Hadron Spectroscopy 8 Mathematical Techniques Suh-Urk Chung BNL/NY/USA α, TUM/Munich/Germany β PNU/Busan/Republic of Korea (South Korea γ http://cern.ch/suchung/ α Senior Scientist Emeritus β Scientific Consultant (part time γ Research Professor (part time p.
The QCD Lagrangian: L = q m q i ψ i q ψ i q Isotopic Spin where i = color index ( 3, q = flavor index ( 6 and m q = current-quark mass. u 5 MeV d 3 9 MeV s 75 70 MeV c.5.35 GeV b 4.0 4.4 GeV t 74.3 ± 5. GeV flavor SU(2 : (u,d flavor SU(3 : (u,d,s flavor SU(4 : (u,d,s,c Constitutent light-quark mass= (m d +m u /2 = 220 MeV Constitutent strange-quark mass= m s = 49 MeV S. Godfrey and N. Isgur, Phys. Rev. D32, 89 (985 p. 2
C- and G-Parity: A new Definition and Applications C- and G-Parity Operations: We shall adopt a notation a to stand for both the baryon number B and hypercharge Y. Anti-particles are denoted ā, so that ( a = (B,Y, ā = ( B,Ȳ = ( B, Y In addition, we shall use y to denote Y/2; (2 y = Y 2 = 2 (B +S, Q = y +ν where S, Q, ν are the strangeness, the charge and the third component of isospin, respectively. p. 3
Let I be the isospin operator. Then, we have (3 [I i,i j ] = iǫ ijk I k We start with a state having an isospin σ and its third component ν which transforms according to the standard jm representation, so that (4 where I ± =I x ± ii y and (5 I z σν = ν σν I ± σν = F ± (ν σν ± I 2 σν = σ(σ + σν F ± (ν = (σ ν(σ ±ν + Note that F ± (ν = F ( ν. We shall require that anti-particle states transform in the same way as their particle states according to the standard representations given above. The C operation changes a state aν to ā ν. (We use a shorthand notation where the isospin σ is omitted from a more complete description of the state aσν. If anti-particle states are to transform in the same way as particle states, it is necessary that one define an anti-particle through the G operation. The key point is that G is defined so that its operation does not perturb the ν quantum number. p. 4
To define the G operator, we need to first introduce a rotation by 80 around the y-axis: (6 U 2 [R y (π] = ( 2σ, U [R y (π] = ( 2σ U[R y (π], U[R y (π] σν = ( σ ν σ ν It will be shown later that R y (π commutes with the C operator. We therefore define the G operator through (7 G = CU[R y (π] = U[R y (π]c, C = ( 2σ U[R y (π]g We are now ready to define an anti-particle state via (8 G aν = g āν G āν = ḡ aν and require that g and ḡ be independent of ν and furthermore that an arbitrary isospin rotation R(α, β, γ commutes with G: (9 [ ] U[R(α,β,γ],G = 0 p. 5
The action of C on particle and anti-particle states is (0 C aν = g( σ+ν ā ν C āν = ḡ( σ+ν a ν It is customary to define C such that C 2 = I, in which case ( gḡ( 2σ = For hadrons, we shall define g and ḡ via (2 while for quarks, (3 g = η( y+σ, ḡ = η( ȳ+σ g = η( B+y+σ, ḡ = η( B+ȳ+σ Note that the exponents in these expressions are always integers. p. 6
The quantity F defined by (4 2 F = B +y = (3B +S 2 will be termed the intrinsic flavor of a particle. Note that the intrinsic flavor is always an integer, as shown in the following table: states u d s π η K p n Λ Σ Ξ Ω F 0 0 0 3 3 2 2 3 0 It is seen that the intrinsic flavor of an anti-particle is the negative of that of the particle, i.e. F = F. With these definitions, we can make η a real number and let it take on values of + or, so that η 2 = +. Then, we have, since C 2 = + and gḡ( 2σ = +, (5 G 2 = ( 2σ conforming to the standard expressions. p. 7
