StaticQuarkModel Malcolm JohnC4: particle physics

Units

natural S.I.

energy MeV ⇒ MeV × 1.6022 · 10−13⇒ J

mass MeV ⇒ MeV/c2×

1.6022 · 10−13

(3 · 108)2 ⇒ kg

time MeV −1⇒ ~/MeV ×

1.0546 · 10−34

1.6022 · 10−13 × 1012⇒ ps

frequency MeV ⇒ MeV/h ×1.6022 · 10−13

1.0546 · 10−34 / 2π ⇒ Hz

length MeV−1⇒ ~c/MeV × 1.0546·10−34

× 3·108

1.6022·10−13 × 1015︸ ︷︷ ︸ ⇒ fm

197

wavelength MeV−1⇒ h/( MeV/c) × 197 × 2π ⇒ fm

cross section MeV−2⇒ ~2c2/MeV2

× (197)2/100 ⇒ barn

(1 b = 100 fm2)In transfering to and from MeV units, ~c = 197 MeV fm is useful.

1 Operators and groups

The mathematics of groups deals with sets of operations withsome underlying symmetry. A set of operational elementsg1, g2, gi...gn (1 ≤ n ≤ ∞) is a group if three conditions hold.

• an identity, e exists such that gie = egi = gi. Though im-portant in the definition, the identity is not part of themgroup.

• operations are associative, g1(g2g3)=(g1g2)g3 though theydo not have to be commutative. Group that are commu-tative (g1g2 = g2g1) are Abelian groups.

• every element of a group must have an inverse that is alsopart of the group such that g1g−1

1 = e.

There are several fundamental operations upon which wewould expect physical behaviour not to depend. Translations inspace; occurrence in time; rotation about an axis; reversal ofspatial coordinates or the swapping of positive charge for nega-tive charge. If such an operator, Q is applied to a wavefunction,| ψ〉 = Q | ψ〉 (the bar labels the transformed wavefunction), theprobabilities of that wavefunction should not change:

〈φ | ψ〉 = 〈φ | Q†Q | ψ〉 = 〈φ | ψ〉 ,

Q−1= Q

†

(unitary)(1)

Similarly, Q must not change the energy of the system,

〈φ | H | ψ〉 = 〈φ | Q†HQ | ψ〉 = 〈φ | H | ψ〉 , (2)

If Q is unitary, we can write, The similarity transformation that

leaves the Hamiltonian unchanged is . Thus,

Q−1HQ = H ⇒ HQ − QH = 0 , (3)

the operator Q must commute with the Hamiltonian. Thesestatements indicate that unitary transformations, and thegroups thereof, have an important role in fundamental physics.Also consider the time variation of the Q expectation value,

d〈Q〉dt

=ddt〈ψ∗ | Q | ψ〉

= 〈∂

∂tψ∗ | Q | ψ〉 + 〈ψ∗ |

∂Q∂t| ψ〉 + 〈ψ∗ | Q |

∂

∂tψ〉

= 〈+i~

Hψ∗ | Q | ψ〉 + 〈ψ∗ | Q | −i~

Hψ〉 +⟨∂Q∂t

⟩=

i~

[〈ψ∗ | [HQ − QH] | ψ〉

]+

⟨∂Q∂t

⟩d〈Q〉

dt=

i~

⟨[H,Q]︸ ︷︷ ︸⟩

+

⟨∂Q∂t

⟩= 0 Q has no explicit time dependence

〈Q〉 = const.

This is Ehrenfest’s theorem. It shows that the expectation val-ues of Q are constants of the motion and thus gives a quantummechanical observable with good quantum numbers.

Parity and charge parity

A simple example are discrete groups that contain just one el-ement in addition to the identity. In this case it has to be itsown inverse. Examples are the reversal of spatial coordinates,

Malcolm John - 2020

P(x) −→ −x, or charge reversal (C(particle) −→ antiparticle).Physics must to be invariant after two applications of such bi-nary operators, so

P2| ψ〉 = I | ψ〉 = λ2

| ψ〉

where | ψ〉 are eigenvectors and λ are eigenvalues λ = ±1.An interesting question is, does P commute with the Hamilto-nian? Due to its simplicity, this could be assumed. Under thisassumption, if a state is an eigenstate of parity with a certaineigenvalue (±1), any transition will preserve this eigenvalue.

Fundamental particles have intrinsic parity, which is defined byconvention as positive for particles and negative for antiparti-cles. The force mediators are vector constructs, defined withnegative parity.

P | f 〉 = +1 | f 〉 for f = e−, µ−, qP | f 〉 = −1 | f 〉 for f = e+, µ+, q (4)

P | b〉 = −1 | b〉 for b = γ, g, Z0, W± .

The charge-parity operator, C, changes particle into antiparti-cle so few particles can be eigenstates of C. Charge parity isa useful conserved quantity when considering neutral bosons.The photon is a fundamental, neutral boson whose intrinsiccharge quantum number is inferred by considering classicalelectromagnetism. The vector potential (A) associated with thephoton is related to the electric field (E) and scalar potential (φ)by ∂A/∂t = −E−∇φ. Upon reversing charge, E and φ flip sign,so must the vector potential. Thus,

C | γ〉 = − | γ〉 . (5)

3D rotation as a fundamental group

Consider a rotation ε, about the y-axis to point a from b:a = ρy(ε)b and its inverse, b = ρ−1

y (ε)a.xayaza

=

cos ε sin ε

1− sin ε cos ε

xbybzb

xbybzb

=

cos ε − sin ε

1sin ε cos ε

xayaza

and impose invariance,

ψ(a) = ψ(b) = ψ(ρ−1y (ε)a)

ψ

xa

ya

za

= ψ

xa cos ε − za sin ε

ya

za cos ε + xa sin ε

Then using the first terms of the [co]sine Taylor expansion,

ψ(a) = ψ(a) + εxa∂ψ(a)∂z

− εza∂ψ(a)∂x

+ ...

≈

[1 + ε

(xa∂

∂z− za

∂

∂x

)]ψ(a)

≈

[1 −

i~εJy

]︸ ︷︷ ︸ψ(a) as: Jy = −i~

[z∂

∂x− x

∂

∂z

]︸ ︷︷ ︸

R(ε) (r ∧ p)y

Successive infinitesimal rotation are applied to perform amacroscopic rotation, This is the property of a Lie group,

R(β) = limn→∞

(1 −

i~

β

nJy

)n

= exp(−

i~βJy

).

(6)

For the general 3D case, this becomes,

R(θ) = exp(−

i~J · θ

). (7)

Any combination of successive rotations can be reduced toa combination of the three Jx, Jy, Jz generators that form aclosed, complete symmetric group of special (det R = 1) or-thogonal (RT

= R−1), S O(3) transformations. Orthogonality isthe real vector space analogue of unitarity.

