Документ взят из кэша поисковой машины. Адрес оригинального документа : http://nuclphys.sinp.msu.ru/thgr/eng/ch2.pdf Дата изменения: Thu Sep 13 16:04:35 2007 Дата индексирования: Tue Oct 2 00:28:01 2012 Кодировка: Поисковые слова: m 2

Chapter 2 Unitary symmetry and quarks
2.1.1 Baryon and meson unitary multiplets

2.1 Eightfold way. Mass formulae in

SU

(3)

.

Let us return to baryons 1=2+ and mesons 0; . As we remember there are 8 particles in each class: 8 baryons: - isodublets of nucleon (proton and neutron) and cascade hyperons 0 ;, isotriplet of -hyperons and isosinglet , and 8 mesons: isotriplet , two isodublets of strange K -mesons and isosinglet . Let us try to write baryons B (1=2+ ) as a 8-vector of real elds B = (B1 ::: B8)= (~ N4 N5 B6 B7 B8), where ~ =(B1 B2 B3)= ( 1 2 3). Then the basis vector of the 8-dimensional representation could be written in the matrix form: 1X B = p 8=1 k Bk = 2k 0 1 1 N4 ; iN5 C 3 + p3 B8 1;i 2 1B p2 B 1 + i 2 ; 3 + p13 B8 B6 ; iB7 C = (2.1) @ A 2 p3 B8 N4 + iN5 B6 + iB7 ;

0 B B @

1 p2

0

+

; ;

1 p6

0

;

+ 1 p 2 0

+
0

1 p

6

0

;

2 p

p n
6

0

1 C C: A

24


At the left upper angle of the matrix we see the previous espression (1.23) from theory of isotopic group SU (2). In a similar way pseudoscalar mesons yield 3 3 matrix

0 B P =B @

1 p2

0

; K;

+

1 p6

;

+ 1 p 2 0

K

0

+

1 p

6

;

1 K+ C K0 C : A 2
p
6

(2.2)

Thus one can see that the classii cation proves to be very impressive: instead of 16 particular particles wehavenow only 2 unitary multiplets. But what corollaries would be? The most important one is a deduction of mass formulae, that is, for the rst time it has been succeded in relating among themselves of the masses of di erent elementary particles of the same spin.

2.1.2 Mass formulae for octet of pseudoscalar mesons
As it is known, mass term in the Lagrangian for the pseudoscalar meson eld described by the wave function r has the form quadratic in mass (to assure that Lagrange-Euler equation for the free point-like meson would yield Gordon equation where meson mass enters quadratically)

LP = m2 P m P

2

and for octet of such mesons with all the masses equal (degenerated):

LP = m2 P P m P
(note that over repeated indices there is a sum), while m = 140 MeV,mK = 490 MeV, m = 548 MeV. Gell-Mann proposed to refute principle that Lagrangian should be scalar of the symmetry group of strong intereractions, here unitary group SU (3), and instead introduce symmetry breaking but in such a way as to conserve isotopic spin and strangeness (or hypercharge Y = S + B where B is the baryon charge equal to zero for mesons). For this purpose the symmetry breaking term should have zero values of isospin and hypercharge. Gell-Mann proposed a simple solution of the problem, that is, the mass term should transform as the 33-component of the octet formed by product of two meson octets. (Note that in either meson or baryon octet 25


33-component of the matrix has zero values of isospin and hypercharge) First it is necessary to extract octet from the product of two octets entering Lagrangian. It is natural to proceed contracting the product P P along upper and sub indices as P P or P P and subtract the trace to obtain regular octets M = P P ; 1P P 3 , N = P P ; 1P P 3 M = 0 N =0 (over repeated indices there is a sum). Components 33 of the octets M33 I N33 would serve us as terms which break symmetry in the mass part of the Lagrangian LP : One should only takeinto accountthat m in the meson octet there are both particles and antiparticles. Therefore in order to assure equal masses for particles and antiparticles, both symmetry breaking terms should enter Lagrangian with qual coe cients. As a result mass term of the Lagrangian can be written in the form
2 LP = mP P P + m2P (M33 + N33)= m 1 2 = m0P P P + m2P (P 3P3 + P3 P 3): 1 Taking together coe cients in front of similar bilinear combinations of the pseudoscalar elds we obtain 4 m2 = m2P m2 = m2P + m2P m2 = m2P + 3 m2P 0 K 0 1 0 1 wherefrom the relation follows immediately

4m2 =3m2 + m K

2

4 0 245 = 3 0 30 + 0 02(g\ w )2 :

The agreement proves to be impressive taking into account clearness and simplicity of the formalism used.

