0 _{= i =1,i}

3 = 0

π− = i =1,i₃ = −1

Recall that p = i = 1

2,i3 = + 1

2 n = i = 1

2,i3 = − 1 2

Pion-Nucleon Scattering π+ = i =1,i₃ = +1

π0 _{= i =1,i}

3 = 0 π− = i =1,i₃ = −1

p = i = 1

2,i3= + 1 2

n = i = 1

2,i3 = − 1 2

π

×

N

Six Possible Elastic Scattering Channels

π+ + p → π+ + p π− + n → π− + n

⎫ ⎬ ⎪ ⎭⎪

π− + p →π− + p

π− + p → π0 + n

π+ + n →π+ + n

π+ + n →π0 + p

⎫

⎬ ⎪ ⎪⎪

⎭ ⎪ ⎪ ⎪

final state must

have i= 3

2

final state will have both i = 3

Clebsch-Gordon Table:

π

×

N

π + + p →π+ + p = π+ p = 3

2, 3 2

π− + n →π− + n = π− n = 3

2,− 3 2

π− + p →π− + p = π− p = 1

3 3 2,−

1

2 −

2 3

1 2,−

1 2

π− + p→π0 + n = π0 n = 2

3 3 2,−

1

2 +

1 3

1 2,−

1 2

π+ + n→π+ +n = π+ n = 1

3 3 2,

1

2 +

2 3

1 2,

1 2

π+ +n→ π0 + p = π0 p = 2

3 3 2,

1

2 −

1 3

1 2,

Isospin Conservation

〈π+ p H_πN π+p = 〈π−n H_πN π−n = 〈3

2 HπN 3

2 = M3

〈π−p| H_πN π−p = 1

3 3 2,−

1 2 −

2 3

1 2,−

1 2

⎛ ⎝⎜

⎞

⎠⎟

|

HπN

|

1 3

3 2,−

1 2 −

2 3

1 2,−

1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 3 3

2 HπN 3 2 +

2 3

1

2 HπN 1 2 =

1

3M3 + 2

3M1

〈π−p| H_πN π0n = 1

3 3 2,−

1 2 −

2 3

1 2,−

1 2

⎛ ⎝⎜

⎞

⎠⎟ |HπN |

2 3

3 2,−

1

2 +

1 3

1 2,−

1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = 2 3 3

2 HπN 3

2 −

2 3

3

2 HπN 1

2 =

2

3 M3− 2

3 M1 〈π+p|H_πN π+p = 〈π−n|H_πN π−n =M₃

〈π−p|H_πN π−p = 1

3M3+ 2 3M1

〈π−p|H_πN π0n = 2

phase-space factor

Agrees with observation! 〈π+p|H_πN π+p = 〈π−n|H_πN π−n = M₃

〈π−p|H_πN π−p = 1

3M3+ 2 3M1

〈π−p|H_πN π0n = 2

Strangeness

1950s: Not all

nuclear interactions conserve isospin!

Strong Nuclear Force Weak Nuclear Force

Isospin conserved Isospin not conserved

Short lifetimes Long lifetimes

Strange particles produced in pairs

Strange particles decay as singletons

Λ

has no charged counterpart

Proposal: Two kinds of nuclear force!

Gell-Mann/Nishijima

(modify Heisenberg’s formula so that it still works)

1960s: Many new particles and interactions observed –

haphazard assignments of isospin and strangeness

Two interesting features of the states of angular momentum 3/2 :

1) All states that had the same total isospin with J=3 / 2 had the same mass

eg. Δ++, Δ+, Δ0, Δ− all had mass 1232 MeV and all have i _{= 3 2}

eg. Ξ*, Ξ*0 both had mass 1533 MeV and both have i _{= 12}

2) States of different strangeness differed by the same amount of mass

Δ++ = i = 3

2,i3= + 3 2

Δ+ = i = 3

2,i3= + 1 2

Δ0 _{= i =} 3

2,i3=− 1 2

Δ− = i = 3

2,i3=− 3 2

strangeness

0

-2

-3 i=3/2

i=1

i=1/2

i=0

-1232 MeV

1384 MeV

1533 MeV

1672 MeV?

strangeness

0

-2

939 MeV

1193 MeV

1318 MeV 1116 MeV

1964! in

detected

1961 in

predicted

:

