InternalSymmetriesofStrongInteractionsusers.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter...24.2 Isospin in Nuclear Physics 1335 Figure 24.4 Mirror nuclei 5B11 and 6C11 with their

In science one tries to tell people something

that no one ever knew before.

But in poetry, it’s the exact opposite.

Paul Dirac (1902–1984)

24

Internal Symmetries of Strong Interactions

If we open up a standard table of the presently known elementary particles, we seea confusing variety of them with many different properties. Before trying to developa more detailed theory of their interactions it is useful to find certain organizingprinciples which correlate their quantum numbers and mass spectra [1].

24.1 Classification of Elementary Particles

The most fundamental distinction concerns their statistical properties. They canbe of the Fermi-Dirac or Bose-Einstein type, in which case the particles are calledfermions or bosons, respectively. Historically, the fermions were subdivided intolight and heavy fermions. The former are called leptons (named after the Greekλεπτoς=“fine, small, thin”), which were historically electrons, muons, and neutrinos.Recently, also heavier particles were found which should be considered as leptons.This is due to the similarity of their interactions with those of the traditional, truelylight leptons. In fact, the characteristic property of leptons is now that they arefermions with no strong interactions. The distinction according to mass is no longersignificant.

The most important of the traditional heavy fermions are those particles whichprovide most of the mass of to atomic nuclei. These are the so-called nucleons, whichcan be protons or neutrons. These particles have strong interactions. They can beexcited in nuclear collisions, thereby producing short-lived resonances. Also theseare classified as elementary heavy fermions. The set of all heavy fermions with stronginteractions are called baryons (particles named after the Greek βαρυς = “heavy”).All baryons can be produced in collision or decay processes, sometimes in the formof antiparticles which, according to the spin-statistics theorem of Section 7.10, haveexactly the same mass and spin as the corresponding particles, except for somequantum numbers such as the charge. They carry an opposite sign.

The particles with Bose-Einstein statistics are separated into those that don’tand those that do possess strong interactions. The former play an important role inthe theoretical description of gravitational, electromagnetic, and weak interactions.

1330

24.1 Classification of Elementary Particles 1331

They are called gravitons, photons, and intermediate vector bosons W and Z. Thebosons with strong interactions are called mesons (middle-heavy particles after theGreek µεσoς). The set of all particles with strong interactions, baryons and mesons,are called hadrons (after the Greek αδρoς = “stout”).

Only few of all these particles are completely stable. The most prominent onesmake up the stable matter of the universe. These are the protons, electrons, photons,neutrinos, and gravitons. The proton is the only stable baryon. The other importantnuclear constituents, the neutrons, live on the average only 898± 16 seconds. Afterthis, a neutron decays into a proton, an electron, and an antineutrino.

Among the leptons, only electrons and neutrinos are stable particles. The muonsare particles very similar to the electrons, but much heavier. They decay in about10−6 seconds into an electron, a neutrino, and an antineutrino.

The proton is not only the only stable heavy fermion, it is the only stable stronglyinteracting particle. All mesons are unstable. The decays proceeding over a “long”time, as in the case of the neutron, are called weak. Even lifetimes of 10−13 secondsare still considered as “long” in elementary particle physics, and the decay is calledweak. A particle with such a lifetime down to 10−11 sec leaves an observable tracein a bubble chamber.

Most of the hadrons which decay via strong interaction processes live only for amuch shorter time than that, so short that they do travel any visible distance in abubble chamber. They are observable only as resonances in scattering cross sections.The most prominent example is the first resonance in pion nucleon scattering shownin Figs. 24.1 and 24.2. The resonance is called ∆(1232). It is found at a pion beammomentum

pπ = 0.34GeV/c. (24.1)

In the center of mass coordinate frame, the energy is equal to the mass of theresonance M ,

ECM =M =√

(Eπ +mp)2 − p2π =√

m2π +m2

p + 2mpEπ (24.2)

where Eπ =√

p2π +m2π is beam energy and mπ, mp are the masses of pion and

proton. From pπ = 0.34GeV/c one finds M = 1232MeV/c2 which is the numbergiven in the parenthesis of ∆(1232). The large width in the total cross section showsthe rather short lifetime. The peak is of the Lorentz shape and can be parametrizedas

σtot ∝ 2πλ2Γ2

(ECM −M)2 + (Γ/2)2, (24.3)

where λ ≡ h/pπ is the Compton wavelength of the incoming pion beam. At theenergies ECM = M ± Γ/2, the cross section has fallen to half its peak value. Theparameter Γ is therefore identified with the width of the resonance. From the ex-perimental curve we extract the width

Γ ≈ 115 MeV. (24.4)

1332 24 Internal Symmetries of Strong Interactions

In the quantum mechanical theory of resonance scattering it is known that such across section indicates an exponentially decaying wave function of the form

Ψ ∝ e−i(M−iΓ/2)t/h. (24.5)

Indeed, such a wave function leads to a scattering amplitude which contains a polein the energy plane below the real axis, proportional to

1

ECM −M + iΓ/2. (24.6)

Its absolute square gives the above Lorentzian cross section. The lifetime of such awave function is

τ = 2h/Γ. (24.7)

This is why the inverse width is sometimes called half-life. In physical units, theright-hand side has to be multiplied with h if Γ is measured in MeV and τ in seconds.The ∆(1231)-resonance has therefore a lifetime of

τ = 2h/Γ = 0.6583× 10−21 MeV

Γsec

≈ 5.72× 10−24 sec. (24.8)

Figure 24.1 Total and elastic π+-proton cross section showing clearly the first nucleon

resonance called ∆(1232).

24.1 Classification of Elementary Particles 1333

Figure 24.2 Total and elastic π−-proton cross section of the proton in a beam of neg-

atively charged pions π−. It shows once more the first nucleon resonance called ∆(1232)

as in the previous Figure 24.1, but with only a third of the height. This agrees with the

isospin-3/2 assignment and the amplitude relations (24.44) if one invokes the optical the-

orem (9.36), according to which the total cross section is proportional to the imaginary

part of the forward scattering amplitude.

It decays mostly into a pion and a nucleon. Such a short-lived decay is called strong

decay.

A small fraction, however, about 0.6% of the decays, goes into a nucleon and aphoton. This is called the radiative decay channel.

Since the ∆(1232)-resonance can decay into a nucleon and a photon, we shouldexpect that it can also be produced in a collision taking place from the final states,i.e., when a photon hits a nucleon. Indeed, the cross sections in Fig. 24.3 clearlyshow this resonance.

In the past twenty years, various simple organizing principles have been foundaccording to which the variety of particles and resonances can be classified and re-membered and their interactions be related. These principles are based on symmetrygroups associated with certain characteristic invariances of the different interactions.


Figure 24.3 Photon-proton and photon-deuteron total cross sections showing clearly the

first nucleon resonance ∆(1232) as well as a second resonance called N(1520).

They have helped in understanding the quantum numbers, the energy spectra, thevarious decay channels, and the different scattering cross sections.

24.2 Isospin in Nuclear Physics

If one compares binding energies of so-called mirror nuclei (which are obtainedfrom each other by the interchange of protons and neutrons), one finds that thedifference may be attributed entirely to the different electromagnetic interactions ofprotons and neutrons. For instance, 3H= nnp and 3He= ppn have binding energiesBH3 = 8.452MeV and B3He = 7.728MeV, respectively, with the difference beingcaused by the proton-neutron mass difference mn −mp = 1.293323± .000016MeVand by the different Coulomb field energy around the nuclei. Indeed, if the Coulombenergy is approximated by that of a uniformly charged sphere of charge Z and radiusR, the field energy is

ECoul = (1/2)(Z − 1)6

5

e2

4πR= (1/2)Z(Z − 1)

6

5

αhc

Rm, (24.9)

24.2 Isospin in Nuclear Physics 1335

Figure 24.4 Mirror nuclei 5B11 and 6C

11 with their excited states (the numbers are

the excitation energies in MeV). The ground state of 6C11 decays into that of 5B

11 by a

positive β-decay (i.e., emission of a positron and a neutrino) as indicated by the arrow. An

estimate of the Coulomb energy via formula (24.9) explains roughly the energy difference

of 1.98MeV between the two level schemes.

with α = e2/4πhc ≈ 1/137 being the fine-structure constant. Using the estimateR ≈ 1.45A1/3 × 10−13 cm for the nuclear radius, where A is the atomic number, wecan write

ECoul ≈ Z(Z − 1)6

5· A−1/3 MeV. (24.10)

For A = 3, Z = 2 we estimate ECoul ≈ 0.57MeV, and find the difference of thebinding energies to be

BH3 − BHe3 ≈ 1.29MeV − 0.57MeV ≈ 0.72MeV, (24.11)

in rough agreement with the observed 0.76MeV.Not only the binding energies, but also the excitation spectra of mirror nuclei

differ merely by a common electromagnetic shift, as seen in the example in Fig. 24.4.This property of mirror nuclei has led nuclear theorists to postulate that, if it werepossible to switch off the electromagnetic interactions, the potentials Vpp and Vnnwould turn out to be exactly equal. This symmetry is called charge symmetry.

The charge symmetry possesses a natural extension. When analyzing the dataof nucleon-nucleon scattering at low energy, it was found that also Vpp ≈ Vpn. Thisled Heisenberg to postulate that within purely nuclear interactions, protons andneutrons should be two indistinguishable states of the same particle, the nucleon N .He described the two states in analogy with the two spin states of a spin 1/2 objectby a two-component object

N =

(

pn

)

, (24.12)


and he called this object an isospinor state. Heisenberg considered the states as arepresentation of a rotation group in a fictitious space, the space of isotopic spin.Its generators are denoted by Ii. They obey the commutation rules of the rotationgroup

[Ii, Ij] = iǫabcIk. (24.13)

The two-component isospinors are rotated by the 2 × 2 representation of Ii, givenby Ii = σi/2. Proton and neutron are eigenstates of the third component I3 = σ3/2,with eigenvalue 1/2 and −1/2.The exchange of protons and neutrons corresponds to a rotation around the 2 axesof isospin by 1800:

eiπI2(

pn

)

= eiπσ2/2

(

pn

)

=

(

0 1− 1 0

)(

pn

)

=

(

n− p

)

. (24.14)

The postulate of indistinguishability of protons and neutrons can then be formu-lated mathematically as the commutativity of the generators of isospin Ii with theHamiltonian of strong interactions

[Ii, Hstrong] = 0. (24.15)

This is called isotopic spin symmetry or isospin invariance of strong interactions.If |A〉 is a nuclear n body wave function, then the isospin acts additively on eachnucleon, just as ordinary angular momentum (compare Appendix 3B)

Ii|A〉 = (Ii × 1× . . .× 1 + . . .+ 1× 1× 1× . . .× Ii)|A〉. (24.16)

Nuclei, which possess a large difference in their number of protons np and neutronsnn, have a third component of isospin

I3|A〉 =1

2(np − nn)|A〉. (24.17)

Their total isospin I must be at least equal to I3. The ground state has I = I3, theexcited states have a higher I. If the Coulomb energies are taken into account, theycan be seen to be members of isospin multiplets formed together with neighboringnuclei of equal total number of nucleons and different np − nn (see Fig. 24.5). Withthe help of isospin rotations, the charge symmetry operation may be represented asa rotation around the second axis by an angle π, namely as exp(iπI2).