2.1.3 Mass formulae for the baryon octet
26

JP

=

1+ 2

Mass term of a baryon B S J P = 1 + in the Lagrangian is usually linear in 2 mass (to assure that Lagrange-Euler equation of the full Lagrangian for free


point-like baryon would be Dirac equation where baryon mass enter linearly)

LB = mB BB : m
For the baryon octet B with the degenerated (all equal) masses the corresponding part of the Lagrangian yields

LB = mB B B m
But real masse are not degenerated at all: mN 940 m 1192 m 1115 m 1320 (in MeV ). Also here Gell-Mann proposed to introduce mass breaking through breaking in a de nite way a symmetry of the Lagrangian: LB = m0B B + m1B 3 B3 + m2B3 B 3 ): m Note that here there are two terms with the 33-component as generally speaking m1 6= m2. (While mesons and antimesons are in the same octet, baryons and antibaryons forms two di erent octets) Then for particular baryons we have: 1 2 p = B3 n = B3 mp = mn = m0 + m1 + 1 ; = B2 m 0 = m = B2 0 1 2 2 ; p6 0 = B33 m = m0 + 3 (m1 + m2) 3 0 3 = B2 m 0 = m0 + m2 ; = B1 The famous Gell-Mann-Okubo mass relation follows immediately:
;

2(mN + m )= m +3m

4520 = 4535:

(Values at the left-hand side (LHS) and right-hand side (RHS) are given in MeV.) The agreement with experiment is outstanding which has been a stimul to further application of the unitary Lie groups in particle physics.

2.1.4 Nonet of the vector meson and mass formulae
Mass formula for the vector meson is the same as that for pseudoscalar ones (in this model unitary space do not depends on spin indices ). 27


But number of vector mesons instead of 8 is 9, therefore we apply this formula taking for granted that it is valid here, to nd mass of the isoscalar vector meson !0 of the octet: m2 0 = 1 (4m2 ; m2) ! 3K wherefrom m!0 = 930 MeV. But there is no such isoscalar vector meson of this mass. Instead there are a meson ! with the mass m! = 783 MeV and a meson with the mass m = 1020 Mev. Okubo was forced to introduce nonet of vector mesons as a direct sum of the octet and the singlet 1 0p 0 p 1 + + 16 !8 K+C 2 B ; V =B ; p12 0 +0 p16 !8 K 0 C + A @ K; K 0 p1 1 0 0C 30 B 1 + B 0 p3 0 0 C (2.3) @ A 1 0 0 p3 0 whichwe write going a little forward as 1 0 p1 0 p1 + + 2! K+C 2 B ; ; p12 0 + p12 ! K 0 C (2.4) V =B A @ ; 0 K K

where 1 is essentially cos ! , ! being the angle of ideal mixing of the 3 octet and singlet states with I =0 S = 0. Let us stress once more that introduction of this angle was caused by discrepancy of the mass formula for vector mesons with experiment. "Nowadays (1969!) there is no serious theoretical basis to treat V9 (vector nonet{V.Z.) as to main quantity never extracting from it !0 (our 0{V.Z.) as SpV9 so Okubo assertion should be seen as curious but not very profound observation." S.Gasiorowicz, "Elementary particles physics", John Wiley & Sons, Inc. NY-London-Sydney, 1968. We will see a little further that the Okubo assertion is not only curious but also very profound. 28

q

! ! 0 q1 q 1 ! @ q q 2 A !8 3 3 2 1

;

3

3

0


2.1.5

Decuplet of baryon resonances with and its mass formulae

JP

=

3+ 2

Up to 1963 nine baryon resonances with J P = 3 + were established: isoquartet 2 ( with I = 3 ) (1232) ! N + isotriplet (1385) ! + + isodublet 2 (1520) ! + . But in SU (3) there is representation of the dimension 3 10, an analogue to IR of the dimension 4 in SU (2) ( with I = 2 ) which is given by symmetric tensor of the 3rd rank (what with symmetric tensor describing the spin state J P = 3 + yields symmetric wave function for the 2 particle of half-integer spin!!!). It was absent a state with strangeness -3 whichwe denote as . With this state decuplet can be written as:
;
222 223

;

0 221

+ 211 0 123 0 133 + 113

++ 111

?;? 333 Mass term of the Lagrangian for resonance decuplet B following GellMann hypothesis about octet character of symmetry breaking of the Lagrangian can be written in a rather simple way:

233

;

L

B M

= M0 B B

+ M1 B3 B 3 :

Really from unitary wave functions of the decuplet of baryon resonances B and corresponding antidecuplet B due to symmetry of indices it is possible to constract an octet in a unique way. The result is:

M =M

0

M = M0 + M1 M = M0 +2M1 M = M0 +3M1 Mass formula of this kind is named equidistant. It is valid with su cient accuracy, the step in mass scale being around 145 MeV. But in this case the predicted state of strangeness -3 and mass (1530 + 145) = 1675 MeV cannot
;

29


be a resonance as the lighest two-particle state of strangeness -3 would be ( (1320)K (490)) with the mass 1810 MeV! It means that if it exists it should be a particle stable relative to strong interactions and should decay through weak interactions in a cascade way loosing strangeness -1 at eachstep. This prediction is based entirely on the octet symmetry breaking of the Lagrangian mass term of the baryon decuplet B . Particle with strangeness -3 ; was found in 1964, its mass turned out to be (1672 5 0 3) MeV coinciding exactly with SU (3) prediction! It was a triumph of unitary symmetry! The most of physicists believed in it from 1964 on. (By the way the spin of the ; hyperon presumably equal to 3/2 has never been measured.)

30


2.2 Praparticles and hypothesis of quarks
Upon comparing isotopic and unitary symmetry of elementary particles one can note that in the case of isotopic symmetry the lowest possible IR of the dimension 2 is often realized which has the basis (1 0)T , (0 1)T along this representation, for example, N K K transform, however at the same time unitary multiplets of hadrons begin from the octet (analogue of isotriplet in SU (2)I ). The problem arises whether in nature the lowest spinor representations are realized. In other words whether more elementary particles exist than hadrons discussed above? For methodical reasons let us return into the times when people was living in caves, used telegraph and vapor locomotives and thought that -mesons were bounded states of nucleons and antinucleons and try to understand in what way one can describe these states in isotopic space. Let us make a product of spinors N a Nb a b =1 2, and then subrtract and add the trace Nc N c c =1 2 3, expanding in this wayaproduct oftwo irreducible representations (IR) (two spinors) into the sum of IR's: 1 (2.5) Nb N a =(NbN a ; 2 baNc N c)+ 1 baNc N c 2 what corresponds to the expansion in terms of isospin 1 1 =1 + 0 or (in 2 2 terms of IR dimensions) 2 2=3 + 1: In matrix form ! ! p = (pp ; nn)=2 np (p n) n pn ;(pp ; nn)=2 + ! (pp + nn)=2 0 + (2.6) 0 (pp + nn)=2 which we identify for the J=0 state of nucleon and antinucleon spins and zero orbital angular momentum with the pion isotriplet and isosinglet : ! ! 1 1 + p0 p0 0 2 2 = 1 1 ; ;p 0 + 0 p2 0 2 while for for the J=1 state of nucleon and antinucleon spins and zero orbital angular momentum with the -meson isotriplet and isosinglet !: ! ! 1 1 + p0 p !0 0 2 2 1 1 ; ;p 0 + 0 p2 !0 : 2 31


This hypothesis was successfully used many times. For example, within this hypothesis the decay 0 into two -quanta was calculated via nucleon loop,

0

p p

p

The answer coincided exactly with experiment what was an astonishing achievement. Really, mass of two nucleons were so terribly larger than the mass of the 0-meson that it should be an enormous bounding energy between nucleons. However the answer was obtained within assumption of quasi-free nucleons (1949, one of the achievements of Feynman diagram technique applied to hadron decays.) With observation of hyperons number of fundamental baryons was increased suddenly. So rst composite models arrived. Very close to model of unitary symmetry was Sakata model with proton, neutron and hyperon as a fundamental triplet. But one was not able to put baryon octet into such model. However an idea to put something into triplet remains very attractive.

32


2.3 Quark model. Mesons in quark model.
Model of quarks was absolutely revolutionary. Gell-Mann and Zweig in 1964 assumed that there exist some praparticles transforming along spinor representation of the dimension 3 of the group SU (3) (and correspondingly antipraparticles transforming along conjugated spinor representation of the dimension 3 ), and all the hadrons are formed from these fundamental particles. These praparticles named quarks should be fermions (in order to form existing baryons), and let it be fermions with J P = 1 + 2 q = 1 2 3 q1 = u q2 = d q3 = s: Note that because one needs at least three quarks in order to form baryon of spin 1/2, electric charge as well as hypercharge turn out to be DROBNYMI non-integer(!!!) which presented 40 years ago as open heresy and for many of us really unacceptable one. Quarks should have the following quantum numbers: Q u d s 2/3 -1/3 -1/3

