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Lectures on symmetries and particle physics
Stefan Floerchinger
Institut fur Theoretische Physik, Philosophenweg 16, D-69120 Heidelberg, Germany
E-mail: [email protected]
Abstract: Notes for lectures that introduce students of physics to symmetries and particle physics.
Prepared for a course at Heidelberg university in the summer term 2017.
Contents
1 Introduction and motivation 3
2 Finite groups 8
3 Lie groups and Lie algebras 20
4 SU(2) 22
5 SO(N) and SU(N) 34
6 SU(3) 39
7 Lorentz and Pioncare group 52
8 Conformal group 61
9 Non-Abelian gauge theories 63
10 Grand unification 65
– 1 –
Suggested literature
The application of group theory in physics is a well established mathematical subject and there are
many good books available. A selection of references that will be particularly useful for this course
is as follows.
• P. Ramond, Group Theory, A Physicist’s Survey
• A. Zee, Group Theory in a Nutshell for Physicists
• J. Fuchs and C. Schweigert, Symmetries, Lie algebras and Representations
• H. Georgi, Lie algebras in Particle Physics
• H. F. Jones, Groups, Representations and Physics
– 2 –
1 Introduction and motivation
1.1 Symmetry transformations
Studying different kinds of symmetries and their consequences is one of the most fruitful ideas in
all branches of physics. This holds especially in high energy and particle physics but not only there.
To make this work, we first define the notion of a symmetry transformation and relate it to the
mathematical concept of a group.
It is natural to demand for symmetry transformations that
• A symmetry transformation followed by another should be a symmetry transformation itself.
• Symmetry transformations should be associative.
• There should be a unique trivial symmetry transformation doing nothing.
• For each symmetry transformation there needs to be a unique symmetry transformation re-
versing it.
With these properties, the set of all symmetry transformations G forms a group in the mathematical
sense. More formally, a group G has the properties:
1) Closure: For all elements f, g ∈ G the composition f · g ∈ G .
2) Associativity: (f · g) · h = f · (g · h) .
3) Identity element: There exists a unique e ∈ G such that e · f = f = f · e for all g ∈ G .
4) Inverse element: For all elements f ∈ G there is a unique f−1 ∈ G such that f · f−1 =
f−1 · f = e .
1.2 Infinitesimal symmetries
In physics the group elements g ∈ G which describe a symmetry can often be parametrised by a
continuous parameter on which the group elements depend in a differentiable way, e.g.
R→ G
α 7→ g(α) .
In this situation it is possible to study infinitesimal symmetries which are characterised by their
action close to the identity element e of the group. Without loss of generality we can choose g(0) = e
. Moreover, if the parametrisation obeys
g(α1) · g(α2) = g(α1 + α2)
for any two group elements g(α1) and g(α2) one speaks of a one-parameter subgroup. Let us now
restrict to very small parameters α1, α2 such that the corresponding group elements are close to
the identity element. In this case one speaks of an infinitesimal symmetry which is described as a
local one-parameter subgroup, and characterised by the derivative
d
dαg(α)
∣∣∣α=0
.
– 3 –
1.3 Classical mechanics
Lagrangian description. In classical mechanics, the equations of motions of a physical system
with action
S =
∫dt L(q(t), q(t), t) ,
where the Lagrangian L is a function of position q and velocity q, can be derived by the principle
of least action
δS =
∫dt
∂L
∂qδq +
∂L
∂qδq
=
∫dt
∂L
∂q− d
dt
∂L
∂q
δq = 0 .
This gives the Euler-Lagrange equation
∂L
∂q− d
dt
∂L
∂q= 0 .
Suppose there is a mapping
hs : q 7→ hs(q) , (1.1)
for s ∈ (−ε, ε) ⊂ R such that
hs=0(q) = q ,
for any position q, and an induced map
hs : q 7→ hs(q) =∂hs(q)
∂qq , (1.2)
for the corresponding velocity q. The Lagrangian is then said to be invariant under (1.1) and (1.2)
if there is a differentiable function Fs(q, q, t) such that
L(hs(q), hs(q), t) = L(q, q, t) +d
dtFs(q, q, t) . (1.3)
The invariance (1.3) gives rise to a conservation law. For any solution to the equations of motion
φ : t 7→ q = φ(t)
we define
Φ(s, t) = hs φ(t)
and find from (1.3)
0 =∂
∂s
(L(Φ, ∂tΦ)− d
dtFs
)=∂L
∂q
∂Φ
∂s+∂L
∂q
∂2Φ
∂s∂t− d
dt
∂Fs∂s
=d
dt
(∂L
∂q
∂Φ
∂s− ∂Fs
∂s︸ ︷︷ ︸=Q
)
where Q is a conserved quantity called Noether charge.
Consider for example as a point particle of mass m with the Lagrangian
L =1
2mx2 − V (x) ,
– 4 –
where x ∈ R3 it the position. Suppose the potential V (x) is translational invariant such that L is
invariant under
hs : R3 → R3
x 7→ x + sa
for a ∈ R3 and s ∈ (−ε, ε) ⊂ R while hs is the identity map and Fs = 0. Then
∂Φ
∂s=∂hs∂s
= a ,
and therefore
Q = mx · a .
We have found momentum conservation in direction a. The close connection between symmetries
and conservation law is truly remarkable.
Hamiltonian description. In the Hamiltonian formulation of classical mechanics the physical
information of a system in encoded in the Hamiltonian H which is a function of position q, momenta
p and time t. The equations of motion are
dp
dt= −∂H
∂q,
dq
dt=∂H
∂p.
For systems which can also be described in the Lagrangian framework, the Hamiltonian is given by
the Legendre transformation
H(q, p, t) = pq − L(q, q, t) ,
where q is defined implicitly by
p =∂L
∂q.
The fundamental tool in Hamiltonian mechanics is the Poisson bracket. It is a map from the space
of pairs of differentiable functions of dynamical variables q and p to a single differential function.
For such functions f, g we define
·, · :(f(q, p), g(q, p)
)7→ f, g(q, p) ,
where
f, g =∂f
∂p
∂g
∂q− ∂f
∂q
∂g
∂p.
The Poisson bracket possesses three properties:
1) Bilinear: λf + µg, h = λf, h+ µg, h ,
2) Antisymmetric: f, g = −g, f ,
3) Jacobi identity: f, g, h+ g, h, f+ h, f, g = 0 ,
for differentiable functions f, g, h and for λ, µ ∈ R. The bilinearity holds in both arguments. By
these properties the Poisson bracket turns the space of functions of q and p into a Lie algebra.
If H does not explicitly depend on time, and for any differentiable function f(q(t), p(t)) which
is on a trajectory that is a solution to the equations of motion, the time derivative can be written
d
dtf(q(t), p(t)) = H, f .
For a conserved quantity we therefore get
H, f = 0 .
– 5 –
Quantum mechanics. Starting from the Hamiltonian description it is most convenient to work
in the Heisenberg picture. The canonical quantisation maps the Poisson bracket of differentiable
functions in q and p to the commutator of the associated operators q and p in some suitable Hilbert
space H,
·, · 7→ i
~[·, ·] ,
where i is the imaginary unit, ~ denotes Planck’s constant and the commutator is defined by
[A,B] = A ·B −B ·A .
In particular one has
p, q = 1 7→ i
~[p, q] = 1 ,
which is the Heisenberg commutation relation. The commutator equips H with a Lie algebra in
the same way the Poisson bracket does in the Lagrangian description. In particular we have the
properties
1) Bilinear: [λA+ µB,C] = λ[A,C] + µ[B,C] ,
2) Antisymmetric: [A,B] = −[B,A] ,
3) Jacobi identity: [A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0 ,
for operators A,B,C and λ, µ ∈ R. Analogous to before the time dependance of the operators is
described byd
dtA(t) =
i
~[H,A] .
Observables which describe conservations laws are characterised by
[H,A] = 0 .
By the Jacobi identity, the space of all conserved quantities forms a closed Lie subalgebra.
As an example we consider angular momentum conservation of the electron in the quantised hydro-
gen atom. Let the eigenstates of the Hamiltonian H be labelled by the principal quantum number
n, total angular momentum l and the projection on the z-axis, m. Since angular momentum is
conserved the operator commutes with the Hamiltonian,
[Lz, H] = 0
and therefore for two eigenstates |n′, l′,m′〉, |n, l,m〉,
0 = 〈n′, l′,m′|[Lz, H]|n, l,m〉 = (m′ −m) 〈n′, l′,m′|H|n, l,m〉 .
Accordingly,
〈n′, l′,m′|H|n, l,m〉
can only be non-zero for m 6= m′, which is the selection rule.
One may also start from a conserved operator and construct the corresponding symmetry transfor-
mation. For a self-adjoint operator A ∈ H with
[H,A] = 0 ,
– 6 –
we define the unitary operator
UA(t) = eitA =
∞∑n=0
(itA)n
n!,
for t ∈ R. They obey the multiplication law
UA(t1) · UA(t2) = UA(t1 + t2) ,
for all t1, t2 ∈ R and form a unitary one-parameter group. As an example consider the momentum
operator in position space defined as
p =~i
d
dx,
which generates infinitesimal translations. For a ∈ R the operator
exp(ipa
~
)= ea
ddx
generates finite translations, e.g.
eaddx f(x) = f(x+ a) ,
for a suitable function f .
– 7 –
2 Finite groups
Now that we discussed the close relation between symmetries and the mathematical concept of a
group it seems natural to start our analysis with simple cases, i.e. finite groups of low order. The
order ord(G) of a group G is defined to be the number of group elements.
2.1 Order 2
The simplest symmetry transformation is a reflection,
P : x→ −x (2.1)
which applied twice is the identity, i.e. PP = e. The parity transformation (2.1) has the simplest
finite group structure involving only two elements. The group is known as the cyclic group Z2 with
the multiplication table
Z2 e P
e e P
P P e
The group Z2 has many manifestations for example in terms of reflections about the symmetry axis
of an isosceles triangle.
Studying higher order group will naturally involve more elements which can manifest in more
symmetries, e.g. the equilateral triangle is symmetric under reflections about any of its three
medians as well as rotations of 2π3 around its center,
which is described by a group of order six.
In general we denote the elements of a group of order n by e, a1, a2, ..., an−1 where e denotes
the identity element.