The postulate of isospin symmetry leads immediately to interesting selectionrules for decay processes by nuclear interactions, such as α decay or emission ofdeuterons. Consider a nucleus |A〉 that has an equal number of protons and neutrons,i.e. a nucleus that is charge symmetric to itself, a so-called self-conjugate nucleus .Since it consists of an even number of isospin 1/2 nucleons, its total isospin isnecessarily integer, and the third component of its isospin is zero. The operationexp(iπI2) must therefore produce the same state, except for a phase factor η:

exp(iπI2)|A〉 = η|A〉. (24.18)

24.2 Isospin in Nuclear Physics 1337

Figure 24.5 Singlets and triplets of isospin in the nuclei 6C14, 7N

14, 8O14 (with the exci-

tation energies in MeV). The dotted lines connect levels that are roughly degenerate after

subtracting the Coulomb energy according to the estimate (24.9).

Applying this operation twice must lead back to the original state. Hence the phasefactor η can only be ±1. It is called the intrinsic charge parity of the nucleus. Everycharge-symmetric nucleus has a definite charge parity.

The charge parity η is calculable uniquely as a function of the total isospin.We use the property of the eigenfunctions of angular momentum Ylm(Θ, ϕ) withm = 0 to pick up a phase (−)l, when rotated around the y-axis by an angle π [seethe matrix elements of the rotation matrix (4.883)]. This is also obvious from therepresentation [(I+I3)!(I−I3)!]1/2a†(I+I3)

p a†(I−I3)n |0〉 of an isospin I, I3 -state in terms

of creation operators a†p, a†n of isospin up and down, as constructed in Eq. (4.871).

Under an isospin rotation around the 2-axis by an angle π, the operators a†p and a†ngo over into a†n and −a†p, respectively. A charge symmetric state with I3 = 0 hastherefore the charge parity

η = (−)I . (24.19)

The charge parity gives rise to interesting section rules for decay processes via nuclearinteractions. If the Hamiltonian commutes with all generators of isospin, the initialand final states must have equal charge parities. Typical forbidden processes are

5B∗10 ր

ց3Li

6 + α,

4Be8 + d,

(24.20)

where 5B∗10 can be the 4.5MeV or the 6.5MeV resonance of 5B

10. Now, the excitednucleus 5B

∗10 of excitation energy 4.5 MeV is a member of an isospin triplet formedtogether with ground states of the isotopes 4Be

10 and 6C10. It therefore has a charge

parity η = −1. On the other hand, the nuclei 3Li6 and 4Be

8 have isospin 0 andtherefore a positive charge parity η = +1. The two processes (24.20) are thusforbidden processes, which have indeed never been observed in the laboratory.


Let us now study the consequences of the larger isotopic spin symmetry in nuclearreactions. A deuteron in its ground state has the orbit in a symmetric s-wave, andthe spin states are given by the three symmetric triplet states

| ↑↓〉, 1√2| ↑↓ + ↓↑〉, | ↓↓〉. (24.21)

The antisymmetry of the wave function requires the isotopic wave function to beantisymmetric, i.e., it must be

1√2|pn− np〉. (24.22)

This is an I = 0 -state of isospin.If a deuteron collides with 4Be

9, which has I = 12, I3 = −1

2, the isospin of the

two-particle system is necessarily I = 12, I3 = −1

2. We can now compare two possible

final states, namely

5B∗101.7MeV + n (24.23)

and

4Be10 + p. (24.24)

If we denote the isospins of the two particles by I(1), I(2), the isospin states are inthe first case

|I(1), I(1)3 ; I(2), I(2)3 〉 = |1, 0; 1

2,− 1

2〉, (24.25)

and in the second|I(1), I(1)3 ; I(2), I

(2)3 〉 = |1,−1; 1, 1

2〉. (24.26)

The strengths, with which these two final states are coupled to the initial state|12,−1

2〉, are proportional to the Clebsch-Gordan coefficients

〈 1

2,− 1

2|1, 0; 1

2,− 1

2〉 = 1√

3, (24.27)

〈 1

2,− 1

2|1,−1; 1, 1

2〉 = −

√

2

3, (24.28)

respectively (recall Table 4.2). Hence the first reaction cross section should be halfas big as the second, a fact borne out by experiment.

24.3 Isospin in Pion Physics

In 1935, Hideki Yukawa [2] postulated the existence of a Bose particle which shouldmediate the strong interactions between nucleons in a similar way as the photonsdo between charges. Since the size of nuclei are of the order of 1013cm, Yukawaconcluded that instead of a Coulomb potential

v(x) =1

4πr, (24.29)

24.3 Isospin in Pion Physics 1339

which solves the field equation

−∂x2v(x) = δ(3)(x), (24.30)

the potential between nucleons should have a finite range and read

vY (x) =e−µr

4πr. (24.31)

This solves the differential equation

(−∂x2 + µ2)vY (x) = δ(3)(x). (24.32)

He identified the inverse range µ with the mass of the particle. This particle wasdiscovered experimentally around 1946, and is now called the pion. It exists in threecharge states π+, π0, π−. The neutral pion has a mass

mπ0 = 134.9642± .0038 MeV, (24.33)

from which the masses of the charged pions differ by

mπ± −mπ0= 4.6043± .00037 MeV. (24.34)

The π-mesons exist only for a small time. The neutral π0 meson decays mainly intotwo photons, with a lifetime

τπ0 ≈ (0.87± .04)× 10−16sec. (24.35)

This relatively fast decay is caused purely by electromagnetic interactions. Thecharged π±-states decay mainly into µ, νµ with a lifetime of

τπ± ≈ (2.6030± 0.0023)× 10−8sec. (24.36)

This decay is so slow that it proceeds via weak interactions.Since nuclear forces are supposed to arise mainly from pion interactions, the

isosymmetry of nuclear forces implies that the pion-nucleon coupling should be sym-metric under isospin rotations. The pion itself is obviously an isospin 1 object. If itinteracts with a nucleon of isospin 1/2, the total isospin can be either 3

2or 1

2. The

Clebsch-Gordan coefficients for the different channels of total isospin |I, I3〉 are thefollowing (see Table 4.2)

|π+p〉 = |1, 1; 12, 12〉 = |3

2, 32〉,

|π+n〉 = |1, 1; 12,−1

2〉 = 1√

3|32, 12〉+

√

23|12, 12〉,

|π0p〉 = |1, 0; 12, 12〉 =

√

23|32, 12〉 − 1√

3|12, 12〉,

|π0n〉 = |1, 0; 12,−1

2〉 =

√

23|32,−1

2〉+ 1√

3|12,−1

2〉,

|π−p〉 = |1,−1; 12, 12〉 = 1√

3|32,−1

2〉 −

√

23|12,−1

2〉,

|π−n〉 = |1,−1; 12,−1

2〉 = |3

2,−3

2〉.

(24.37)


From these we find the amplitudes for scattering processes to be all given in termsof two amplitudes, one for total isospin I = 3

2and one for total isospin I = 1

2. The

amplitudes are given by the so-called scattering matrix, whose matrix elements areobtained by evaluating the scattering operator S between incoming and outgoingstates. The detailed theory of the scattering matrix has been given in Chapter 9.From that theory it follows that if all generators Ii of isospin rotations commute withthe Hamiltonian operator of strong interactions Hstrong, they also commute with thescattering operator S. Then the amplitudes do not depend on the I3 components ofthe total isospin:

[Ii, S] = 0. (24.38)

We therefore can use I+ inside the matrix elements 〈I, I3|S|I, I3〉 to change I3 toany allowed value |I3| ≤ I. From

〈I, I3|I+S|I, I3 − 1〉 = 〈I, I3|SI+|I, I3 − 1〉 (24.39)

it follows that

〈I, I3 − 1|S|I, I3 − 1〉√

(I + I3)(I − I3 + 1)

= 〈I, I3|S|I, I3〉√

(I + I3)(I − I3 + 1) (24.40)

[see Eqs. (4.370), (4.374)]. Thus the diagonal matrix elements 〈I, I3|S|I, I3〉 areindependent of I3. Similarly one sees that the non-diagonal elements vanish. Wetherefore define the reduced scattering amplitude SI of a given isospin I by theequation

〈I, I ′3|S|I, I3〉 ≡ δI′3I3S

I . (24.41)

This statement is a special case of the Wigner-Eckart theorem (4.899) for thematrix elements of an arbitrary spherical tensor operator TI,I3 of isospin I. Such atensor operator is defined by the transformation law [compare (4.895)]

[Ii, TI,I3] = TI,I′3DI(Ii)

I′3I3, (24.42)

where DI(Ii)I3I′3 are the representation matrices of the generators Ii of the isospin

rotation group. According to the Wigner-Eckart theorem, the matrix elements ofa tensor operator between the representation states |I, I3〉 are related by Clebsch-Gordan coefficients:

〈I ′′, I ′′3 |TI′,I′3|I, I3〉 = 〈I ′′, I ′′3 |I ′, I ′3; I, I3〉〈I ′′||T ||I〉. (24.43)

These account for the dependence on the quantum numbers I3, I′3, I

′′3 . The matrix

elements depend only on a few reduced matrix elements 〈I ′′||T ||I〉 which are nonzeroif the isospins I and I ′ can be coupled to I ′′. Within this definition, the scatteringoperator is a trivial tensor operator of isospin zero, in which (24.43) reduces to(24.41).

24.4 SU(3)-Symmetry 1341

Note that (24.41) is also a manifestation of Schur’s Lemma, according to whicha matrix commuting with an irreducible set of matrices must be proportional to theunit matrix.

As an application of (24.41), the scattering amplitudes for various pion-nucleonscattering processes are given by the following combinations of only two amplitudes,S3/2 and S1/2:

Sπ+p→π+p = S3/2,

Sπ+n→π+n = (1/3)S3/2 + (2/3)S1/2,

Sπ0p→π0p = (2/3)S3/2 + (1/3)S1/2,

Sπ+n↔π0p = (√2/3)S3/2 − (

√2/3)S1/2,

Sπ0n→π0n = (2/3)S3/2 + (1/3)S1/2,

Sπ−p→π−p = (1/3)S3/2 + (2/3)S1/2,

Sπ0n↔π−p = (√2/3)S3/2 − (

√2/3)S1/2,

Sπ−n→π−n = S3/2. (24.44)

The ↔ symbols in the charge-exchange reactions exhibit time-reversal invariance.By taking the absolute squares of the amplitudes, we see that the cross sectionsmust satisfy

2σπ0p→π0p = σπ+p→π+p + σπ−p→π−p. (24.45)

This is borne out by experiment.

24.4 SU(3)-Symmetry

In 1944, the set of fundamental particles was enriched by a particle which was aheavy neutral meson of mass ≈ 500 MeV [3], with a lifetime of τ ≈ 10−10 sec. Itbecame known as a strange particle [4] under the name K0. It is observed to decaymainly into π+, π− and has a mass and a lifetime

mK0 ≈ 497.72± 0.07 MeV ,

τK0 ≈ (0.8323± 0.0022)× 10−10sec. (24.46)

In 1951, a charged partner of this particle was discovered which decays mostly intoa µ+ and a neutrino of the meson type, namely the K+ meson. It has a mass anda lifetime

mK+ ≈ 493.667± 0.014 MeV ,

τK+ ≈ (1.2371± 0.0026)× 10−8sec. (24.47)

Later also a K− meson was found, that decays into a negatively charged meson anda neutrino with the same lifetime. In 1955 one discovered, moreover, that there


are actually two K0-mesons, now called K0 and K0. These mesons could soon beproduced abundantly in particle accelerators and it became possible to study theirinteractions with nucleons. In the course of this it was found that not only themesons but also the nucleons possess strange partners. Moreover, if one assigned tothe strange mesons a quantum number S = 1, called strangeness, and to the strangepartner of the baryons in the associate production a quantum number S = −1, thenstrangeness is a conserved quantum number in a nuclear scattering process. If anordinary non-strange meson such as a pion hits a nucleon, the result is either a pionand a nucleon, both of strangeness zero, or a strange meson with S = 1 togetherwith a strange partner of the nucleon with S = −1. For example

π−p→ π−p (24.48)

or

π−p րց

K0Λ0,K+Σ−.