I
1/2 1/2 0

I

3

Y=S+B 1/3 1/3 -2/3

B 1/3 1/3 1/3

1/2 -1/2 0

in order to assure the right quantum numbers of all 8 known baryons of spin 1/2 (2.1): p(uud), n(ddu), + (uus), 0(uds), ; (dds), (uds), 0(ssu), ; (ssd). In more details we shall discuss baryons a little further in another section in order to maintain the continuity of this talk. First let us discuss meson states. We can try to form meson states out of quarks in complete analogy with our previous discussion on nucleonantinucleon states and Eqs.(2.5,2.6): 1 q q =(q q ; 3 q q )+ 1 q q ) 3

0 (u d s) B @

1 uC d A= s

0 1 uu du su C B ud dd sd A = @ us ds ss
33


0 1 D1 du su B ud D2 sd C + 1 (uu + dd + ss)I @ A3 us ds D3
where

(2.7)

1 D1 = uu ; 1 qq = 2 (uu ; dd)+ 1 (uu + dd ; 2ss)= 3 6 1q q + p q q 1 =2 3 23 8 1 D2 = dd ; 1 qq = ; 2 (uu ; dd)+ 1 (uu + dd ; 2ss)= 3 6 1q q + p q q 1 = ;2 3 23 8 2 1 D3 = ss ; 1 qq = ; 6 (uu + dd ; 2ss)= ; p q 8q: 3 3 We see that the traceless matrix obtained here could be identi ed with the meson octet J P =0; (S -state), the quark structure of mesons being: ; =(ud) + =(du) K ; =(us) K + =(su) 1 0 = p (uu ; dd) 2 0 K =(sd) K 0 =(ds) 1 = p (uu + dd ; 2ss): 6 In a similar way nonet of vector mesons Eq.(2.10) can be constructed. But as they are in 9, we take straightforwardly the rst expression in Eq.(2.7) with the spins of quarks forming J =1 (always S -state of quarks): 1 010 uu du su u (u d s)" B d C = B ud dd sd C = A @A@ us ds ss J =1 s" 010 1 1 + p2 + p2 ! K+C B ; ; p12 0 + p12 ! K 0 C =B (2.8) @ A ; 0 K K 34


This construction shows immediately a particular structure of the meson : it contains only strange quarks!!! Immediately it becomes clear more than strange character of its decay channels. Namely while ! meson decays predominantly into 3 pions, the meson practically does not decay in this way (2 5 09)% although energetically it is very 'pro table' and, instead, decays into the pair of kaons ( 00 (49 1 0 9)% into the pair K +K ; and (34 3 0 7)% into the pair KLKS ). This strange experimental fact becomes understandable if we expose quark diagrams at the simplest level:

s ud s

K K

ud s ud s ud
Thus wehaveconvinced ourselves that Okubo note on nonet was not only curious but also very profound. 35


We see also that experimental data on mesons seem to support existence of three quarks. But is it possible to estimate e ective masses of quarks? Let us assume that -meson is just made of two strange quarks, that is, ms = m( )=2 510 MeV. An e ective mass of twolight quarks let us estimate from nucleon mass as mu = md = Mp=s 310 MeV. These masses are called constituent ones. Now let us look howit works.

M

p(uud

)

=M

n(ddu)

=930 Mev

(input)( 940)exp

M (qq s) = 1130 Mev ( 1192)exp M (uds) =1130 Mev ( 1115)exp M (ssq) =1330 Mev ( 1320)exp (There is one more very radical question: Whether quarks exist at all? From the very beginning this question has been the ob ject of hot discussions. Initially Gell-Mann seemed to consider quarks as some suitable mathematical ob ject for particle physics. Nowadays it is believed that quarks are as real as any other elementary particles. In more detail we discuss it a little later.)

36


"Till Parisina's fatal charms Again attracted every eye..."{ Lord Byron, 'Parisina'

37


2.4 Charm and its arrival in particle physics.
Thus, there exist three quarks! During 10 years everybody was thinking in this way. But then something unhappened happened. In november 1974 on Brookehaven proton accelerator with max proton energy 28 GeV (USA) and at the electron-positron rings SPEAR(SLAC,USA) it was found a new particle{ J= vector meson decaying in pions in the hadron channel at surprisingly large mass 3100 MeV and surprisingly long mean life and, correspondingly, small width at the level of 100 KeV although for hadrons characteristic widths oscillate between 150 MeV for rho meson, 8 Mev for ! and 4 MeV for meson. Taking analogy with suppression of the 3-pion decay of the vector meson which as assumed is mainly (ss) state the conclusion was done that the most simple solution would be hypothesis of existence of the 4th quark with the new quantum number "charm". In this case J= (3100) is the (cc)vector state with hidden charm.