From the actions of the C on aν and āν as defined previously, it is easy to work out the commutation relations between C and I; (6 {C,I x } = {C,I z } = 0, [C,I y ] = 0 ln other words, C anti-commutes with I x and I z while it commutes with I y. This gives a ready justification of the definition of G-parity. We can further deduce that, since rotations commute with G, i.e. GU[R(α,β,γ] = U[R(α,β,γ]G, (7 CU[R(α,β,γ]C = U[R y (π]u[r(α,β,γ]u [R y (π] This shows that the actions of I-spin rotation under charge-conjugation can be expressed in terms of I-spin 90 rotations. Recapitulate: G aν = η( y+σ āν (8 G āν = η( ȳ+σ aν C aν = η( y ν ā ν C āν = η( ȳ ν a ν For quarks, replace y = (B +S/2 B +y = (3B +S/2 and ȳ = ( B + S/2 B +ȳ = (3 B + S/2. Conclude: η is the charge conjugation of the nonstrange neutral members of any meson family of SU(3. [Note G = C( I ] p. 8
As an example, Consider the members of the pion SU(3 family, i.e. {π}. We set η = + and find (9 Cπ ± = π, Cπ 0 = +π 0, Cη = +η, Cη = +η Gπ = π, Gη = +η and (20 C G ( K + K 0 ( K + K 0 = = ( +K K 0 ( K 0 K, C, G ( ( K0 K 0 K = +K + ( K 0 ( +K + = +K 0 K Note that C 2 = I and G 2 = I, consistent with the usual results as applied to the states with I = /2. For the {ρ} SU(3 family, we must set η =, so that (2 Cρ ± = +ρ, Cρ 0 = ρ 0, Cω = ω, Cφ = φ Gρ = +ρ, Gω = ω and (22 C G ( K + K 0 ( K + K 0 = = ( K + K 0, C ( + K0 +K, G ( K 0 ( +K 0 = K K K + ( K 0 ( K + = K 0 p. 9
Two-Particle States: We shall work out here the effect of C and G operations on a particle-antiparticle system in an eigenstate of total isospin, total intrinsic spin, orbital angular momentum and total spin. We use the notations I,S,l and J for these quantum numbers. (Note that I was used as an isospin operator and S denoted strangeness in section2. Each single-particle state in the two-particle center-of-mass(cm system will be given a shorthand notation, (23 a,+ k,ν,m = a,+ k,σ ν,s m ā, k,ν 2,m 2 = ā, k,σ 2 ν 2,s 2 m 2 where k is the 3-momentum of the particle in the CM system, and σ and s are isospin and spin of the particles σ = σ 2 = σ and s = s 2 = s. The two-particle system in a given state of Iν and lsjm is given by (24 aāν = (σ ν σ 2 ν 2 Iν(s m s 2 m 2 Sm s (Sm s lm JM ν ν 2 m m 2 d k Y l m( k a,+ k,ν,m ā, k,ν 2,m 2 where Y l m( k is the usual spherical harmonics. p. 0
We note C a,+ k,ν,m ā, k,ν 2,m 2 = ( ν +ν 2 ā,+ k, ν,m a, k, ν 2,m 2 = ( ν +ν 2 +2s a, k, ν 2,m 2 ā,+ k, ν,m G a,+ k,ν,m ā, k,ν 2,m 2 = ( 2σ ā,+ k,ν,m a, k,ν 2,m 2 = ( 2σ+2s a, k,ν 2,m 2 ā,+ k,ν,m where the second lines have been derived by interchanging two wave functions, which brings in a factor ( 2s, positive for mesons and negative for fermions. The effect of C and G on the two-particle states can now be worked out. By interchanging the subscripts and 2 and by the operation k k, we obtain (25 C aāν = ( l+s+ν aā ν G aāν = ( l+s+i aāν where we have used the relationship Y l m ( k = ( l Y l m ( k and the following formulas for the Clebsch-Gordan coefficient (σ 2 ν 2 σ ν Iν = (σ ν σ 2 ν 2 Iν (σ 2 ν 2 σ ν Iν = ( I 2σ (σ ν σ 2 ν 2 Iν, σ = σ 2 = σ (s 2 m 2 s m Sm s = ( S 2s (s m s 2 m 2 Sm s, s = s 2 = s p.