(24.49)

There are also associate production processes where two K’s are produced in addi-tion to a nucleon:

π−p→ K−K0p, (24.50)

π+p→ K+K0p. (24.51)

These allow to deduce that the strangeness of K+, K− is S = 1,−1, respectively.By scattering further a strange meson K− on a proton,

K−p→ K+Ξ−, (24.52)

one is able to produce a new particle Ξ− of strangeness S = −2. This particle iscalled cascade since it decays in a cascade-like process with the first step

Ξ− → Λ0π− (24.53)

and the second steps

Λ0 → pπ−, (24.54)

π− → µ−νµ. (24.55)

Note that in contrast to the production process, the decay of the strange particlesviolates strangeness. For instance:

Σ− → π− nS : −1 0 0.

(24.56)

However, this violation can be blamed on another interaction, the weak interaction,since the decay proceeds quite slowly, with a lifetime of roughly 10−10 sec.


Figure 24.6 Pseudoscalar meson octet states associated with the triplet of pions. The

same picture holds for the vector meson octet states with the replacement (24.62).

While non-strange particles have a charge

Q = N/2 + I3, (24.57)

where N is the total nucleon-number, the charge of strange particles is given by

Q = Y/2 + I3, (24.58)

with Y being defined as a convenient combination of N , and S, called hypercharge

Y ≡ N + S. (24.59)

Murray Gell-Mann plotted the known ordinary and strange pseudoscalar mesonstates in the I3, Y plane and found the multiplet of eight particles shown in Fig. 24.6(there and in subsequent similar plots the ket symbols | . . .〉 of the states are omit-ted). The center contains, beside the π0-meson, another pseudoscalar neutral mesoncalled the η meson. Its mass is

mη = 548.8± 0.6 MeV (24.60)

and it decays with a width of

Γ ≈ 1.05± 0.15keV, (24.61)

principally into γγ and 3π0. A ninth much heavier pseudoscalar meson η′(958) ofwidth 8.5MeV, known at that time, did not fit into the scheme by having a largermass, and no partners.

Gell-Mann applied the same extension to the spin 1 meson resonances of negativeparity to nine pseudoscalar particles, replacing

K → K∗(892),π → ρ(770),η → φ(1020),η′ → ω(783).

(24.62)


He continued to organize the nucleons and their strange partners in this way andfound again an octet, shown in Fig. 24.7, with the masses and lifetimes given inTable 24.1.

Figure 24.7 Baryon octet states associated with isodoublet of nucleons.

Table 24.1 Masses and lifetimes of the octet states associated with the isodoublet of

nucleons.

m( MeV) τ (10−10 sec)p 938.2796± 0.0027 stablen 939.5731± 0.0027 (898 ± 16) × 1010

Σ+ 1189.37 ± 0.06 0.8 ± 0.004Σ− 1197.34 ± 0.05 1.482± 0.011Σ0 1192± 0.08 (5.8± 1.3) × 10−10

Λ0 1115.60 ± 0.05 2.632 ± 0.020Ξ0 1314.9 ± 0.6 1.642 ± 0.015Ξ− 1321.32 ± 0.13 1.642 ± 0.015

Gell-Mann further noticed that there exists no similar regularity for the most promi-nent excited states of the nucleon, the resonance ∆(1232) of mass 1232MeV andwidth 115±5MeV. They exist with charges −, 0,+,++ and possess strange partnersΣ(1385) of width ≈ 37MeV with charges −, 0,+, and Ξ(1530) of width ≈ 10MeVwith charges −, 0. When he looked at the masses of these excited states, he observedan about equal spacing on the Y -axis. When plotted in an I3−Y plane, the particlesfill an equilateral triangle except for a missing lower corner. See Fig. 24.8. Startingfrom these observations he put forward the hypothesis that just as nuclear forcesare independent under isospin rotations, which form the group of unitary matri-ces in two dimensions, they should also be approximately invariant under the mostdirect extension of this group, which includes the additional quantum number ofstrangeness. So he postulated an approximate invariance of the strong interactionsunder the group of unitary matrices in three dimensions, SU(3). Since the masssplittings are much larger than in the former SU(2)-multiplets, this symmetry ismuch more broken in nature than the isotopic symmetry.


Figure 24.8 Baryon decuplet states associated with the first resonance of nucleons.

In field theory, the group theoretic implications of SU(3)-symmetry are studiedby giving the relativistic fields an extra SU(3)-subscript. One introduces a funda-mental Dirac field with subscripts α = 1, 2, 3:

ψ(x) → ψα(x) =

ψ1(x)ψ2(x)ψ3(x)

, (24.63)

which changes under SU(3)-transformations by group elements from the fundamen-tal three-dimensional representation of SU(3), called 3. Such a Dirac field is calleda quark field, and will be denoted by qα(x). Omitting the spacetime arguments, thetransformation law is:

3 : qα → Uαβqβ , U ∈ SU(3). (24.64)

The Hermitian adjoint of a quark field, (qα)† ≡ (q†)α. It transforms according to

the complex conjugate representation, called 3∗. To emphasize the different trans-formation behavior we write [Uα

β]† as (U∗)αβ so that

3∗ : q†α → U∗αβq

†β, U ∈ SU(3). (24.65)

The unitary scalar product q†q ≡ (q†)αqα is an invariant, likewise the contraction ofqα with any other particle field, which transforms like (q†)α. For Dirac fields it ismore convenient to work with the fields q(x) = q†(x)γ0 rather than q†(x) becauseof their more favorable Lorentz transformation properties, in particular the Lorentzinvariance of the scalar product q(x)q(x). Thus we shall work with conjugate quarkfields

qα = (q1, q2, q3). (24.66)


This field is said to transform according to the representation 3 of SU(3). Thecontraction qαqα is obviously invariant under Lorentz- and SU(3)-transformations:

qαqα = invariant. (24.67)

We have seen in Eq. (7.308) that the charge-conjugate Dirac field is directly relatedto q by a similarity transformation of the Dirac indices, which for quark fields reads

qC(x) = CqT (x). (24.68)

The fields q annihilate antiparticles, and will therefore be referred to as antiquark

fields , ignoring at this point the similarity transformation.The traceless tensor field obtained from the direct product of qα and qβ,

Mαβ ≡ qαq

β − 1

3δα

βqγ qγ, (24.69)

is in general a bilocal field containing qα(x, t) and qβ(x′, t′) at different spacetimepoints which are irrelevant to the present discussion and therefore omitted. Thisfield forms an invariant 8-dimensional representation space of SU(3), which Gell-Mann identified with the 8 meson states described above. This octet representationis irreducible. This means that every state of it can be reached by performing groupoperations on an arbitrary single fixed state. Gell-Mann called the fields up, down,and strange quark fields and denoted them by u, d, s :

qα =

uds

. (24.70)

Similarly for the antiquarks:qα = (u, d, s). (24.71)

He associated the octet fields (24.69) with the pseudoscalar mesons as shown inFig. 24.9. The field Mα

β annihilates the corresponding particle, for example π+ ≡M1

2 = du annihilates the particle π+. In contrast to isospin, the SU(3)-symmetryis strongly broken. The pion and the K mesons have quite different masses, namely135MeV and 490MeV.

The operator which measures the third component of isospin, I3, is the numberoperator

I3 =1

2[(Nu − Nd)− (Nu − Nd)]. (24.72)

Here and in the sequel we denote by Nu,d,s the total number operator of the u, d, s-quarks and their antiparticles, respectively. Thus we have the commutation rules

−[Nu,d, qα] = nu,dqα, α(u, d, s) (24.73)

with the eigenvaluesnu = (1, 0, 0), nd = (0, 1, 0), (24.74)


Figure 24.9 Quark content of the pseudoscalar meson octet. The particle- and quark-

symbols denote the annihilation parts of the corresponding fields.

and similar relations between the number operators and the antiquarks qα = (u, d, s).The hypercharge of the mesons is given by

Y =1

3[(Nu + Nd − 2Ns)− (Nu + Nd − 2Ns)], (24.75)

where Ns counts the number of strange quarks, i.e., it is defined to have, for (u, d, s)and (u, d, s), the eigenvalues ns = (0, 0, 1) and (0, 0,−1), respectively, with corre-sponding eigenvalues of Ns for the antiquarks.

The group SU(3) is a 3 · 3 − 1 = 8-parameter group. Its Lie algebra possesses8 traceless Hermitian generators which are conventionally denoted by λa/2 (a =1, · · · , 8) with the 3× 3 traceless Hermitian matrices

λ1 =

0 1 01 0 00 0 0

, λ2 =

0 −i 0i 0 00 0 0

, λ3 =

1 0 00 −1 00 0 0

,

λ4 =

0 0 10 0 01 0 0

, λ5 =

0 0 −i0 0 0i 0 0

,

λ6 =

0 0 00 0 10 1 0

, λ7 =

0 0 00 0 −i0 i 0

, λ8 = 1√3

1 0 00 1 00 0 −2

.

(24.76)

Note that they have the same trace orthonormality relations as the Pauli matrices:

tr(λaλb) = 2δab. (24.77)

It is easy to see that the 3×3 -matrices U = eiαaλa/2 span the space to unitary 3×3-matrices with unit determinants. The first three λa’s are the direct extension ofthe σi-matrices of isospin into three dimensions.


Table 24.2 Structure constants of SU(3). All elements not given in the table follow from

the total antisymmetry.

abc fabc123 1147 1/2156 −1/2246 1/2257 1/2345 1/2367 −1/2

458√3/2

678√3/2

The group structure is specified by the commutation rules of the generators

[λa, λb] = 2ifabcλc, (24.78)

where the structure constants fabc are totally antisymmetric in abc, due to the Jacobiidentity satisfied by the commutators. Their values are given in Table 24.2. Let Ga

be the operators in Hilbert space associated with these Lie algebra elements. Theseact on the three quark field operators of the fundamental representation as follows:

−[

Ga, qα]

= (λa/2)αβqβ. (24.79)

An explicit operator expression for the generators can be given by assigning canonicalcommutation or anticommutation relations to the q-fields:

[qα, q†β] = δα

β (24.80)

Then we can express the generators Gi simply as follows [recall Section 2.5]

Ga = q†α(λa/2)αβqβ. (24.81)

It is easy to see that these operators Ga have indeed the same commutation relationas the matrices λa/2:

[Ga, Gb] = ifabcGc. (24.82)

Besides these commutators, some calculations will involve the anticommutators ofthe λa-matrices. They are given by

λa, λb

= 2dabcλc + (4/3)δab, (24.83)

where dabc are completely symmetric in abc and have the nonzero matrix elementsshown in Table 24.3.


