Lie Theory in Particle Physics

Tim Roethlisberger

May 15, 2018

Abstract

In this report we look at the representation theory of the Lie algebra of SU(3). We constructthe general finite dimensional irreducible representation of the complexified Lie algebra.Using tensor product representations we find the structure behind the baryons and mesonsobserved in experiments in the early 1960s, which leads us to the idea of the quark model.

1 Introduction

In the 1950s and 1960s the increase in the energies of particle accelerators led to the discoveryof many unknown particles. There were so many different particles this became known asthe particle zoo. The big question was whether there was an underlying structure to thisparticle zoo. To answer this question people tried to order the particles depending on theirproperties.

For example the spin 3/2 particles could be ordered by plotting their hypercharge Yagainst their isospin I3. The hypercharge of a particle is the sum of its strangeness S andbaryon number B, so Y = S + B. The strangeness S, baryon number B and isospin I3 areall quantum numbers that can be measured by experiment.

In the early 1960s nine spin 3/2 particles were known. In figure 1 we have plotted theirhypercharge against their isospin. This plot already exhibits some structure, as did othersuch plots for particles with different spin. There was however no known particle withI3 = 0, Y = −2, which would complete the triangle in figure 1. In this report we will discusshow to find a structure behind this pattern. For this we will consider representations ofthe Lie algebra of SU(3) which will lead us to the quark model. This model describes theexperimental results and led people to predict the missing particle with I3 = 0, Y = −2.

All figures are taken from Jan B. Gutowski’s lecture notes ”Symmetry and ParticlePhysics” [1]

2 SU(3) Lie algebra representations

We consider the Lie Group SU(3) = U ∈ Mat3×3|UU † = 1, det(U) = 1 of the specialunitary 3 × 3 matrices. Its Lie algebra, which is the tangent space at the identity, is then

1

Figure 1: The spin 3/2 baryons known in the early 1960s. The “missing” particle at I3 =0, Y = −2 had not been discovered yet.

the set of all antihermitian and traceless 3 × 3 matrices denoted by L (SU(3)) = su(3) =A ∈Mat3×3|A† = −A, tr(A) = 0.

We now allow for complex linear combinations of the elements of su(3), that is thecomplexification of su(3). Since we allow complex linear combinations the new elements willnot be necessarily antihermitian, but they will however still be traceless, which is exactlythe defining property of the Lie algebra of SL(3) = U ∈ Mat3×3|det(U) = 1. We choosea basis of this complexified Lie algebra as follows:

h1 =

12

0 00 1

20

0 0 0

h2 =

1

2√3

0 0

0 12√3

0

0 0 1√3

e1+ =

0 1√2

0

0 0 00 0 0

e2+ =

0 0 00 0 1√

2

0 0 0

e3+ =

0 0 1√2

0 0 00 0 0

(1)

e1− =

0 0 01√2

0 0

0 0 0

e2− =

0 0 00 0 00 1√

20

e3− =

0 0 00 0 01√2

0 0

It is important to note that the elements h1 and h2 are both diagonal and they commute.

The maximal commuting and diagonalisable subalgebra of a Lie algebra is called the Cartansubalgebra. In our case of sl(3) the Cartan subalgebra is spanned by h1 and h2. This Cartansubalgebra is important because if d is a representation of the Lie algebra then it follows thatd(h1) = H1 and d(h2) = H2 commute and thus they are simultaneously diagonalisable. This

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allows us to find a common eigenbasis of the vector space V on which the representation actsand in turn we can label the basis states of the vector space V with the eigenvalues withrespect to H1 and H2. If v is such a basis state and H1v = λ1v and H2v = λ2v then we call(λ1, λ2) the weight of the state. We are only interested in finite-dimensional representations.The first representation we look at is the adjoint representation, which will play an importantrole for all the other representations.