−

Ω

Gell-Mann / Zweig quarks / aces

J = 3 2

J = 1 2

i3

i3 Δ++

Δ+

Δ0

Δ−

Σ+

Σ0

Σ−

Ξ* _Ξ*0

Ω−

p n

Σ+

Σ0

Σ−

Ξ Ξ0

Λ

Aces or Quarks

•  Perhaps there is a symmetry between isospin and strangeness

•  Isospin & Strangeness  Flavour

•  SU(2) isospin symmetry  SU(3) flavour symmetry

•  SU(3)-flavour is not an exact symmetry due to greater mass

of the strange quark

•  Proposal: all baryons are bound states of 3 quarks in some

combination

•  SU(3) symmetry implies that a baryon is unchanged if we

interchange any two quark flavours

Ground state

0

=



2 1 2

1 2

3

2 1 2

1 2

1

⊗ ⊕

=

⊗

The irreps of SU(3)

Decuplet

10

χ

Singlet

_χ

A

Octet-12

Antisymmetric on the first pair

8 12

χ

χ

₁₃8

∝

χ

12

8

+

χ

23 8

uuu

ddd

sss

uss+sus+ssu

3

dss+sds+ssd

3

dds+dsd+sdd 3

uus+usu+suu

3

uud+udu+duu

3

udd+dud+ddu

3

uds+usd+dsu+dus+sdu+sud

6

uds−usd+dsu−dus+sud−sdu 6

(2(ud−du)s+(us−su)d−(ds−sd)u) / 12

((us−su)d+(ds−sd)u) / 2

(ud−du)d 2

(ud−du)u

2

(ds−sd)d

2

(us−su)d 2

(ds−sd)s

2

(us−su)s

2

χ

₂₃8 (2s(ud−du)+d(us−su)−u(ds−sd)) / 12

(d(us−su)+u(ds−sd)) / 2

d(ud−du) 2

u(ud−du) 2

d(ds−sd) 2

u(us−su) 2

s(ds−sd) 2

s(us−su) 2

Octet-23

Antisymmetric on the last pair

There is also an Octet-13

(antisymmetric on the first-last)

uuu

ddd

sss

uss+sus+ssu

3

dss+sds+ssd

3

dds+dsd+sdd 3

uus+usu+suu

3

uud+udu+duu

3

udd+dud+ddu

3

uds+usd+dsu+dus+sdu+sud

6 strangeness 0 -2 -3 i=3/2 i=1 i=1/2 i=0 -1232 MeV 1384 MeV 1533 MeV 1672 MeV

J = 3 2 i3 Δ++ Δ+ Δ0 Δ− Σ+ Σ0 Σ−

Ξ* _Ξ*0

Ω−

(2(ud−du)s+(us−su)d−(ds−sd)u) / 12

((us−su)d+(ds−sd)u) / 2

(ud−du)d 2

(ud−du)u

2

(ds−sd)d

2

(us−su)d 2

(ds−sd)s

2

(us−su)s

2

(2s(ud−du)+d(us−su)−u(ds−sd)) / 12

(d(us−su)+u(ds−sd)) / 2

d(ud−du) 2

u(ud−du) 2

d(ds−sd) 2

u(us−su) 2

s(ds−sd) 2

s(us−su) 2

⊕

strangeness 0 -2 939 MeV 1193 MeV 1318 MeV 1116 MeV

J = 1 2 i3 p n Σ+ Σ0 Σ−

Ξ Ξ0

The Spin part

Use this for the decuplet

Use this for the octet Each bound state has three spin- 1

C

ol

o

u

r

•  Solution? Add in a new

quantum number – colour!

•  Need 3 different colours

•  Postulate: Every naturally

occurring bound state of quarks is a colour singlet!

Ground state

symmetric

0

=



2 1 2

1 2

3

2 1 2

1 2

1

⊗ ⊕

=

⊗

⊗ _SU(3)

flavour

irreps

symmetric

SU(3)_colour irreps! antisymmetric!

violates Pauli principle?!?!?

strangeness

0

-2

-3 i=3/2

i=1

i=1/2

i=0

-1232 MeV

1384 MeV

1533 MeV

1672 MeV

J = 3 2

i3 Δ++

Δ+

Δ0

Δ−

Σ+

Σ0

Σ−

Ξ* _Ξ*0

Ω−

ϕ_A = 1