Table 24.3 The symmetric couplings dabc. All values not given explicitly are obtained

from the total symmetry in abc.

abc dabc abc dabc118 1/

√3 355 1/2

146 1/2 366 -1/2157 1/2 377 -1/2

228 1/√3 448 -1/2

√3

247 -1/2 558 -1/2√3

256 1/2 668 -1/2√3

338 1/√3 778 -1/2

√3

344 1/2 888 -1/√3

Consider now the antiquark fields, which transform according to the represen-tation 3∗ of SU(3). Under commutation with Ga, the antiquark fields behave asfollows

−[Ga, qα] = −qβ(λa/2)βα = −(λa∗/2)αβ q

β. (24.84)

The antiquark fields have the same commutation rules as the Hermitian conjugateof the quark fields, i.e., they commute like (q†)α ≡ (qα)†:

[qβ, q†α] = δβα. (24.85)

Thus, the generators of SU(3) have, for quark and antiquark fields, the operatorrepresentation:

Ga = q†(λa/2)q − q†(λa∗/2)q. (24.86)

The generator of the third component of isospin is

I3 = G3 = q†(λ3/2)q − q†(λ3/2)q

=1

2(u†u− d†d)− 1

2(u†u− d†d). (24.87)

Similarly, the hypercharge is given by

Y =2√3G8 =

2√3

[

q†(λ8/2)q − q†(λ8/2)q]

=1

3

[

(u†u+ d†d− 2s†s)− (u†u+ d†d− 2s†s)]

, (24.88)

and the charge by

Q = Y/2 + I3

=1

3

[

(2u†u− d†d− s†s)− (2u†u− d†d− s†s)]

. (24.89)


The raising and lowering of the third component is done with the operators

I± = 1

2

[

q†λ1±i2q − q†(λ1±i2)∗q]

=

u†d − d†ud†u − u†d

, (24.90)

where λ1±i2 is a short notation for λ1 ± iλ2. A pair of operators which raise andlower the hypercharge by a unit (thereby also raising and lowering I3 by half a unit)is1

V± = 1

2

[

q†λ4±i5q − q†(λ4±i5)∗q]

=

u†s − u†ss†u − s†u

. (24.91)

The effect of these operators on the quark and antiquark states |qα〉 (created by theoperators (q†)α when acting on the vacuum |0〉 ) is illustrated in Fig. 24.10 [thisbeing the analog to the illustration in Fig. 4.3]. Note that there exists a related pair

Figure 24.10 Effect of raising and lowering operators on quark and antiquark states.

Here and in the subsequent weight diagrams the ket symbols of the states are omitted.

of operators which raise and lower hypercharge by a unit while acting oppositely toV± in the isospin:

U± = 1

2

[

q†λ6±i7q − q†(λ6±i7)∗q]

=

d†s − d†ss†d − s†d

. (24.92)

1Note that there exists a third component to the two operators V±, namely V3 = 12 (u

†u −s†s)− 1

2 (u†u− s†s), which commutes with V± in the same way as the isospin operators. The three

operators form the so-called V -spin subgroup of SU(3). A similar set of commutation rules holdsfor the U± operators of (24.92) and U3 = 1

2 (d†d− s†s)− 1

2 (d†d− s†s).


These, however, will play no role in the future discussion.

Consider now the composite states associated with the product representation3 × 3 of a quark and an antiquark assigned to the pseudoscalar mesons in (24.69).The quantum numbers I3 and Y are additive so that the quantum numbers of themeson octet can be found by vector addition, i.e. by placing the 3 diagram withthe (I3, Y ) origin once on each state of the 3 diagram and by adding up the vectorsas shown in Fig. 24.11. In the theory of group representations the vectors |I3, Y 〉of quarks and antiquarks are referred to as the fundamental weights. Using the

susd

duud

us ds

dd uu

ss

Figure 24.11 Addition of the fundamental weights in product representation space of 3

and 3 vectors.

raising and lowering operators I±, V±, it is easy to find all states of an irreduciblerepresentation starting out from any fixed state. For instance, the operator I+ actsupon the particle state |π−〉 = |du〉 as follows:

I+|π−〉 = |uu− dd〉 ≡√2|π0〉, (24.93)

whereas

I+|π0〉 = −√2|du〉 ≡ −

√2|π†〉. (24.94)

Similarly, the application of V+ to |π−〉 = |ud〉 yields

V+|π−〉 = −|sd〉 = −|K0〉, (24.95)

whereas applying V+ to |K0〉 we find

V+|K0〉 = |uu− ss〉. (24.96)

This state has the same (I3, Y )-quantum numbers as |π0〉, but it is not an eigenstateof total isospin I2. Such a state is given by

|η〉 =√

1

6|uu+ dd− 2ss〉, (24.97)


so that V+|K0〉 can be decomposed into a π0- and an η-state as follows:

V+|K0〉 = 1

2|uu− dd〉+ 1

2|uu+ dd− 2ss〉

=√

1

2|π0〉+

√

3

2|η〉. (24.98)

It has become customary (de Swart-convention2) to change the phases to the q-states so that the matrix elements of I± and V± are all positive. This is achieved bychanging the phases of the states of the representation 3∗ as shown in Fig. 24.12. For

0 + 12

− 12

23

− 13

Y

T3

−|s〉

−|d〉I±

|u〉

V±1

1

Figure 24.12 States of the 3-representation with phases in the de Swart-convention.

the SU(3)-representation matrices U∗, these phase-changes on the states correspondto a similarity transformation. The new matrices will be denoted by U :

U∗ → U . (24.99)

Correspondingly, we shall say that the antiquark fields q, after the appropriate phasechanges, transform according to the 3-representation.

The phases of all higher representations are adjusted accordingly. The mesonoctet states associated with the product of 3 and 3-states are shown in Fig. 24.13.As a check we calculate a couple of phases:

V+|π0〉 = V+√

1

2|uu− dd〉

= −√

1

2|su〉 =

√

1

2|K+〉

(24.100)

and

V+|K0〉 = V+| − ds〉 = −|du〉 = |π+〉, (24.101)

2J.J. de Swart, Rev. Mod. Phys. 35 , 916 (1963).


−1 0 +1

+1

0

−1

K+ = −suK0 = −sd

π+ = −duπ− = udπ0 = 1√

2(uu− dd)

η = 1√6(uu+ dd− 2ss)

K− = us K0 = −ds

T3

S

Figure 24.13 Quark-antiquark content of the meson octet states with phases in the

de Swart-convention.

which are indeed positive.Note that these quark-antiquark states can also be obtained by forming certain

combinations of λa-matrices and sandwiching them between the creation operatorparts of the quark and antiquark fields

q†α = (u†, d†, s†), (q†)α = (u†, d†, s†), (24.102)

and applying them to the vacuum. The creation operators of the meson octet statesare

M †a = q†(λa/

√2)q†, (24.103)

with the combinations indicated in Fig. 24.14. These combinations of λa indices areoften used to specify octet states instead of the quark content. Any Hermitian octetoperator Oa (for instance Ga itself and M †

a) with the indices −(1 + i2)/√2 adds to

a given state the quantum numbers of a π+. Let us verify the de Swart-phases ofthe assignments in Fig. 24.14 by applying I±, V± to the meson states |Ma〉 =M †

a|0〉with the above index combinations. Using the commutation rules

[Ga,M†b] = ifabcM

†c (24.104)

we have, for example,

[I+,M†π−] =

√2f123M

†3 =

√2M †

π0 ,

[I+,M†π0 ] = if123(−M †

2 + iM †1) = −

√2M †

π+ , (24.105)

[I+,M†K0] = (f174M

†4 − f165M

†5) =M †

K+.

There is another simple way of constructing an octet representation of SU(3)from quark states without the use of antiquarks. By applying the product of threefields q†αq

†βq

†γ at three different places3 to the vacuum one obtains 27-states. If these

3This assumption of different places is necessary to distinguish all 27-states, for otherwise somestates will coincide due to particle identity.


−1 0 +1

+1

0

−1

K+ = −(4 + i5)/√2K0 = −(6 + i7)/

√2

π+ = −(1 + i2)/√2π− = (1 − i2)/

√2

π0 = 3

η = 8

K− = (4 − i5)/√2 K0 = −(6− i7)/

√2

T3

S

Figure 24.14 Combination of indices a in the pseudoscalar octet field M †a of (24.103).

Here −(1± i2)/√2-stands for −(M †

1 ± iM †2)/

√2.

are decomposed into the irreducible contents of SU(3), they lead to the followingmultiplets

3× 3× 3 = 10 + 8 + 8 + 1. (24.106)

The octet states can be identified with the nucleon octet states. In the decuplet thereare doubly charged non-strange states which can be identified with the I3 = 3/2component of the first nucleon resonance ∆(1236). It is therefore suggestive toidentify the other 10-states with the strange partners of this resonance. At the timewhen Gell-Mann proposed this identification, he could only match 9 of the decupletstates with known particles and resonances. There was one more state of strangeness3 (at the bottom of the weight diagram to be constructed in Fig. 24.17) which closesit to a triangle. Since the masses within the two rows are almost degenerate andsince they grow from row to row by about the same amount

m∆ −mΣ∗ ≈ −150,

mΣ∗ −mΞ∗ ≈ −145,

with the masses given in units MeV, Gell-Mann extrapolated the mass of the tenthparticle to be

M ≈ M∆ + 3(MΣ∗ −M∆)

≈ 1232 + 3 · 1567 ≈ 1682. (24.107)

He therefore postulated the existence of a particle with this mass, a negative charge,and a hypercharge −2 as it resulted from the three quark content of the states. He


called this particle Ω−. It was indeed found in 1964 at a mass of 1675 ± 8 Mev.This was celebrated as a triumph of the approximate SU(3)-symmetry hypothesisand the 10-assignment of the excited nucleons and their strange partners.

Let us explicitly construct the irreducible representations contained in the pro-duct 3 × 3 × 3 and assign them to physical states. We perform the multiplicationsuccessively

3× 3× 3 = 3× (3× 3). (24.108)

The product of two states 3× 3 is easily reduced to

3× 3 = 6 + 3. (24.109)

To see this we write down the symmetric and antisymmetric combinations as shownin Fig. 24.15. The first is a 6-representation, the second transforms in the same way

d u

s

d u

s

dd ud, du uu

ss

us, suds, sd=×

dd uu

ss

(ds+ sd)/√2

(ud+ du)/√2

(us+ su)/√2 +

(ds− sd)/√2 (us− su)/

√2

−(ud− du)/√2

=

3 × 3

6 + 3

Figure 24.15 Quark content in the reduction of the product 3× 3 = 6 + 3.

as an antiquark representation 3. As before we have assigned the phases to satisfy thede Swart-convention. After forming these products we continue the multiplication:

3× 3× 3 = 3× (6 + 3) = 3× 6 + 3× 3. (24.110)

Let us now form 3× 6 and 3× 3. The product 3× 3 has been calculated before interms of quark-antiquark states:

3× 3 = 8 + 1. (24.111)

Now the octet states contain three quarks, as shown in Fig. 24.16. The octet statescan be identified with the nucleons and their strange partners.


− u(ud− du)/√2

u(us− su)/√2

s(us− su)/√2

−d(ud− du)/√2

d(ds− sd)/√2

s(ds− sd)/√2

3 × 3

ud

s

−(ud− du)/√2

(us− su)/√2(ds− sd)/

√2

×

8

=

+

+

1

1√6

+uds−usd+dsu−dus+sud−sdu

Figure 24.16 Octet and singlet states obtained from 3× 3 in the product space of three-

quarks.