2.1 The adjoint representation

The adjoint representation of an element x in the Lie algebra g acts on the Lie algebra itselfand is defined as:

ad(x) :

g→ g

y 7→ [x, y]

The representation property is satisfied as a consequence of the Jacobi identity of the com-mutator. So for su(3) we have an eight dimensional representation. The common eigenstatesof H1 and H2 are just the basis states of the complexified Lie algebra we defined earlier. Wecan calculate their weights by evaluating the commutator. Doing so we find:

State v [h1, v] [h2, v] Weight (λ1, λ2)h1 0 0 (0, 0)h2 0 0 (0, 0)e1+ e1+ 0 (1, 0)e1− −e1− 0 (-1,0)

e2+ −12e2+

√32e2+ (−1

2,√32

)

e2−12e2− −

√32e2− (1

2,−√32

)

e3+12e3+

√32e3+ (1

2,√32

)

e3− −12e3− −

√32e3− (−1

2,−√32

)

We can plot these states on the so called weight-plane as shown in figure 2. The non-zeroweights of the adjoint representation are called roots of the Lie algebra and will play animportant role for general representations. We will call the adjoint representation the 8representation.

2.2 The su(2) subalgebras

In preparation for considering the general irreducible finite dimensional representation, letus consider the following commutation relations:

[h1, e1±] = ±e1± [e11, e

1−] = h1

[

√3

2h2 −

1

2h1, e

2±] = ±e2± [e2+, e

2−] =

√3

2h2 −

1

2h1

[

√3

2h2 +

1

2h1, e

3±] = ±e3± [e3+, e

3−] =

√3

2h2 +

1

2h1

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Figure 2: The eigenstates of the adjoint representation plotted on the weight-plane. Thestates h1 and h2 lie at the origin since they commute with each other and themselves.

These are exactly the commutation relations of the complexified su(2) Lie algebra. The

generators em± play the role of raising and lowering operators and h1,√32h2− 1

2h1 and

√32h2 +

12h1 are three linearly dependent matrices that play the role of the Cartan generator ofsu(2). The eigenstates of the su(3) representation are thus also eigenstates of the threesu(2) subalgebras. This puts constraints on the possible weights of the representation sincewe know that the eigenvalues of a finite-dimensional su(2) representation are half-integers.Thus for a weight (λ1, λ2) we find

2λ1 ∈ Z,√

3λ2 − λ1 ∈ Z,√

3λ2 + λ1 ∈ Z.

Adding the latter two equations we find 2√

3λ2 ∈ Z. Using the fact that in our normalizationthe eigenvalues of an su(2) representation are spaced by 1 all the weights of an irreduciblesu(3) representation are constrained to lie on a lattice of equilateral triangles with length 1as depicted in figure 2.

Now let v be an eigenstate of H1 and H1 with weight (λ1, λ2). We can then look at theeffect of Em

± = d(em± ) acting on v:

(H1

H2

)Em± v =

([H1, E

m± ]

[H2, Em± ]

)v + Em

±

(H1

H2

)v

= (root(em± ))tEm± v +

(λ1λ2

)Em± v (2)

So either Em± v is zero, that is Em

± annihilates v, or it is again an eigenstate of H1 and H2

with its weight shifted by the root of em± . This can be depicted in the weight-plane as shownin figure 3.

The blue lines in figure 3 represent all the points on the weight plane which have zeroeigenvalue with respect to one of the three su(2) subalgebras. Since representations of su(2)are symmetric about the zero eigenvalue the su(3) representations have to be symmetric

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Figure 3: Effect of the operators Em± acting on the state in the center of the arrows

about the three blue axes. Therefore it is enough to know the structure of the representationin one of the sectors confined by the blue lines.

2.3 The general irreducible finite-dimensional representation

We define the highest weight state of an irreducible finite-dimensional representation as thestate w that is an eigenstate of H1 and H2 and is annihilated by all the raising operators Em

+ .Such a state has to exist, since otherwise one could construct infinitely many new states byrepeated application of the raising operators. The highest weight state has to be unique aswell, since otherwise one can find two subspaces V1 and V2 of V , by applying the differentraising and lowering operators, on each of which the representation would be irreducible andthus on V the representation would not be irreducible.

Since the highest weight state is a highest weight state of all three su(2) subalgebras weknow it has non-negative eigenvalues with respect to all three of the subalgebras. Geometri-cally this means the highest weight has to lie in the red region of figure 4 or on its boundary.We will consider three different cases, (i) the highest weight state lies on the vertical edge of

Figure 4: The highest weight state of an irreducible finite-dimensional su(3) representationhas to lie in the red region of the weight-plane or on its boundary.

the red region, (ii) it lies on the diagonal edge of the red region or (iii) it lies inbetween.