2.2 Order 3
There is only one group of order three which is Z3 with the multiplication table
Z3 e a1 a2
e e · e = e e · a1 = a1 e · a2 = a2
a1 a1 · e = a1 a1 · a1 = a2 a1 · a2 = e
a2 a2 · e = a2 a2 · a1 = e a2 · a2 = a1
written compactly for clarity
– 8 –
Z3 e a1 a2
e e a1 a2
a1 a1 a2 e
a2 a2 e a1
The group elements can be written as e, a1 = a, a2 = a2, with the law that a3 = e. One says that
a is an order three element. The structure of the group is completely fixed by the multiplication
table. However, this get rather cumbersome, in particular for larger groups and it is convenient
to use another characterisation of the group elements and their multiplication laws, the so-called
presentation. For example Z3 has a presentation
〈a | a3 = e〉 .
The first part of the presentation lists the group elements from which all others can be constructed –
the so-called generators – while the second part gives the rules needed to construct the multiplication
table.
This generalises to the cyclic group of order n ∈ N which has the group elements e, a, a2, ..., an−1and a presentation is given by
〈a | an = e〉 . (2.2)
One distinguishes between a group G which is an abstract entity defined by the set of its group
elements and their multiplication table (or, equivalently, a presentation) and a representation of a
group.
A representation can be seen as a manifestation of the group multiplication laws in a concrete
system or, in other words, a kind of incarnation of the group.
For example, the group Z3 has a representation in terms of rotations by 2π3 around the center of
an equilateral triangle. Zn has a one dimensional representation
1, ei 2πn , e2i 2πn , ..., e(n−1)i 2πn (2.3)
where the group elements are represented equidistantly on the unit circle in the complex plane and
the action of the generator a is represented by a rotation of 2πn around the centre.
.
....
.
.
.. . .
.
C
Often one works with representations as operations in a vector space:
A matrix representation R of a group G on a vector space V is a group homomorphism
R : G→ GL(V )
onto the general linear group GL(V ) on V , i.e. a map from a group element g to a matrix R(g)
such that
R(g1 · g2) = R(g1) · R(g2)
for all g1, g2 ∈ G. We define the dimension of the representation R by
dim(R) = dim(V )
and denote the identity element of a group by e and of a representation by 1.
– 9 –
2.3 Order 4
At order four there are two different groups. Again there is Z4 which in the representations (2.3)
reads 1, i,−1,−i. The other group is the dihedral group D2 with the multiplication table
D2 e a1 a2 a3
e e a1 a2 a3
a1 a1 e a3 a2
a2 a2 a3 e a1
a3 a3 a2 a1 e
The fact that the multiplication table is symmetric shows that D2 is Abelian, ai ·aj = aj ·ai . Two
representations of D2 are: (1 0
0 1
),
(1 0
0 −1
),
(−1 0
0 1
),
(−1 0
0 −1
),
f1(x) = x, f2(x) = −x, f3(x) =
1
x, f4(x) = − 1
x
.
For both groups of order four elements form a subgroup, namely Z2. It is straight forward to see
from the presentation (2.2) for Z4,
〈a | a4 = e〉 ,
that the group generated by a2 is Z2. The same goes for the group generated by ai ∈ D2 for
i ∈ 1, 2, 3. For D2 one can actually go on and write it as a direct product of two factors Z2,
D2 = Z2 ⊗ Z2 .
We use here the notion of a direct product.
Let (G, ) and (K, ∗) be two groups with elements ga ∈ G, a ∈ 1, ..., nG and ki ∈ K, i ∈1, ..., nK respectively. The direct product is a group (G ⊗ K, ?) of order nGnK with elements
(ga, ki) and the multiplication rule
(ga, ki) ? (gb, kj) = (ga gb, ki ∗ kj) .
Since and ∗ act on different sets, they can be taken to be commuting subgroups and we simply
write gaki = kiga for the elements of G⊗K.
As long as it is clear we denote for notational simplicity the group (G, ) by G and the group
operation as g1 g2 = g1g2.
Exercise: Show that D2 = Z2 ⊗ Z2 but Z4 6= Z2 ⊗ Z2.
2.4 Lagrange’s theorem
If a group G of order N has a subgroup H of order n, then N is necessarily an integer multiple of n.
Consider a group G = ga | a ∈ 1, ..., N with subgroup H = hi | i ∈ 1, ..., n ⊂ G. Take a
group element ga ∈ G but not in the subgroup, ga /∈ H. Then it follows that also gahi is not in H,
gahi /∈ H, because if there would exist hj ∈ H such that gahi = hj then ga = hjh−1i ∈ H which
would object our assumption. Continuing in this way we can construct the disjoint cosets
gaH = gahi | i ∈ 1, ..., n
– 10 –
and write G as a (right) coset decomposition
G = H ∪ g1H ∪ ... ∪ gkH ,
with k ∈ N. Therefore the order of G can only be a multiple of its subgroup’s order: N = nk.
As an immediate consequence, groups of prime order cannot have subgroups of smaller order.
Hence for a group of prime order p all elements are of order one or order p and thus G = Zp.
2.5 Order 5
By Lagrange’s theorem there is only Z5.
2.6 Order 6
We now know that there are at least two groups of order six: Z6 and Z2 ⊗ Z3 but the question is
whether they actually differ. The generator a of Z3 is an order-three element, the generator b of
Z2 an order-two element and thus a3 = e = b2 and ab = ba. Now (ab)6 = a6b6 = e is a order-six
element und therefore
Z6 = Z2 ⊗ Z3 .
To construct another order six group we again take an order-three element a and an order-two
element b. The set
e, a, a2, b, ab, a2b
together with the relation ba = a2b 6= ab form the dihedral group D3 which is non-Abelian. A
presentation is given by
〈a, b | a3 = e, b2 = e, bab−1 = a−1〉 .
D3 is the symmetry group of the equilateral triangle
A
C
B
where a is a rotation by 2π3 (ABC → BCA) and b is a reflection (ABC → BAC). Higher dihedral
groups have higher polygon symmetries:
D4 D5
The group can also be represented by permutations as indicated above. The element a corresponds
to a three-cycle A→ B → C → A (A B C
B C A
)(2.4)
whereas the element b is a two-cycle: A→ B → A(A B
B A
). (2.5)
– 11 –
In general the n! permutations of n objects form the symmetric group Sn. From (2.4) and (2.5) we
see
D3 = S3 .
The group operation for a k-cycle can be represented by k × k matrices, e.g. a three-cycle can be
represented by (3× 3) matrices 0 1 0
0 0 1
1 0 0
ABC
=
BCA
.
2.7 Order 8
From what we know already we can immediately write down four order eight groups which are not
isomorphic to each other since they contain elements of different order: Z8, Z2⊗Z2⊗Z2, Z4⊗Z2.
Additionally there is the dihedral group D4 and a new group Q called the quaternion group.
An element q of the deduced quaternion vector space H generalises the complex numbers,
q = x0 + e1x1 + e2x2 + e3x3
q = x0 − e1x1 − e2x2 − e3x3
with xi ∈ R, i ∈ 1, 2, 3 and the relation
e2i = −1
for the imaginary units ei as well as
e1e2 = −e2e1 = e3
plus cyclic permutations. For q ∈ HN(q) =
defines a norm with
N(qq′) = N(q)N(q′) .
A finite group of order eight can be taken to be the set
1, e1, e2, e3,−1,−e1,−e2,−e3
and is called the quaternion group Q.
Exercise: Convince yourself that Q is a closed group, indeed.
A matrix representation of the quaternion group Q is given by the Pauli matrices
ej = −iσj , σjσk = δjk + iεjklσl ,
with
σ1 =
(0 1
1 0
), σ2 =
(0 −ii 0
), σ3 =
(1 0
0 −1
).
– 12 –
2.8 Permutations
The permutations of n objects can be represented by the symbol(1 2 3 ... n
a1 a2 a3 ... an
)and form the group Sn. Every permutation can uniquely be resolved into cycles, e.g.(
1 2 3 4
2 3 4 1
)∼ (1234)(
1 2 3 4
2 1 3 4
)∼ (12)(3)(4)(
1 2 3 4
3 1 2 4
)∼ (132)(4) .
Any permutation can also be decomposed into a product of two-cycles,
(a1a2a3...an) ∼ (a1a2)(a1a3)...(a1an) .
The symmetric group Sn has a subgroup of even permutations An ⊂ Sn of order n!2 . Notice that
the odd permutations do not form a subgroup since there is no unit element.
2.9 Cayley’s theorem
Every group of finite order n is isomorphic to a subgroup of Sn.
Let G = ga | a ∈ 1, ..., n be a group and associate the group elements with the permuta-
tions
ga 7→ Pa =
(g1 g2 ... gng1ga g2ga ... gnga
).
This construction leaves the multiplication table invariant,
gagb = gc 7→ PaPb = Pc
and Pa | a ∈ 1, ..., n is called the regular representation of G and it is by construction a subgroup
of Sn.
2.10 Concepts
Conjugacy. For a group G = ga | a ∈ 1, ..., n we define the conjugate to ga ∈ G with respect
to the element g ∈ G as
ga = ggag−1 .
Then
gagb = gc 7→ gagb = gc
since
gga g−1g︸ ︷︷ ︸e
gbg−1 = ggagbg
−1 = ggcg−1 .
Note if [g, ga] = 0 then ga and g are self-conjugate with respect to each other. For a permutation
P =(k l m p q
)conjugacy leaves the cycle structure invariant, that is
gPg−1 =(g(k) g(l) g(m) g(p) g(q)
),
since gPg−1g(k) = gP (k) = g(l) .
– 13 –
Classes. For a group G we define the class Ca to consist of all ga that are conjugate to ga,
Ca = ga = ggag−1 | g ∈ G .
Now take another group element gb ∈ G but such that gb /∈ Ca. The set of all conjugates forms the
class Cb and so on,Cb = gb = ggbg
−1 | g ∈ G...
Note that the classes are disjoint, Ci∩Cj = ∅, for i 6= j. In this way we can decompose G in classes
G = C1 ∪ C2 ∪ ... ∪ Ck .
Normal subgroup. Let G be an group and H ⊂ G a subgroup such that for all g ∈ G and
hi ∈ H the conjugates stay in H, that is ghig−1 ∈ H. In this case H is called a normal subgroup.
Quotient group. Let G,K be groups and consider a map:
G→ K
ga 7→ ka
such that gc = gagb 7→ kakb = kc. Set H to be the kernel of that map,
H = ga ∈ G | ga 7→ e ⊂ G .
H then is a normal subgroup because for all hi ∈ H
ghig−1 7→ kek−1 = e .
