48. SU(n) Multiplets and Young Diagrams1

48. SU(n) Multiplets and Young Diagrams
Revised August 2019 by C.G. Wohl (LBNL).
This note tells (1) how SU(n) particle multiplets are identified or labeled, (2) how to find thenumber of particles in a multiplet from its label, (3) how to draw the Young diagram for a multiplet,and (4) how to use Young diagrams to determine the overall multiplet structure of a composite
system, such as a 3-quark or a meson-baryon system.
In much of the literature, the word “representation” is used where we use “multiplet,” and“tableau” is used where we use “diagram.”48.1 Multiplet labels
An SU(n) multiplet is uniquely identified by a string of (n−1) nonnegative integers: (α, β, γ, . . .).
Any such set of integers specifies a multiplet. For an SU(2) multiplet such as an isospin multiplet,the single integer α is the number of steps from one end of the multiplet to the other (i.e., it is one
fewer than the number of particles in the multiplet). In SU(3), the two integers α and β are thenumbers of steps across the top and bottom levels of the multiplet diagram. Thus the labels for
the SU(3) octet and decupletare (1,1) and (3,0). For larger n, the interpretation of the integers in terms of the geometry of themultiplets, which exist in an (n−1)-dimensional space, is not so readily apparent.
The label for the SU(n) singlet is (0, 0, . . . , 0). In a flavor SU(n), the n quarks together form a(1, 0, . . . , 0) multiplet, and the n antiquarks belong to a (0, . . . , 0, 1) multiplet. These two multiplets
are conjugate to one another, which means their labels are related by (α, β, . . .) ↔ (. . . , β, α).

48.2 Number of particles
The number of particles in a multiplet, N = N (α, β, . . .), is given as follows (note the patternof the equations). In SU(2), N = N (α) is(α + 1).
1(48.1)(α + 1) (β + 1) (α + β + 2)··.
1
12(48.2)(α+1) (β+1) (γ+1) (α+β+2) (β+γ+2) (α+β+γ+3)·····.
111223(48.3)

N=In SU(3), N = N (α, β) is
N=In SU(4), N = N (α, β, γ) is
N=S. Navas et al. (Particle Data Group), Phys. Rev. D 110, 030001 (2024)
31st May, 2024 10:21am

47. SU(n) multiplets and Young diagrams
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48. SU(n) Multiplets and Young Diagrams