Properties of the J generators

Consider the rotation 2π around axis i. The S O(3) transforma-tion should leave a wavefunction unchanged,

R(2π) | ψ〉 = I | ψ〉 = exp

− i~

2πJi︸ ︷︷ ︸ | ψ〉

= 0

(8)

which requires the eigenvalues of Ji to be integers. Writing thegenerator Ji = −i~ εi jk x j∂k in matrix form,

Jx = −i~

0 0 00 0 10 −1 0

Jy = −i~

0 0 −10 0 01 0 0

Jz = −i~

0 1 0−1 0 00 0 0

(9)

reveals the commutation relation,

[Ji, J j] = i εi jk Jk . (10)

Forming the eigenvalue equation for the Eq. 9 matrices con-firms three integer eigenvalues of Ji: m = [−1, 0,+1]~. The Ji

matrices are Hermitian (observable quantum numbers) but thenon-zero commutation of Eq. 10 means the three projectionsare not simultaneously observable. This non-commutivity is anexpression of the fact that Ji transforms as a vector. On theother hand, the Casimir operation,

J2= J2

x + J2y + J2

z (11)

= 2~2I

is a scalar so [J2, Ji] = 0. The operator J2 corresponds to thetotal angular momentum, which is simultaneously observablewith one projection along an axis. By convention, the z axisis chosen. The generator matrices of Eq. 9 are the irreducibleS O(3) representation of the rotation group for j = 1.Next we calculate the eigenvalues of J2. For this, it is useful to

Malcolm John - 2020

define raising(+) and lowering(−) operators,

J± = Jx ± iJy , (12)

such that

[Jz, J±] = ±J± ,

or JzJ± = J±(Jz + 1) .

The J± operations raise or lower the m quantum numberthrough the quantised eigenvalues from the minimum to maxi-mum values: − j,− j + 1, ... , j − 1, j, where j is identified at thetotal angular momentum quantum number. Then from Eq. 11,

J2= J2

x + J2y + J2

z

=︷ ︸︸ ︷J−J+ − i(JxJy − JyJx︸ ︷︷ ︸) +J2

z

iJz

= J−J+ + Jz(Jz + 1) (13)

Applying this result to the state, ψ where the angular momen-tum projection along the z-axis is at it maximal value (m = j)such that J+ | ψ〉 = 0,

J2| ψ〉 = j( j + 1) | ψ〉 . (14)

The same conclusion is reached from applying the loweringfirst to the state m = − j, where J− | ψ〉 = 0,

J2= J+J− + i(JxJy − JyJx) + J2

z

= J+J− + Jz (Jz − 1) (15)

= +(− j)(− j − 1)

= j( j + 1)

Clebsch Gordan coefficients

In the above, raising is applied to the m = j state and lower-ing to the m = − j state. For the general case the action ofraising/lowering is not null.

J± | j,m〉 = C±( j,m) | j,m ± 1〉 (16)

with the boundary condition that C± is required to be 0 forJ− | j,−j 〉 and J+ | j,+j 〉.

To derive these constants two observations are made.Eq. 13&15 show effect of a raising, then lowering operationis a real number that is not in general, unity.

J−J+ = j( j + 1) − m(m + 1)

J+J− = j( j + 1) − m(m − 1)(17)

Second, that Jx and Jy are Hermitian so J+ is the Hermitian-conjugate of J− (and vice-versa),∣∣∣∣∣ J− | j,m〉

∣∣∣∣∣2 = 〈 j,m | J+J− | j,m〉

= C2−〈 j,m | j,m〉 = C2

−

where C2− is real and defined in Eq. 17. Similarly,∣∣∣J+ | j,m〉

∣∣∣2 = 〈 j,m | J−J+ | j,m〉 = C2+

The C± are Clebsch-Gordan coefficients, defined as,

C±( j,m) =√

j( j + 1) − m(m ± 1) , (18)

and are used extensively is to calculate the relative intensity(probability) of combined states.

Example: the (j1 = 1) × (j2 = 1) combination

Imagine the combination of two spin-1 particles, each with pos-sible states: m = −1, 0, 1, into a system with total j = 2. Totalj = 1, 0 is also possible but neglected in this example. Thecombination has five possible states, m = −2,−1, 0, 1, 2 but thecomposition of each state must be found by applying the low-ering operator from the fully-stretch out state, calculating theClebsch-Gordan coefficients from Eq. 18. First calculate vari-ous C-G coefficients,

J− | 2,+2〉 = 2 | 2,+1〉

J− | 2,+1〉 =√

6 | 2, 0〉

J− | 1,+1〉 =√

2 | 1, 0〉

J− | 1,−1〉 = 0

Then start from the “stretched-out" state successively applyingthe lowering operator to both sides by calculating the C-G co-eficients dividing-through by any factor appearing on the LHC.

| 2,+2〉 =| 1,+1〉 | 1,+1〉 (“stretched-out" state)

| 2,+1〉 = 1√

2| 1, 0〉 | 1,+1〉 + 1

√2| 1, 1〉 | 1, 0〉

| 2, 0 〉 = 1√

6| 1,−1〉 | 1,+1〉 +

√23 | 1, 0〉 | 1, 0〉

+ 1√

6| 1, 1〉 | 1,−1〉

| 2,−1〉 = 1√

2| 1,−1〉 | 1, 0〉 + 1

√2| 1, 0〉 | 1,−1〉

| 2,−2〉 =| 1,−1〉 | 1,−1〉

Clebsch-Gordan coefficients for several common combinationsare published in several places, notably in the PDG.

The d j

m′,mmatrix

Generic rotation can be parameterised by three Euler anglesdefining three successive rotations. Most usefully :1. rotation by angle α about the original z-axis.2. rotation by angle β about the intermediate y-axis.3. rotation by angle γ about the final z-axis. For a wavefunc-tion with angular momentum j being rotated such that its z-projection changes from m to m′, the probability could be writ-ten,

P jm′,m

(αβγ) =∣∣∣∣ 〈 j,m′ | e− i

~ γJz e−i~ βJy e−

i~αJz | j,m〉

∣∣∣∣2=

∣∣∣∣ e−i~ (γm′+αm)

〈 j,m′ | e−i~ βJy | j,m〉

∣∣∣∣2=

∣∣∣∣ d jm′,m

(β)∣∣∣∣2 ,

Malcolm John - 2020

where,d j

m′m(β) = 〈 j,m′ | e−

i~ βJy | j,m〉 (19)

The change of the z-projection (m → m′) only occurs duringthe rotation around y. The two z-rotations lead just to a phaseshift that is unobservable in the modulus-square. The state| j,m〉 rotates into a linear combination on the (2 j + 1) | j,m′〉states.

We calculate the d jm′,m

matrix the specific case of j = 1 andapply it to the realistic case of e+e− → µ+µ−. There are 3 z-projections (−1, 0, 1) so there are 3 × 3 matrix elements in thiscase.

1. From inspecting the expansion of e−iβJy (with ~ = 1):

e−iβJy = 1 − iβJy −12!β2J2

y +i

3!β3J3

y +14!β4J4

y + ...

2. Look separately for solutions of J2n+1y | j,m〉 and J2n

y |

j,m〉

3. From the raising/lowering operators J± = Jx ± iJy and

Clebsch-Gordan coefficient, C−(1, 1) =√

2.