We now use H to build set of group elements and define a multiplication between these sets. More
specific, let gaH, gbH be two cosets and H the trivial coset. We define the multiplication of cosets,
(gahi)︸ ︷︷ ︸∈gaH
(gbhj)︸ ︷︷ ︸∈gbH
= gagb g−1b higb︸ ︷︷ ︸hi
hj
= gagb hihj︸︷︷︸∈H
∈ gagbH .
This shows that the cosets can be multiplied in the same way as the original group elements ga and
gb.
One then easily verifies that the cosets gaH have a group structure. This group is called the
quotient group G/H. One can show that the quotient group G/H has no normal subgroup if H is
the maximal normal subgroup. The systematic classification of groups in form of a decomposition
into simple groups is based on this.
Simple group. A simple group is a group without (non-trivial) normal subgroups. One can
classify simple groups:
- Zp for p prime
- An, n ≥ 5
- Infinite families of groups of Lie type
- 26 sporadic groups
However, a detailed discussion of this goes beyond the scope of these lectures.
– 14 –
2.11 Representations
Let V be a N dimensional vector space with an orthonormal basis |i〉,N∑i=1
|i〉 〈i| = 1 , 〈i|j〉 = δij ,
where 1 is the identity element in the space of N -dimensional matrices GL(V ). Let G be a order
n group with representation R on V such that the action of g ∈ G is represented by
|i〉 7→ |i(g)〉 = Mij(g) |j〉
where M(g) ∈ GL(V ). Then for ga, gb, gc, g ∈ G and gagb = gc,
M(ga) ·M(gb) = M(gc)
M(g−1)ij = (M(g)−1)ij .
The representation is called trivial if M(g) = 1 for all g ∈ G. The representation R is called
reducible if one can arrange M(g) in the form
M(g) =
(M [1](g) 0
N(g) M [⊥](g)
),
where M [1](g) ∈ GL(V1) acts on the subspace V1 ⊂ V of dimension d1, M [⊥](g) ∈ GL(V ⊥1 ) on its
orthogonal complement V ⊥1 ⊂ V of dimension N − d1 and N(g) is a (N − d1) × d1 matrix. Then
for g, g′ ∈ G
M [1](gg′) = M [1](g) ·M [1](g′)
M [⊥](gg′) = M [⊥](g) ·M [⊥](g′)
as well as
N(gg′) = N(g) ·M [1](g′) +M [⊥](g) ·N(g′) .
One can in fact simplify this further. With
W =1
ord(G)
∑g∈G
M [⊥](g−1) ·N(g)
we diagonalise the representation(1 0
W 1
)·(M [1] 0
N M [⊥]
)·(
1 0
−W 1
)=
(M [1] 0
0 M [⊥]
)by
W ·M [1](g1) =1
ord(G)
∑g∈G
M [⊥](g−1) ·N(g) ·M [1](g1)
= −N(g1) +1
ord(G)
∑g′∈G
M [⊥](g1g′−1) ·N(g′)
= −N(g1) +M [⊥](g1) ·W
with g′ = gg1. In this basis we say R is completely reducible,
R = R1 ⊕R⊥1 .
If R⊥1 is reducible we can further reduce until there are no subspaces left,
R = R1 ⊕R2 ⊕ ...⊕Rk ,
with k ∈ N.
– 15 –
2.12 Schur’s lemmas
From the notation established before we write the action of a g ∈ G in the subspace V1 ⊂ V as
|a〉 7→ |a(g)〉 = M[1]ab (g) |b〉
where |a〉 is a orthonormal basis of V1. Since V1 is a subspace of V we can write
|a〉 = Sai |i〉
with S a d1 ×N matrix. Now we write the group action
|a〉 7→ |a(g)〉 = M[1]ab (g) |b〉 = M
[1]ab (g) Sbi |i〉
or equivalently
|a〉 = Sai |i〉 7→ Sai |i(g)〉 = Sai Mij(g) |j〉
Therefore
R reducible⇒ Sai Mij(g) = M[1]ab (g) Sbj for all g ∈ G . (2.6)
Schur’s first lemma: If matrices of two irreducible representations of different dimension can
be related as in (2.6), then S = 0.
If now d1 = N,S is a N ×N matrix and if |a〉 and |j〉 span the same space the representations Rand R1 are related by a similarity transformation
M [1] = S ·M · S−1 .
Thus for S 6= 0⇒ R is reducible or there is a similarity relation.
Schur’s second lemma: Let R be an irreducible representation. Any matrix S with
M(g) · S = S ·M(g)
for all g ∈ G is proportional to 1.
If |i〉 is an eigenket of S then |i(g)〉 is also an eigenket:
M(g)S |i〉︸︷︷︸λ|i〉︸ ︷︷ ︸
λ|i(g)〉
= SM(g) |i〉︸ ︷︷ ︸|i(g)〉
⇒ S = λ1 .
Let Ra,Rb be two irreducible representations of dimension da and db respectively. Construct
the mapping
S =∑g∈G
M [a](g) ·N ·M [b](g−1) ,
where N is any da × db matrix. Then for g ∈ G
M [a](g) · S = S ·M [b](g) .
– 16 –
For Ra 6= Rb by Schur’s first lemma
1
ord(G)
∑g∈G
M[a]ij (g) M [b]
pq (g−1) = 0 .
On the other hand for Ra = Rb by Schur’s second lemma
1
ord(G)
∑g∈G
M[a]ij (g) M [a]
pq (g−1) =1
daδiqδjp .
Combining these two results leaves us with
1
ord(G)
∑g∈G
M[a]ij (g) M [b]
pq (g−1) =1
daδiqδjpδ
ab .
Let us define the scalar product
(i, j) =1
ord(G)
∑g∈G〈i(g)|j(g)〉 ,
where |j(g)〉 = M(g) |j〉. It is invariant under the group action
(i(ga), j(ga)) =1
ord(G)
∑g∈G〈i(gag)|j(gag)〉
=1
ord(G)
∑g′∈G
〈i(g′)|j(g′)〉
= (i, j)
where g′ = gag runs over all group elements. Therefore for v,u ∈ V
(M(g−1)v,u) = (v,M(g)u) .
On the other hand
(M†(g)v,u) = (v,M(g)u) ,
where M† is the Hermitian conjugate of M and therefore
M†(g) = M(g−1) = M−1(g) .
Therefore all representations of finite groups are unitary with respect to the scalar product (·, ·).
2.13 Crystals
A crystal is defined as a lattice which is invariant under translations,
T = n1u1 + n2u2 + n3u3 , (2.7)
with ui ∈ R3, ni ∈ N, i ∈ 1, 2, 3. Specifying additional symmetries of the crystal one can consider
two symmetry groups:
• Space group: translations, rotations, reflections and possibly inversion.
• Point group: rotations, reflections and possibly inversion, e.g. Zn cyclic group, Dn dihedral
group.
– 17 –
Crystallographic restriction theorem: Consider a crystal invariant under rotations through2πn around an axis. Then n is restricted to n ∈ 1, 2, 3, 4, 6.
z
T1
T2
T3
T4
t21
Consider a translation vector T1. By rotations it gets transformed to T2,T3, ...,Tn . By the
group property of the space group, also the differences tij = Ti − Tj are translation vectors and
by construction they are orthogonal to the rotations axis, which we may take to coincide with the
z-axis. Take now the minimum of the differences,
|t| = mini,j
(|tij |)
and without loss of generality normalise |t| = 1. Take now a point A on the rotational symmetry axis
which by t gets translated to another symmetry point B. Rotation by an angle ϕ around A brings
B to B′. Similar, rotation by ϕ around B brings A to A′. Because A′ and B′ are also symmetry
points the difference A′B′ must be a translation vector and so we can infer that A′B′ = p ∈ N0.
A B
ϕ
|t|
A′B′
• •
••
With
A′B′ = 1 + 2 sin(ϕ− π
2
)= 1− 2 cos(ϕ)
we conclude
cos(ϕ) =1− p
2
and thus
p ∈ 0, 1, 2, 3
⇒ ϕ ∈π
3,π
2,
2π
3, π
⇒ n ∈ 6, 4, 3, 2 .
– 18 –
This closes the prove of the restriction theorem above.
Quasicrystals can show five-fold symmetries but have no periodic structure (2.7), e.g. Penrose
tiling.
– 19 –
3 Lie groups and Lie algebras
3.1 Lie groups
Lets assume that for the group G the group elements g ∈ G depend smoothly on a set of N
continuous parameters αA ∈ R and A ∈ 1, ..., N such that
g(α)∣∣∣α=0
= e ,
is the identity element of the group. Thus for a representation of the group
M(α)∣∣∣α=0
= 1 , (3.1)
where M(α) is the action of g(α) represented on the vector space V and 1 the identity element of
the representation. Expanding (3.1) in a small neighbourhood of the identity element up to first
order yields
M(α) = 1+ iαATA ,
with the so-called generators of the representation of the group
TA = −i ∂
∂αAM(α)
∣∣∣α=0
. (3.2)
By (3.2) the group generators are hermitian for unitary representations. Groups of this type are
called Lie groups. Because of the closure of the group
M(α) = limk→∞
(1+
iαATA
k
)k= eiαAT
A
(3.3)
is well-defined and called the exponential parametrisation of the representation. It can reach all
elements at least in some neighbourhood of the unit element.
3.2 Lie algebras
In the exponential parametrisation there is a one-parameter subgroup of the form
U(λ) = eiλαATA
, λ ∈ R ,
for which for λ1, λ2 ∈ RU(λ1) · U(λ2) = U(λ1 + λ2) .
In general for two different generators
eiαATA
eiβBTB
6= ei(αA+βA)TA ,
for continuous parameters αA, βA ∈ R, A ∈ 1, ..., N, but because the exponential parametrisation
forms a representation of the group close to the identity element, there needs to be δA ∈ R, A ∈1, ..., N such that
eiαATA
eiβBTB
= eiδCTC
.
To see which conditions have to be met we write
iδCTC = ln(1 + eiαAT
A
eiβBTB
− 1︸ ︷︷ ︸K
) ,
– 20 –
and expand K to leading order
K =(1+ iαAT
A − 1
2(αAT
A)2 + ...)(1+ iβBT
B − 1
2(βBT
B)2 + ...)− 1
= iαATA + iβBT
B − αATAβBTB −1
2(αAT
A)2 − 1
2(βBT
B)2 + ... ,
as well as the logarithm
ln(1+K) = K − 1
2K2 + ... .
Then
iδCTC = K − 1
2K2 + ...
= iαATA + iβBT
B − αATAβBTB −1
2(αAT
A)2 − 1
2(βBT
B)2 +1
2(αAT
A + βBTB)2 + ...