We next work out the effect of the parity operation(π on the two-particle states. Since antifermions have opposite intrinsic parities to those of their fermion partners, the Π operation brings in the factor ( 2s. In addition, the 3-momentum k changes sign under the Π operation. Therefore, we have (26 Π a,+ k,ν,m ā, k,ν 2,m 2 = ( 2s a, k,ν,m ā,+ k,ν 2,m 2 So, again by using the operation k k, we obtain the familiar result (27 Π aāν = ( l+2s aāν It follows from (3 that a particle-antiparticle with ν = 0 is in an eigenstate of C with its eigenvalue ( l+s. This result applies to all neutral N N, q q, K K and ππ systems, with S = 0 for dikaon and dipion systems. For all ν, a particle-antiparticle system has the G-parity equal to ( l+s+i. Charged N N, q q, K K systems have I =, so that their G-parity is ( l+s+ (again S = 0 for dikaons. Since the G-parity is + for dipions, one has l+i = even for any ππ system. For all ν, the intrinsic parity of a particle-antiparticle system is given by ( l+2s. p. 2
(K Kπ 0 Systems: This case represents an example of a nontrivial application of the C- and G-parity operators introduced thus far. We start with the K intermediate systems. A K decays into a πk. For K s with positive strangeness, one has (28 K + = K 0 = 2 3 π+ K 0 3 π0 K + 2 3 π0 K 0 3 π K + and for negative strangeness (29 K 0 = K = 2 3 π+ K 3 π0 K0 2 3 π0 K 3 π K0 One uses a convention in which ordering of particles signifies different momenta, so that one must keep track of it with care. p. 3
It is seen that the C and G operators act on K s in the following way: (30 C ( K + K 0 = ( K + K 0, C ( K 0 K = ( +K 0 K + and (3 G ( K + K 0 = ( + K 0 +K, G ( K 0 K = ( K + K 0 Let A g I (K stand for the decay amplitude X 0 (K Kπ 0 where I is the isospin of the X and g its G-parity (32 A g I (K = 2 [ (K + K +g K 0 K 0 ( I ( K 0 K0 +gk K +] and (33 GA g I (K = ga g I (K, CA g I (K = g( I A g I (K p. 4
Introducing the K decays, one sees that { A g [ I (K = (π + K 0 K +g( I (π K0 K +] 6 [ +( I (π K + K0 +g( I (π + K K 0]} (34 { [ (π 0 K + K +g( I (π 0 K K +] 2 +( I [ (π 0 K 0 K0 +g( I (π 0 K0 K 0]} We next consider two different intermediate states involving K K. Let a s refer to a 0 (980, a 2 (320 and other I G = objects, and f s stand for either f 0 (980, f 2 (270 or other I G = 0 + states. They are given by (35 a 0 = [ K + K +K 0 K0 +( K 0 K 0 +K K + ] 2 a [ = K 0 K +K K 0], a + [ = K + K0 + K 0 K +] 2 2 where Ga = a and Ca 0 = +a 0 as it should be. p. 5
and (36 f = 2 [ K + K K 0 K0 ( K 0 K 0 K K + ] so that Gf = +f and Cf = +f. Again, let A g I (a and Ag I (f refer to the decay amplitude for X 0 a+π and X 0 f +π A + 0 (a = 3 (π + a π 0 a 0 +π a + (37 = [π + (K 0 K a +π + (K K 0 a +π (K + K0 a +π ( K ] 0 K + a 6 [π 0 (K + K a +π 0 (K 0 K0 a +π 0 ( K ] 0 K 0 a +π 0 (K K + a 2 (π + a π a + A + (a = 2 = [π + (K 0 K a +π + (K K 0 a π (K + K0 a π ( K ] 0 K + a 2 and (38 A (f = 2 [ π 0 (K + K f π 0 (K 0 K0 f π 0 ( K0 K 0 f +π 0 (K K + f ] One sees that CA + 0 (a = +A+ 0 (a, CA+ (a = A+ (a and CA (f = +A (f. p. 6
