Application of Group Theory in Particle Physics using the Young Tableaux Method 2006 PASSHE-MA CONFERENCE (March 31 – April 1) Akhtar Mahmood ( Assistant Professor of Physics) Jack Dougherty ( Undergraduate Research Assistant) Edinboro University of Pennsylvania

Maury Gell-Mann ( at CalTech ) ( at CalTech ) Proposed the existence of Quarks as the fundamental building blocks of matter in the late 1960s. Awarded Nobel Prize in Physics in 1969 for the development of the Quark model, and the classification of elementary particles. FlavorQ/eu+2/3 d-1/3 s-1/3

Three Families of Quarks Generations IIIIII Charge = -1/3 d (down) s (strange) b (bottom) Charge = +2/3 u (up) c (charm) t (top) Also, each quark has a corresponding antiquark. The antiquarks have opposite charge to the quarks Increasing mass

Hadrons- the composites of Quarks Baryons are a composites of three quarks Baryons are a composites of three quarks Mesons are a composites of a quark-antiquark pair Mesons are a composites of a quark-antiquark pair

Let’s make some more baryons ! s u d Lambda (  ) Q = 0 M=1116 MeV/c 2 s u u Sigma (   ) Q = +1 M=1189 MeV/c 2 s u d Sigma (   ) Q = 0 M=1192 MeV/c 2 s d d Sigma (   ) Q = -1 M=1197 MeV/c 2 uds Q Mass +2/3-1/3 Quarkupdownstrange uuddss ~5 [MeV/c 2 ]~10 [MeV/c 2 ]~200 [MeV/c 2 ]

The Quark Configuration of the Charmed-Strange Baryon The Quark Configuration of the Charmed-Strange Baryon us c ds c

Summary of Young Tableaux Method – The SU(N) Notation Here N = 2 (for Spin up or down) N = 1 to 5 (for quark flavors – up, down, strange, charm, beauty)

Young Tableaux (Young Diagrams) In the SU(N) Notation, N is denoted by a box.

(N) - Total Number of Spins or Quark Flavors

The conjugate representation N for Anti-quarks. Denoted by a column of N-1 boxes. If N = 4, then the conjugate representation of N is with 3 boxes.

RULES FOR SU(N) REPRESENTATION

No row is longer than any above it. Descending rows are always shorter than the ones above them. No row is longer than any above it. Descending rows are always shorter than the ones above them. Allowed Not Allowed

Box values always increase from left to right in a row Box values always increase from left to right in a row

NOT PERMITTED NOT PERMITTED

Going from left to right, no box- columns can be longer than the previous one. Allowed Not Allowed Allowed

Box values always decrease from top to bottom in a column 3 2 1

NOT PERMITTED NOT PERMITTED 3 4 5

PERMITTED

Fully Symmetric Configuration S =S =S =S =

Mixed Symmetric Configuration M =M =M =M =

Fully Anti-symmetric Configuration A =A =A =A =

WHAT IS THE NET VALUE OF A VALID CONFIGURATION ? A RATIO OF….. Product of Box Values (n) V = Product of Possible “Hook” Values (h)

A numerator (n) is defined as the product of the actual value each consecutive box. n = 3  4  5 = FOR N = 3

A denominator is defined as the product of all of the possible consecutive “hooks” Hooks Hooks h = 3  2  1 = 6

FOR N = 3, IN THIS PARTICULAR CONFIGURATION FOR N = 3, IN THIS PARTICULAR CONFIGURATION 3  4  5 3  4  5 V = = 10 V = = 10 3  2  1 3  2  1

LET’S CALCULATE THE VALUE OF THIS CONFIGURATION FOR ANY “N” ! LET’S CALCULATE THE VALUE OF THIS CONFIGURATION FOR ANY “N” ! NN+1 NN-1

“n” is defined as the product of the actual value of each consecutive box. n =  ( each box values ) (product) n = ( N )( N + 1 )( N – 1 )( N )

FOR N = 3 FOR N = n = 3  4  2  3 = 72

“h” is defined as the product of the possible “hook” values of each consecutive box- configuration n = (“hook” value of each box-configuration) (product)

h = 3  2  2  1 = 12

FOR N = 3, IN THIS PARTICULAR CONFIGURATION FOR N = 3, IN THIS PARTICULAR CONFIGURATION 3  4  2  3 3  4  2  3 V = = 6 V = = 6 3  2  2  1 3  2  2  1