= iαATA + iβBT
B − 1
2[αAT
A, βBTB ] + ... ,
and therefore up to second order
[αATA, βBT
B ] = −2i(δC − αC − βC)TC = iγCTC , (3.4)
for continuous parameters γA ∈ R, A ∈ 1, ..., N. Therefore there is a relation
γC = αAβBfAB
C ,
where the fABC ∈ R and A,B,C ∈ 1, ..., N are called structure constants. Thus from (3.4)
[TA, TB ] = ifABCTC , (3.5)
is again a generator and (3.5) is the Lie algebra. From the commutator property in (3.5) we see
that the structure constants are anti-symmetric,
fABC = −fBAC .
Moreover, for unitary representations with TA = (TA)† one has
−i(fABC
)∗TC = [TA, TB ]† = [TB , TA] = ifBACT
C = −ifABCTC
and thus the structure constants are real,
fABC =(fABC
)∗.
The generators also satisfy the Jacobi identity
[TA, [TB , TC ]] + [TB , [TC , TA]] + [TC , [TA, TB ]] = 0 .
– 21 –
4 SU(2)
4.1 Algebras
The Lie algebra of SU(2) is the smallest non-trivial Lie algebra. It plays an important role in
physics not only because it is isomorphic to the Lie algebra of rotations SO(3) but also because
we are interested in the group SU(2). We will study representations of the Lie algebra of SU(2) in
Hilbert spaces since a lot of physics can be described there.
In the simplest non-trivial case the Lie algebra of SU(2) is represented in a two-dimensional
Hilbert space with orthonormal basis |i〉 , i ∈ 1, 2 by the three generators
T+ = |1〉 〈2| , T− = |2〉 〈1| , T 3 =1
2
(|1〉 〈1| − |2〉 〈2|
).
They satisfy the algebra
[T+, T−] = 2T 3 , [T 3, T±] = ±T± , (4.1)
and by defining the hermitian operators
T 1 =1
2
(T− + T+
), T 2 =
i
2
(T− − T+
),
the commutator algebra reads
[TA, TB ] = iεABCTC , (4.2)
for A,B,C ∈ 1, 2, 3. The algebra (4.2) is closed under commutation and satisfies the Jacobi
identity and is therefore a Lie algebra. This explicit construction is an irreducible representation
of the Lie algebra of SU(2) of dimension two, denoted 2. It is called the fundamental or spinor
representation.
To study other representations it is useful to introduce Casimir operators. These are operators
that commute with the Lie algebra and in the case of SU(2) there is only one,
C2 = (T 1)2 + (T 2)2 + (T 3)2 , [C2, TA] = 0 .
Since TA = (TA)† by construction we also have C2 = C†2 .
The states of the Hilbert space can be labelled by the eigenvalues of the maximal number of
commuting operators. That is C2 and by choice T 3 and thus the algebra will be represented on
eigenstates of these two operators,
C2 |c,m〉 = c |c,m〉 , T 3 |c,m〉 = m |c,m〉 ,
with c,m ∈ R since the operators are hermitian. Because the Casimir operator C2 is positive
definite, we know the spectrum of T 3 is bounded and therefore expect the maximal and minimal
values of T 3 to be of the order ±√C2. From the commutation relations (4.1) we infer
T+ |c,m〉 ∝ |c,m+ 1〉 ,
because the states are uniquely labelled by the eigenvalues of C2 and T 3. Then
T 3 T+ |c,m〉 = (m+ 1) T+ |c,m〉
= (m+ 1) d(+)m |c,m+ 1〉
for some d(+)m ∈ R, m ∈ R and
C2 T+ |c,m〉 = c T+ |c,m〉
= c d(+)m |c,m+ 1〉 .
– 22 –
Since we know T 3 is bounded, there needs to be a so-called highest weight state |c, j〉 of the
representation for which j ∈ R is the maximal value of T 3 subject to,
T+ |c, j〉 = 0 , T 3 |c, j〉 = j |c, j〉 .
Similarly we can infer
T− |c,m〉 = d(−)m |c,m− 1〉
from the commutation relations (4.1) for some d(−)m ∈ R, m ∈ R and find a lowest weight state
|c, k〉 , k ∈ R,
T− |c, k〉 = 0 , T 3 |c, k〉 = k |c, k〉 .
Then the Casimir operator is determined by the highest weight state,
C2 |c, j〉 =
((T 3)2 +
1
2
(T+T− + T−T+
))|c, j〉
=(
(T 3)2 +1
2
[T+, T−
])|c, j〉
=(
(T 3)2 + T 3)|c, j〉
= (j2 + j) |c, j〉 ,
and therefore c = j(j + 1). Analogously for the lowest weight state
C2 |c, k〉 = (k2 − k) |c, k〉 ,
and thus
k(k − 1) = j(j + 1) . (4.3)
Under the assumption that j is the maximal value of T 3 there is only the solution k = −j to (4.3).
Thus there are (2j + 1) states,
|j, k〉 , k ∈ −j, ..., j ,
such that 2j ∈ N and where we have relabelled c by j since it is determined by it. We denote the
representation by the number of states, 2j + 1. Mathematicians usually use 2j for the classification,
the so-called Dynkin label. Using
T± |j,m〉 = d(±)m |j,m± 1〉 ,
we find from
[T+, T−] |j,m〉 = 2T 3 |j,m〉 ,
that
d(+)m−1d
(−)m − d(+)
m d(−)m+1 = 2m .
One then finds that
d(+)m =
√(j −m)(j +m+ 1) ,
d(−)m =
√(j +m)(j −m+ 1) ,
must hold and the eigenstates form an orthonormal set
〈j,m|j,m′〉 = δmm′ ,
j∑m=−j
|j,m〉 〈j,m| = 1 .
– 23 –
Fundamental representation 2: In the smallest representation 2 with j = 12 , the generators
are represented by the Pauli matrices,
TA =σA
2,
which satisfy
σAσB = δAB1+ iεABCσC , σA∗
= −σ2σAσ2 .
Adjoint representation 3: For j = 1 the generators can be represented by the three hermitian
matrices
T 1 =
0 0 0
0 0 i
0 −i 0
, T 2 =
0 0 i
0 0 0
−i 0 0
, T 3 =
0 i 0
−i 0 0
0 0 0
.
They can be written using the Levi-Civita-symbol,
(TA)bc = −iεAbc .
This generalises to an infinite number of irreducible representations such that if the values of
T 3 are on an axis
T 3
−2 −1 0 1 2
• •
•• • •• • •
• • • • •
j = 12
j = 1
j = 32
j = 2
then a point represents an element of a representation, with different representations aligned parallel
to the T 3-axis. As shown above, the Lie algebra of SU(2) has one Casimir operator and the states
are uniquely determined by one label. In general Lie algebras will have as many labels as Casimir
operators. This is called the rank of the Lie algebra and hence the Lie algebra of SU(2) has rank
one.
Another way of generating all irreducible representations is by taking direct products of the
smallest representation. Lets suppose TA(1) and TA(2) are two copies of the Lie algebra of SU(2) each
satisfying [TA(a), T
B(a)
]= iεABCTC(a) ,
for a ∈ 1, 2, A,B,C ∈ 1, 2, 3 and [TA(1), T
B(2)
]= 0 .
Then the sum of the generators[TA(1) + TA(2), T
B(1) + TB(2)
]= iεABC(TC(1) + TC(2)) ,
generate the same algebra acting on the direct product states, denoted |·〉(1) |·〉(2). Since the sum
of the generators satisfies the same commutation relations one is able to represent their action
in terms of the previously derived representations. Consider the direct product of representation
2j + 1⊗ 2k + 1. The highest weight state |j〉(1) |k〉(2) is uniquely determined by its values of T 3(a),
a ∈ 1, 2 since the Lie algebra of SU(2) is of rank one and therefore the highest weight state
satisfies
T 3 |j〉(1) |k〉(2) =(T 3
(1) + T 3(2)
)|j〉(1) |k〉(2) = (j + k) |j〉(1) |k〉(2) ,
– 24 –
which must also be the highest weight state of the representation 2(j + k) + 1. To generate the
rest of the states we apply the sum of the lowering operators T− = T−(1) +T−(2) to the highest weight
state,
T− |j〉(1) |k〉(2) ∝ |j − 1〉(1) |k〉(2) + |j〉(1) |k − 1〉(2) .
The orthogonal combination
|j − 1〉(1) |k〉(2) − |j〉(1) |k − 1〉(2) ,
is the highest weight state of the 2(j + k− 1) + 1 representation because
T 3(|j − 1〉(1) |k〉(2) − |j〉(1) |k − 1〉(2)
)= (j + k − 1)
(|j − 1〉(1) |k〉(2) − |j〉(1) |k − 1〉(2)
),
and
T+(|j − 1〉(1) |k〉(2) − |j〉(1) |k − 1〉(2)
)= 0 .
Subsequently applying the sum of lowering operators decomposes the direct product representation,[2j + 1
]⊗[2k + 1
]=[2(j + k) + 1
]⊕[2(j + k− 1) + 1
]⊕ ...⊕
[2(j− k) + 1
]where we assumed without loss of generality j ≥ k.
As an example consider the direct product 2⊗2 of two spinor representations of the Lie algebra
of SU(2). Denote the highest weight state by
|↑↑〉 = |12,
1
2〉(1)|12,
1
2〉(2)
,
and generate the state
|↓↑〉+ |↑↓〉 = |12
;−1
2〉(1)|12
;1
2〉(2)
+ |12
;1
2〉(1)|12
;−1
2〉(2)
by applying the sum of lowering operators. Doing so again will give the lowest weight state
|↓↓〉 = |12
;−1
2〉(1)|12
;−1
2〉(2)
.
These three states form the three-dimensional representation 3 of the Lie algebra. The linear
combination
|↓↑〉 − |↑↓〉 = |12
;−1
2〉(1)|12
;1
2〉(2)− |1
2;
1
2〉(1)|12
;−1
2〉(2)
is a singlet state as it is annihilated by either the sum of lowering or raising operators and therefore
forms the representation 1. In summary we have confirmed 2⊗ 2 = 3⊕ 1.