J+ − J− = Jx + iJy − Jx + iJy

Jy = −i2

[J+ − J−]

so, Jy | 1, 1〉 = −i2

[C+(1, 1) · 0 −C−(1, 1) | 1, 0〉

]=

i√

2| 1, 0〉

4. Operate again with Jy; C+(1, 0) = C−(1, 0) =√

2,

J2y | 1, 1〉 =

i√

2Jy | 1, 0〉

=1

2√

2(C+(1, 0) | 1, 1〉 −C−(1, 0) | 1,−1〉)

=12

(| 1, 1〉 − | 1,−1〉)

5. And again:

J3y | 1, 1〉 =

12

Jy(| 1, 1〉− | 1,−1〉)

=12

(−i2

[J+ − J−] | 1, 1〉 −

−i2

[J+ − J−] | 1,−1〉)

=i4

(√

2 | 1, 0〉 +√

2 | 1, 0〉)

=i√

2| 1, 0〉

= Jy | 1, 1〉

6. Note the cyclical pattern and conclude:

J2n+1y | 1, 1〉 =

i√

2| 1, 0〉

J2ny | 1, 1〉 =

12

(| 1, 1〉 − | 1,−1〉)

thus

〈1, 1 | J2n+1y | 1, 1〉 = 0

〈1,−1 | J2n+1y | 1, 1〉 = 0

〈1, 0 | J2n+1y | 1, 1〉 = i

√2

〈1, 1 | J2ny | 1, 1〉 = 1

2

〈1,−1 | J2ny | 1, 1〉 = − 1

2

〈1, 0 | J2ny | 1, 1〉 = 0

7. Using these relations the d jm′m

elements are,

d111 = 〈1, 1 | e−iβJy | 1, 1〉

= 〈1, 1 | 1 +(−iβJy) +12!

(−iβJy)2+

i3!

(−iβJy)3+

14!

(−iβJy)4+ ... | 1, 1〉

= 1 −12!β2

2+

i4!β4

2− ... ...

=12

(1 + (1 −12!β2

+14!β4− ...))

=12

(1 + cos β)

d1−11 = 〈1,−1 | e−iβJy | 1, 1〉

=12

(1 − (1 −12!β2

+14!β4− ...))

=12

(1 − cos β)

8. Elements are related by,

d1m′,m = (−1)m−m′d1

m,m′ = d1−m,−m′ .

The complete d1m′,m matrix is,

12

(1 + cos β) −1√

2sin β

12

(1 − cos β)

1√

2sin β cos β −

1√

2sin β

12

(1 − cos β)1√

2sin β

12

(1 + cos β)

. (20)

This result is used, for example, to determine the probabilityof rotating the photon of an electromagnetic interaction from, a| 1,+1〉 state to a combination of | 1,+1〉 and | 1,−1〉 states, asoccurs in e+e− → γ → µ+µ− collisions.

Pe+e−→µ+µ− ∝∣∣∣d1

1,1

∣∣∣2 +∣∣∣d1−1,1

∣∣∣2∝

∣∣∣ 12 (1 + cos θ)

∣∣∣2 +∣∣∣ 1

2 (1 − cos θ)∣∣∣2

∝1 + cos2 θ

4, (21)

where θ is the scattering angle in the zero-momentum frame.

Malcolm John - 2020

2 SU(2)

The S O(3) commutation relation, Eq. 10 can be realised us-ing the 2 × 2, special (det U), unitary (U† = U−1) matrices ofS U(2). To find these matrices remember that Lie operators aredescribed by exponentiated generators, e.g. Eq.6. Here, thegenerator X has eigenvalues X = Ix (x is a vector of eigenval-ues) and U has eigenvalues u where u = ex,

eTr X= eTr(Ix)

= e∑

i xi = Πiexi = det(eX) = det(U)

as the determinant of a matrix is equal to the product of itseigenvalues. So for operator matrices with unit determinant,exp(Tr X) = 1, Tr X = 0, their generators must be traceless.

Applying the unitary condition to a matrix, X =

(a bc d

),(

d −b−c a

)=

(a∗ c∗

b∗ d∗

)gives d = a∗, c = −b∗. Writing out X with explicit real andimaginary parts,

X =

(a b−b∗ a∗

)=

(xa + iya xb + iyb

−xb + iyb xa − iya

), (22)

shows that the real part of a must be zero, otherwise X wouldhave a non-zero trace. A natural expansion,

X = iyb

(0 11 0

)︸ ︷︷ ︸ +ixb

(0 −ii 0

)︸ ︷︷ ︸ +iya

(1 00 −1

)︸ ︷︷ ︸ . (23)

σx σy σz

reveals three, traceless, Hermitian matrices known as the Paulispin matrices. This is the expected dimensionality of S U(2).Complex n × n matrices have 2n2 real free parameters butthe unitarity requirements imposes n2 constraints. The S pecialconstraint of requiring det(U) = 1 imposes one extra constraintso,

dim [S U(n)] = n2− 1 , (24)

which is indeed three in the case n = 2.

Rotations with SU(2)

With 2× 2 matrices, we expect two eigenvalues for each of thethree Pauli matrices: m = − 1

2 ,+12 for j = 1

2 , which by conven-tional choice is defined along σz. Thus the generators, Xi ofrotations in S U(2)-space are defined,

si = 12 σi , (25)

with commutation relation,

[si, s j] = iεi jk sk , (26)

and for which a raising and lowering operator is

s± =sx ± isy

s+ =

(0 10 0

)s− =

(0 01 0

) (27)

s+

(01

)=

(0 10 0

) (01

)=

(10

)s+

(10

)=

(0 10 0

) (10

)=

(00

).

Working backwards,

sx =12

(s+ + s−) and sy =i2

(s− − s+)

and sz is derived from the requirement:

sz

(10

)=

12

(10

)and sz

(01

)= −

12

(01

)confirms the S U(2) generators as the Pauli spin matrices:

sx = 12

(0 11 0

)︸ ︷︷ ︸ sy = 1

2

(0 −ii 0

)︸ ︷︷ ︸ sz = 1

2

(1 00 −1

)︸ ︷︷ ︸

σx σy σz

The algebra of S U(2) transformation [almost] identical to S O(3)rotations but here, si = 1

2 σi generate rotational operators,

U = exp(−i 12 σ · θ) . (28)

An expansion of the exponential shows the importance of thefactor, 1

2 . Using the y-component as an example,

(σy)2=

(0 −ii 0

)(0 −ii 0

)=

(1 00 1

),

Uy =1 +(−i 1

2 σyθy)

1!+

(−i 12 σyθy)2

2!+

(−i 12 σyθy)3

3!

+(−i 1

2 σyθy)4

4!+

(−i 12 σyθy)5

5!+

(−i 12 σyθy)6

6!+ ...

=

1 −(θy

2

)2

2!+

(θy

2

)4

4!+

(θy

2

)6

6!+ ...

− iσy

(θy

2

)1!−

(θy

2

)3

3!+

(θy

2

)5

5!+ ...