The completely antisymmetric combination of uds-states is a singlet state underSU(3)-transformations:

|Λ〉 = 1√3!(|uds〉+ |dsu〉+ |sud〉+ |dus〉+ |sdu〉+ |usd〉). (24.112)

The proton state is

|p〉 = 1√2|u(ud− du)〉. (24.113)

The state |Σ0〉 in the octet is found by applying I− to the right most state |Σ†〉 =−|u(us− su)〉, thereby obtaining

|Σ0〉 = −1

2|d(us− su) + u(ds− sd)〉. (24.114)

The isosinglet state |Λ0〉 is obtained by applying V − to |p〉 = |u(ud− du)〉/√2 and

by separating the result into a linear combination of |Σ0〉 and an orthogonal state,which must be |Λ0〉. First we have

V −|p〉 =√21

2|sud+ usd− sdu− uds〉. (24.115)

The scalar product with |Σ0〉 gives

〈Σ0|V −|p〉 = 1√2. (24.116)


uuuddd

sss

suu, u(us+ su)/√2

uss, s(us+ su)/√2

3 × 6

ud

s

uu(ud+ du)/√2dd

ss

(us+ su)/√2(ds+ sd)/

√2×

10 + 8

=

Figure 24.17 Irreducible three-quark states 10 and 8 in the product 3 × 6 (the symbol

(. . .)s denotes complete symmetrization).

Hence we rewrite

V −|p〉 =√2(

1

2|Σ0〉+ √

3

2|Λ0〉

)

(24.117)

and identify

|Λ0〉 ≡ −√

1/12|u(ds− sd)− d(us− su)− 2s(ud− du)〉. (24.118)

Consider now the product 3× 6. It gives the states shown in Fig. 24.17. The state|uuu〉 is identified with the 1232MeV resonance ∆++(1232) of isospin and parity3/2†. By applying the operators I± and V± to |uuu〉 we find the ten decuplet states

|∆++〉 = |uuu〉,|∆+〉 =

√1

3|uud+ udu+ duu〉,

|∆0〉 =√

1

3|udd+ dud+ udd〉,

|∆−〉 = |ddd〉,|Σ+〉 =

√1

3|uus+ usu+ suu〉,

|Σ0〉 =√

1

6|uds+ usd+ dsu+ dus+ sud+ sdu〉,

|Σ−〉 =√

1

3|dds+ dsd+ sdd〉,

|Ξ0〉 =√

1

3|uss+ sus+ ssu〉,

|Ξ−〉 =√

1

6|dss+ sds+ ssd〉,

|Ω−〉 = |sss〉

(24.119)


In addition to this decuplet, there is a further octet of particle states which canalternatively be assigned to the nucleon octet. Its quark states are found by con-structing one state, say the proton, and applying the operators I±, V± to it. Theproton state is found by linearly composing |uud〉 and |u(ud+ du)〉/

√2 so that the

result is orthogonal to |∆+〉 in (24.119) and to |p〉 of (24.113):

|p′〉 =√

1

6|u(ud+ du)− 2duu〉. (24.120)

The prime in |p′〉 indicates that the quark wave function differs from (24.113). Allother octet states associated with |p′〉 are then obtained by applying I±, V± to |p′〉,for instance

|n′〉 = I−|p′〉 = −√

1

6|d(ud+ du)− 2udd〉. (24.121)

As a first test of the octet assignment we calculate the mass differences withinthe nucleon octet assuming that the mass operator transforms like an SU(3) singlet,plus a term which behaves like a neutral isosinglet member of an octet, i.e., like η.Fortunately, the Clebsch-Gordan coefficients for the matrix elements

〈8a|8b|8c〉 (24.122)

are already in our position. They are of two types, related to the tracestr([λa, λb]λc) = fabc and tr(λaλbλc) = dabc. The first is antisymmetric, the secondsymmetric in abc. We therefore expand

〈8a|8b|8c〉 = F fabc +Ddabc, (24.123)

where F and D are two unknown irreducible matrix elements. To relate these toparticles, we only have to multiply them with the SU(3)-polarization vectors ǫa(m)of the particles whose assignments are clear from Fig. (24.14). For instance

ǫa(π+) = − 1√

2(1, i, 0, 0, 0, 0, 0, 0). (24.124)

Then we have between octet states:

〈8m2|8m|8m1〉 = ǫ∗a(m2)ǫ∗b(m)(F fabc +Ddabc)ǫc(m1). (24.125)

We may also write

〈8m2|8m|8m1〉 = F fm2∗m∗m1+Ddm∗

2m∗m1

, (24.126)

with obvious notation. The matrix associated with the b = 8 = η component isassumed to explain three mass differences. Two of them may be used to determinethe two irreducible matrix elements F and D. The symmetry breaking will thenallows us to derive one relation between the masses. Using the values of fabc and


dabc with b = 8 from Table 24.3, we have for the differences of the masses from somecommon SU(3) singlet mass msg:

∆mN = 〈p|8η|p〉 = Ff485 +Dd484 = F√3/2−D/2

√3,

∆mΣ = 〈Σ+|8η|Σ+〉 = Dd181 = D/√3,

∆mΛ = 〈Λ0|8η|Λ0〉 = Dd888 = −D/√3,

∆mΞ = 〈Ξ0|8η|Ξ0〉 = Ff687 +Dd686 = −F√3/2−D/2

√3.

(24.127)

Using these equations we readily find that the common singlet mass is

msg = 1

2(mΣ +mΛ) = 1

2(mN +mΞ) + 1

4(mΣ −mΛ), (24.128)

so that the masses have to satisfy the famous Gell-Mann–Okubo relation

1

2(mN +mΞ) = 1

4(3mΛ +mΣ). (24.129)

It holds reasonably well, the left-hand side being approximately equal to 1.128GeV,the right-hand side equal to 1.134GeV.

For the decuplet, the same ansatz reproduces the equal mass splittings betweenthe rows that had led Gell-Mann to his prediction of the Ω-meson.

For the pseudoscalar meson octet the relative mass splitting is very large andthe mass relation cannot be expected to be as good as for the baryon octet. If weuse, however, the square masses, it is nevertheless in surprisingly good agreementwith experiment:

m2K = 1

4(3m2

η +m2π). (24.130)

The left-hand side is equal to 0.25GeV2, the right-hand side to 0.23 GeV2. Anargument for using square-masses is that, in contrast to the fermion Lagrangian,the boson Lagrangian has a mass term proportional to m2.

The vector meson octet has a smaller relative mass splitting and the mass relationshould be good. If we insert, however, the experimental masses

m2K∗ = 1

4(3m2

ω +m2ρ), (24.131)

there is a bad surprise: The left-hand side is equal to 0.80GeV2, considerably largerthan the value 0.61 GeV2 on the right hand. This disagreement can be resolvedby postulating that the symmetry breaking part of the Lagrangian mixes the iso-singlet octet state with the SU(3) singlet state [5]. This is always possible with thesymmetry breaking transforming like an η-meson. Specifically, if ω0, φ0 denote theSU(3)-states, the physical particles can be

ω = cos θω0 − sin θφ0,

φ = sin θω0 + cos θφ0, (24.132)

with some unknown mixing angle θ. The symmetry breaking Lagrangian can containthe mass terms

Lsymm br = −m28ω

20 −m2

1φ20 −m2

18φ0ω0 + . . . , (24.133)


with some SU(3) octet, singlet, and mixing mass parameters and the associatedSU(3) fields. The octet mass parameter is determined from the Gell-Mann–Okuborelation (24.131) as

m8 = 930 MeV. (24.134)

Between the physical states (24.132) this becomes diagonal, if the mixing angle θ isgiven by

θ =1

2arctan

m218

m29 −m2

1

. (24.135)

The diagonal masses are

m2ω = 1

2(m2

8 +m21)− 1

2[(m2

8 −m21)

2 −m418], (24.136)

m2φ = 1

2(m2

8 +m21) +

1

2[(m2

8 −m21)

2 −m418]. (24.137)

From the experimental numbers we determine the parameters

m18 ≈ 0.65 MeV, cos θ ≈ 0.77. (24.138)

The mixing angle can be tested by experiment, since it determines the decay rateof the processes

φ→ K+K−, (24.139)

which occurs with a width 2.2MeV. Since the vector meson has spin 1, the pseu-doscalar kaons must come out in a p-wave. Their orbital wave function is thereforeantisymmetric. But for bosons the total wave function must be symmetric, whichimplies that the SU(3)-quantum numbers have to be coupled antisymmetrically.This, however, excludes the SU(3)-singlet part of φ from the decay (since its cou-pling would be proportional to δab in the SU(3)-indices of the pseudoscalar mesons).The amplitude for the decay is therefore proportional to the SU(3)-Clebsch-Gordancoefficient cos θ(8ω0

|8K+|8K−)∝cos θ〈0, 0| 12, 1

2,− 1

2〉f845 = cos θ × (1/

√2) × (

√3/2).

This should be compared with the decay ρ0 → π+π−, whose amplitude is propor-tional to (8ρ0|8π−|8π+)∝〈1, 0|1, 1; 1,−1〉f123 = (1/

√2) × 1. Taking the squares, we

find the ratio∣

∣

∣

∣

∣

S(φ→ K+K−)

S(ρ0 → π+π−)

∣

∣

∣

∣

∣

2

=3

4cos2 θ ≈ 0.44. (24.140)

The decay rates contain a phase space factor

Γ ∝ q3

M2, (24.141)

where q is the momentum of the emerging pseudoscalar mesons, and M the massof the decaying vector meson. In the processes at hand we have qK ≈ 127,Mφ ≈1020, qπ ≈ 359,Mρ ≈ 770MeV, so that the phase space factors are 0.0020MeV and0.078MeV, respectively. Together with (24.140) this gives the ratio of decay rates

∣

∣

∣

∣

∣

γ(φ → K+K−)

γ(ρ0 → π+π−)

∣

∣

∣

∣

∣

2

≈ 0.011. (24.142)

24.5 Newer Quarks 1361

The experimental ratio is

∣

∣

∣

∣

∣

γ(φ→ K+K−)

γ(ρ0 → π+π−)

∣

∣

∣

∣

∣

2

exp

≈ 0.014, (24.143)

in reasonable agreement with SU(3)-symmetry.

For the pseudoscalar mesons the same type of mixing can occur between η- andη′- mesons. However, since their mass difference is relatively large compared to thatbetween ω and φ, their mixing angle is much smaller, which explains why no mixingwas needed to satisfy the Gell-Mann–Okubo relation (24.130).

As a further test of the octet assignment we may calculate branching ratios forthe decay of the decuplet resonances in the nucleon and the meson octets. Also herethe agreement with experimental data is reasonably good.

24.5 Newer Quarks

Particles discovered in the 1970s and the theoretical attempts to explain their in-teractions and decay properties suggested the existence of more quarks. They werenamed charmed quark, with the symbol c, top quark, with the symbol t, and bottom

quark, with the symbol b. In Table 24.4 we have listed the masses and quantumnumbers of the quarks discovered so far.

Table 24.4 List of Quarks and their properties taken from the Particle Data Group [11].

Quark (I, I3) Charge Strangeness Mass (MeV)

u 1

2, 1

2

2

30 2.3+0.7

−0.5

d 1

2, - 1

2− 1

30 4.8+0.5

−0.3

s 0,0 − 1

3−1 95± 5

c 0,0 2

30 1275± 25

t 0,0 2

30 173070± 0.890

b 0,0 − 1

30 4180± 30

It was tempting to incorporate these quarks into the above broken symmetryconsiderations and extend the basic quark antiquark SU(3)-multiplet by these ad-ditional quarks and antiquarks to higher multiplets, with an even larger brokensymmetry group. This has led indeed to a reasonable particle classification if weadd the lightest of the additional quarks, the charmed quark c, to the other threeand forming a quartet u, d, s, c. This quartet can then be treated very approximatelyas a fundamental representation 4 of an internal symmetry group SU(4). Clearly,the corresponding antiquarks must be transformed according to the representation4. The weight diagram of these representations is three-dimensional and containsan additional “charm” axis labeled by C in Fig. 24.18.