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If the highest weight state lies on the vertical axis it is in the j = 0 representation withrespect to the first su(2) subalgebra, thus E1

−v = 0 and there cannot be another state toits left. We can repeatedly act on the highest weight state with one of the other loweringoperators (namely m = 2, 3) until we reach the state that we get by mirroring the highestweight state in the diagonal blue lines. In this manner we can find all the states on two edgesof a triangle (see figure 5), which all have multiplicity one. These states are all highest weight

Figure 5: A 15-dimensional representation with the highest weight state on the vertical axis.All states have multiplicity one.

states of representations of one (or two) of the su(2) subalgebras. We can then proceed inthe same way as before and construct the remaining states in the triangle. These states alsohave multiplicity one which can be shown using the commutation relations following fromthe matrices in equation (1), and the fact that E1

−v = 0.The case with the highest weight state lying on the diagonal blue line is very similar.

This time the highest weight state is the j = 0 representation with respect to the secondsu(2) subalgebra. Again we can find the states on two edges of the triangle by repeatedlyapplying E1

− and E3− until we reach the mirrored states. The remaining states can be reached

by applying E1− and E3

− in combination. Again all the states have multiplicity 1. This caseis depicted in figure 6.

We now consider the case with the highest weight state lying inbetween the blue lines,that is in the red region of figure 4. This time we have Em

−w 6= 0, so none of the loweringoperators annihilate the highest weight state w. This will result in higher multiplicities forthe states in the inner layers.

For the outer layers we can proceed with the same analysis as before. The outer layerof the resulting hexagon (as in figure 7) can be produced by applying the different loweringoperators and using the symmetry of the su(2) representations. Thus all the states of

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Figure 6: A 21-dimensional representation with the highest weight state on the diagonalaxis. All states have multiplicity one.

the outer layer have multiplicity one. If we proceed to the next layer of the hexagon themultiplicity of the states increases by one. This happens for every layer until we reach thefirst triangular layer. Then all the following triangular layers will have the same multiplicityas the first triangular layer. This is illustrated in figure 7. The outer layer is colored in blackand has multiplicity one. The next layer, colored in red, has multiplicity two. This continuesuntil the first triangular layer, colored in bright blue, is reached. This layer has multiplicity5. The next triangular layer then also has multiplicity 5.

Once again we will not give the full proof for the multiplicities of states. The ideahowever is again to use the commutation relations between the different raising and loweringoperators and derive upper and lower bounds for the multiplicities. Interested readers canfind the proof in the lecture notes “Symmetry and Particle Physics” by Jan B. Gutowski onpages 74 to 76.

To find the dimension of the general representation we can now just count up the numberof states. We will call the length of the upper edge of the hexagon m and the length of theother edge n, as depicted in figure 7. Without loss of generality we assume m > n. Thelength of the edge of the first triangular layer will thus be (m − n + 1). So there are1 + 2 + · · ·+ (m− n+ 1) = 1

2(m− n+ 1)(m− n+ 2) states in the interior triangle. Each of

these states has multiplicity (n+ 1).There are n hexagonal layers labelled from k = 1 to k = n. Then in the k-th layer there

are 3((m + 1 − (k − 1)) + (n + 1 − (k − 1) − 2)) = 3(m + n − 2k + 2) states, each of them

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Figure 7: A 315-dimensional representation representing the general case. The differentcolors stand for different multiplicities of states. The length m of the upper edge of thehexagon would be 8 in this case. The length n of the other edge would be 4 in this case.

with multiplicity k. Summing up we find:

1

2(n+ 1)(m− n+ 1)(m− n+ 2) +

n∑k=1

3k(m+ n− 2k + 2) =1

2(m+ 1)(n+ 1)(m+ n+ 2)

Taking m = 0 or n = 0, this formula works for the triangular representations as well. Thelowest dimensional irreducible representations have dimension d = 1, 3, 6, 8, 10 and so on.Here the 8-dimensional representation is the adjoint representation we saw earlier.