Uy = cos

θy

2− iσy sin

θy

2. (29)

The periodicity of the S U(2) rotations has a peculiar minus signcomes out of a 360 rotation, U(2π) | ξ〉 = − | ξ〉 and it takestwo rotations to achieved an null rotation, U(4π) | ξ〉 =| ξ〉. Theobjects, | ξ〉, that are rotated by the S U(2) rotations, and spin-12 by construction, are clearly not vectors (for which we expect360 rotation to be null) but are known as spinors. Thoughthere is no classical equivalent, a Möbius strip can visualise ofhow a 720 rotation can be necessary to achieve a null rotation.

SU(3)

This group describes transformations between three eigenval-ues. The Gell-Mann basis is common, it is built on (and con-tains) the Pauli spin matrices. It extends the idea of raisingand lowering between “spin 1

2 " and “spin − 12 " with a third de-

gree. Imagine moving around a triangle rather than just up anddown. The Gell-Mann basis is defined as Ti = 1

2 λi where, the

Malcolm John - 2020

first three S U(3) basis matrices are those of S U(2),

λ1 =

0 1 01 0 00 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

1 0 00 −1 00 0 0

(30)

The S U(2) basis can be applied between the 1st and 3rd ele-ment,

λ4 =

0 0 10 0 01 0 0

λ5 =

0 0 −i0 0 0i 0 0

λa =

1 0 00 0 00 0 −1

(31)

And again between the 2nd and 3rd elements,

λ6 =

0 0 00 0 10 1 0

λ7 =

0 0 00 0 −i0 i 0

λb =

0 0 00 1 00 0 −1

(32)

We note that the three diagonal matrices, λ3, λa and λb arenot independent, λ3 = λa − λb. So choose to keep λ3 as-is andmake an orthogonal combination of the other two,

λ8 = 1√3

[λa + λb

]= 1√

3

1 0 00 1 00 0 −2

(33)

We have found n2− 1 = 8 traceless matrices as required for a

basis of S U(3).

Combined representations

Consider two spinors with angular momentum ja and jb. Thecombination

| j,m〉 =| ja,ma〉 | jb,mb〉

has total angular momentum in a range of discrete values,

j = jmin, ( jmin + 1), ... , ( jmax − 1), jmax,

where jmin = | jA − jB| and jmax = | jA + jB|. For each value ofj there are 2 j + 1 states in an irreducible representation (mul-tiplet). Consider for example, the combination of two spin- 1

2spinors:

(2 12 +1)(2 1

2 +1) = 4 combinations,

jmin = 12 −

12 = 0,

jmax = 12 + 1

2 = 1,

Nstates( jmin) = (2 jmin + 1) = 1,

Nstates( jmax) = (2 jmax + 1) = 3 .

Hence to describe the multiplets arising from the combinationof two spin- 1

2 spinors, we write,

2 ⊗ 2 = 3 ⊕ 1 .

Combining two spin- 12 spinors by-hand

Combinations of two spin- 12 spinors can have total spin j = 0

or j = 1,m ∈ −1, 0, 1.

| 1, 1〉 =| 12 ,

12 〉 |

12 ,

12 〉

=|↑, ↑〉

| 1, 0〉 =

√12 |

12 ,

12 〉 |

12 ,−

12 〉 +

√12 |

12 ,−

12 〉 |

12 ,

12 〉

=

√12 (|↑, ↓〉 + |↓, ↑〉)

| 1,−1〉 =| 12 ,−

12 〉 |

12 ,−

12 〉

=|↓, ↓〉

| 0, 0〉 =

√12 |

12 ,

12 〉 |

12 ,−

12 〉 −

√12 |

12 ,−

12 〉 |

12 ,

12 〉

=

√12 (|↑, ↓〉 − |↓, ↑〉)

The j = 1 triplet is symmetric under the interchange of spinors.The singlet contains one j = 0 state which is deduced fromrequiring orthogonality to the |1, 0〉 state. The singlet state ismanifestly antisymmetric, meaning that a factor of −1 dropsout upon the interchange of spinor 1 and 2. This deconstruc-tion explicitly demonstrates 2 ⊗ 2 = 3 ⊕ 1.

Young’s Tableaux

Young’s Tableaux is a tool of group mathematics used to cal-culate the number, dimension and symmetry of the irreduciblerepresentations. It is quicker than the ‘by-hand’ approach ex-emplified above for 2 ⊗ 2. Tableaux are constructed using thefollowing rules:

1. A tableau represents one SU(n) object and can take npossible labels,

for S U(3), ≡ 1 , 2 or 3

2. Objects are combined by adding tableaux to the right orbottom forming a top-left justified stack ensuring no rowis longer than the one above it.

OK OK OK no

3. A single “horizontal" row configuration represents a mul-tiplet of symmetric combinations. A single “vertical" col-umn configuration represents an antisymmetric multiplet.All other configurations are of mixed symmetry.

symmetric antisym. mixed

4. When populating the tableaux with labels (1,2,3 etc...) toinvestigate the dimensionality of the multiplet, two morerules must be respected to avoid double-counting per-mutations:a. Moving right across a row of tableaux, labelsmay notdecrease (it can stay the same);b. Moving down any column of tableaux, labels must in-crease (it can not stay the same).This last rule implied that there can be at-most n rows oftableaux in the SU(n) group.

1 1 OK 11

no 1 23

OK 2 13

not

Applying these rule to the example of two S U(2) spinors recov-ers the by-hand result.

⊗ = ⊕

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= 1 1 1 2 2 2︸ ︷︷ ︸ ⊕ 12︸︷︷︸

sym. asym.

2 ⊗ 2 = 3 1

Combining three spinors

When adding the third spinor to the previous result, one notesthat there is two ways that the third tableau may be added tothe symmetric triplet, but only one way to add it to the asym-metric singlet without violating rule 4. The result is symmetricquadruplet and two doublets of mixed symmetry. They retainseparate labels MS and MA indicating the symmetry of thefirst two spinor in the combination.

⊗ ⊗ =

(⊕

)⊗

= ︸ ︷︷ ︸ ⊕ ︸ ︷︷ ︸ ⊕︸ ︷︷ ︸sym. MS MA

= 1 1 1 1 1 2 1 2 2 2 2 2

⊕1 12

1 22

⊕1 12

1 22

Thus, (2 ⊗ 2) ⊗ 2 = 4 ⊕ 2 ⊕ 2 (34)

Combining three spinors by hand

Start by adding a spin-up spinor to the |1, 1〉 state, then workdown using the Clebsch-Gordan recipe.

| 32 ,

32 〉 = |↑↑↑〉

| 32 ,

12 〉 = 1

√3(|↓↑↑〉+ |↑↓↑〉+ |↑↑↓〉)

| 32 ,−

12 〉 = 1

√3(|↓↓↑〉+ |↑↓↓〉+ |↓↑↓〉)

| 32 ,−

32 〉 = |↓↓↓〉

Two more states come from subtracting angular momentumfrom |1, 1〉 and requiring orthogonality to the | 3

2 ,+12 〉 states:

| 12 ,

12 〉 = 1

√6(2 |↑↑↓〉− |↓↑↑〉− |↑↓↑〉)

| 12 ,−

12 〉 = −1

√6(2 |↓↓↑〉− |↑↓↓〉− |↓↑↓〉)

These states are of mixed symmetry. Finally, two states arederived from adding a spinor to the two-spinor singlet:

| 12 ,

12 〉 = 1

√2(|↑↓↑〉− |↓↑↑〉)

| 12 ,−

12 〉 = −1

√2(|↑↓↓〉− |↓↑↓〉)

which is also of a mixed symmetry.