Another Example (For N=3) n = 3  4  2 = 24 h = 3  1  1 = 3 V = 24  3 = 8 M 34 2

|Baryon(Qq 1 q 2 )  A =|Space  S  |Color  A  Spin  S,MS,MA  Flavor  S,MS,MA COMPLETE BARYON WAVE FUNCTION q2q2 l´ l´ l q1q1 q3q3 SIZE = 1 fm

Space is described by the Parity (P) of the Baryon which is defined as the state of the particle P ≡ (-1 )(l + l´) In the ground state l = 0 and l´=0  P = (-1) 0 = + P ≡ (-1 )(l + l´) In the ground state l = 0 and l´=0  P = (-1) 0 = +

(+)  Ground state (l = 0 and l’ = 0) (+)  Ground state (l = 0 and l’ = 0) (-) Excited state (l = 1 and (-) Excited state (l = 1 and l’ = 0 or (l = 0 and l’ = 1) l’ = 0 or (l = 0 and l’ = 1)

Color The strong force that binds the quarks with gluons carry “color charge” i.e. red, green and blue. Since a baryon is a Fermion, it must obey Pauli’s Exclusion Principle, and hence the color charge combination must be anti-symmetric for all baryons.

COLOR: From the SU(3) c Color Symmetry Group Only the color combination is valid - Color Singlet (Colorless) - Color Singlet (Colorless)

The 3 quark color are Red, Green and Blue 3  3  3  35   34 2   4  5 3  2  1   3  4  2 3  1  1 3  4  2 3  1  1 3  2  1 10 S 8 MS 8 MA 1A1A 

1 A  Color Singlet  Colorless R + G + B = White (Colorless) 16RGB-GRB+BRG-RBG+GBR-BGR A

Total Angular Momentum ( J P ) J  L + S where L = l +l’ and P  (-1) (l +l´) Each quark has a Spin (S) of ½ and a J P of ½ +. WHAT ARE POSSIBLE J P VALUES OF A BARYON ?

q2q2 l´ l´ l q1q1 q3q3 In Ground State l = 0 and l’ =0

q 1 (½) +  q 2 (½) + {q 1 q 2 } (1) +  q 3 (½) + {q 1 q 2 q 3 } (3/2) + If q 1  q 2  q 3 (3 Distinct J P Values) {q 1 q 2 }q 3 (1/2) +/ [q 1 q 2 ]q 3 (1/2) + q 1 or q 2 = u, d, s q 3 = c, b (Symmetric) (Mixed-Symmetric) (Mixed-Antisymmetric) [q 1 q 2 ] (0) +  q 3 (½) +

The Quark Configuration of the Charmed-Strange Baryon in the Ground State The Quark Configuration of the Charmed-Strange Baryon in the Ground State us c ds c

q 1 (½) +  q 2 (½) + {q 1 q 2 } (1) +  q 3 (½) + {q 1 q 2 q 3 } (3/2) + If q 1 = q 2  q 3 (2 Distinct J P Values) {q 1 q 2 }q 3 (1/2) +/ q 1 or q 2 = u, d, s q 3 = c, b (Symmetric) (Mixed-Symmetric)

q 1 (½) +  q 2 (½) + {q 1 q 2 } (1) +  q 3 (½) + {q 1 q 2 q 3 } (3/2) + If q 1 = q 2 = q 3 (1 Distinct J P Value) q 1 or q 2 = u, d, s q 3 = c, b (Symmetric)

SPIN: From the SU(2) S Spin Symmetry Group Three distinct spin states for each of the Three distinct J p values. J p (3/2) +  4 S J p (1/2) +/  2 MS J p (1/2) +  2 MA

SPIN CONFIGURATION USING YOUNG’S TABLEAUX 2  2  2  24   23 1   3  4 3  2  1   2  3  1 3  1  1 2  1  3 3  1  1 2  1  0 3  2  1 4S4S 2 MS 2 MA 0A0A 

4 separate Spin orientations (S, S z ) for S = 3/2 SzSz +3/2 +1/2 -1/2 -3/2 (S, S z ) or (J, J z ) for S z = 4 S and J = (3/2) + (3/2, +3/2) (3/2, +1/2) (3/2, -1/2) (3/2, -3/2)