47. SU(n) Multiplets and Young Diagrams

Note that in Eq. (48.3) there
is no factorwith(α + γ + 2): only a consecutive sequence of the label
47.3.
Young
diagrams
integers appears in any factor. One more example should make the pattern clear for any SU(n).
n) particle multiplets
are identified
In SU(5),N = N (α,orβ, γ, δ) isA Young diagram consists of an array of boxes (or some othermber of particles in a multiplet from itssymbol) arranged in one or more left-justified rows, with each row
g diagram for a multiplet, and (4) how (α+1)(δ+1)(α+β+2)
being(β+1)
at least (γ+1)
as long as
the rowbeneath. (β+γ+2)
The correspondence between
N=·····
mine the overall multiplet structureof 1a diagramandamultiplet
label
is:Thetoprow111
2
2 juts out α boxes to-quark or a meson-baryon system.
therightpast
theendofthesecondrow,thesecond row juts out β(γ+δ+2) (α+β+γ+3) (β+γ+δ+3) (α+β+γ+δ+4)
×···. A diagram (48.4)
boxes
to the 3right past the 3end of the third4 row, etc.
ine word “representation” is used where
2SU(n) has at most n rows. There can be any number of “completed”
u” is used where we use “diagram.”
From the symmetry of theseequations,is clearthat multipletsare conjugate
one affectanothercolumnsof n itboxes
buttressing
the left thatof a diagram;thesetodon’t
have the same number of particles,
butsocanothermultiplets.
For
example,theSU(4)multipletsthe label. Thus in SU(3) the diagrams(3,0,0) and (1,1,0) each have 20 particles. Try the equations and see.
,,,,uely identified bya stringof (n−1)
48.3
Young
diagrams. .). Any such set ofAintegers
specifies consists of an array of boxes (or some other symbol) arranged in one or more
Young diagram
iplet such as anleft-justified
isospin multiplet,the eachrows, withrow being
least as (1,0),long as(0,1),
the rowbeneath.
The(3,0).
correspondence
represent
the atmultiplets(0,0),(1,1), andIn anyof steps from one
end of athemultiplet
betweendiagramand a multiplet
label
is:
The
toprowjuts
out
αboxestotheright
pastthethe endSU(n), the quark multiplet is represented by a single box,
r than the numberof
particlesintheof the second row, the second
row juts
out β boxes
the right
past theend andof thethird row,
antiquarkmultipletby a tocolumnof (n−1)
boxes,
a singletby aetc.
ntegers α and βA are
the numbers
diagramin SU(n)ofhas atcompleted
most n rows.
There
be any number of “completed” columns of ncolumnof ncanboxes.
m levels of the multiplet
diagram. Thus
boxes buttressing
the left of a diagram; these don’t affect the label. Thus in SU(3) the diagramsnd decuplet
47.4., Coupling
multiplets
together,,,
The following recipe tells how to find the multiplets that occur
in coupling two multiplets together. To couple together more thanrepresent the multiplets (1,0),(0,1), (0,0),first(1,1),
and two,(3,0).then
In anySU(n),thewithquark
multiplet
two multiplets,
couplecouplea thirdeachof the isrepresented by a single box,
theantiquark
multipletbyacolumnof(n−1)
boxes,
andasingletbymultiplets obtained from the first two, etc.
a completed column of n boxes.
First a definition: A sequence of the letters a, b, c, . . . is admissible
ifatpoint in the sequence at least as many a’s have occurred as
48.4 Coupling multiplets any
togetherb’s,
atleastmany
have occurred
as c’s, inetc.couplingThus abcd
aabcb
The following recipe tells how to asfindthe b’smultipletsthat occur
twoandmultiplets
are more
admissible
sequences
and abb
acb areNow
the recipe:
together. To couple together
than two
multiplets,
firstandcoupletwo,not.
thencouple
a third with(a)from
Drawthethefirst
Young
for the two multiplets, but in one ofeach of the multiplets obtained
two,diagramsetc.
the diagrams
replacea,thein the firstifrow
withpointa’s, inthetheboxesinn, the interpretationof atheintegers A sequenceFirstdefinition:of the lettersb, c,boxes. . . is admissible
at anysequencethesecondrow
withb’s,etc.
Thus,
tocouple
twoSU(3)octets(such
the multiplets, atwhichexist
inanleast as many a’s have occurred as b’s, at least as many b’s have occurred as c’s, etc. Thus abcd
so readily apparent.
as the π-meson
the not.
baryonoctet),
we start withandand aabcb are admissible sequences
and abboctetand andacb areNow
the recipe:
Young diagramsfor
thetwomultiplets,butinoneof
thediagrams
replacethelet is (0, 0, . . . , 0). (a)In aDraw
flavorthe
SU(n),
a a.
The
unlettereddiagramforms
theupperleft-handcornerofall(1, 0, . . . , 0) multiplet,
andfirst
the rown with a’s,boxes in theb the boxes in the second row with b’s, etc. Thus, to couple two SU(3)enlargeddiagrams
below.
0, 1) multiplet. octets
These (suchtwo multipletsas the π-mesontheoctet
and thebaryonconstructedoctet), we start
withand ba a . The unlettered(b) Add
the a’sfromlettereddiagramconstructedto the right-handhich means theirlabels are
relatedby left-handdiagramforms
the upper
cornerof allthetheenlargeddiagrams
below. endsof the rows of the unlettered diagram to form all possible legitimate(b) Add the a’s from the lettered diagram to the right-hand ends of the rows of the unlettered
Young diagrams that have no more than one a per column. In general,
diagram to form all possible legitimate Young diagrams that have no more than one a per column.
there will be several distinct diagrams, and all the a’s appear in eachles
In general, there will be several distinct diagrams, and all the a’s appear in each diagram. At thisdiagram. At this stage, for the coupling of the two SU(3) octets, wetwo SU(3) octets, we have:multiplet, N =stage,
N (α,forβ, .the. .), couplingis given of thehave:he equations).
a a,a,a,.
aaa(α + 1).
1aUse theto furtherenlarge
the diagrams
already(c) Use the b’sto further(c)enlargetheb’sdiagrams
alreadyobtained,
subject
to theobtained,
same rules.
(47.1)
subject
tothesamerules.
Thenthrowaway
any
diagramin whichtheleft
Then throw away any diagram in which the full sequence of letters formed by reading
right tofullsequenceoflettersformedbyreadingrighttoleftinthefirstrow,in the first row, then the second row, etc., is not admissible.
then the second row, etc., is not admissible.
(d) Proceed as in (c) with the c’s (if any), etc.
(d) Proceed as in (c) with the c’s (if any), etc.
β + 1) (α + β + 2)·.
(47.2)
1
2
31st ofMay,
The final result
the2024coupling of the two SU(3) octets is:=⊗a ab=

+ β + 2).
2a(47.1)

3(47.2)a(c) Use the b’s to further enlarge the diagrams already obtained,
subject to the same rules. Then throw away any diagram in which thefull sequence of letters formed by reading right to left in the first row,
48. SU(n) Multiplets and Young Diagrams
then the second row, etc., is not admissible.
(d) Proceed as in (c) with the c’s (if any), etc.

The final result
theresultcouplingof coupling
the two ofSU(3)octetsis: octets is:
Theoffinalof thethe twoSU(3)(β+γ+2) (α+β+γ+3)·.
23(47.3)⊗a a ⊕ba a
b=a a ⊕a ⊕a ba ⊕aa ⊕baa b.

bbawith (α + γ + 2): only a
Here only the diagrams with admissible sequences of a’s and b’s andappears in any factor. Onewith
fewer thanrows (sincen = 3) have
In terms
Hereonlythediagrams
withfouradmissible
sequencesof a’sbeen
andkept.
b’s andwith offewer than four rows
ar for any SU(n). In SU(5),be written(since n = 3) multiplet
have beenlabels,kept. theIn above
terms mayof multipletlabels, the above may be written(1,2)2)⊕⊕(3,(3,0) 0)⊕ (0,3) ⊕⊕1)(0,⊕0)(0,. 0) .
α+β+2) (β+γ+2)(1,1)1)⊗⊗(1,(1,1)1)==(2,(2,⊕ (0,3)(1,⊕ 1)(1,⊕1)(1,⊕1)(1,·22
In termsofofparticles,
numbers ofparticles,
it may be written+3) (α+β+γ+δ+4)
In terms of numbers
it maybe written·(47.4).
48 8⊗⊗
88==2727⊕ 10⊕ 10⊕ 8⊕⊕88⊕⊕81 ⊕. 1.
⊕ 10⊕ 10s clear that multiplets that
Theof thenumberson theleft hereis equal
ona useful check. (Seee number of particles,but
The productoftheproductnumbers
on theleft hereis equal
to thesum toon thethe sumright,
theright,
ausefulcheck.
(SeealsoSec.
15ontheQuarkModel.)U(4) multiplets (3,0,0)
andalso Sec. 15 on the Quark Model.)ations and see.

31st May, 2024