Application to SU(3)

The strong force has three charges of identical magnitudeand is thought to exhibit perfect S U(3) symmetry. The threecharges are imagined as colour,

= 1 2 3 = Red Blue Green

Unlike, electric charge, colour charge cannot be observed di-rectly because all hadrons are neutral “colourless" combina-tions. It is the study of the patterns of colour combinationsthat much of the nature of the underlying force is inferred. Thecombination of two S U(3) spinors is,

⊗ = ︸︷︷︸ ⊕ ︸︷︷︸sym. asym.

= 1 1 1 2 1 3 2 2 2 3 3 3

⊕12

13

23

Thus, 3 ⊗ 3 = 6 ⊕ 3 (35)

Extend to three spinors as per the Young’s Tableaux rules:

⊗ ⊗ =

(⊕

)⊗

= ⊕ ⊕ ⊕

sym. MS MA asym.

Thus, (3 ⊗ 3) ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 (36)

For colour S U(3), the colour singlet plays an important roleensuring the baryon wavefunction is antisymmetric.

RGB

=1√

6

(+ | RGB 〉 − | GRB 〉 + | BRG 〉

− | RBG 〉 + | GBR 〉 − | BGR 〉).

(37)

3 Isospin

In the 1930s, the theoretical success of S U(2) and the similar-ity in mass of the proton and neutron inspired postulation thatthey were the same fundamental object, just with a differentprojection. Though this is long since known not to be the case,it is still a useful symmetry when dealing with hadronic physics.The concept is:

• |u〉 and |d〉 form a S U(2) doublet like spin- 12

I3 | u〉 = 12 | u〉, I3 | d〉 = − 1

2 | d〉

• Raising & lowering operators exist:

I+ | d〉 =| u〉, I− | u〉 =| d〉, I− | d〉 = 0

• Total isospin is conserved in strong interactions.

Isospin conservation stems from the underlying symmetry ofthe S U(2) group. For the group properties of symmetry to ap-ply, two assumptions are imposed:

1. the u and d quarks have an identical [strong] coupling.

2. the u and d quarks have [essentially] identical mass.

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The pattern of u, d flavour multiplets for baryons is identicalto the 2 ⊗ (2 ⊗ 2) = 4 ⊕ 2 ⊕ 2 structure of 12 ,−

12 .

| I = 32 , I3 = 3

2 〉 = | uuu〉

| I = 32 , I3 = 1

2 〉 = 1√

3(| duu〉 + | udu〉 + | uud〉)

| I = 32 , I3 = − 1

2 〉 = 1√

3(| ddu〉 + | udd〉 + | dud〉)

| I = 32 , I3 = − 3

2 〉 = | ddd〉

| I = 12 , I3 = 1

2 〉 = 1√

6(2 | uud〉 − | duu〉 − | udu〉)

| I = 12 , I3 = − 1

2 〉 = −1√

6(2 | ddu〉 − | udd〉 − | dud〉)

| I = 12 , I3 = 1

2 〉 = 1√

2(| udu〉 − | duu〉)

| I = 12 , I3 = − 1

2 〉 = −1√

2(| udd〉 − | dud〉)

Spin-flavour wavefunction of the proton

The proton is the lowest-lying stable baryon; as L = 0, the spa-tial part of the wavefunction, ψspace is symmetric. The colourwavefunction, ψcolour is always antisymmetric. Hence for the to-tal wavefunction to a antisymmetric (as the proton is a fermion),ψspinψ f lavour must be symmetric. The proton has S = 1

2 so mustbe built from the mixed symmetry wavefunctions because thesymmetric quadruplet is for S = 3

2 only. The correct combina-tion is symmetric,

|p↑〉spin, f lavour ∝(ψspin(MS ) ψ f lavour(MS )

+ψspin(MA) ψ f lavour(MA)) (38)

where,

ψspin(MS ) = | 12 ,

12 〉 = 1

√6(2 |↑↑↓〉 − |↑↓↑〉 − |↓↑↑〉)

ψspin(MA) = | 12 ,

12 〉 = 1

√2(|↑↓↑〉 − |↓↑↑〉)

and with different notation, but identical logic:

ψ f lavour(MS ) = 1√

6(2 | uud〉 − | udu〉 − | duu〉)

ψ f lavour(MA) = 1√

2(| udu〉 − | duu〉)

Multiplying all these terms together and normalising givesthe proton (spin, f lavour) wavefunction,

| p↑〉spin, f lavour = 1√

18( 2 | u↑u↑d↓〉 − | u↑u↓d↑〉 − | u↓u↑d↑〉

+2 | u↑d↓u↑〉 − | u↑d↑u↓〉 − | u↓d↑u↑〉

+2 | d↓u↑u↑〉 − | d↑u↓u↑〉 − | d↑u↑u↓〉 ) . .

To confirm the symmetry, notice that each line is related to an-other by the swapping the position of any two spinors.

Isospin of antiquarks

When dealing with antiquarks care must be taken of howcharge conjugation is applied. We choose a conventionwhich leave the Clebsch-Gordan coefficients the same as withquarks, though this introduces a sign convention (the Condon-

Shortley convention),

I− | d〉 → − | u〉,

I+ | u〉 → − | d〉,

I+ | d〉 = I− | u〉 = 0

(39)

For example, consider the SU(2) isospin combinations of 2 ⊗2 = 1 ⊕ 3. The singlet is symmetric,

1√

2(|dd〉 + |uu〉)=|0, 0〉

and the antisymmetric triplet identifies three pions (+, 0,−),

|1, 1〉 = |ud〉 = | π+〉

|1, 0〉 = 1√

2(|dd〉 − |uu〉) = | π0

〉

|1,−1〉 = − |ud〉 = | π−〉 .

Experimentally, the use of S U(2) seems justified by the tripletof pions being [almost] degenerate in mass:

m(π+) = m(π−) = 140 MeV/c2, m(π0) = 135 MeV/c2.

Example use of isospin: analysis of ∆ decays

• The ∆(1232) baryons contain only u and d with total spinS =

3/2 and isospin I =3/2.

• Baryon number conservation requires the ∆ to decay top or n, higher mass baryons are not accessible.