2 separate Spin orientations (S, S z ) for S = 1/2 SzSz +1/2 -1/2 (S, S z ) or (J, J z ) for S z = 2 MS OR S z = 2 MA and J = (1/2) +/ OR J =(1/2) + (1/2, +1/2) (1/2, -1/2) (1/2, +1/2) (1/2, -1/2) (S, S z ) or (J, J z ) for S z = 2 MS OR S z = 2 MA and J = (1/2) +/ OR J =(1/2) + (1/2, +1/2) (1/2, -1/2) (1/2, +1/2) (1/2, -1/2)

|Spin S  Flavor S |Spin S  Flavor S | Spin MS  Flavor MS | Spin MS  Flavor MS | Spin MA  Flavor MA | Spin MA  Flavor MA

J P = (3/2) + |Spin S  |Flavor S  |4 S  |(3/2) +  S | S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,+3/2) |1/√3( +  + ) S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,+1/2) |1/√3( +  + ) S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,-1/2) | S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,-3/2) J P = (3/2) + |Spin S  |Flavor S  |4 S  |(3/2) +  S | S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,+3/2) |1/√3( +  + ) S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,+1/2) |1/√3( +  + ) S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,-1/2) | S  |{q 3 1/√2 {q 1 q 2 + q 2 q 1 }} S (3/2,-3/2)

J P = (1/2) +/ |Spin MS  |Flavor MS  |2 MS  |(1/2) +/  MS |1/√6(- -  + 2 ) MS  |q 3 1/√2 {q 1 q 2 + q 2 q 1 } MS (1/2,+1/2) |1/√6(+ +  - 2 ) MS  |q 3 1/√2 {q 1 q 2 + q 2 q 1 } MS (1/2,- 1/2) J P = (1/2) +/ |Spin MS  |Flavor MS  |2 MS  |(1/2) +/  MS |1/√6(- -  + 2 ) MS  |q 3 1/√2 {q 1 q 2 + q 2 q 1 } MS (1/2,+1/2) |1/√6(+ +  - 2 ) MS  |q 3 1/√2 {q 1 q 2 + q 2 q 1 } MS (1/2,- 1/2)

J P = (1/2) + |Spin MA  |Flavor MA  |2 MA  |(1/2) +/  MA |1/√2 ( - ) MA  |q 3 1/√2 {q 1 q 2 - q 2 q 1 } MA (1/2,+1/2) |1/√2 ( - ) MA  |q 3 1/√2 {q 1 q 2 - q 2 q 1 } MA (1/2,-1/2) J P = (1/2) + |Spin MA  |Flavor MA  |2 MA  |(1/2) +/  MA |1/√2 ( - ) MA  |q 3 1/√2 {q 1 q 2 - q 2 q 1 } MA (1/2,+1/2) |1/√2 ( - ) MA  |q 3 1/√2 {q 1 q 2 - q 2 q 1 } MA (1/2,-1/2)

Quark Mass and Charge FLAVOR MASS (GeV) ELECTRIC CHARGE FLAVOR MASS (GeV) ELECTRIC CHARGE UP /3 UP /3 DOWN /3 DOWN /3 STRANGE /3 STRANGE /3 CHARM /3 CHARM /3 BOTTOM /3 BOTTOM /3 TOP /3 TOP /3

QUANTUM PROPERTIES OF BARYONS Baryon #: B = 1 (Each Quark has a Baryon # of 1/3) Isospin: I 3 = Q -½(B+s+c+b+t) Hypercharge: Y = 2(Q - I 3 ) - (c  b  t)

Ordinary matter – SU(2) F Symmetry Group - only up and down quarks (N F =2) 2  2  2  24   23 1   3  4 3  2  1   2  3  1 3  1  1 2  1  3 3  1  1 2  1  0 3  2  1 4S4S 2 MS 2 MA 0A0A 

For N F = 2 (u and d Quarks), We can have 4 Baryons with J P (3/2) + and 2 Baryons with J P (1/2) +/

u (½) +  u (½) + {uu} (1) +  d (½) +  + (3/2) + If q 1 = q 2  q 3 (2 Distinct J P Values) p (1/2) +/ q 1 or q 2 = u, u q 3 = d {uud}(Symmetric) u{ud}(Mixed-Symmetric)

d (½) +  d (½) + {dd} (1) +  u (½) +  0 (3/2) + If q 1 = q 2  q 3 (2 Distinct J P Values) n (1/2) +/ q 1 or q 2 = d, d q 3 = d {ddu}(Symmetric) u{dd}(Mixed-Symmetric)