Similarly consider the direct product 2⊗3 of the a spinor and adjoint representation of the Lie
algebra of SU(2). The highest weight state
|32
;3
2〉 = |1
2,
1
2〉(1)|1, 1〉(2)
is uniquely determined by its values of T 3. Applying the sum of lowering operators generates the
other states, e.g.
|32
;1
2〉 =
1√3T− |3
2;
3
2〉
=
√1
3|12,−1
2〉 |1, 1〉+
√2
3|12,
1
2〉 |1, 0〉
where the coefficients of the states are called Clebsch-Gordan coefficients. Continuing analogous to
before we decompose the direct product representation to the sum of the representation 4 and 2,
i.e. 2⊗ 3 = 4⊕ 2.
– 25 –
4.2 Groups
Let TA, A ∈ 1, 2, 3 denote the hermitian generators of the Lie algebra of SU(2). From (3.3) we
know that the exponential parametrisation
U(θ) = eiθATA
, θ =
θ1
θ2
θ3
∈ R3 , (4.4)
satisfy the group axioms where we showed closure under multiplication up to leading order.
Fundamental representation: The group generated by (4.4) in the fundamental representation
is SU(2). The group elements read
U(θ) = exp
(iθA
σA
2
)= cos
(θ
2
)12 + iθAσ
A sin
(θ
2
)(4.5)
where
θ =√θAθA , θA =
θAθ,
and σA are the Pauli matrices. From (4.5) it is easy to see that the group elements transform under
θ 7→ θ + 2π like
U(θ) 7→ −U(θ) .
Because the group elements are unitary, and have unit determinant their general form is given by(α β
−β∗ α∗
),
where α, β ∈ C and |α|2 + |β|2 = 1. Writing
α = α1 + iα2 ,
β = β1 + iβ2 ,
such that αi, βi ∈ R we can represent the group elements as points on the surface of the three-sphere
S3,
α21 + α2
2 + β21 + β2
2 = 1 .
SU(2) is isomorphic to S3 and the group manifold of SU(2) is simply connected.
Adjoint representation: In the adjoint representation the group SO(3) is generated by (4.4)
(remember as mentioned before that the Lie algebra of SU(2) is isomorphic to the Lie algebra of
SO(3)). In the exponential parametrisation the group elements read(R(θ)
)bc
= exp(θAεAbc) , (4.6)
generating rotations in three-dimensional Euclidean space such that for v ∈ R3 transforms under
the group action as
v 7→ v′ = R(θ)v .
Because the scalar product is invariant under these transformations,
v · v 7→ v′ · v′ = v · v ,
we have
R(θ)TR(θ) = 13 , det(R(θ)) = 1 .
– 26 –
The group elements (4.6) can be written(R(θ)
)bc
= δbc cos(θ) + εbcAθA sin(θ) + θbθc(1− cos(θ)) ,
which is symmetric under θ 7→ θ + 2π. Limiting the range to θ ∈ (−π, π) we can represent the
group elements as points in the closed three-ball where the antipodal points of the surface are to be
identified because θ = π and θ = −π represent the same group element. From this consideration we
can already infer that the group manifold of SO(3) is not simply connected. A loop in the group
manifold extending to the boundary and closing by running through the antipodal point can not be
contracted to a single point. Indeed SO(3) is double connected and SU(2) is its double universal
cover.
π
•
•
4.3 Three-dimensional harmonic oscillator
The quantum harmonic oscillators time evolution is given by the Hamiltonian
H =p · p2m
+1
2kx · x
= ~ω(A† ·A +3
2)
where x, p are position and momentum operators respectively, m is the particles mass and k = mω2
with ω the angular frequency of the oscillator. The Hamiltonian is then rewritten in terms of the
creation and annihilation operators A† respectively A which obey the commutation relations
[Ai, A†j ] = δij , [A†i , A
†j ] = 0 = [Ai, Aj ] .
Then the eigenstates are
H |n1, n2, n3〉 = ~ω(n1 + n2 + n3 +
3
2
)|n1, n2, n3〉
for ni ∈ N, i ∈ 1, 2, 3 where
|n1, n2, n3〉 =
3∏i=1
(A†i )ni
√ni!|0〉 .
Define transition operators for the first exited states such that
P12 |1, 0, 0〉 = |0, 1, 0〉P13 |1, 0, 0〉 = |0, 0, 1〉P23 |0, 1, 0〉 = |0, 0, 1〉 .
Since these have the same energy they commute with the Hamiltonian
[H,Pij ] = 0 (4.7)
– 27 –
for all Pij ∈ P12, P13, P23 and can be written in terms of creation and annihilation operators
Pij = i(A†iAj −A†jAi)
satisfying the Lie algebra of SU(2). We identify them with the angular momentum operators
L1 = P23 = x2p3 − x3p2
L2 = P13 = x1p3 − x3p1
L3 = P12 = x1p2 − x2p1
and from (4.7) we know
[H,Li] = 0
for all i ∈ 1, 2, 3 and as for any vector operator
[Li, A†j ] = −iεijkA†k .
Thus they span the 3 representation of the Lie algebra of SU(2). The Nth excited state transforms
as the N -fold symmetric direct product
(3⊗ ...⊗ 3︸ ︷︷ ︸N times
)sym .
Therefore we can decompose any exited level, e.g. the second excited state,
3⊗ 3 = 5⊕ 1
where the quadrupole is a symmetric traceless tensor,
[A†iA†j +A†jA
†i −
2
3δijA
† ·A†] |0〉
and the singlet
(A† ·A†) |0〉 .
In general we can decompose
3⊗ 3 = 5⊕ 3⊕ 1
Tij = T sym., tracelssij + T anti-sym
ij︸ ︷︷ ︸εijktk
+δijt .
4.4 Bohr atom
As another application we can consider the Bohr’s atoms Hamiltonian
H =p · p2m
− e2
r, r =
√x · k .
The spectrum of the energy states is given by
En = −Ryn2
where n is the principal quantum number, l = 0, 1, ..., n − 1 is the angular momentum quantum
number, m = −l, ..., 0, ..., l. The states for n = 2 are in the representationl = 0 : 0
l = 1 : 3
– 28 –
and have the same energy. There is a symmetry which for which we need a transition operator from
0 to 3 that commutes with the Hamiltonian. In classical mechanics there is the Laplace-Runge-Lenz
vector which precisely meets there requirements,
Aclassicali = εijkpjLk −me2xi
r.
For quantum theory we define the hermitian form
Ai =1
2εijk(pjLk + Lkpj)−me2xi
r
which obeys
L ·A = 0 = A · L ,
[Li, Aj ] = iεijkAk
as well as
[H,Aj ] = 0 .
This invariance is a hidden symmetry for 1r -potentials. Starting with the commutation relation
[Ai, Aj ] = iεijkLk(−2mH)
and defining the operator
Ai =Ai√−2mH
to give the simple commutation relation
[Ai, Aj ] = iεijkLk
as well as
[Li, Aj ] = iεijkAk
and additionally we know
[Li, Lj ] = iεijkLk .
Defining the linear combinations
X(+)j ≡ 1
2(Lj + Aj)
X(−)j ≡ 1
2(Lj − Aj)
which commute
[X(+)i , X
(−)j ] = 0
we are left with two copies of the Lie algebra of SU(2)
[X(+)i , X
(+)j ] = iεijkX
(+)k
[X(−)i , X
(−)j ] = iεijkX
(−)k .
Now the Casimir operators are the same
C(+)2 =
1
4(Li + Ai)(Li + Ai)
=1
4(Li − Ai)(Li − Ai) = C
(−)2
– 29 –
because of A · L = 0 = L ·A and therefore
C(+)2 = j(j + 1) = C
(−)2
where j1 = j = j2. To express the Hamiltonian in terms of the Casimir operator we calculate
AiAi = (−2mH)AiAi
= (p · p− 2me2
r)︸ ︷︷ ︸
−2mH
(LiLi + 1) +m2e4
and arrive at
H = −12me
4
L · L + A · A + 1
= −12me
4
4C(+)2 + 1
= − me4
2(2j + 1)2.
Therefore we get the well-known result of the principal quantum number
n = 2j + 1
and the angular momentum
Lj = X(+)j +X
(−)j
which degeneracy of levels n = 2j + 1 amounts to the direct product 2j + 1× 2j + 1:
l = 2j = n− 1
l = 2j − 1 = n− 2
...
l = 0
4.5 Isospin
Fermi-Yang model: Even though protons (mass mp = 938 MeV) carry electromagnetic charge
and neutrons (mass mn = 939 MeV) do not, the small mass difference of the two particles led to
the assumption that they share a symmetry which leaves the strong interaction invariant. Assume
the nucleons and antinucleons are Fermi oscillator states
|p〉 = b†1 |0〉 , |n〉 = b†2 |0〉
|p〉 = b†1 |0〉 , |n〉 = b†2 |0〉
where |0〉 denotes the vacuum state and the creation and annihilation operators satisfy the anti-
commutation relations
bi, b†j = δij = bi, b†j
which all other anticommutators vanishing. From these operators one can construct the generator
of the Lie algebra of SU(2)
Ij =1
2b†α(σj)αβbβ −
1
2b†α(σ∗j )αβ bβ
– 30 –
as well as the generator of the Lie algebra of U(1)
I0 =1
2(b†αbα − b†αbα) .
The direct product of the isospin representations 2(p
n
),
(p
n
)then decomposes such that the 3 representation furnishes three particles
|π+〉 = b†1b†2 |0〉
|π0〉 = (b†1b†1 − b
†2b†2) |0〉
|π−〉 = b†2b†1 |0〉
of masses: mπ± = 139 MeV, mπ0 = 135 MeV. As for the proton and neutron the mass difference is
explained by the symmetry breaking when weak and electromagnetic forces are taken into account.
The electro magnetic charge is given by
Q = I3 + I0 .
Wigner supermultiplet model: Extending the isospin symmetry by combining it with spin,
SU(4) ⊃ SU(2)I ⊗ SU(2)spin
leads to nucleons N and antinucleons N transforming as isospin and spin spinors
|N〉 ∼ 4 = (2I ,2spin)
|N〉 ∼ 4 = (2I ,2spin)
such that we can decompose
4⊗ 4 = 15⊕ 1
and the pions belong the 15-dimensional representation
15 = (3I ,3spin)︸ ︷︷ ︸ρ+,ρ0,ρ−
ρ-meson (vector)
1−,149 MeV
⊕ (1I ,3spin)︸ ︷︷ ︸ω
ω-meson (vector)
1−,783 MeV
⊕ (3I ,1spin)︸ ︷︷ ︸π+,π0,π−
pion (scalar)
0−
and the singlet state corresponds to the scalar η-meson (0−, 538 MeV). Considering the quark
content for π+
u u d d u d ∼ ud ∼ π+ .