Figure 24.18 The four lowest quarks u, d, s, c and their position in the three-dimensional

weight space with the quantum number “charm”. By combining a quark and an antiquark

one finds the 16 states on the right-hand side, of which 15 form an irreducible representa-

tion of SU(4). The new states D0,D+, F+ and their antiparticles have been found in the

laboratory.

24.6 Tensor Representations and Young Tableaux

A systematic construction of the tensor representation in direct product spaces suchas 3× 3× 3 is possible with the help of Young tableaux. The irreducible represen-tations are obtained by forming tensors which transform irreducibly under permu-tations of the indices. They are of a definite symmetry type specified by the Youngtableau introduced in Appendix 2A. The tableau 1 2 3 indicates a tensor of com-

plete symmetry transforming like a 10-representation, while

1

2

3

corresponds to a

complete antisymmetric tensor transforming like an SU(3) singlet. For the mixed

tableau1 2

3we adopt the convention to first symmetrize the state |qαqβqγ〉 in

the first two indices, then antisymmetrize in the first and the third index. Equi-

valently we can use another tableau, say1 2

3, or

2 1

3. Each gives an octet

representation. For instance, the proton wave function (24.113) arises from the

symmetry operations of the Young tableau2 1

3applied to the state |uud〉. The

symmetrization in the first two indices gives

|uud〉 → 2|uud〉 (symmetry in 21), (24.144)

the antisymmetrization in the second and third index (of the original tensor |qαqβqγ〉)|uud〉 → 2|u(ud− du)〉 (antisymmetry in 23). (24.145)

Hence:

2 1

3|uud〉 = 2|u(ud− du)〉 = 2

√2|p〉. (24.146)

24.6 Tensor Representations and Young Tableaux 1363

Equivalent proton states could have been obtained from the tableau

1 2

3|uud〉 = 2|uud− duu〉, 1 2

3|udu〉 = 2|udu− duu〉, (24.147)

or the sum of the two which gives

|u(ud+ du)− 2duu〉. (24.148)

This is once more the proton state |p′〉 of Eq. (24.120) found in the product 6× 3.It cannot be decided by SU(3)-symmetry alone, which mixture of the two inde-

pendent three-quark wave functions corresponds to the proton in nature. This hashas to emerge from a dynamical model [6, 7]. We shall see later that the largersymmetry group SU(6) makes definite predictions on this.

In general, the dimensionality of the SU(n)-representation of a tableau is ob-tained from the following formula

dSU(n) =D(l1, l2, . . . , ln)

D(n− 1, n− 2, . . . , 0), (24.149)

whereD(x1, . . . , xn) ≡

∏

i<j

(xi − xj), (24.150)

andli ≡ n+mi − i, i = 1, 2, 3, . . . , n, (24.151)

with mi being the number of boxes in the ith row of the tableau. Note that alsorows with no boxes have to be counted, up to i = n.

A useful alternative formula for the dimension is

dSU(n) =l1!l2! · · · ln!

0!1! · · · (n− 1)!

1∏

i,j hij, (24.152)

where hij is the sum of the number of boxes to the right of the element ij in thetableau plus the number of boxes below ij plus 1.

Take some examples. The trivial Young tableau has m1 = 1, m2 = 0, m3 =0, . . . , mn = 0 and thus l1 = n, l2 = n− 2, . . . , ln = 0. Hence

d = n. (24.153)

The first non-trivial one has m1 = 2, m2 = 0, . . . , mn = 0 and thus l1 =n+ 1, l2 = n− 2, l3 = n− 3, . . . , ln = 0. Hence

d = n(n+ 1)/2, (24.154)

which gives 6 for SU(3). Further

d = n(n− 1)/2, (24.155)


which gives 3 for SU(3), the dimension of the representation 3.For 3-quark states, the dimensions are

d = n(n+ 1)(n + 2)/6, (24.156)

d = (n− 1)n(n+ 1)/3, (24.157)

d = n(n− 1)(n− 2)/6, (24.158)

which take the values 10, 8, 1, respectively, for SU(3). Somewhat more generally, thedimensions of some simple tableaux are

d1 2 3 ··· k

=

(

n + k − 1k

)

, (24.159)

d1 2 3 ··· k

=

(

n+ k − 1k + 1

)

k, (24.160)

d1

2...

k

=

(

n

k

)

. (24.161)

A column with n boxes is completely antisymmetric in the corresponding indicesand thus proportional to the invariant antisymmetric tensor ǫi1i2,...,in. This is whyits dimensionality is 1. In constructing all Young tableaux any such column can beomitted. For instance, in SU(3),

= . (24.162)

One often indicates this cancellation by crossing out any complete column with avertical line:

→ . (24.163)

The representation associated with each Young tableaux occurs as often as dimen-sionality of the associated representation of the permutation group. This follows

24.6 Tensor Representations and Young Tableaux 1365

from the fact that the representation matrices of the permutation group commutewith those of SU(3). They can therefore be brought simultaneously to a block form

D1

D1

D2

. . .

. (24.164)

Along the diagonal there are ni identical irreducible representation blocks Di of di-mensionality di, etc. Now, according to Schur’s lemma, any matrix commuting withsuch a block matrix must be itself a block matrix with the dual form, that consistsof di identical blocks of dimensionality ni. The formulas for the dimensionality ofthe irreducible representations of the permutation group were given in App. 2A. Ifthe Young tableau has r boxes, the dimensionality is given by

dSr =r!

∏

ij hij, (24.165)

where the numbers hij are the same as in (24.149).

Using this formula we find, for instance, that the product of three-quark states

× × decompose into one SU(3) decuplet , two octets ,

and one singlet :

3× 3× 3 = 10 + 8 + 8 + 1. (24.166)

The Young tableau can also be used to find the irreducible contents in a productof two or more irreducible representations. For instance, suppose we want to know

the irreducible contents of 8×8. We take the two Young tableaux ×and distinguish the rows of boxes in the second factor by a, b, c, . . ., in the examplea a

b. Then we add the lettered boxes to the first tableau in the following way:

a) add all a’s in such a way that one obtains all proper tableaux(i.e., with mi+1 ≥ mi) which have no more than one a in each column;

b) add the b’s, following the same rule;c) add the c’s, following the same rule.


In our example we thus obtain the expansion

× a a

b=

a a

b+

a a

b8× 8 27 10

+a

a b+

aa

b10 8

+

a

ba

+ a

a b8 1

(24.167)

In accordance with the rule given earlier we have dropped all complete columns.Note that due to the antisymmetry of any complete column, an incomplete columnis equivalent to the complex conjugate representation, i.e., a tableau that would beobtained from the missing boxes necessary to complete it (indicated by a box witha circle), fo instance

=

3

. (24.168)

This is how we can see immediately that the representation associated with the

tableau is a 10

=

10

. (24.169)

The octet representation, being the adjoint representation of SU(3) is real [its gen-erators are (Ga)bc = −ifabc with real antisymmetric matrices (fa)bc; recall (4.83)].

From the Young tableau of the octet this is obvious since the missing boxes

in

form once more the same tableau , so the representation is equal

to the complex conjugate of itself and thus real.

24.7 Effective Interactions among Hadrons 1367

24.7 Effective Interactions among Hadrons

There exists a great variety of strongly interacting particles with many possibleinteractions between them. Let us study a few of them.

24.7.1 The Pion-Nucleon Interaction

The most important strong interaction is the one between pions and nucleons sinceit gives rise to the dominant part of the forces which keep the nuclei together. If weneglect electromagnetic interactions, this interaction exhibits isospin symmetry ofpions and nucleons, so that a system of fields πa and N has the following effectiveaction

A = Aπ +AN +AπNN , (24.170)

with

Aπ =∫

d4x1

2

[∂πa(x)]2 −m2

ππ2a(x)

, (24.171)

AN =∫

d4x N(x)(i/∂ −MN )N(x), (24.172)

AπNN = gπNN

∫

d4x N(x)iγ5τaN(x)πa(x). (24.173)

Here πa(x) is an isovector field and N(x) a Dirac spinor field with an additionalisospinor index, which is not explicitly written down. The matrices τa are 2×2 Paulimatrices acting on the isospinor indices of N(x). Studies of pion nucleon scatteringamplitudes determine the size of the pseudoscalar coupling constant gπNN to be

g2πNN

4π≈ 14.4± .4 . (24.174)

In some analyses, the pion nucleon interaction (24.173) is parametrized in the axialvector form

AπNN =fπNN

Mπ

∫

d4xN(x)γµγ5τaN(x)∂µπa. (24.175)

For physical nucleons on the mass shell, the two interactions are the same with arelation between the two coupling constants

fπNN = − Mπ

2MNg. (24.176)

Thus

f 2πNN = 0.081± 0.002. (24.177)

In either form, the coupling strength is too large to perform perturbative calculationswith the interaction AπNN .


It is useful to find the SU(3)-symmetric extension of the pion nucleon interaction.In SU(3), there are two ways of coupling three octets with each other, an antisym-metric one fijk and a symmetric one, dijk. The easiest way to do the calculation isby writing the nucleon octet as a 3× 3 -matrix.

B =

1√2Σ0 + 1√

6Λ0 Σ+ p

Σ− − 1√2Σ0 + 1√

6Λ0 n

−Ξ− Ξ0 − 2√6Λ0

, (24.178)

and the pseudoscalar octet as

P =

1√2π0 + 1√

6η0 π+ K+

π− − 1√2π0 + 1√

6η0 K0

K− K0 − 2√6η0

. (24.179)

In this context the phases of π± are chosen as π± ≡ (π1 ± iπ2)/√2 so that P is a

Hermitian matrix. Then the interaction Lagrangian has a simple explicit form

LPBB = gF

F +Dtr([B, B]P ) + g

D

F +Dtr(B, BP ), (24.180)

where g ≡ gπNN is the pion nucleon coupling constant and F/D the so-called “F -over-D -ratio”, which gives the relative strength of antisymmetric and symmetriccoupling. Working out the different components, this yields the expansion in termsof isospinors

Lint = g

(NN) · + (1− 2α)(ΞΞ) · +3− 4α√

3NNη0

−3− 2α√3

ΞΞη0 − (1− 2α)(NK) ·− (ΞKc) ·

−(1− 2α) · (K†N)− · (Kc†

Ξ)− 3− 2α√3

NΛK

−3− 2α√3

K†ΛN +3− 4α√

3ΞΛKc +

3− 4α√3

Kc†ΛΞ

+i(1− α)(×) · +2α√3(Λ) · +

2α√3(Λ) ·

+2α√3( ·)η0 − 2α√

3ΛΛη0

. (24.181)

Here we have used the isospinors

N ≡(

pn

)

, Ξ ≡(

Ξ0

Ξ−

)

, (24.182)

and

K ≡(

K+

K0

)

, Kc ≡(

K0

−K−

)

. (24.183)


Note that Kc is the charge conjugate spinor CK∗. The parameter α is defined by

α ≡ D

F +D(24.184)

so that

F/D = (1− α)/α. (24.185)

The two numbers g = gπNN and F/D parametrize all couplings.The pion nucleon coupling was written down in Eq. (24.174). The F/D-ratio is

determined experimentally from an analysis of the production process

π−p→ Σ0K+ (24.186)

which contains the coupling

gnΣ−K+ = −√2g(1− 2α)

= −√2gF −D

F +D. (24.187)

From the experimental data one finds

F

D≈ 2

3. (24.188)

For the coupling between pions and Σ-particles, the interaction Lagrangian(24.181) yields

gΣΣπ = g(1− α) = gF

F +D. (24.189)

The SU(3)-relations between coupling constants in the Lagrangian (24.181) are,of course, in agreement with standard SU(3)-Clebsch-Gordan coefficients. The co-efficients of the SU(2)-subgroup can be taken from Table 4.2 in Chapter 4.1, Theisoscalar factors from Table 24.5 in Appendix 24A on p. 1378.