2.4 Complex conjugate representation

If we look at a finite dimensional representation d then d(v) is a matrix and we can defined(v) = d(v)∗ as the complex conjugate of d(v). Since d is a representation it satisfies

d([v, w]) = d(v)d(w)− d(w)d(v).

If we take the complex conjugate of this equation we get

d([v, w]) = d(v)d(w)− d(w)d(v)

which shows that d is again a representation. For our case of su(3) we can look at ih1:

d(ih1) = (d(ih1))∗ = (iH1)

∗ = −iH∗1 .

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Using the linearity of d we can also write this as

d(ih1) = id(h1) = iH1.

This means H1 = −H∗1 and in the same way we find H2 = −H∗2 . This shows that if v isan eigenstate of H1 and H2 with weight (λ1, λ2) then v∗ is an eigenstate of H1 and H2 withweight (−λ∗1,−λ∗2). Since the weights of our representations are real this means that theweight diagram of the complex conjugate representation is just the weight diagram of theoriginal representation inverted in the origin.

2.5 The 1, the 3 and the 3

The lowest dimensional irreducible representation of su(3) is the trivial representation de-noted by 1. It is called the singlet and is one-dimensional; m = n = 0 in figure 7. For allelements v ∈ su(3) d(v) = 0 and hence there is just one eigenstate with weight (0, 0).

The next representation we look at is the 3. It is the fundamental representation withd(v) = v and it is 3-dimensional; m = 1 and n = 0 in figure 7. We can choose a basis ofeigenstates:

state weight

u =

100

(12, 12√3)

d =

010

(−12, 12√3)

s =

001

(0,− 1√3)

already anticipating the quark model we will see later. The weight diagram of the 3 is shownin figure 8. The highest weight state is u.

The next representation we consider is the complex conjugate of the 3 which is the 3.Since the states in the 3 were real we can just take the same states and, as we saw in theprevious section, the weights are just inverted:

state weight

u =

100

(−12,− 1

2√3)

d =

010

(12,− 1

2√3)

s =

001

(0, 1√3)

We can again plot the weight diagram which is just the inverted weight diagram of the 3 asshown in figure 9. This time s is the highest weight state.

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Figure 8: The weight diagram of the 3 representation. The states are labeled as in the quarkmodel.

2.6 Tensor product representations

We will now consider tensor product representations, which will in general not be irreducible.Consider the irreducible representations d1 and d2 on V1 and V2 respectively. Then

d = d1 ⊗ 1 + 1⊗ d2 is the corresponding representation on the tensor product V = V1 ⊗ V2.We saw above that we can choose an eigenbasis of V1 and an eigenbasis of V2. The tensorproduct of these basis states is then a basis of V . Let v1 ∈ V1 and v2 ∈ V2 be states withweights (p1, q1) and (p2, q2) respectively. Then

d(h1)v1 ⊗ v2 = (d1(h1)⊗ 1 + 1⊗ d2(h1))v1 ⊗ v2= p1v1 ⊗ v2 + p2v1 ⊗ v2= (p1 + p2)v1 ⊗ v2.

We can do the same with d(h2) to find

d(h2)v1 ⊗ v2 = (q1 + q2)v1 ⊗ v2.

So the weight of a state in the tensor product is the sum of the individual weights, thatis v1 ⊗ v2 has weight (p1 + p2, q1 + q2). Starting with a highest weight state, which is notnecessarily unique, we can find the associated set of states on which the representation isirreducible using the analysis of section 2.3. We can then choose a new highest weight statefrom the remaining states and do the same again. We end up with a decomposition of thetensor product V into invariant subspaces V = W1 ⊕W2 ⊕ · · · ⊕Wn, where each Wi is anirreducible representation. The same can be done for a triple tensor product V = V1⊗V2⊗V3where the weights are just the sum of the three individual weights. We now consider thetensor product representations that will bring us to the quark model.

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Figure 9: The weight diagram of the 3 representation. The states are labeled as in the quarkmodel.