Modified Lie algebra of SU(2) We create a new Lie algebra with three elements
L1 = iT 1
L2 = iT 2
L3 = T 3
where T 1, T 2, T 3 are generators of the Lie algebra of SU(2). These are no longer hermitian and the
new algebra is
[L1, L2] = −iL3
[L2, L3] = iL1
[L3, L1] = iL2 .
– 31 –
The Casimir operator
Q = (L1)2 + (L2)2 − (L3)2
obeys
[Q,Lj ] = 0
But is no longer bounded. In the adjoint representation the group generated is SU(1, 1) which
elements are of the form
eiθALA
= exp
(1
2
(iθ3 −θ1 + iθ2
−θ1 − iθ2 −iθ3
)).
They are not unitary and of the general form(u v
v∗ u∗
)with unit determinant
uu∗ − vv∗ = 1 .
Consider the commutation relation for a bosonic harmonic oscillator with creation and annihilation
operators a† and a respectively satisfying
[a, a†] = 1 .
Consider the canonical Bogoliubov transformation(a
a†
)7→(u v
v∗ u∗
)(a
a†
)=:
(b
b†
)where u, v ∈ C and the operators b, b† satisfy
[b, b†] = 1 .
Then
[b, b†] = [ua+ va†, v∗a+ u∗a†] = (|u|2 − |v|2)[a, a†]
and thus
1 = |u|2 − |v|2 .
Therefore there set of canonical Bogoliubov transformations forms the group SU(1, 1). Further con-
sider the canonical commutation relation of quantum mechanics on some suitable complex Hilbert
space
[x, p] = i
for position and momentum operators x and p respectively. With
q =
(x
p
)these can be rewritten as
[qi, qj ] = iΩij
with the symplectic matrix
Ω =
(0 1
−1 0
).
– 32 –
The set of all transformations S such that
S · Ω · ST = Ω
form the symplectic group Sp(2,R). One can verify that there is a one-to-one correspondence
SU(1, 1)→ Sp(2,R)
by a change of basis from (a, a†) to (x, p). To study the representations of SU(1, 1) consider the
operators made up of one bosonic harmonic oscillator with creation and annihilation operators a†
and a respectively,
L+ = L1 + iL2 =1
2a†a† , L− = L1 − iL2 =
1
2aa
with the commutator
[L+, L−] = −(1
2+ a†a) =: −2L3 .
Then
[L3, L±] = ±L±
but in contrast to SU(2) there is no highest weight state since the Casimir operator is not bounded.
Starting from the vacuum state |0〉 we get
L− |0〉 = 0
and
(L+)n |0〉 ∝ |2n〉 .
and therefore generate all states of even occupation number
|0〉 , |2〉 , |4〉 , ... .
Again there also is another representation of SU(1, 1) generating all states of odd occupation number
|1〉 , |3〉 , |5〉 , ... .
Harmonic
oscillator
potential
|0〉
|1〉
|2〉
– 33 –
5 SO(N) and SU(N)
5.1 SO(N)
The group of rotations in N -dimensional Euclidean space is SO(N) where the group elements are
N ×N matrices R (fundamental representation) satisfying
RT ·R = 1 (5.1)
and
det(R) = 1 (5.2)
(demanding only det(R))2 = 1 generates the group O(N)). A vector is defined by the transformation
under rotation
vi 7→ Rijvj
where i, j = 1, ..., N . Analogously we define tensors be to objects transforming like
T ij 7→ RimRjnTmn
W ijn 7→ RisRjtRnuW stu
exemplary for second- and third-rank tensors, which can be generalised to arbitrary indexed tensors.
To be complete a scalar does not transform at all,
s 7→ s .
The anti-symmetric combination
Aij =1
2(T ij − T ji) = −Aji
again is a tensor and transforms to an anti-symmetric one
Aij 7→ RimRjnAmn .
There are N(N−1)2 independent components for anti-symmetric second-rank tensor. Analogously we
can do for a symmetric second-rank tensor
Sij =1
2(T ij + T ji)
with
Sij 7→ RimRjnSmn
and 12N(N + 1) independent components because of the additional trace. Now the trace is a scalar
since
Sii 7→ RikRil︸ ︷︷ ︸(RT )kiRil=δkl
Skl = Sii
Constructing the symmetric trace-less second-rank tensor
Sij = Sij − δij Skk
N
we are left with N(N+1)2 − 1 components. Therefore we decomposed the representation
N⊗N =
[N(N− 1)
2
]⊕[
N(N + 1)
2− 1
]⊕ 1 .
– 34 –
For example for SO(3): 3⊗ 3 = 5⊕ 3⊕ 1. For a general treatment of T ijk...m Young tableaux was
invented to keep take of patterns. For rotations we can use (5.1) to write
RikRilδkl = Rik(RT )kj = δij
and thus δij is an invariant symbol. Using the N -dimensional Levi-Civita symbol εijk...n which is
totally antisymmetric
ε...k...m... = −ε...m...k...
and
ε123...N = 1
we are able to rewrite the determinant of (5.2) to give
RipRjq...Rnsεpq...s = (det(R))︸ ︷︷ ︸=1
εij...n
and therefore εijk...n is also an invariant symbol.
Dual tensors Let Aij be an anti-symmetric second-rank tensor. We define its dual to be the
totally anti-symmetric (N − 2)-rank tensor
Bk...n = εijk...nAij .
For different values of N we get
N = 3 : Bk = εijkAij (vector)
N = 2 : B = εijAij (scalar) .
In the case of N = 4 we again get a anti-symmetric second-rank tensor which is used in electro-
magnetics, namely the fields strength tensor and its dual in 4-dimensional Euclidean space (for the
treatment in Minkowski space we are dealing with SO(3, 1) not SO(4) anymore),
F kl = εijklAij .
Self-dual / Anti-self-dual For SO(2n) we can construct two irreducible representations with an
additional feature. Consider an totally anti-symmetric n-rank tensorAi1i2...in which has (2n)(2n−1)...(n+1)n! =
(2n)!(n!)2 components. Constructing the dual tensor
Bi1i2...in =1
n!εi1i2...inj1j2...jnAj1j2...jn
and its dual
Ai1i2...in =1
n!εi1i2...inj1j2...jnBj1j2...jn
we see that the are dual to each other. Therefore the tensors
T i1...in± =1
2(Ai1...in ±Bi1...in)
are self-dual and anti-self-dual with (2n)!2(n!)2 independent components. which can be schematically
seen by considering
εT± =1
2(εA± εB) = ±1
2(A±B) = ±T± .
– 35 –
Lie algebra of SO(N) The expansion around the identity element for SO(2) is
R = 1+ dθ
(0 1
−1 0
)whereas for SO(3)
R = 1+ idθAJA
with the already introduces matrices
J1 = −i
0 0 0
0 0 1
0 −1 0
, J2 = −i
0 0 −1
0 0 0
1 0 0
, J3 = −i
0 1 0
−1 0 1
0 0 0
.
In general we can infer for SO(N)
R = 1+A
RT ·R = 1
⇒ A = −AT .
The generators then are
(J(mn))(ij) = −i(δmiδnj − δmjδni)
which obey the algebra
[J(mn), J(pq)] = i(δmpJ(nq) + δnpJ(mp) − δnpJ(mp) − δmqJ(np)) .
Finally let us count the entries for the above considered tensors. Assume we have T ijk... = −T jik... =
εijsT sk... antisymmetric in the first index, then we need not take anti-symmetric pair into account.
For a symmetric traceless tensor, Sij ⇒ Sii = 0, Sijk ⇒ Siik = 0. In general Si1i2...in a symmetric
traceless tensor we will have
εijSiji3...in = 0
and counting the components by the routine
S11...1, S11...2, S11...22, ..., S22...2
will give (n+ 1) entries, an continuing by
S33...3x, S33...xx, ..., Sxx...xx
where the x denotes 1 or 2 will yield
n∑k=0
(k + 1) =1
2(n+ 1)(n+ 2)
components without the traceless constraint. Therefore we need to take δijSijm1...Mn−2 = 0 into
account and get1
2(n− 2 + 1)(n− 2 + 2) =
1
2(n− 1)n
traceless constraints. Finally we have
d = 2n+ 1 = 2j + 1, j ∈ N
dimension of representation as Si1...in .
– 36 –
5.2 SU(N)
Now that we have discussed SO(N) we are interested in how to get to SU(N). We have already
seen that SU(2) is the universal cover of SO(3) for any 2j + 1, j ∈ N. To start off we have
O(N) : OT ·O = 1⇒ (det(O))2 = 1⇒ det(O) = ±1
U(N) : U†U = 1⇒ det(U)∗ det(U) = 1⇒ det(U) = eiϕ .
For U(N) we know
- elements eiϕ1 form U(1)
- elements with det(U) = 1⇒ SU(N) .
As a side remark: eikN 2n1N has det = 1, k ∈ N and forms ZN group and therefore
U(N) = U(1)× (SU(N)/ZN ) .
Now for U(N) the quadratic form
ψi 7→ U ijψj
ξi 7→ U ijξj
ξ†ψ = ξ∗Tψ =
N∑j=1
ξ∗jψj
is invariant. For the conjugate spinors we have
ξi∗ 7→ U∗ijξj
∗= ξ∗j(U†)ji
We will now use the notation ψi = ψ∗i for conjugate spinors. The transformation laws then take
the form
ψi 7→ ψ′i = U ijψj
ψi 7→ ψ′i = ψj(U†)ji
and therefore
ξiψi 7→ ξj (U†)jiU
ik︸ ︷︷ ︸
δjk
ψk = ξiψi .
From this we can infer that δij exists as an invariant symbol, whereas δij , δij do not. We can now
write the tensors in SU(N) to have two indices ϕi1...imj1...jn. The transformation law is then
ϕijk 7→ ϕ′ijk = U ilU
jm(U†)nkϕ
lmn
and contractions are only allowed between upper and lower indices,
ϕijj = δkjϕijk
SU(N)→ U ilϕljj .
To lower or raise indices we use the antisymmetric symbol
εi1i2...in(U†)i1j1 ...(U†)injn = (det(U))∗︸ ︷︷ ︸
=1
εj1...jn
U i1j1 ...Uinjnεj1...jn = det(U)︸ ︷︷ ︸
=1
εi1...in
– 37 –
making εi1...in and εi1...in invariant symbols. For example for SU(4) we can lower indices: ϕkpq =
εijpqϕijk . In case of SU(2) the speciality is that it needs only upper indices with the invariant
symbols εij , εij , δij . We can upper all indices
ϕij...kmn...s = εmnεnb...εstϕab...tij...k
and moreover only need symmetric tensors. In general this does not work,
ψmnp = εmnpjψj
but for SU(2)
ψm = εmjψj .