The complete decay amplitude for X 0 (K Kπ 0 may now be written (39 where (40 2A = A + 0 +A+ +A 0 +A A + 0 = x+ 0 A+ 0 (K +y + 0 A+ 0 (a A + = x+ A+ (K +y + A+ (a A 0 = x 0 A 0 (K A = x A (K +y A (f where the superscripts ± once again specifies g = ± and the subscripts 0 or stand for I. The variables x g I and yg I are the unknown parameters in the problem. Note that an isoscalar X 0 cannot couple to π 0 +f, so that one must set y 0 = 0. p. 7
Consider next the amplitude corresponding to π K S K +. The complete amplitude is A = { 2 x + 0 6 + 4 +x 0 { 2 3 y+ 0 [ (π K + K S +(π K S K +] 0 x+ [ (π K + K S (π K S K +] 0 x [ π (K + K S a +π (K S K + ] a 0 y+ [ (π K + K S +(π K S K +] [ (π K + K S (π K S K +] } [ π (K + K S a +π (K S K + ] a } Similarly one finds, for the π + K S K amplitude, A = { 2 x + 0 6 + 4 x 0 { 2 3 y+ 0 [ (π + K K S +(π + K S K ] 0 +x+ [ (π + K K S (π + K S K ] 0 x [ π + (K K S a +π + (K S K ] a 0 +y+ [ (π + K K S +(π + K S K ] [ (π + K K S (π + K S K ] } [ π + (K K S a +π + (K S K ] a } C π K S K + = π + K S K {x + 0, x, y+ 0 {x +, x 0, y+ } = C = + eigenstates } = C = eigenstates p. 8
Flavor SU(3 Irreducible Representations: J. J. de Swart, Rev. Mod. Phys. 35, 96 (963. G. E. Baird and L. C. Biedenharn, J. math. Phys. 4, 449 (963; 5, 723 (964. S. U. Chung, E. Klempt, and J. K. Körner, Eur. Phys. J. A. 5, 539 (2002 Let D(p, q be an irreducible representation characterized by two integers p and q. For a physically realizable representation, one must have p q = 3 n, where n = 0, ±, ±2, ±3,... The number of basis vectors in an irreducible representation is given by the dimensionalityn of the representation (4 N = (+p(+q [+ 2 (p+q ] There are two Casimir operators F 2 and G 3 with the eigenvaluesf 2 and g 3. They are given by f 2 = [ p 2 +q 2 +pq +3(p+q ] 3 (42 g 3 = 8 (p q(2p+q +3(2q +p+3 So an irreducible representation can be equivalently characterized by D(f 2,g 3 corresponding to the two Casimir eigenvalues. p. 9
See Table I for a few examples of practical importance. Table I: Irreducible Representations of SU(3 N 8 0 0 27 (p,q (0,0 (, (3,0 (0,3 (2,2 (f 2,g 3 (0,0 (3,0 (6,9 (6, 9 (8,0 An eigenstate (or a wave function belonging to an irreducible representation is given by the eigenvalues corresponding to a set of five commuting operators (43 {F 2, G 3, Y, I 2, I 3 } where I is the isotopic spin and Y is the hypercharge. It is conventional to use I and Y for both operators and eigenvalues. Thus, the eigenvalue for the SU(2 Casimir operator I 2 is I(I +, and that for Y is just Y = B +S, but the eigenvalue for I 3 is denoted m here. Introduce new notations for convenience: (44 µ = {f 2,g 3 }, σ = {Y, I}, and ν = {Y, I, m} Then, the eigenstate can be given a compact notation φ (µ ν. p. 20