• As the strong interaction dominates, use isospin to un-derstand relative rates remembering:

|p〉 = | 12 ,

12 〉 |n〉 = | 1

2 ,−12 〉

|π+〉 = |1, 1〉 |π0

〉 = |1, 0〉 |π−〉 = |1,−1〉

• Decompose I = 32 state into possible combinations I = 1

2and I = 1

∆++

= |p〉|π+〉

∆+

= 1√

3|n〉|π+

〉 +

√23 |p〉|π

0〉

∆0

= 1√

3|p〉|π−〉 +

√23 |n〉|π

0〉

∆−

= |n〉|π−〉

• One can deduce the ratio of branching fractions:

B(∆+→ π0 p)

B(∆+→ π+n)

=|〈π0 p | ∆+

〉|2

|〈π+n | ∆+〉|

2 = 2

• or the relative ∆ cross sections of π− and π+ scatteringoff a proton target at

√s ∼ m(∆):

σ(π+ p)σ(π−p)

=|〈∆

++| π+ p〉|2

|〈 ∆0| π−p〉|2

= 3 .

J PC of hadrons

Parity and, where applicable, charge quantum numbers areoften quoted, for a particular state, with its total angular mo-mentum, J = L + S , as a JPC number. The JP of a particle

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is closely related to the spatial transformation properties of thestate. 0− are labelled pseudoscalar particles and 0+ scalar. 1−

are labelled vector particles and 1+ states are axial vectors.

• Mesons are qq bound states. As these have opposite par-ity, a ground state mesons will always have P = −1. Ex-cited states bring additional parity factors according to(−1)L.

Pmeson = (−1)L+1

C is defined (for light neutral mesons only) by interchang-ing q↔ q and swapping their positions and spin.

Cmeson = (−1)L+S

• Baryons contain three quarks and as such cannot betheir own antiparticle; C is undefined. The calculationof baryon parity is more complex than for mesons be-cause one must consider the angular momentum of athree-body system. The intrinsic parity of a baryon is(+1)3

= +1; similarly it is −1 for an antibaryon. In full:

Pbaryon = B (−1)L12 (−1)L3

where B is baryon number, L12 is the angular momentumbetween quark 1 and 2; L3 is the angular momentum ofthe third relative to the 1-2 pair.

Situations where consideration of JPC prove insightful are,

• JPC of a π0

Ground state: L=S=J=0. From the formula: P = −1,C = +1. So JPC

= 0−+. It is a pseudoscaler. Thedominant decay mode is π0

→ γγ. The conservationof charge conjugation is straight forward. Parity conser-vation requires that he overall |γγ〉 is antisymmetric.

• JPC of a ρ(770)Vector state with the quark spins aligned: L=0, S=1, J=1.From the formula: P = −1, C = −1. So JPC

= 1−−.

• Is ρ0→ π+π− an allowed decay?

Excited initial state, S = J = 1. L=0 so P = −1. Also fromabove, C = −1 so JPC

= 1−−. The pair of pions have in-trinsic parity +1 from (−1)π+ (−1)π− and each J = 0. Soproducing them in a L = 1 state will simultaneously con-serve J, P,C.

• What about ρ0→ π0π0?

This would be similar to the previous example, requiringthe pion pair to be in a L = 1 state. However, applyingC to the final state has no effect: C | π0π0

〉 =| π0π0〉.

C = +1. JPC(π0π0) = 1−+. So this process is forbiddenby charge violation with the strong or EM force. Indeed,it has never been observed.

• ρ0→ l+l−

Similar to ρ0→ π+π− except l+l− system has intrin-

sic parity −1 from (+1)l− (−1)l+ . Producing as final statewith the spins aligned, S = 1 simultaneously conservesJ, P,C. Note that, this process only proceeds via theelectromagnetic force and is therefore only 10−4 as likelyas the π+π− mode.

• ρ0→ π0γ

The initial state is JPC= 1−−. The photon: JPC

= 1−− andthe π0 0−+. But, the photon carry parity (−1)L away froma system.So this EM decay is allowed.

Strangeness

The hope that up-ness and down-ness was sufficient to de-scribed hadronic matter, died when a new particle, originallyknown as the V0, was observed in 1948 (in Manchester). Mostevidently, the displaced π+π−vertex meant this new particlewas neutral and long-lived. By adding together the momen-tum of the daughter pions, the invisible particle was measuredto have a mass ∼ 1

2 that of the proton, ruling out it being abaryon. Particles with this new long-lived property were as-signed a new quantum number: strangeness.

τ−θ puzzle Soon after, two long-lived, strange mesons wereobserved with indistinguishable masses and lifetimes.

• The “τ±" decayed isotropically to π±π0. By angular mo-mentum conservation, this implies the J(τ) = 0 and as-suming parity conservation, P(τ) = +1.

• The “θ±" decayed to π±ππ. P(π) = −1, (−1)3= −1. Again

assuming parity conservation, the parity of the θ seemedin opposition to the tau. Suspicion that J(θ) = 1 wastested by examination of the decay kinematics (in Oxfordby Dick Dalitz) but they showed L12 = L3 = 0, confirmingJ(θ) = 0 and P(θ) = −1.

Parity of an ensemble of pions

• The intrinsic parity of a quark is +1, an antiquark, −1.Hence a J=0 meson like the pion has P = −1.

• The parity of the two pion state is the product of theirintrinsic parities and any angular momentum be-tween them. However, by conservation of angularmomentum from the initial “τ", L = 0:

P(ππ) = (−1)(−1)(−1)L= +1.

• The three pion system is more complicated becausethere are two angular momenta:

1. the relative motion of the first two pions, L12

2. the motion of the third relative to the axis of thefirst two, L3.

• However, because the overall angular momentummust be conserved from the “θ" (J = 0), L12 ≡ L3.

P(πππ) = (−1)(−1)(−1)(−1)L12 (−1)L3 = −1.

In a 1956 literature review Lee and Yang concluded therewas no evidence of parity conservation in weak processes.They correctly postulated in a paper entitled, Question of ParityConservation in Weak Interactions that the “τ" and “θ" mesonswere the same particle (now called the K±) and that parity is

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not a conserved quantity of the weak force, in stark contrast tothe strong force, electromagnetism and common sense.

The confirmation of parity violation came from a dedicated,and renown experiment by Madame Wu and collaborators.The essential details are,

• 6027Co is an artificial isotope that beta-decays to 60

28Ni with ahalf-life of 5.27 years.

• It has spin=5 whereas 60Ni has spin=4 hence the electronand antineutrino of the beta decay have J=1.

• The cobalt sample was cooled to 0.01K (!) in a magneticfield such that all their spins were aligned.

• This spin alignment is vital because in order to inspectproperties of parity, a reference direction is needed, andthe only “intrinsic" direction for the initial state is the spinaxis.

• An electron counter was mounted above the sample anddepending on the polarity of the magnetic field, it recordsthe electron flux in, or opposite to, the direction of theCobalt-60’s spin.

• The data is shown below. It is seen that there is a deficit inelectrons in the direction of the spin.

4 S U(3) flavour

With a new flavour quantum number, the applicability of SU(3)is tested for three eigenvectors,

| u 〉 =

100

| d 〉 =

010

| s 〉 =

001

The Gell-Mann matrices (Eq. 30-33) combine to form three

raising/lowering operators that navigate the S U(3) multiplet,

I± = 12 (λ1 ± iλ2)

V± = 12 (λ4 ± iλ5)

U± = 12 (λ6 ± iλ7)

I3 = 12 λ3 S = 1

√3λ8

where I3 gives the projection along the “horizontal" and the Sdefines the “vertical" position in the triangle plots.