In general an element of SU(2)
U = exp(i
2θ · σ)
where
σ1 =
(0 1
1 0
), σ2 =
(0 −ii 0
), σ3 =
(1 0
0 −1
)are the Pauli matrices, is not real. With
σ2σ∗aσ2 = −σa
we get
σ2[ei2θ·σ]∗σ2 = e
i2θ·σ
thus there exists a similarity transformation
S[ei2θ·σ]∗S−1 = e
i2θ·σ
where
S = −iσ2 =
(0 1
−1 0
), S−1 = iσ2 =
(0 −1
1 0
).
Now for general SU(N) we write the elements
U = eiM =∞∑k=0
1
k!(iM)k
where M = M† is hermitian and therefore U†U = 1. Due to ln(det(M)) = tr(ln(M)) we get
det(U) = eitr(M) = 1⇒ tr(M) = 0 .
As seen for SU(2) we therefore have
M = θ · σ2
= θATA
– 38 –
6 SU(3)
For SU(3) the analogous to the Pauli matrices of SU(2) as generators are the Gell Mann matrices
λ1 =
0 1 0
1 0 0
0 0 0
, λ2 =
0 −i 0
i 0 0
0 0 0
, λ3 =
1 0 0
0 −1 0
0 0 0
,
λ4 =
0 0 1
0 0 0
1 0 0
, λ5 =
0 0 −i0 0 0
i 0 0
, λ6 =
0 0 0
0 0 1
0 1 0
,
λ7 =
0 0 0
0 0 −i0 i 0
, λ8 =1√3
1 0 0
0 1 0
0 0 −2
which are normalised to tr(λaλb) = 2δab. The first three Gell-Mann matrices contain the Pauli
matrices (λ1, λ2, λ3) ∼ (σ1, σ2, σ3) acting on a subspace. We can now look at commutators and
since λ3 and λ8 are diagonal,
[λ3, λ8] = 0 .
Moreover
[λ4, λ5] = 2i
1 0 0
0 0 0
0 0 −1
=: 2iλ[4,5] = i(λ3 +√
3λ8)
and therefore λ4, λ5, λ[4,5] form the Lie algebra of SU(2). The same we can do for
[λ6, λ7] = 2i
0 0 0
0 1 0
0 0 −1
=: 2iλ[6,7] = i(−λ3 +√
3λ8)
and are then left with three copies of the Lie algebra of SU(2) which overlap. The Lie algebra of
SU(3) is generated in the fundamental representation by
TA =λA2
.
In the light of the above we define I± = T 1 ± iT 2, U± = T 6 ± iT 7, V± = T 4 ± iT 5 which form a
basis together with T 3 and T 8 for which
[T 3, T 8] = 0 .
– 39 –
This generates a subalgebra,
[T 3, I±] = ±I±
[T 3, U±] = ∓1
2U±
[T3, V±] = ±1
2V±
[T 8, I±] = 0
[T 8, U±] = ±√
3
2U±
[T 8, V±] = ±√
3
2V±
[I+, I−] = 2T 3
[U+, U−] =√
3T 8 − T 3
[V+, V−] =√
3T 8 + T 3
[I+, V−] = −U−[I+, U+] = V+
[U+, V−] = I−
[I+, V+] = 0
[I+, U−] = 0
[U+, V+] = 0
with all the hermitian conjugate relations. Since there are two Casimir operators we can label the
states in every irreducible representation by eigenstates of T 3 and T 8,
|i3, i8〉
such that
T 3 |i3, i8〉 = i3 |i3, i8〉T 8 |i3, i8〉 = i8 |i3, i8〉 .
One can then verify that
T 3I± |i3, i8〉 = (I±T3 ± I±) |i3, i8〉 = (i3 ± 1)I± |i3, i8〉
T 8I± |i3, i8〉 = I±T8 |i3, i8〉 = i8I± |i3, i8〉
and thus
I± |i3, i8〉 ∝ |i3 ± 1, i8〉 .
Analogous
T 3U± |i3, i8〉 = (i3 ∓1
2)U± |i3, i8〉
T 3U± |i3, i8〉 = (i8 ±√
3
2)U± |i3, i8〉
giving
U± |i3, i8〉 ∝ |i3 ∓1
2, i8 ±
√3
2〉
– 40 –
and finally in the same manner
V± |i3, i8〉 ∝ |i3 ±1
2, i8 ±
√3
2〉 .
Therefore considering the lattice spanned by i3 and i8 the operators U±, V± and I± can be repre-
sented by
i3
i8
I+I−
U+
U−
V+
V−
and are called root vectors. Correctly normalized they read
I± = (±1, 0)
U± = (∓1
2,±√
3
2)
V± = (±1
2,±√
3
2)
Associating root vectors to T 3 and T 8 we would get (0, 0) for both. We can now make a few
observations:
- Root vectors have equal lengths or vanish
- Angles between them are π3 ,
2π3 and so on
- sometimes the sum of two roots give another root sometimes not
- the sum of a root vector with itself is not a root vector
- positive roots and simple roots
Positive roots are characterised by i3 > 0. Therefore V+, I+, U− are positive and V−, I−, U+ are
negative. Moreover
I+ = V+ + U−
and V+,U− are the simple roots.
Weight diagrams and representations LetR be some (irreducible or reducible) representation
of SU(3) and plot the states |i3, i8〉 in the i3, i8 plane. For the fundamental representation 3 we
identify the states with quark states
|u〉 := |12,
1
2√
3〉
|d〉 := |−1
2,
1
2√
3〉
|s〉 := |0,− 1√3〉
– 41 –
such that the weight diagram is
i3
i8
strangness
− 12
12
12√
3
− 1√3
0
1
•u•d
•s
Using the root vectors we see that
V− |u〉 ∝ |s〉I− |u〉 ∝ |d〉U+ |s〉 ∝ |d〉V+ |s〉 ∝ |u〉
as well as
V+ |u〉 = I+ |u〉 = U+ |U〉 = U− |u〉 = 0 .
We can then consider the representation under charge conjugation such that the generators
TAC = −(TA)† = −(TA)T
with
[TA, TB ] = ifABCTC
and therefore
[TAC , TBC ] = ifABCTCC .
We are then left with
TAC = TA
for A = 2, 5, 7 and
TAC = −TA
for A = 1, 3, 4, 6, 8. Thus we conclude
C |i3, i8〉 ∼ |−i3,−i8〉 .
We are now ready to look at the charge conjugate representation 3∗ where we identify the states
as the antiparticles of the 3 states, namely
|u〉 = |−1
2,− 1
2√
3〉
|d〉 = |12,− 1
2√
3〉
|s〉 = |0, 1√3〉
and the weight diagram reads
– 42 –
i3
i8
− 12
12
1√3
− 12√
3
•u
•d
•s
For the singlet representation 1 the weight diagram is almost trivial with only one state (i3, i8) =
(0, 0)
i3
i8
•
In the adjoint representation we have
adj(TA)X = [TA, X]
with X = XCTC . This satisfies the algebra since
[adj(TA), adj(TB)]X?= adj([TA, TB ])X ⇔ [TA, [TB , X]]− [TB , [TA, X]]
?= [[TA, TB ], X]
⇔ [TA, [TB , TC ]] + [TB , [TC , TA]] + [TC , [TA, TB ]]?= 0
is the Jacobi identity. The weight table for 8 can then be computed using the commutation relations,
state [T 3, V ] [T 8, V ] weight (i3, i8)
T 3 0 0 (0, 0)
T 8 0 0 (0, 0)
I+ +1 0 (+1, 0)
I− −1 0 (−1, 0)
U+ − 12 +
√3
2 (− 12 ,+
√3
2 )
U− + 12 −
√3
2 (+ 12 ,−
√3
2 )
V+ + 12 +
√3
2 (+ 12 ,+
√3
2 )
V− − 12 −
√3
2 (− 12 ,−
√3
2 )
and has the weight diagram
– 43 –
i3
i8
•T 3
•T8
•I+
•V+•
U+
•I−
•V−
•U−
Analogous to the SU(2) case we can look at the tensor product representation with two represen-
tations TA(1) and TA(2) with states |i3, i8〉(1) and |j3, j8〉(2) respectively. Define
TA := TA(1) + TA(2)
and as before we have
T 3 |i3, i8〉(1) |j3, j8〉(2) = (TA(1) + TA(2)) |i3, i8〉(1) |j3, j8〉(2) = (i3 + j3) |i3, i8〉(1) |j3, j8〉(2)
where the difference to SU(2) is that we have two labels for SU(3) since its Lie algebra is of rank
three. Let us denote
I+, U+, V+ 3 Em+I−, U−, V− 3 Em−
m = 1, 2, 3 and study 3⊗ 3. The weight table is then
state weight (i3, i8)
u⊗ u (1, 1√3)
d⊗ d (−1, 1√3)
s⊗ s (0,− 2√3)
u⊗ d, d⊗ u (0, 1√3)
u⊗ s, s⊗ u ( 12 ,−
12√
3)
d⊗ s, s⊗ d (− 12 ,−
12√
3)
and the corresponding weight diagram
– 44 –
i3
i8
•dd •du •ud •uu
•ds •sd •su •us
•ss
For the highest weight state we need Em+ |s〉 = 0 and therefore it is u ⊗ u. Applying Em− |u⊗ u〉generates the representation 6 with weight table
state weight (i3, i8)
u⊗ u (1, 1√3)
d⊗ d (−1, 1√3)
s⊗ s (0,− 2√3)
1√2(u⊗ d+ d⊗ u) (0, 1√
3)
1√2(u⊗ s+ s⊗ u) ( 1
2 ,−1
2√
3)
1√2(d⊗ s+ s⊗ d) (− 1
2 ,−1
2√
3)
and weight diagram
i3
i8
• • •
• •
•
The remainder of the decomposition is 3∗ with the weight table
state weight (i3, i8)1√2(d⊗ u− u⊗ d) (0, 1√
3)
1√2(d⊗ s− s⊗ d) (− 1
2 ,−1
2√
3)
1√2(s⊗ u− u⊗ s) ( 1
2 ,−1
2√
3)
– 45 –
and weight diagram
i3
i8
•
• •
We have thus decomposed 3⊗ 3 = 6⊕ 3∗ where for 3∗
|i3, i8〉∗ = − |i3, i8〉
denotes the charge conjugation. Similar one can derive the weight tables and diagrams for 3∗⊗3∗ =
6∗ ⊕ 3. Now for 3⊗ 3∗. The weight table is
state weight (i3, i8)
u⊗ u, d⊗ d, s⊗ s (0, 0)
u⊗ s ( 12 ,√
32 )
u⊗ d (1, 0)
d⊗ s (− 12 ,√
32 )
d⊗ u (−1, 0)
s⊗ u (− 12 ,−
√3
2 )
s⊗ d ( 12 ,−
√3
2 )
and the weight diagram
i3
i8
•ds •us
•du •uu •dd•ss
•ud
•su •sd
– 46 –
Applying lowering operators to the highest weight state u⊗ s give the states
u⊗ s , u⊗ d , d⊗ s , d⊗ u , s⊗ u , s⊗ d , 1√2
(d⊗ d−u⊗ u) ,1
2(d⊗ d+u⊗ u−2s⊗ s)
which furnish the 8 representation with weight diagram
i3
i8
• •
• •• •
• •
There is a singlet representation left with the state 1√3(u⊗ u+ d⊗ d+ s⊗ s) which is evident since
it is annihilated by Em+ . Thus we have decomposed 3⊗ 3∗ = 8⊕ 1. Finally considering 3⊗ 3⊗ 3
the weight table can be derived analogously and the weight diagram is
i3
i8
•ddd •••ddu •••duu •uuu
•••dds ••••••dus
•••uus
•••dss •••uss
•sss
In the exact same way as before applying lowering operators to the highest weight state (uuu) we
can decompose 3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1. The first irreducible representation are the symmetric
combinations
u⊗ u⊗ u , d⊗ d⊗ d , s⊗ s⊗ s , 1√3
(u⊗ u⊗ s+ u⊗ s⊗ u+ s⊗ u⊗ u) , ...