Take, for instance, the matrix element 〈p|π0|p〉. The SU(2) Clebsch-Gordon co-efficient is 1/

√3. The antisymmetric and symmetric isoscalar factors are

√3/√12 =

1/2 and√9/√20 = 3/2

√5, respectively. Hence, with the normalization factors of

(24A.13), we have

〈p|π0|p〉 = F +D

2. (24.190)

The matrix element 〈Σ+|π0|Σ+〉 has an SU(2)-Clebsch-Gordan 1/√2 and the

isoscalar factors√

2/3 and 0, respectively, so that

〈Σ+|π0|Σ+〉 = F, (24.191)

just as before in (24.189).


Let us also calculate the coupling gnΣ−K+ for which the Clebsch-Gordan coeffi-

cient is −√

2/3 and the isoscalar factors are 1/2 and −3/2√5, respectively. Thus

〈n|K+|Σ−〉 = 1√2(F −D), (24.192)

and

gnΣ−K+ = −g√2F −D

F +D, (24.193)

as in (24.187).In Appendix 24A we have collected a few useful formulas for SU(3)-calculations.

24.7.2 The Decay ∆(1232)→ Nπ

The interactions between the various members of the nucleon octet with pseudoscalarmesons and the decuplet resonances of spin-parity JP = 3/2+ are all determined byonly one SU(3)-invariant matrix element. This, in turn, may be chosen to coincidewith the coupling between the proton, the π+, and the ∆++(1232)-resonance. TheLagrangian for this specific interaction is conventionally written as

L∆++pn+ =g∗

MN

(

ψ++µ ψ+∂µπ

+ + c.c.)

, (24.194)

where ψµ is a Rarita-Schwinger spinor for the spin 3/2 particle ∆++(1232). Somepeople write 4h/Mπ instead of g∗/MN .

The isospin structure can be incorporated by multiplying the Lagrangian by theClebsch-Gordan factor

〈32c|1

2a1b〉

〈3232|121211〉 , (24.195)

where c, a, b are the isospin orientations of ∆, N and π. Alternatively, we can useRarita-Schwinger isospinors ∆a for the ∆-particle (each of the three components∆1,∆2,∆3 is a two-component isospinor), and we can write in isospace (suppressingthe Lorentz indices):

L =g∗

MN

(

N − N)

· . (24.196)

Indeed, the isospinor for the ∆++-resonance has only a + -component with isospinup, so that N contains the particles ∆++, p, and π+ in the form ∆++pπ+, as in(24.194).

For calculations it is useful to record the completeness relation of Rarita-Schwinger isospinors

∑

s3

∆a∆b = δab 1−1

3τaτb =

2

3δab 1−

1

6[τa, τb], (24.197)


where 1 stands for the 2× 2 unit matrix in isospace.The coupling g∗ can be determined by an analysis of the strength of the ∆++

resonance in the π+p scattering amplitude. This gives

g∗ ≈ 12.43 g ≈ 0.92 g (24.198)

corresponding to

h ≈ 0.457 g. (24.199)

If we assume the ∆ resonance to be very narrow, we can immediately calculate thedecay rate from the interaction (24.194). The decay amplitude is found as follows:We consider a ∆++-resonance in its rest frame and use helicity amplitudes along thedirection of the momentum of the emerging proton, which we assume to run in the3-direction. The ∆ helicity can be ±3/2 or ±1/2. Only the latter two orientationsallow to conserve angular momentum with the helicities of the proton being ±1/2.The two orientations decay with the same rate, so we only need to study one case, saythe helicity +1/2. The Rarita-Schwinger spinor is a Clebsch-Gordan combinationof spin J,M = 1, 0 with 1/2, 1/2 and of J,M = 1, 1 with 1/2− 1/2:

uµ(0; 3

2, 1

2) =

√

2

3

0001

u(0, 1

2)− 1√

6

01i0

u(0,− 1

2). (24.200)

Between the state

a†π+a†p|0〉 (24.201)

and a ∆-state at rest 〈0|a∆ with unit normalization in a finite volume V , the wavefunctions for the pion and the nucleon are

1√2EπV

e−iqπx1√Vu(pN , s

′3), (24.202)

and for the ∆-resonance

1√Vuµ(p∆, s3). (24.203)

We therefore find the decay amplitude [matrix element of the t-matrix (9.290)]:

t =g∗

MN

uµu qµ 1√V

3

1√2Eπ

1√

EN/MN

(24.204)

=g∗

MN

√

2

3cosh

ζ

2pN

1√2Eπ

1√V

3

1√

EN/MN

, (24.205)


where

u(pN , 1

2)u(0, 1

2) = cosh

ζ

2=

1

2MN(EN +MN ), (24.206)

with EN , Eπ being the energies of proton and pion in the rest frame of the decaying∆-resonance, and pN the momentum of the nucleon emerging in the z-direction. Thedecay rate is found from formula (9.338), yielding for the helicity h = 1/2 -state:

Γ∆h=1/2→Nπ =1

(2π)32πENEπ

M∆

∫

dΩ|t|2

=4π

(2π)32π

pNM∆

g∗2

M2N

2

3p2N

1

4(EN +MN )

=g∗2

4π

2

3

p3NM2

NM∆

(EN +MN). (24.207)

Since the resonance ∆ is, with probability 1/2, in the helicity-3/2 state, where itcannot decay, the final decay rate is only one half of the value (24.207), i.e.,

Γ =g∗2

4π

1

3

p3NM2

NM∆

(EN +MN). (24.208)

In terms of the three particle masses involved, the energy EN is given by

EN =M∆

2 +M2N −M2

π

2M∆

, (24.209)

and the momentum pN of the outcoming proton (equal to that of the pion) is

pN =√

[M2∆ − (MN +Mπ)2][M2

∆ − (MN −Mπ)2]/2M∆. (24.210)

Using the experimental width

Γ∆ ≈ 115MeV, (24.211)

we obtain

g∗ ≈ 14.43 ≈ 1.07g , (24.212)

corresponding to

h ≈ 0.53, (24.213)

in reasonable agreement with the independent determination (24.199) from the pion-nucleon scattering.


24.7.3 Vector Meson Decay ρ(770)→ππ

Among mesons, the most directly observable coupling is the ρππ-coupling. It governsthe strength of the most prominent resonance in the p-wave pion-pion scatteringamplitude. Since ρ is a vector meson of isospin 1, it is coupled to the two pions ina p-wave. The Lagrangian interaction is given by

L =gρππ2

µ ·(

×↔∂µ

)

. (24.214)

For the decay ρ+ → π0π−, the invariant t-matrix element [defined in (9.290)] is givenby

〈π0π+|t|ρ+〉 = −gρππ1

√

2V Epi

2

1√

2VMρ

2pµπǫµ(pρ, s3), (24.215)

where ǫµ(pρ, s3) is the polarization vector of the ρ meson and the decay rate is,according to formula (9.339),

Γρ+→π0π+ = g2ρππ4π

(2π)2pCM

M2ρ

1

234|pµπǫµ|2. (24.216)

The average over the initial polarization gives

1

3pµπp

νπ

∑

s3

ǫµ(pρ, s3)ǫν(pρ, s3) = −1

3pµπp

νπ

(

gµν −pµρp

νρ

p2ρ

)

=1

3p2CMπ, (24.217)

where pCM π is the momentum of the pions in the rest frame of the decaying ρ-meson:

pCM π =1

2

(

M2ρ − 4m2

π

)1/2. (24.218)

This brings (24.216) to the form

Γρ+→π0π+ =g2ρππ4π

2

3

p3CM π

M2ρ

=g2ρππ4π

2

3

(

M2ρ − 4m2

π

)3/2

8M2ρ

. (24.219)

The experimental decay width is

Γρππ ≈ 153MeV. (24.220)

From this one finds the coupling constant

g2ρππ4π

≈ 2.85. (24.221)


24.7.4 Vector Meson Decays ω(783)→ρπ and ω(783)→ πππ

Another important coupling is that between ω, ρ, and π:

Lωρπ = gωρπǫµνλκ∂µων∂λκ · . (24.222)

It is responsible for the decay

ω → πππ, (24.223)

which proceeds at a rate

Γω→πππ = 9.8± 0.3MeV. (24.224)

The ππ -channels are dominated by the ρ-resonance. One can therefore assume asequential decay, finding

Γω→3π = (Mω − 3Mπ)4(

M2ρ − 4M2

π

)−2MωM

2π

1√27

1

4

g2ρππ4π

g2ωρπ4π

× 3.56

≈ 10.3MeVg2ρππ4π

g2ωρπ4π

, (24.225)

where the number 3.56 is due to the numeric integration of a phase space integral[8]. The experimental rate gives

g2ρππ4π

≈ 0.95

M2π

, (24.226)

and hence

g2ωρπgρππ2

≈ 0.117

M2π

≈ 6.141

GeV2 . (24.227)

24.7.5 Vector Meson Decays K∗(892)→Kπ

In πK -scattering one observes the SU(3)-partner of the ρ-meson, the strange vectormeson K∗ (892). Its coupling is written as

LK∗Kπ =1

2gK∗Kπ(K

†µK)i

↔∂µ. (24.228)

Its decay width is

ΓK∗Kπ = 3

(

12gK∗Kπ

)2

4π

2

3

p3CMK∗

M2K

, (24.229)

where

pCMK∗ =1

2MK∗

√

[

M2K∗ − (MK∗ +Mπ)

2] [

M2K∗ − (MK∗ −Mπ)

2]

(24.230)


is the momentum of the decay products in the rest frame of K∗. This implies a ratiowith respect to the ρππ -width:

ΓK∗Kπ

Γρππ=

3

4

g2K∗Kπ

g2ρππ

M2ρ

M2K∗

p3CMK∗

p3CM ρ

≈ 3

4

g2K∗Kπ

g2ρππ× 0.385. (24.231)

The experimental width of K∗ is

ΓK∗Kπ = (51.8± 0.8)MeV, (24.232)

so that its ratio with respect to the ρ-width (24.220) is 0.3406. Inserting this into(24.231), we extract a ratio of coupling constants

gK∗Kπ

gρππ≈ 1.086. (24.233)

By SU(3)-symmetry, this ratio is predicted to be unity, in good agreement withexperiments.