ud c J= c ud
And what is the mass of charm quark? Let us make a bold assumption that as in the case of the (1020) meson where mass of a meson is just double value of the ('constituent') strange quark mass (500 MeV) ( so called 'constituent' masses of quarks u and d are around 300 Mev as wehavesee) mass of the charm quark is just half of that of the J= (3100) particle that is around 1500 MeV (more than 1.5 times proton mass!). But introduction of a new quark is not so innocue. One assumes with this hypothesis existence of the whole family of new particles as mesons with 38

ud


the charm with quark content (uc), (cu), (dc), (cd), with masses around (1500+300=1800) MeV at least and also (sc) I (cs) with masses around (1500+500=2000) MeV. As it is naturally to assume that charm is conserved in strong interaction as strangeness does, these mesons should decaydue to weak interaction loosing their charm. For simplicitywe assume that masses of these mesons are just sums of the corresponding quark masses. Let us once more use analogy with the vector (1020) meson main decay channel of which is the decay K (490) K (490) and the production of this meson with the following decay due to this channel dominates processes at the electron-positron rings at the total energy of 1020 MeV. If analogue of J= (3100) with larger mass exists, namely, in the mass region 2 (1500+300)=3600 MeV, in this case such meson should decay mostly to pairs of charm mesons. But suchvector meson (3770) was really found at the mass 3770 MeV and the main decaychannel of it is the decayto twonew particles{ pairs of charm mesons D0 (1865)D0 (1865) or D+ (1870)D; (1870) and the corresponding width is more than 20 MeV!!!

c
(3770)

D ud D

c

New-found charm mesons decayas it was expected due to weak interaction what is seen from a characteristic mean life at the level of 10;12 ; 10;13 s. Unitary symmetry group for particle classi cation grows to SU (4). It is to note that it would be hardly possible to use it to construct mass formulae as SU (3) because masses are too di erent in 4-quark model. In any case a problem needs a study. 39


In the model of 4 quarks (4 avors as is said today) mesons would transform along representations of the group SU (4) contained in the direct product of the 4-dimensional spinors 4 and 4 : 4 4= 15 + 1 or 1 q q =(q q ; 4 q q )+ 1 q q ) 4 where now =1 2 3 4:

0 (2980) D0 (1865) B D0 (1865) p1 0 + p1 B P = B D; (1870) 2 ; 6 B @ F ;(1969) K; 0 (3100) D 0 (2007) B D 0(2007) p1 0 + p1 ! B V = B D ; (2010) 2 ; 2 B @ F ;(2112) K;

D+ (1870)

; ;

+

1 p

2

0

K

0

+

1 p

6

F +(1969) K+ K0

;

2 p

1 C C C C A

(2.9)

6

D + (2010)
1 p2

1 F +(2112) C + K+ C C 1 0+p ! K0 C A 2 K0

(2.10)

2.5 The fth quark. Beauty.
Per la bel lezza delle donne molti sono periti... Eccl.Sacra Bibbia Because of beauty of women many men have perished... Eccl. Bible

Victorious trend for simplicity became even more clear after another important discouvery: when in 1977 a narrow vector resonance was found at the mass around 10 GeV, (1S )(9460),; 50Kev, everybody decided that there was nothing to think about, it should be just state of two new, 5th quarks of the type ss or cc. This new quark was named b from beauty or bottom and was ascribed the mass around 5 GeV M =2, a half of the mass of a new meson with the hidden "beauty" (1S )(9460) = (bb). 40


ud b ud b ud
Immediately searches for next excited states was begun which should similarly to (3770) have had essentially wider width and decayinto mesons with the quantum number of "beauty". Surely it was found. It was (4S )(10580),; 10 MeV. It decays almost fully to meson pair BB and these mesons B have mass 5300 = (5000 + 300) MeVw and mean life around 1.5 ns.

b
(10000)

B ud B

b

41


2.6 Truth or top?
Truth is not to be sought in the good fortune of any particular conjuncture of time, which is uncertain, but in the light of nature and experience, which is eternal. Francis Bacon But it was not all the story as to this time there were already 6 leptons ; e, ; , ; e (and 6 antileptons) and only 5 quarks: c u with the electric charge +2/3e and d s b with the electric charge -1/3e. Leaving apart theoretical foundations (though they are and serious) we see that a symmetry between quarks and leptons and between the quarks of di erent charge is broken. Does it mean that there exist one more, 6th quark with the new ' avour' named "truth" or "top" ? For 20 years its existence was taking for granted by almost all the physicists although there were also many attempts to construct models without the 6th quark. Only in 1996 t-quark was discouvered at the mass close to the nucleus of 71Lu175, mt 175GeV. We have up to now rather little to say about it and particles containing t-quark but the assertion that t-quark seems to be uncapable to form particles.