I3

Y

I±

U± V±

|d〉 |u〉

|s〉

I±

V± U±

|u〉 |d〉

|s〉

The isospin (I-spin) transformation is still valid, but it is joinedby U-spin and V-spin which navigate the (s, d) and (s, u) dou-blets respectively. The complete set of raising and loweringoperators are:

I+ | d〉 = | u〉 I− | u〉 = | d〉

V+ | u〉 = | s〉 V− | s〉 = | u〉

U+ | s〉 = | d〉 U− | d〉 = | s〉

I+ | u〉 = − | d〉 I− | d〉 = − | u〉

U+ | d〉 = | s〉 U− | s〉 = | d〉

V+ | s〉 = − | u〉 V− | u〉 = − | s〉

The Condon-Shortley convention is used for the antiquarkswhich gives rise to the ‘extra’ minus signs.

SU(3) mesons

Young’s tableau for the SU(3) antiquark triplet is the antisym-metric reciprocal of the quark triplet defined according to

qi = εi jkqkq j (40)

where a negative εi jk indicates the antiparticle. Explicitly,

SU(3) quarks = SU(3) antiquarks =

= u , d , s =sd

, us

, du

= u d s

Thus Young’s tableaux of quark-antiquark combinations are,

⊗ = ︸︷︷︸ ⊕ ︸︷︷︸3 ⊗ 3 = 1 ⊕ 8

We now have the information to extend beyond thepreviously-observed π+, π0, π− triplet of SU(2) and identify allthe SU(3) qq states. We start with the simplest case of J = 0pseudoscalar mesons.

Malcolm John - 2020

−

−

The six states with non-zero I and non-zero S are have un-ambiguous quark content. The other three need thought. Theflavourless, symmetric singlet, known as the η′ is

η′ = 1√

3(|uu〉 + |dd〉 + |ss〉)

The remaining two are part of the octet and are deduced withrotations within the group,

I− | ud〉 = | dd〉 − | uu〉 + X flavourless

U− | ds〉 = | ss〉 + | dd〉 + Y flavourless

V− | us〉 = | ss〉 − | uu〉 + Z flavourless

These equations are not linearly independent so the conven-tion is to choose one to be the established, isospin-triplet part-ner of the π+ and π−,

π0= 1√

2(|dd〉 − |uu〉) . (41)

The last member of the octet is deduced by requiring its wave-function be orthogonal to both the η′ the π0, and normalised to1.

η · π0= 0

α|uu〉β|dd〉γ|ss〉

· 1√

2

|uu〉−|dd〉

0

= 0 ⇒ β = α

η · η′ = 0α|uu〉β|dd〉γ|ss〉

· 1√

3

|uu〉|dd〉|ss〉

= 0 ⇒ γ = −2α

η · η = 1 ⇒ α2+ α2

+ (−2α)2= 1

α = 1√

6

Thus, |η〉 = 1√

6

(|uu〉 + |dd〉 − 2|ss〉

)(42)

The J = 1 vector mesons exhibit a similar pattern of stateswith the notable difference that “octet-singlet" mixing occursamongst the I = 0, S = 0 states. Experimentally it is seen thatthe singlet state (φ(1020)) decays largely to kaons the other(ω(782)) to pions (i.e. it has lost its ss content).

|ρ0〉 = 1

√2(|dd〉 − |uu〉) B(ρ0

→ π+π−) = 100%

|ω〉 ≈ 1√

2(|dd〉 + |uu〉) B(ω→ π+π−π0) = 90%

|φ〉 ≈ |ss〉 B(φ→ KK) = 84%

B(φ→ π+π−π0) = 15%

This is further breaking of the pattern anticipated by the S U(3)symmetry and more evidence that the s quark mass is signifi-cantly different to that of the u and d quarks.

−

−

SU(3) baryons

The baryon wavefunction is made of four parts:

Ψ = ψspaceψspinψ f lavourψcolour .

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We consider ground state baryons that have no orbital angularmomentum (L = 0) such that ψspace is symmetric. The colourwavefunction ψcolour is identical for all baryons and always anti-symmetric. Hence, for the overall fermionic wavefunction to beantisymmetric, ψspinψ f lavour must be symmetric. The possiblecombinations for ψspin are:

2 ⊗ (2 ⊗ 2) = 8 = 3 ⊕ 3 ⊕ 1 ⊕ 1 = 4︸︷︷︸⊕ 2︸︷︷︸⊕ 2︸︷︷︸sym. MS MA

where we recall the four symmetric states are spin- 32 states,

whereas the other are spin- 12 . The S U(3) irreducible represen-

tation for three particle combinations is:

3 ⊗ 3 ⊗ 3 = 10︸︷︷︸⊕ 8︸︷︷︸⊕ 8︸︷︷︸⊕ 1︸︷︷︸sym. MS MA asym.

So the symmetric (SU(3),SU(2)) combinations must be,

1√

2[( 8 , 2 ) + ( 8 , 2 )] and ( 10 , 4 )

MA MA MS MS sym. sym.

The J = 12 triplet and J = 3

2 decuplet are

Note that the ∆++ is manifestly symmetric in ψ f lavourψspinψspace.

This observation was the driving force for an additional degreeor freedom : colour charge. Also, the quark model was used topredict the S = −3Ω−. It was verified in 1964 when a new par-ticle was observed to decay via three weak decays. The weak

decay is identified by particles “flying" an appreciable distance;this implying longer lifetimes than the EM or strong interactionscould describe. The production and decay chain is:

K−p→Ω−K+K0S

Ω− → Ξ0π− .... Ξ0→ Λ0π0 .... Λ0

→ pπ−

where each decay is a strangeness-changing weak decay.

5 The November Revolution

By the mid-1960s, the quark model had established itself as anexcellent explanation for the patterns of discovered hadronicstates. Though it was still not clear if a “quark", which hadnever been observed in isolation, was real or just a convenientmathematical construct.

GIM suppression

It was known that the eigenstates of the charged weak interac-tion were not the same as the mass eigenstates. Cabibbo pa-rameterised this by introducing an ‘angle’ that described howmuch the d and s mass eigenstates contributed to the weakinteraction with the u:(

u)← weak force→

(d cos θc + s sin θc

)However, this could not explain why dimuon decays were sorare; experiment measured B(K0

L → µ+µ−) ≈ 10−8 when theFeynman diagram was straightforward,

s

d

µ+

µ−

u ν

W+

W−

A ∝ sin θc cos θc

K0L

In 1970 the “GIM mechanism" (Glashow, Illiopoulos, Maiani) in-troduced a + 2

3 partner to the s-quark. A second up-type quarkgive a second diagram that contributes with an [almost] equaland opposite contribution.

s

d

µ+

µ−

c ν

W+

W−

A ∝ − sin θc cos θc

K0L

The new model, with two generation of doublets becomes:(uc

)← weak force→

(cos θc sin θc

− sin θc cos θc

) (ds

)The reason this process happens at all is because thecancelling-out of these two diagrams is not exact because theu and c quarks have different mass. And from a considerationof kaon mixing, mc was estimated at ≈ 1.5 GeV/c2.