24.7.6 Axial Vector Meson Decay a1(1270)→ ρπ

A more involved coupling governs the decay of the axial vector meson a1(1270). Itis seen as a resonance of mass 1270 MeV and width

Γa,ρπ ≈ 316± 45MeV (24.234)

in the ρπ-scattering amplitude. Angular momentum and parity allow s- and d-wave interactions, so that there exist two independent coupling constants with aLagrangian density

L = ga1ρπ (aµ × µ) · + ha1ρπ (∂µaν × ∂ν

µ) · . (24.235)

Let ǫ(p), ǫ(q) denote the polarization vectors of the a1 and ρ meson, respectively,and k the momentum of the outgoing pion. Then the invariant t-matrix element forA+ → ρ−π+ reads

tba =1

√2V

3

1√p0q0k0

[

ga1ρπǫ∗µ(p)ǫ

µ(q)− ha1ρπqνǫ∗ν(p)pµǫµ(q)

]

. (24.236)

For the calculation of decay rates it is better to use a more complicated-lookingdecomposition into a longitudinal and a transverse helicity amplitude

tL =1

√2V

3

1√p0q0k0

gLPµqν

(

gµλ −qµqλq2

)(

gνκ −pνpκp2

)

ǫ∗ ∗ κ(p)ǫλ(q),

tT = − gTm2

a

ǫµα′β′

γǫλαβγqβpαqβ′pα′ǫ∗λ(p)ǫµ(q). (24.237)


The relation between them is

−ha1ρπ = gL + gTm2

a1 +m2ρ −m2

π

2m2a1

,

−ga1ρπ =gTm2

a1

(

m2a1+m2

ρ −m2π

2

)2

−m2a1m2

ρ

. (24.238)

The advantage of these amplitudes is, that they are decoupled in the decay process.A somewhat tedious calculation (see Appendix 24B) gives

Γa1ρπ = ΓLa1ρπ

+ ΓTa1ρπ

, (24.239)

where

ΓLa1ρπ

=g2Lnπ

1

3

p5ρm2

ρ

, ΓTa1ρπ

=g2Tnπ

2

3

p5ρm2

a1

. (24.240)

Since the experimental masses have the ratio

m2a1

m2ρ

≈ 2.72, (24.241)

we get

Γa1ρπ ≈ 5.5× 10−5(

g2L + 2.735 g2T)

GeV, (24.242)

and find for the coupling constants the relation

g2L +2m2

ρ

m2a1

g2T ≈ g2L + 0.68g2T ≈ 75.1. (24.243)

The ratio between the coupling constants gL and gT has to be determined by someother method.

24.7.7 Coupling of ρ(770) to Nucleons

The coupling is defined by the non-minimal Lagrangian density

L = gρNN

[

µψγµψ +

κρNN

2M(∂µν − ∂νµ)ψσ

µνψ]

. (24.244)

An analysis of the phase shifts in ππ → NN yields the coupling strengths [9]

g2ρNN

4π= 0.6± 0.1, κρNN = 6.6± 1.0. (24.245)

Appendix 24A Useful SU(3)-Formulas 1377

Appendix 24A Useful SU(3)-Formulas

The matrix elements of an octet operator between two octets labeled by the real indices i =1, 2, 3, . . . , 8 of the adjoint representation 8 can be written as follows:

〈8i|8k8j〉 = F (−ifkij) +Ddkij = F (Fk)ij +D(Dk)ij , (24A.1)

where fijk are the structure constants of SU(3), and dijk those of the anticommutators

λi, λj = dijkλk +4

3δij .

The SU(3)-vector ei associated with a specific particle is found from the assignment of the nucleonoctet to the 3× 3 -matrix elements

N =

1√2Σ0 + 1√

6Λ Σ+ p

Σ− − 1√2Σ0 + 1√

6Λ n

Ξ− Ξ0 − 1√62Λ

, (24A.2)

by contraction with the λ-matrices:

ei =1√2tr(

λiN)

. (24A.3)

The proton, for example, corresponds to the octet vector

ei(p) =1√2tr

λi

0 0 10 0 00 0 0

. (24A.4)

It has the non-zero components

e4(p) =1√2, e5(p) =

i√2. (24A.5)

For the π0-couplings we calculate the matrix elements F0ij and D0ij between two proton states asfollows:

〈p|F0|p〉 =1

2, 〈p|D0|p〉 =

1

2. (24A.6)

Hence

〈p|π0|p〉 =F +D

2. (24A.7)

Similarly, the SU(3)-vector of the Σ+-particle is

ei(Σ+) =1√2tr

λi

0 1 00 0 00 0 0

. (24A.8)

It has the non-zero components

e1(

Σ+)

=1√2, e2

(

Σ+)

=i√2. (24A.9)

For the coupling of π0 to Σ-particles we need the matrix elements (F0)ij and (D0)ij betweenΣ+-states:

〈Σ+|F0|Σ+〉 = 1, 〈Σ+|D0|Σ+〉 = 0. (24A.10)


Hence

〈Σ+|π0|Σ+〉 = F. (24A.11)

The matrix elements of π+ between Σ+Σ+ and pp have therefore the ratio

Σ+π+Σ+

pπp=

2F

F +D, (24A.12)

just as in (24.189), (24.191), and (25.100).The SU(3)-matrix elements may also be calculated by using tables of the SU(3)-Clebsch-Gordan

coefficients. The matrices Fi Di correspond to the coefficients

Fi =√3〈8a|88〉, Di =

√

5

3〈8s|8 8〉. (24A.13)

The Clebsch-Gordan coefficients can be decomposed into a product of an isospin SU(2)-Clebsch-Gordan coefficient and a so-called isoscalar factor

〈8ai|8j; 8k〉 = 〈〈8aYi|8Yj; 8Yk〉〉(T i, T i3|T j, T j

3 ;Tk, T k

3 ), (24A.14)

where T i, T i3 is the isospin content of the SU(3)-index i. The isoscalar factors are listed in Table

24.5.

Table 24.5 Isoscalar factors of SU(3). We have omitted the square roots of the number

in the arrays on the right-hand sides, i.e., −3 stands for −√3. This is indicated by the

superscript 1/2 on each array. The table is from the Particle Data Properties Booklet [10].

1 → 8⊗ 8

(Λ) → (NK Σπ Λη ΞK) =1√8

(

2 3 − 1 − 2)1/2

81 → 8⊗ 8

NΣΛΣ

→

Nπ Nη ΣK ΛKNK Σπ Λπ Ση ΞKNK Σπ Λη ΞKΣK ΛK Ξπ Ξη

=1√20

9 − 1 − 9 − 1− 6 0 4 4 − 62 − 12 − 4 − 29 − 1 − 9 − 1

1/2

83 → 8⊗ 8

NΣΛΞ

→

Nπ Nη ΣK ΛKNK Σπ Λπ Ση ΞKNK Σπ Λη ΞKΣK ΛK Ξπ Ξη

=1√12

3 3 3 −32 8 0 0 − 26 0 0 63 3 3 −3

1/2

10 → 8⊗ 8

∆ΣΞΩ

→

Nπ NηNK Σπ Λπ Ση ΞK

ΣK ΛK Ξπ ΞηΞK

=1√12

− 6 6− 2 2 − 3 3 3

3 − 3 3 312

1/2

8 → 10⊗ 8

NΣΛΞ

→

∆π ΣK∆K Σπ Ση ΞK

Σπ ΞKΣK Ξπ Ξη ΩK

=1√15

−12 38 − 2 − 3 2

−9 63 − 3 − 3 6

1/2

10 → 10⊗ 8

∆ΣΞΩ

→

∆π ∆π ΣK∆K Σπ Ση ΞKΣK Ξπ Ξη ΩK

Ξ KΩη

=1√24

15 3 − 68 8 0 − 8

12 3 − 3 − 612 − 12

1/2

Appendix 24B Decay Rate for a1 → ρπ 1379

Appendix 24B Decay Rate for a1 →ρπ

Denoting the polarization vectors ǫ(p) and ǫ(q) of a1 and ρ mesons by ǫ and ǫ′, and using thetransversality properties pµǫ

µ = 0, qµǫ′µ = 0, respectively, the amplitudes are (dropping the

normalization factors 1/√2Nq0 associated with the three particles)

t = gL(pǫ′∗)(qǫ) +

gTM2

a1

∣

∣

∣

∣

∣

∣

ǫ′∗ǫ ǫ′∗q ǫ′∗pǫq q2 qpǫp qp p2

∣

∣

∣

∣

∣

∣

(24B.1)

= gL(pǫ′)(qǫ) +

gTM2

a1

[

M2a1M2

ρ (ǫ′∗ǫ) + (pǫ′∗)qp− (ǫ′∗ǫ)(qp)2

]

.

This expression shows directly the relation of the couplings gL and gT with the couplings ga1,ρπ

and ha1ρπ . We now put a1 in its rest frame, so that pµ = (Ma1, 0, 0, 0), and we find the scalar

products

ǫ′∗(q, 1)ǫ∗(p, 1) = −1, ǫ′∗(q, 2)ǫ∗(p, 2) = −1, ǫ′∗(q, 3)ǫ(p, 3) = − q0Mρ

,

and further

(qǫ(p, 1)) = 0, (pǫ′∗(q, 1)) = 0,

(qǫ(p, 2)) = 0, (pǫ′∗(q, 2)) = 0,

(qǫ(p, 3)) = −pCM, (pǫ′∗(q, 3)) =pCM

Mρ

Ma1, (24B.2)

where

pCM =1

2Ma1

√

[M2a1

− (Mρ −Mπ)2][Ma2

1

− (Mρ −Mπ)2] (24B.3)

is the center-of-mass momentum of the ρ-meson. We therefore find the helicity amplitudes (writingthe helicities as superscripts in parentheses)

〈ρ(0)π1|t|a1(0)〉 = −gLp2CM

Ma1

Mρ

, (24B.4)

〈ρ 1

2π|t|a12

1 〉 =gTM2

a1

(

−M2a1M2

ρ + E2CMM2

a1

)

= gT p2CM,

where

ECM =M2

a1+M2

ρ −M2π

2Ma1

(24B.5)

is the energy of the ρ meson in the center-of-mass frame. From this we obtain directly the longi-tudinal width

ΓLa1→ρπ =

g2L4π

2

6

p5CM

M2ρ

, (24B.6)

having inserted a factor 2 due to isospin, and a transversal width

ΓTa1→ρπ =

g2T4π

2

6· 2p

5CM

M2a1

. (24B.7)

Notes and References

For discussion of internal symmetries see the book byS. Gasiorowicz, Elementary Particle Physics, John Wiley & Sons, N.Y. 1966.

The individual citations refer to:


[1] For an introduction seeH. Fritzsch, Elementary Particles, Building Blocks of Matter, World Scientific, Singapore,2005.

[2] H. Yukawa, Proc. Phys. Math. Soc., Japan 17, 48 (1935).

[3] For a brief early history of the discovery of the strange mesons seeH.S. Bridge, in Progress in Cosmic Ray Physics, North-Holland, Amsterdam, 1956.

[4] M. Gell-Mann, Phys. Rev. 92, 833 (1952);T. Nakano and K. Nishijima, Progr. Theor. Phys. 10, 581 (1953).

[5] J.J. Sakurai, Phys. Rev. Letters 9, 472 (1962).

[6] An example for such a model isR.P. Feynman, M. Kislinger, and F. Ravndal, Phys. Rev. D 3 , 2706 (1971).

[7] Other quark models are discussed inM. Danilov, R. Mizuk, (arXiv:0704.3531);K. Carter, Symmetry Magazine, September 2006 (http://www.symmetrymagazine.org/cms/?pid=1000377);F. Buccella, P. Sorba, Mod. Phys. Lett. A 19, 1547 (2004);F. Buccella, H. Hogaasen, J.M. Richard, and P. Sorba, Eur. Phys. J. C 49, 743 (2007).

[8] For details see M. Gell-Mann, D. Sharp, W.G. Wagner, Phys. Rev. Lett. 8, 261 (1962).

[9] G.E. Brown, R. Machleidt, Phys. Rev. 50, 1731 (1994).

[10] This booklet can be downloaded from the internet address http://pdg.lbl.gov/2013/reviews/rpp2013-rev-su3-isoscalar-factors.pdf.

[11] J. Beringer at al. Phys. Rev. D 86, 010001 (20014) (http://pdg.lbl.gov/2013/tables/rpp2013-sum-quarks.pdf).

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