42


2.7 Baryons in quark model
Up to nowwehave considered in some details mesonic states in frameworks of unitary symmetry model and quark model up to 5 quark avours. Let us return now to SU (3) and 3- avour quark model with quarks q1 = u q2 = d q3 = s and let us construct baryons in this model. One needs at least three quarks to form baryons so let us make a product of three 3-spinors of SU (3) and search for octet in the expansion of the triple product of the IR's: 3 3 3 = 10 +8+80 +1: As it could be seen it is even two octet IR's in this product so we can proceed to construct baryon octet of quarks. Expanding of the product of three 3-spinors into the sum of IR's is more complicate than for the meson case. We should symmetrize and antisymmetrize all indices to get the result: q q q= 1 (q q q + q q q + q q q + q q q + q q q + q q q )+ 4 1 (q q q + q q q ; q q q ; q q q ; q q q ; q q q )+ 4 1 (q q q ; q q q + q q q ; q q q + q q q ; q q q )+ 4 1 (q q q ; q q q ; q q q + q q q ; q q q + q q q )= 4

T f g + T f g + T ] + T ]: All indices go from 1 to 3. Symmetrical tensor of the 3rd rank has the dimenSSS sion Nn =(n3 +3n2 +2n)=6 and for n=3 N3SSS = 10. Antisymmetrical tenAAA sor of the 3rd rank has the dimension Nn =(n3 ; 3n2 +2n)=6 and for n=3 mix N3AAA =1. Tensors of mixed symmetry of the dimension Nn = n(n2 ; 1)=3 only for n=3 (N3mix=8) could be rewritten in a more suitable form as T upon using absolutely antisymmetric tensor (or Levi-Civita tensor) of the 3rd rank which transforms as the singlet IR of the group SU (3): Really,
0 0 0

=u u u
0 0

0

:
2 3 132

123

= u1u2u 12

3 3 123

+ u2u3u1 231 + u3u1u2 312 + u1u3u 123 123 12 +u3u2u1 321 + u2u1u3 213 = 123 123 43


=(u1u2u3 + u2u3u1 + u3u1u2 ; u1u3u2 ; u3u2u1 ; u2u1u3) 123 = 123 123 123 123 123 123 = DetU 123 = 123 as DetU = 1. (The same for 213 etc.) For example, partly antisymmetric 8-dimensional tensor T ] could be reduced to B jAs (3) = T ] SU 1 wewould have and for the proton p = B3

p2jpi

As SU

(3)

1 = 2B3

p jAs SU
(3)

(3)

= ;judui + jduui:

(2.11)

Instead the baryon octet based on partly symmetric 8-dimensional tensor T could be written in terms of quark wave functions as
fg

p6B jSy SU
1 3

=

f

q q gq :

1 For a proton B3 wehave

p6jpi

Sy SU

(3)

= 6B

p jSy SU

(3)

=2juudi; judui;jduui:

(2.12)

In order to construct fully symmetric spin-unitary spin wave function of the octet baryons in terms of quarks of de nite avour and de nite spin projection we should cure only to obtain the overall functions being symmetric under permutations of quarks of all the avours and of all the spin pro jections inside the given baryon. (As to the overall asymmetry of the fermion wave function we let colour degree of freedom to assure it.) Multiplying spin wave functions of Eqs.(1.19,1.21) and unitary spin wave functions of Eqs.(2.11,2.12) one gets:

p18B j = p18(B j "

As SU

(3)

tjA + B

j

Sy SU

(3)

TSj ):

(2.13) (2.14)

1 Taking the proton B3 as an example one has

+2u"d

p18jpi = j; udu + uudi j; "#" + ""#i " j2 uud ; udu ; duui j2 ""# ; "#" ; #""i = j2u"u"d# ; u"d"u# ; d"u"u# # u" ; u"u# d" ; d" u#u" +2d# u" u" ; u# u"d" ;
44

=

u#d"u"i:


Wehave used here that

juudi j #""i = ju
2u"u"d

#u"d"

i
(2.15)

andsoon. Important note. One could safely use instead of the Eq.(2.14) the shorter version but where one already cannot change the order of spinors at all!

p6jpi = p6jB i = j "
1 3 3 3

#

;

u"d"u# ; d"u"u#i:

The wave function of the isosinglet oneself upon calcolating 2j

has another structure as one can see
#

i" = ;p6jB i" = jd

" s" u#

+ s"d"u# ; u"s"d

;

s"u"d#i:

(2.16)