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Charmonium

Direct evidence of charm arrived in November 1974 from twoexperiments. One collaboration called it the J; the other the ψ.

• e+ e− collisions at centre of mass energy√

s ≈ 3 GeV/c2

at Stanford, CA. Left diagram. (Burton Richter et al.)

• 28 GeV protons on a beryllium target at Brookhaven, NY.Right diagram. (Sam Ting et al.)

Quantum numbers

First question: how much u-ness and d-ness is there in thisnew state? The J/ψ isospin assignment comes from observing:

B (J/ψ → ρ+π−) = B (J/ψ → ρ0π0) = B (J/ψ → ρ−π+)

Clebsch Gordan coefficients shows the J/ψ has zero isospin.

Clebsch-Gordan coefficents| J,M〉 =

m1 m2 | 2, 0〉 | 1, 0〉 | 0, 0〉

| ρ+π−〉 = | 1, 1〉 | 1,−1〉 1 −1√

1/6

√1/2

√1/3

| ρ0π0〉 = | 1, 0〉 | 1, 0〉 0 0

√2/3 0 −

√1/3

| ρ−π+〉 =| 1,−1〉 | 1, 1〉 −1 1

√1/6 −

√1/2

√1/3

At Stanford, they examined e+e− → µ+µ− with√

s ≈ m(J/ψ )so the final state is produced via either a J/ψ resonance decay-ing electromagnetically or via a virtual photon. The data showsa ‘dip’ on the low-mass side of the resonance which implies thephoton and the J/ψ process are interfering, which means theJ/ψ has the same JPC

= 1−− as the photon.

Width

Consider Breit-Wigner formula:

σ(E) = σmax

Γ2f /4

(E − ER)2+ Γ

2/4

where σmax = 4πo2 (2J + 1)(2s1 + 1)(2s2 + 1)

.

(43)

For the e+e− final state. J = 1, s1 = s2 = 12 :

σ(E)J/ψ→e+e− = 3πo2 Γ2e+e−/4

(E − ER)2+ Γ

2/4

This can be integrated to find total cross section using thesubstitution, tan θ = 2(E − ER)/Γ.

σ(E)J/ψ→e+e− = 3πo2 Γ2e+e−/4

(tan2 θ)Γ2/4 + Γ2/4

= 3πo2

(Γe+e−

Γ

)2

tan2 θ + 1

Then,

d(E − ER) = Γ2 d tan θ , d(tan θ)

dθ = tan2 θ + 1

and∫ ∞

−∞

d(E − ER) 7→∫ π/2

−π/2dθ

σTOTJ/ψ→e+e− = 3πo2

(Γe+e−

Γ

)2Γ

2

∫ π/2

−π/2dθ =

3π2o2

2

(Γe+e−

Γ

)2

Γ

From the measurements made by Richter et al. we can take:

• the area under the e+e− → J/ψ → e+e− peak is ≈ 400 nbMeV (approx. a triangle, height: 100 nb, base: 8 MeV)

• the selection efficiency of | cos θ| < 0.6 on a (1 + cos2 θ)distribution ∼ 50%;

• Γe+e−/Γ = B(J/ψ → e+e−) ≈ 6%

Thus:

Γ =800 nb MeV

32π

2o2 (6%)−2

using : o =~cpc

=197 MeV fm1548 MeV

and taking care of the area units, Γ ≈ 92.7 keV .

This is much narrower than the established J = 1 mesons. Forexample, Γ(K∗) ≈ 50 MeV and Γ(ρ) ≈ 150 MeV. This energy isfar below the experimental resolution; hence the width cannotbe measured directly and this calculation is necessary to de-termine the width. Such a narrow width (“long lifetime" ) meansthe J/ψ can not to be interpreted as some high-mass combina-tion of u,d,s. Instead, the J/ψ heralded the discovery of the 4thquark, “charm”, in a bound cc state.

Charmonium spectrum

Soon after, more JPC= 1−− cc states was found:

• ψ(2S ) Γ = 320 keV. B(ψ(3686)→ J/ψ π+ π−) ≈ 50%

• ψ(3S ) Γ = 27.3 MeV. B(ψ(3770)→ D0 D0) ≈ 85%

Decays of the J/ψ must be via the annihilation of the cc. In thecase of a strong-force, the transition must involve a minimumof three gluons because:

• a single gluon is not colourless so not colour-conserving.A colourless exchange is possible with two gluons (e.g.RB + RB) but this is forbidden for the following reason.

• a gluon is not a C eigenstate. It does not conserve thecharge quantum number C|J/ψ 〉 = −1, C|gg〉 = +1. Athree gluon exchange is the minimum number of gluonsthat can propagate a colourless C-odd transition.

This implies the decays depends on α6s . The suppression of

the strong decay of qq states is known as the "OZI-rule". Alsoseen with the φ (ss). Running of the strong coupling constantmeans OZI-suppression is more significant than with the lightermesons.

c

c

c

c

u

u

D0

D0

Malcolm John - 2020

c

c

u

u

dd

uu

π+

π−

π0

Charmonium decays are governed by kinematics: The J/ψis below the kinematic threshold for production of the lowest-mass open-charm mesons (the D mesons),

m(J/ψ ) < m(ψ′) < 2m(D0) < m(ψ′′)

The ψ(3770) falls apart rapidly (large width) to a D0D0 pair isvia a single gluon exchange. (∝ α2

s)

In considering the whole charmonium spectrum, a comparisonto positronium (bound states of e+e−) can be made,

• Their energy levels are predicted by the non-relativisticSchrodinger equation in analogous fashion to and hy-drogen atom but with reduced mass, me/2.

• Singlets 1S0 and 3S1 split by spin-spin interactions(aligned/not-aligned).

• The three P states have L = 1 and split by L⊕S align-ment.

Inspecting the energy levels of positronium and charmoniumshows striking similarity providing further evidence that the res-onances above m > 3 GeV/c2 are bound qq states. Differencesto positronium are:

• strength of αs leads to larger splitting from spin-spin andspin-orbit interactions

• difference in the relative energies comes from the funda-mental difference in the potential. Unlike the the αEM/rdependence in positronium, charmonium is proportionalto −α/r but αs/r + κr. (At short distances a coulomb-likepotential, at large distances the confinement becomesdominant)

In 1977, an excess of events were observed around10 GeV. Using the e+e− collider at SLAC, three narrow reso-nances were observed. Υ(1S ), Υ(2S ), Υ(3S ). In a strong par-allel to charmonium, a broader resonance, Υ(4S ), also foundto decay to B0B0 pairs, each with a mass around 5.3 GeV/c2.

Until the early 80s, searches for tt resonances were con-ducted, but none were found. The top quark was later discov-ered at Fermilab with a mass close to 175 GeV. At such highmass, so much phase space is available to the quark that itdecays before hadronising into tt.

Malcolm John - 2020