2.6.1 The 3⊗ 3 representation

We will use the basis for the 3 and 3 introduced in section 2.5. With the analysis above wefind the weights of the 3⊗ 3 to be:

state weight

u⊗ s (12,√32

)

u⊗ d (1, 0)

d⊗ s (−12,√32

)

u⊗ u, d⊗ d, s⊗ s (0, 0)d⊗ u (−1, 0)

s⊗ u (−12,−√32

)

s⊗ d (12,−√32

)

In figure 10 the weight diagram is plotted. The shape of the 8-dimensional adjoint repre-sentation is already clearly visible. However the weight (0, 0) has multiplicity three and nottwo. If we act on the highest weight state u ⊗ s with all the lowering operators we find an8 and we are left with a single state with weight (0, 0), a singlet. Explicitly this state is

1√3

(u⊗ u+ d⊗ d+ s⊗ s).

This means 3⊗ 3 is reducible and decomposes as 3⊗ 3 = 8⊕ 1.

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Figure 10: The weight diagram of the 9-dimensional reducible 3 ⊗ 3 representation. Thestates are labeled as in the quark model.

2.6.2 The 3⊗ 3⊗ 3 representation

In the 3⊗ 3⊗ 3 representation we have 33 = 27 states. As discussed above, the weights ofthese states are the sum of the individual weights of their constituents:

state weight

u⊗ u⊗ u (32,√32

)

s⊗ s⊗ s (0,−√

3)

d⊗ d⊗ d (−32,√32

)u⊗ u⊗ s, u⊗ s⊗ u, s⊗ u⊗ u (1, 0)

u⊗ u⊗ d, u⊗ d⊗ u, d⊗ u⊗ u (12,√32

)

s⊗ s⊗ u, s⊗ u⊗ s, u⊗ s⊗ s (12,−√32

)

s⊗ s⊗ d, s⊗ d⊗ s, d⊗ s⊗ s (−12,−√32

)d⊗ d⊗ s, d⊗ s⊗ d, s⊗ d⊗ d (−1, 0)

d⊗ d⊗ u, d⊗ u⊗ d, u⊗ d⊗ d (−12,√32

)u⊗ d⊗ s, u⊗ s⊗ d, d⊗ u⊗ s, d⊗ s⊗ u, s⊗ u⊗ d, s⊗ d⊗ u (0, 0)

The highest weight state of the 3 ⊗ 3 ⊗ 3 representation is u ⊗ u ⊗ u and lies on thediagonal blue line as seen in figure 11. We saw in section 2.3 that acting on this state withthe different lowering operators gives us an irreducible 10-dimensional representation. Wewill call it the 10. The remaining states are plotted in figure 12.

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Figure 11: The weight diagram of the 27-dimensional reducible 3 ⊗ 3 ⊗ 3 representation.The states are labeled as in the quark model.

Figure 12: The weight diagram of the 3⊗3⊗3 representation after removing the 10. Thereare two linearly independent highest weight states.

There are now two linearly independent highest weight states. We can choose one ofthem and act on it with the lowering operators. This gives us an irreducible 8 representation.Removing this invariant subspace as well we can take the other highest weight state and dothe same thing. Again we get an 8. Finally we are left with a single state with weight (0, 0),the singlet. This state is

1√6

(s⊗ d⊗ u− s⊗ u⊗ d+ d⊗ u⊗ s− d⊗ s⊗ u+ u⊗ s⊗ d− u⊗ d⊗ s).

Indeed, it is the only state that is annihilated by all the raising and lowering operators. We

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have now found that the 3⊗3⊗3 representation is reducible and decomposes to 3⊗3⊗3 =10⊕ 8⊕ 8⊕ 1.

3 The quark model

Having constructed representations of su(3) and studied their tensor products, let us nowuse this to understand the structure behind the particle zoo.

The spin 0 and the spin 1 particles can be plotted with their hypercharge Y on the y-axisand the isospin component I3 on the x-axis as in figures 13 and 14. There are two moreparticles with hypercharge 0 and isospin 0, one with spin 0 (denoted η′) and one with spin1 (denoted ϕ). The spin 0 and spin 1 particles each can be identified with the basis states

of the 8 and the 1 in the 3⊗ 3 decomposition if we set (λ1, λ2) = (I3,√32Y ).