Consider a product representation D(p,q D(p 2,q 2. It can be expanded as a direct sum of irreducible representations. The eigenstates of each irreducible representation in the expansion are given by (45 ψ ( µ µ 2 µ γ ν = ν,ν 2 ( µ µ 2 µ γ ν ν 2 ν φ (µ ν φ (µ 2 ν 2 following the notations used previously. The subscript γ is a label which distinguishes two irreducible representations with the same (p,q or (f 2,g 3, e.g. 8 and 8 2. The transformation matrix is real and orthogonal and given by (46 ( µ µ 2 µ γ ν ν 2 ν = ( µ µ 2 µ γ σ σ 2 σ (I m I 2 m 2 Im where the first element on the right-hand side is the SU(3 isoscalar factor and the second element is the usual SU(2 Clebsch-Gordan coefficient. p. 2
Four-quark (q q + q q Vector Mesons: Consider a decay process X a +a 2, where X is a nonstrange (q q +q q meson with J P = and I G = or I G = +. So X is an isovector (I = meson, with both J PC = + or J PC = allowed. The decay products a and a 2 belong to the ground-state S 0 octet, i.e. {π} = {π, K, K, η}. We assume here that the η is a pure SU(3 octet and the η is a pure SU(3 singlet. The following expansion gives relevant irreducible representations (47 8 8 = 8 8 2 0 0 27 The Bose symmetrization requires that a P -wave meson couple only to antisymmetric wave functions of SU(3, i.e. 8 2, 0 and 0, as, 8 and 27 are symmetric under the interchange of a and a 2. Antisymmetric Octet (8 2 : I G = + = J PC = Y I Q wave functions 0 + ( 3 π + π 0 π 0 π + ( K0 K 6 + K + K0 0 ( 3 π + π π π + ( K0 K 0 K 0 K0 ( K K + K + K 3 2 ( π 0 π π π 0 6 2 ( K K 0 K 0 K p. 22
An example of a self-conjugate representation: {π} η = +, C π 0 = η π 0 and C aν = η( S/2 ν ā ν, where I 3 = {ν} S K 0 K + π π 0 η π + I 3 K K 0 p. 23
Eigenstates in the irreducible representations of 0 0: S +2 + 2 0 + +2 I 3 2 : 0 : 0 p. 24
Antisymmetric: 0 Y I Q wave functions 0 + ( 2 π + π 0 π 0 π + + ( K0 K + K + K0 + ( 6 2 π + η ηπ + 0 ( 2 π + π π π + + ( 2 K 0 K 0 K 0 K 0 + ( 2 K K + K + K + ( 2 π 0 η ηπ 0 ( 2 π 0 π π π 0 + ( K K 0 K 0 K + ( 6 2 π η ηπ Antisymmetric: 0 Y I Q wave functions 0 ( 2 π 0 π π π 0 ( K K 0 K 0 K + ( 6 2 π η ηπ 0 ( 2 π + π π π + ( K0 K 2 0 K 0 K0 ( 2 K K + K + K + ( 2 π 0 η ηπ 0 + ( 2 π + π 0 π 0 π + ( K0 K + K + K0 + ( 6 2 π + η ηπ + p. 25
Let φ be the wave function for {ππ} systems. One concludes (48 ] [φ(0+φ(0 = ( π + η ηπ + = I G (J PC = ( + 2 + 2 ] [φ(0 φ(0 = ( π + π 0 π 0 π + + ( K 0 K + K + K 0 2 + 6 3 = I G (J PC = + ( Summarize: Consider a nonstrange isovector X(q q + q q with the quantum numbers of a vector meson J P =. Its decay into {π}+{π} should occur in a P wave. If SU(3 is conserved in the decay, (49 ρ(8 2 : I G (J PC = + ( {ππ} +{K K} ρ x (0 0 : I G (J PC = + ( {ππ}+{k K} π (0+0 : I G (J PC = ( + πη π (8 : IG (J PC = ( + πη p. 26