Light Baryons – SU(3) F Symmetry Group - up, down, and strange quarks (N F = 3) 3  3  3  35   4  5 3  2  1   3  4  2 3  1  1 10 S 8M8M 

N F = 3, in the SU(3) Symmetry Group (u, d and s Quarks) We can have 10 Baryons with J P (3/2) + and 8 Baryons with J P (1/2) + and (1/2) +/ TOTAL # OF BARYONS THAT CAN BE CONSTRUCTED WITH u, d AND s QUARKS = 18

u (½) +  d (½) + {ud} (1) +  s (½) + If q 1  q 2  q 3 (3 Distinct J P Values)  0 (1/2) + q 1 or q 2 = u, d q 3 = s [ud] (0) +  s (½) + [sud] (Mixed-AntiSymmetric) s{ud} (Mixed-Symmetric)  0 (1/2) +/ {sud} (Symmetric) * 0 (3/2) +

J P (1/2) + 8 Baryons Y Axis – Y (Hypercharge) X Axis – I 3 or I z (z – Component of Isospin)

J P (3/2) + 10 Baryons Y Axis – Y (Hypercharge) X Axis – I 3 or I z (z – Component of Isospin)

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J P (1/2) + 8 Baryons - 7 with J P (1/2) +/ and 1 with J P (1/2) + Y Axis – Y (Hypercharge) X Axis – I 3 or I z (z – Component of Isospin)

N F = 3, in the SU(3) Symmetry Group (u, d and s Quarks) From SU(3) we can have a total of 18 Baryons – 10 with J P (3/2) + and 7 with J P (1/2) +/ and 1 with J P (1/2) +

J P (3/2) + 10 Baryons with J P (3/2) + Y Axis – Y (Hypercharge) X Axis – I 3 or I z (z – Component of Isospin)

Charmed Baryons – SU(4) F Symmetry Group - up, down, strange and charm quarks (N F = 4) 4  4  4  46   5  6 3  2  1   4  5  3 3  1  1 20 / S 20 M 

N F = 4, in the SU(4) Symmetry Group (u, d and s Quarks) We can have 20 Baryons with J P (3/2) + and 20 Baryons with J P (1/2) + and (1/2) +/ TOTAL # OF BARYONS THAT CAN BE CONSTRUCTED WITH u, d, s AND c QUARKS = 40

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BUT HOW MANY BARYONS WAS CARRIED OVER FROM THE SU(3) TO THE SU(4) SYMMETRY GROUP? RECALL: From SU(3) we can have a total of 18 Baryons – 10 with J P (3/2) + and 8 with J P (1/2) + [7 with J P (1/2) +/ and 1 with J P (1/2) + ]. IN SU(4): A Total of 40 Baryons - 20 with J P (3/2) +, and 20 with J P (1/2) +/ and 20 J P (1/2) +. NEW – Actually, 10 Charmed Baryons with J P (3/2) + and 12 Charmed Baryons with J P (1/2) +. BUT How many Charmed Baryons with J P (1/2) +/ and with J P (1/2) + ??

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How many Actual Charmed Baryons with J P (1/2) +/ and with J P (1/2) + ? 9 with J P (1/2) +/ and 3 with J P (1/2) + TOTAL = 12

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The Quark Configuration of the Charmed-Strange Baryon in the Ground State The Quark Configuration of the Charmed-Strange Baryon in the Ground State us c ds c

u (½) +  s (½) + {us} (1) +  c (½) + If q 1  q 2  q 3 (3 Distinct J P Values) (1/2) + q 1 or q 2 = u, s q 3 = c [us] (0) +  c (½) + [csu] (Mixed-AntiSymmetric) c{su} (Mixed-Symmetric) (1/2) +/ {csu} (Symmetric) (3/2) +

J P (1/2) + [csu] (Mixed- AntiSymmetric) Mass:  1.3 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

J P (1/2) +/ c{su} (Mixed-Symmetric) Mass:  2.0 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