Figure 13: The spin 0 particles plotted with respect to their hypercharge and isospin com-ponent. These particles are known as pseudoscalar mesons.

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Figure 14: The spin 1 particles plotted with respect to their hypercharge and isospin com-ponent. These particles are known as vector mesons.

We can do the same thing with the spin 12

and spin 32

particles as depicted in figures 15and 16.

Figure 15: The spin 32

particles plotted with respect to their hypercharge and isospin compo-nent. These particles are known as the baryon decuplet (including the missing Ω− particle).

15

Figure 16: The spin 12

particles plotted with respect to their hypercharge and isospin com-ponent. These particles are known as the baryon octet.

The spin 32

particles match the basis states of the 10 and the spin 12

particles match an

8 of the 3⊗ 3⊗ 3 decomposition, again making the identification (λ1, λ2) = (I3,√32Y ).

This motivates the idea that the baryons and mesons consist of smaller particles, calledthe quarks. Here we have encountered three different quarks, the up, the down and thestrange quarks which are the basis states in the 3 representation and have spin 1

2. The

anti-quarks are the basis states in the complex conjugate representation 3. Baryons areparticles that consist of three quarks, so mathematically they are multiplets in the 3⊗3⊗3representation. Mesons consist of one quark and one antiquark, mathematically they aremultiplets in the 3⊗ 3 representation.

Were this SU(3)flavour-symmetry exact, it would imply that all the particles in the samemultiplet have the same mass. Experimentally this is not the case and hence the SU(3)flavour-symmetry is only approximate. While the up (mu = 2.2+0.6

−0.4 MeV) and down (md = 4.7+0.5−0.4

MeV) quarks have similar masses, the mass of the strange quark (ms = 96+8−4 MeV) is

considerably higher [2]. Therefore the SU(2)-symmetry of the up and down quarks is moreaccurate. This is called the SU(2) isospin symmetry, of which I3 is the third component.

With the emergence of the quark model there were new questions to answer. Why dotwo quark states 3 ⊗ 3 not exist? Why does the ∆++ particle exist? In the quark modelthe ∆++ paricle should consist of three up quarks with aligned spin (since ∆++ is a spin 3

2

particle). That is all the quarks should be in the same state, which is not allowed by thePauli exclusion principle. To circumvent this problem a new “color” label for the quarks wasintroduced. There are three different color charges, red, blue and green. Each quark carriesone of the three colors, while the anti-quark carries an anti-color. Now the three quarks inthe ∆++ particle can each carry a different color and the problem is solved. It is observedexperimentally that color charged particles do not exist. It appears that particles have tobe color-neutral. An isolated quark and also a two-quark particle cannot be color-neutralthus they should not exist. This gives rise to a new symmetry, the SU(3)color-symmetry,which is an exact symmetry. The requirement of color-neutrality implies that particles

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should transform in the singlet representation of this SU(3)color-symmetry. This SU(3)color-symmetry is the gauge-symmetry of quantum chromodynamics.

4 The discovery of the missing particle

Murray Gell-Mann and Yuval Ne’eman were the first to organize the baryons and mesonsin this way in 1961. This led to the prediction of the missing particle in figure 15 withisospin 0, hypercharge -2 and spin 3

2. The mass differences between the different isospin

multiplets in the baryon decuplet also allowed them to predict the mass of the missingparticle to be around 1680 MeV. In 1964 such a particle was discovered with almost exactlythe properties predicted. This particle is the Ω− particle and has mass 1672.45±0.29 MeV [2].In 1969 Murray Gell-Mann received the Nobel price in Physics for this prediction and theclassification of elementary particles.

There are three more quark flavours known today, the charm, top and bottom quark.They are however much heavier than the three “light” quarks and thus the SU(3)flavour-symmetry cannot be extended to an SU(6)flavour-symmetry in a meaningful way.

References

[1] Jan B. Gutowski. Symmetry and Particle Physicshttp://personal.maths.surrey.ac.uk/st/jg0033/Resources/lectnotes(master).pdf

[2] C. Patrignani et al. (Particle Data Group), Chin. Phys. C, 40, 100001 (2016) and 2017update

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