3 Instead decuplet of baryon resonances T f g with J P = 2 + would be written in the form of a so called weight diagram (it gives a convenient gra c image of the SU (3) IR's on the 2-parameter plane whichischaracterized by basic elements, in our case the 3rd pro jection of isospin I3 as an absciss and hypercharge Y as a ordinate) which for decuplet has the form of a triangle:

; ;

0 0

+

++ + 0

; ;

and has the following quark content:

ddd udd uud uuu sdd sud suu ssd ssu sss In the SU (3) group all the weight diagram are either hexagones or triangles and often are convevient in applications.
45


For the octet the weight diagram is a hexagone with the 2 elements in center: np ; (0 ) +
;
0

or in terms of quark content:

udd uud sdd sud suu ssd ssu The discouvery of 'charm' put a problem of searching of charm baryons. And indeed they were found! Nowwe already know +(2285 1 0 6m\ w ), c + (2625 6 0 8m\ w ), ++ + 0 (2455), + 0(2465). Let us try to classify them c c c along the IR's of groups SU (4) and SU (3). Nowwemake a product of three 4-spinors q = 1 2 3 4. Tensor structure is the same as for SU (3). But dimensions of the IR's are certainly others: 4 4 4= 204 +2004 +2004 + 4. A SSS symmetrical tensor of the 3rd rank of the dimension Nn =(n3 +3n2 +2n)=6 in SU (4) has the dimension 20 and is denoted usually as 204 . Reduction of theIR204 in IR's of the group SU (3) has the form 204 =103 +63 +33 +13.

Thereisaneasy way to obtain the reduction in terms of the corresponding dimensions. Really, as it is almost obvious, 4- spinor of SU (4) reduces to SU (3) IR's as 44 = 33 +13. So the product Eqs.(1.29,1.30) for n = 4 44 44 =104 +64 would reduce as

(33 +13) (33 +13)=33 33 +33 13 +13 33 +13 13 = 63 + 33 +33 +33 +11: AA As antisymmetric tensor of the 2nd rank T ] of the dimension Nn = n(n ; 1)=2 equal to 6 at n=4 and denoted by us as 64 should be equal to its conjugate T ] due to the absolutely anisymmetric tensor and so it has the following SU (3) content: 64 =33 + 33 . The remaining symmetric tensor SS of the 2nd rank T f g of the dimension Nn = n(n +1)=2 equal to 10 at n=4 and denoted by us as 104 would have then SU (3) content 104 =63 +33 +13 . The next step would beto construct reduction to SU (3) for products 64 44 = 46


2004 + 44 (see Eq.(1.18) at n = 4) and 104 44 = 2004 +204 (Eq.(1.20) at n =4). Their sum would give us the answer for 4 4 4. 64 44 =(33 + 33) (33 +13)= 44 +2004 = =33 33 + 33 33 +33 + 33 = =(33 +13)+(83 +63 + 33 +33:) As 2004 is now known it easy to obtain 204 : 204 =104 44 ; 2004 =103 +63 +33 +13: Tensor calculus with q = a qa + 4 q4 would give the same results. So 204 contains as a part 10-plet of baryon resonances 3=2+ . As to the charm baryons 3=2+ there are two candidate: c (2520) and c(2645) (J P has not been measured 3=2+ is the quark model prediction) Note by the way that J P of the ; which manifested triumph of SU (3) has not been measured since 1964 that is for more than 40 years! Nevertheless everybody takes for granted that its spin-parityis3=2+ . Instead baryons 1=2+ enter 200-plet described by traceless tensor of the 3rd rank of mixed symmetry B ] antisymmetric in two indices in square brackets: p6B = fq q gq ] 0 -plet as wehaveshown is reduced to the sum =1 2 3 4: This 20 of the SU (3) IR's as 2004 =83 +63 +33 +33: It is convenienttochoose reduction along the multiplets with the de nite valueofcharm. Into 8-plet with C =0 the usual baryon octet of the quarks u,d,s (2.1) is naturally placed. The triplet should contain not yet discouvered in a de nite way doubly-charmed baryons:
+

cc

++

+

cc

with the quark content

cc

ccd ccu ccs and, for example, the wave function of + in terms of quarks reads p6j + i" = j2c"c"u# ; cc"u"c# ; u"c"c#i: c (2.17) cc
47