J P (3/2) + {csu} (Symmetric) Mass:  2.1 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

d (½) +  s (½) + {ds} (1) +  c (½) + If q 1  q 2  q 3 (3 Distinct J P Values) (1/2) + q 1 or q 2 = d, s q 3 = c [ds] (0) +  c (½) + [csd] (Mixed-AntiSymmetric) c{sd} (Mixed-Symmetric) (1/2) +/ {csd} (Symmetric) (3/2) +

J P (1/2) + [csd] (Mixed- AntiSymmetric) Mass:  1.2 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

J P (1/2) +/ c{sd} (Mixed-Symmetric) Mass:  2.1 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

J P (3/2) + {csd} (Symmetric) Mass:  1.7 MeV/c 2 CLEO Experiment at CESR (Cornell Electron Storage Ring)

Beauty Baryons – SU(5) F Symmetry Group - up, down, strange, charm, and beauty quarks (N F = 5) 5  5  5  57   6  7 3  2  1   5  6  4 3  1  1 35 S 40 M 

BUT HOW MANY BARYONS WAS CARRIED OVER FROM THE SU(3) TO SU(4) TO THE SU(5) SYMMETRY GROUP? RECALL: From SU(3) we can have a total of 18 Baryons – 10 with J P (3/2) + and 8 with J P (1/2) + [Actually, 7 with J P (1/2) +/ and 1 with J P (1/2) + ]. In SU(4): Total of 40 Baryons - 20 with J P (3/2) +, and 20 with J P (1/2) +/ and 20 J P (1/2) +. But, 10 Charmed Baryons with J P (3/2) + and 12 Actual Charmed Baryons with J P (1/2) + [Actually, 9 with J P (1/2) +/ and 3 with J P (1/2) + ]. In SU(5): Total of 75 Baryons - 35 with J P (3/2)+, and 40 with J P (1/2) +.

NEW: How many actual Beauty Baryons - 15 with J P (3/2) + and 20 with J P (1/2) + ? TOTAL # OF BEAUTY BARYONS = 35 How many Beauty Baryons with J P (1/2) +/ and with J P (1/2) + ??

How many actual Beauty Baryons with J P (1/2) +/ and with J P (1/2) + ? 14 with J P (1/2) +/ and 6 with J P (1/2) + TOTAL = 20

Beauty Baryons – SU(5) F Symmetry Group - up, down, strange, charm, and beauty quarks (N F = 5) AXIS PROBLEM !! Need a 4 th Axis ??

No Possible SU(5) Representations can be added to the Existing Scheme. We can’t physically add a 4th Beauty axis to the SU(4) representation diagram!

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AS OF TODAY NO SU(5) F QUARK REPRESENTATION EXITS !!

My Solution ! My Solution !

CONSIDER A NEW TYPE OF AXIS INSTEAD OF THE TRADITIONAL “CHARM” AXIS IN THE Z-DIRECTION. BUT HOW ?? FLAVOR AXIS !! (A NEW QUANTUM NUMBER !) F = 1 FOR EACH HEAVY QUARK (CHARM AND BEAUTY) F = 0 FOR EACH LIGHT QUARK (UP, DOWN AND STRANGE)

HOW DOES THIS FLAVOR AXIS ACTUALLY WORK – THE FLAVOR QUANTUM # F = 3 (ccc) or (bbb) or (bcc) or (bbc) etc… F = 2 (ccu) or (ccs) or (ccd) or (bbu) or (bbs) or (bbd) or (bcu) or (bcs) etc… F = 1 (cuu) or (css) or (cdd) or (csu) or (bbu) or (bss) or (bsu) or (bdd) etc… F = 0 (uuu) or (sss) or (ddd) or (sud) or (uud) or (ddu) or (suu) or (ssu) etc…

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SU(3) Light Baryons (u, d and s) I3I3I3I3 Y 8 Baryons

SU(3) Light Baryons (u, d and s) I3I3I3I3 Y 10 Baryons

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SU(4) Charmed Baryons

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Beauty Baryons – SU(5) F Symmetry Group - up, down, strange, charm, and beauty quarks (N F = 5) 5  5  5  57   6  7 3  2  1   5  6  4 3  1  1 35 S 40 M 

SU(5) Beauty Baryons

Finally A Solution That Actually Works! Paper in progress for publication in PRL(Physical Review Letters)