Vector mesons (J P = in q q +q q systems: (50 {q q} = 8, {qq} = 3 6, { q q} = 3 6 and hence we must have (5 {q qq q} = ( 8 ( 8 = (3 6 (3 6 = 2 4 8 0 0 27 for a total of 8 states. There are two families of 8-plets, designated V ζ (ζ = ±, given by {V } = 6 self-conjugate members +4 strange members of 0 and 0 + 3 members of 0 and 0 (52 + 3 egregious + members of 0 and 0 {V + } = 6 + self-conjugatemembers +4 strange members of 0 and 0 + 3 + members of 0 and 0 + 3 egregious members of 0 and 0 p. 27
Let χ be the wave function for any state in V ζ, and let χ 0 be its charge-neutral wave function. Then, we have (53 Cχ 0 (0 = ζ χ 0 (0, Cχ 0 (0 = ζχ 0 (0 Gχ(0 = ζχ(0, Gχ(0 = ζχ(0 where we require ζ = ±. The G-parity eigenstates are (54 χ ± = 2 [ χ(0 ζχ(0 ], Gχ± = ±χ ± Define H to be the Hamiltonian that gives rise to the masses in the limit of exact SU(3. Then, one sees that (55 We see that (56 [F 2,H] = 0, [G 3,H] = 0, H χ ± = M ± χ ± G 3 χ ± = 9 χ = H χ = M ± χ So we conclude M + = M. p. 28
Recall that φ is the wave function defined for {ππ} systems. Then we have (57 φ ± = 2 [ φ(0 φ(0 ], G φ± = ± φ ± Consider now the decay X {ππ} in the limit of exact SU(3 (58 2A ± = 2 φ ± M χ ± = [ φ(0 φ(0 ] M [ χ(0 ζχ(0 ] = φ(0 M χ(0 + ζ φ(0 M χ(0 (G G = I, ζ 2 = = 2 φ(0 M χ(0 So we conclude A + = A or φ + M χ + = φ M χ. Summarizing, in the limit of exact SU(3, the observation of I G (J PC = ( + π (400 πη implies that the π (400 must belong to the 0 0 representation of V ζ. And there must exist its partner I G (J PC = ( ρ x (400 {ππ}+{k K} at the same mass and with the same decay strength. For example, we must have g 2( π + (400 π+ η = 3 g2( ρ + x (400 π+ π 0 + 2 3 g2( ρ + x (400 K+ K0 where g 2 is the coupling constant squared. p. 29
Classification of a general decay X {π} +{π}: SU(3 Multiplet J PC Composition Singlet ( even ++ q q, q q +gluon, q q +q q Symmetric Octet (8 even ++ q q, q q +gluon, q q +q q Antisymmetric Octet (8 2 odd q q, q q +gluon, q q +q q Multiplet 20 (0+0 odd + q q +q q Multiplet 20 (0 0 odd q q +q q Multiplet 27 even ++ q q +q q Mulitplet 20: Quantum Numbers Multiplicity I S J P = 8 3/2 ± J P = 4 /2 ± J P = 2 0 ±2 J PC = 3 0 J PC = + 3 0 p. 30
Gluons in QCD Quantum chromodynamics (QCD, the field theory of strong interactions for colored quarks and gluons, is based on the group SU(3 of color. The Lagrangian is (59 L QCD = 4 Fa µν F aµν +i ψ i k γµ (D µ ij ψ j k m k ψi k ψ i k (8π 2 θ QCD ǫµναβ F a µν F a αβ where (60 F a µν = µ A a ν ν A a µ +g s f abc A b µa c ν (D µ ij = µ δ ij i 2 g sλ a ij Aa µ Here the ψk i is the 4-component Dirac spinor for each quark with color index i = {,2,3} and flavor index k = {,6}, and A a µ is the gluon field with the index a = {,8}. g s is the QCD coupling constant, i.e. α s = gs/(4π; 2 the f abc is the usual structure constant of the SU(3 algebra, and λ a ij is the generator of the corresponding SU(3 transformations, so it is a 3 3 matrix with a = {,8} and i,j = {,2,3}; and m k, again with k = {,6}, is the current-quark mass introduced in the previous section. QCD is a gauge field theory with just two parameters g s and θ QCD. p. 3