Antitriplet contains already discouvered baryons with s =1
+

c
0

with the quark content

c

+

c

udc dsc usc and, for example, the wave function of + in terms of quarks reads c 2j +i" = js"c"u# + c"s"u# ; u"c"s# ; c"u"s#i: (2.18) c Sextet contains discouvered baryons with s =1
0

c

+

c

++

c

00 c
0

0+ c
c

(Note that their quantum numbers are not yet measured!) with the quark content ddc udc uuc dsc usc ssc and, for example, the wave function of 0c+ in terms of quarks reads

s"c# +2c" u"d# (2.19) s"c"u# ; c"s"u#i: Note also that in the SU (4) group absolutely antisymmetric tensor of the 4th rank transforms as singlet IR and because of that antisymmetric tensor of the 3th rank in SU (4) T ] =1 2 3 4 transforms not as a singlet IR as in SU (3) (what is proved in SU (3) by reduction with the tensor =1 2 3) but instead along the conjugated spinor representation 4(which also can be proved by the reduction of it with the tensor =1 2 3 4):
+

p12j 0 i = j2u " c" ;u"c"s# ; c"u"s# ;

48


We return here to the problem of reduction of the IR of some group into the IR's of minor group or, in particular, to IR's of a production of two minor groups. Well-known example is given by the group SU (6) SU (3) SU (2)S which in nonrelativistic case has uni ed unitary model group SU (3) and spin group SU (2)S . In the framework of SU (6) quarks belong to the spinor of dimension 66 which in the space of SU (3) SU (2)S could be written as 66 =(3 2) where the rst symbol in brackets means 3-spinor of SU (3) while the 2nd symbol just states for 2-dimensional spinor of SU (2). Let us try now to form a product of 6-spinor and corresponding 6-antispinor and reduce it to IR's of the product SU (3) SU (2)S :

66 =356 +16 =(3 2) (3 2) = (3 3 2 2) = =(8 + 1 3 + 1) = (8 1) + (8 3) + (1 3)] + (1 1) that is, in 356 there are exactly eight mesons of spin zero and 9=8+1 vector mesons, while there is also zero spin meson as a SU (6) singlet. This result suits nicely experimental observations for light (of quarks u d s) mesons. In order to proceed further we form product of two 6-spinors rst according to our formulae, just dividing the product into symmetric and antisymmetric IR's of the rank 2: 66 66 =216 +156 =(3 2) (3 2) = (3 3 2 2) = =(63 + 33 3 + 1) = f(63 3) + (33 1)g21 + (63 1) + (33 3)]15 dimensions of symmetric tensors of the 2nd rank being n(n +1)=2 while of those antisymmetric n(n ; 1)=2. Now we go to product 156 66 which should results as already we have seen in the sum of two IR's of the 3rd rank , one of them being antisymmetric of the 3rdrank (N AAA = n(n2 ; 3n +2)=6) and the other being of mixed symmetry (Nmix = n(n2 ; 1)=3): 156 66 =206 +706 = (63 1) + (33 3)] (3 2) = =(6 3 2) + (3 3 3 2) = = (8 2)+(1 4)]20 + (8 4) + (10 2) + (8 2) + (1 2)]70: Instead the product 216 66 should result as already we have seen into the sum of two IR's of the 3rd rank , one of them being symmetric of the 3rd 6
6

49


rank (N SSS = n(n2 +3n +2)=6) and the other being again of mixed symmetry (Nmix = n(n2 ; 1)=3) and we used previous result to extract the reduction of the 566 -plet:

216 66 =566 +706 = f(63 3) + (33 1)g (3 2) =

=(63 33 3 2) + (33 3 1 2) = = f((8 2) + (10 4))g56 + (8 2) + (1 4)]20+ +(8 4) + (10 2) + (8 2) + (1 2)]70: We now see eminent result of SU (6) that is that in one IR 566 there are octet of baryons of spin 1/2 and decuplet of baryonic resonances of spin 3/2! Note that quark model with al l the masses, magnetons etc equal just reproduces SU (6) model as it should be. In some sense 3-quark model gives the possibility of calculations alternative to tensor calculus of SU (6) group. Some words also on reduction of IR of some group to IR's of the sum of minor groups. We have seen an example of reduction of the IR of SU (4) into those of SU (3). For future purposes let us consider some examples of the reduction of the IR's of SU (5) into those of the direct sum SU (3)+SU (2). Here we just write 5-spinor of SU (5) as a direct sum: 55 =(33 1) + (13 2). Forming the product of two 5-spinors we get: 5
5

55 =155 +105 =(33 1) + (13 2) (33 1) + (13 2) = 33 1) + (33 2) + (33 2) + (13 2 2) = 2) + (1 3)g15 + (33 2) + (33 1) + (1 1)]10 SU (5) group IR's of dimensions 5 and 10 have the the sum SU (3)+SU (2): 55 =(33 1) + (13 2)

=(33 f(6 1) + (33 that is, important for fol lowing reduction to

105 =(33 2) + (33 1) + (1 1): In this case it is rather easy an exercise to proceed alsowithtensor calculus.

50