The Flux-Tube Model N. Isgur and J. Paton, Phys. Rev. D 3 (985 290; N. Isgur, R. Kokoski and J. Paton, Phys. Rev. Lett. 54 (985 869; F. E. Close and P. R. Page, Nucl. Phys. B 443 (995 233; T. Barnes, F. E. Close and E. S. Swanson, Phys. Rev. D 52 (995 5242. Let L and S be the orbital angular momentum and the total intrinsic spin of a q q system. Let s and s 2 be the spins of q and q, i.e. s = s 2 = /2. Then, one readily finds (6 J = L+ S and S = s + s 2 and (62 P = ( L+, C = ( L+S, PC = ( S+ and the wave function in a state of J, L and S is given by (63 JMLS = 2L+ 4π m m 2 ms m (s m s 2 m 2 Sm s (Sm s LM L JM dωd L M L 0 (φ,θ,0 Ω,s m s 2 m 2 where M is the z-component of spin J in a coordinate system given in the rest frame of q q and Ω = (θ,φ specifies the direction of the breakup momentum in this rest frame. p. 32
In the flux-tube model pioneered by Isgur and Paton and an excited gluon in q q +g is described by two transverse polarization states of a string, which may be taken to be clockwise and anticlockwise about the q q axis. Let n (+ m [n ( m ] be the number of clockwise (anticlockwise phonons in the mth mode of string excitation. Two important quantities are defined from these (64 Λ = m [ ] n (+ m n ( m, N = m m [ ] n (+ m +n ( m It is clear from the definition that Λ is the helicity of the flux tube along the q q axis. The number N is a new quantity; it will be given the name phonon number as it represents the sum of all the phonons in the problem weighted by the mode number m. The wave function in the flux-tube model is given by JMLS{n (+ m,n ( m }N Λ = (65 2L+ 4π (s m s 2 m 2 Sm s (Sm s LM L JM m m 2 ms M L dωd L M L Λ (φ,θ,0 Ω,s m ; s 2 m 2 ; {n (+ m,n ( m }N Λ There are a set of five rotational invariants specifying a state in the flux-tube model, i.e. J, L, S, N and Λ. Note Λ L. p. 33
The relevant quantum numbers are (66 P = η 0 ( L+Λ+, C = η 0 ( L+S+Λ+N, PC = ( S+N+ where η 0 = ±. Remark: if n (+ m = n ( m, then Λ = 0 and η 0 = +. Examples are given in tabular form. The lowest-order gluonic hybrid mesons are n (+ = and n ( = 0; N = and Λ = L S 2S+ L J (q q J PC (q q J PC (q q +g 0 P + ++ 3 P J 0 ++, ++, 2 ++ 0 +, +, 2 + 2 0 D 2 2 + 2 ++ 0 +, +, 2 + 2 2 3 D J, 2, 3 +, 2 +, 3 + +, 2 +, 3 + p. 34
An example of n (+ m = n ( m for m = is given below: n (+ = and n ( = ; N = 2 and Λ = 0 L S 2S+ L J (q q J PC (q q J PC (q q +g 0 0 S 0 0 + 0 + 0 3 S 0 P + + 3 P J 0 ++, ++, 2 ++ 0 ++, ++, 2 ++ Another case of interest: n (+ 2 = and n ( 2 = 0; N = 2 and Λ = L S 2S+ L J (q q J PC (q q J PC (q q +g 0 P + + + 3 P J 0 ++, ++, 2 ++ 0 ++, ++, 2 ++ 0,, 2 p. 35
To Be Continuned... p. 36