forming 10 whereas the singlet is the totally anti-symmetric combination
1√6
(s⊗ d⊗ u− s⊗ u⊗ d+ d⊗ u⊗ s− d⊗ s⊗ u+ u⊗ s⊗ d− u⊗ d⊗ s)
– 47 –
forming 1. The octet states have mixed symmetry. One can now assign particle states to the
derived decomposed representations, namely mesons to the decomposition
3⊗ 3∗ = 8⊕ 1
and baryons to
3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1 .
We denote
- B: baryon number
- S: strangeness (number of s quarks − number of s quarks)
- J : spin
- I3: isospin (third component)
- Y = B + S: hypercharge .
We then identify the pseudoscalar mesons (B = 0, J = 0) with the weight diagram 8
I3
Y = B + S
•η•π
0
•π+
•K+
•K0
•π−
•K−
•K0
495 MeV
137 MeV549 MeV
495 MeV
and the additional singlet with the η′-particle. Analogous for the vector mesons (B = 0, J = 1)
with weight diagram
I3
Y = B + S
•ω•ρ0
•ρ+
•K+∗
•K0∗
•ρ−
•K−
∗ •K0∗
892 MeV
770 MeV783 MeV
892 MeV
– 48 –
and the additional singlet is the φ-particle. Now the baryons (B = 1, J = 32 ) are in the 10
representation
I3
Y = B + S
•∆−
•∆0
•∆+
•∆++
•Σ−∗
•Σ0∗•Σ+∗
•Ξ−∗
•Ξ0∗
•Ω−
1235 MeV
1385 MeV
1530 MeV
1670 MeV
and the baryons (B = 1, J = 12 ) in 8
I3
Y = B + S
•Σ0
•Λ0
•Σ+
•p
•n
•Σ−
•Ξ−
•Ξ+
939 MeV
1193 MeV1116 MeV
1318 MeV
Now the question might arise why there are no 3 ⊗ 3 or 3 ⊗ 3 ⊗ 3∗ state. The reason is that one
assumes that all asymptotic states are singlets with respect to a SU(3)C colour symmetry and since
3C ⊗ 3∗C = 8C ⊕ 1C
as well as
3C ⊗ 3C ⊗ 3C = 10C ⊕ 8C ⊕ 8C ⊕ 1C
decompose into a singlet representation and e.g.
3C ⊗ 3C = 6C ⊕ 3∗C
does not there are no 3C ⊗ 3C states.
– 49 –
Let us turn again to the Isospin model of the Lie algebra of SU(2) in the fundamental repre-
sentation we already encountered and denote the nucleons and antinucleons
N i =
(p
n
), Ni =
(p
n
).
Their transformation in the notation established for general SU(N) then is
N i 7→ U ijNj , Ni 7→ Nj(U
†)ji
Turning to the adjoint representation we can write
φ = π · σ = π1σ1 + π2σ2 + π3σ3
=
(π3 π1 − iπ2
π1 + iπ2 −π3
)=
(π0
√2π+
√2π− −π0
)which transforms like
φij 7→ φ′ij = U ilφ
lm(U†)mj .
For Lagrangian density which has a term of the form
L = ...+ gNiφijN
j
is then SU(2) invariant. Explicitly
g(p n)·(
π0√
2π+√
2π− −π0
)·(p
n
)= g(pπ0p− nπ0n+
√2(pπ+n+ nπ−p)
)with Feynman diagrams, e.g.
t
p
•gp
π0
p
•g
p
n
•−g
nπ0
n
•−g n
p
•√2g
nπ+
n
•√
2g p
Now in the same manner for Lie algebra of SU(3) in the adjoint representation,
φ =1√2
8∑a=1
φaλa
=
1√2π0 + 1√
6η π+ K+
π− − 1√2π0 + 1√
6η K0
K− K0 − 2√6η
Start by assuming a Lagrangian with SU(3) symmetry
L0 =1
2gµνtr
((∂µφ)(∂νφ)
)− 1
2m2
0tr(φ2) .
– 50 –
Here all mesons masses would be equal, which clearly is not the case and therefore we have to break
the symmetry. Suppose we have a symmetric Hamiltonian H0 and introduce a H1 which breaks
the symmetry,
H = H0 + αH1 .
As long as α stays small we are in the perturbative regime and know that we just shift the eigenvalue
of H0 by a small amount
〈s|H0|s〉 = Es
〈s|αH1|s〉 = ∆E
and thus write
E = Es + ∆E +O(α2) .
Thus we are interested in introducing a symmetry breaking term in the Lagrangian. Since φ lives
in the 8 representation and L0 ∝ φ2 we look at the decomposition
8⊗ 8 = 27⊕ 10⊕ 10∗ ⊕ 8⊕ 8⊕ 1
where the 1 is the mass term in L0. Since φ2 is symmetric if view as direct product of representations
only the symmetric part is interesting and since
(8⊗ 8)s = 27⊕ 8⊕ 1
(8⊗ 8)a = 10⊕ 10∗ ⊕ 8
one is able to choose 27 or 8 for the breaking term. Since it is always natural to go the simpler way
first we pick 8 and add tr(φ · φ · λ8) as the breaking term. Note that we need to preserve SU(2) in
the upper part of φ to keep the pion structure and do not break isospin. Now write
λ8 = c13 + d
0 0 0
0 0 0
0 0 1
such that the first term adds to the mass term in the Lagrangian and the second is new to break
the SU(3) symmetry in the right way, namely
tr
(φ · φ ·
0 0 0
0 0 0
0 0 1
) = K−K+ +K0K0 +2
3η2
and therefore
L = L0 −1
2κ tr
(φ · φ ·
0 0 0
0 0 0
0 0 1
)corrects the mass term of the K’s and η but not the pion, comparing to
tr(φ · φ) = (π0)2 + η2 + 2K−K+ + ... .
To first order in perturbation theory around the symmetric situation one gets
m2π = m2
0
m2η = m2
0 +2
3κ
m2K = m2
0 +1
2κ
which yields the Gell-Mann Okubo mass formula
⇒ 4m2K = 3m2
η +m2π .
– 51 –
7 Lorentz and Pioncare group
7.1 Real rotations and Lorentz transformations
We use here conventions where the metric in four dimensional Minkowski space is given by
ηµν = ηµν = diag(−1, 1, 1, 1) .
Infinitesimal Lorentz transformations and rotations in Minkowski space are of the form
Λµν = δµν + δωµν (7.1)
with Λµν ∈ R such that the metric ηµν is invariant.
Exercise: Show that this implies
δωµν = −δωνµ .
The spatial-spatial components describe rotations the three dimensional subspace and the
spatial-temporal components Lorentz boost in Minkowski space or rotations around a particular
three-dimensional direction in Euclidian space. Representations of the Lorentz group with
U(Λ′Λ) = U(Λ′)U(Λ)
can be written in infinitesimal form as
U(Λ) = 14 +i
2δωµνM
µν ,
where Mµν = −Mνµ are the generators of the Lorentz algebra
[Mµν ,Mρσ] = i (ηµρMνσ − ηµσMνρ − ηνρMµσ + ηνσMµρ) . (7.2)
The fundamental representation (7.1) has the generators
(MµνF )αβ = −i(ηµαδνβ − ηναδ
µβ ) .
It acts on the space of four-dimensional vectors xα and the infinitesimal transformation in (7.1)
induces the infinitesimal change
δxα =i
2δωµν(Mµν
F )αβ xβ .
One can decompose the generators into the spatial-spatial part
Ji =1
2εijkM
jk, (7.3)
and a spatial-temporal part,
Kj = M j0. (7.4)
Equation (7.2) implies the commutation relations
[Ji, Jj ] = i εijkJk,
[Ji,Kj ] = i εijkKk,
[Ki,Kj ] = −i εijkJk.
In the fundamental representation one has
(JFi )j k = −iεijk
– 52 –
where j, k are spatial indices. All other components vanish, (JFi )00 = (JFi )0
j = (JFi )j 0 = 0. Note
that JFi is hermitian, (JFi )† = JFi . The generator Kj has the fundamental representation
(KFj )0
m = −iδjm, (KFj )m0 = −iδjm
and all other components vanish, (KFj )0
0 = (KFj )mn = 0. From these expression one finds that the
conjugate of the fundamental representation of the Lorentz algebra has the generators
JCj = (JFj )† = JFj , KCj = (KF
j )† = −KFj . (7.5)
This implies that KFj is anti-hermitian,
(KFj )† = −KF
j .
One can define the linear combinations of generators
Nj =1
2(Jj − iKj), Nj =
1
2(Jj + iKj),
for which the commutation relations become
[Ni, Nj ] = iεijkNk,
[Ni, Nj ] = iεijkNk,
[Ni, Nj ] = 0.
This shows that the representations of the Lorentz algebra can be decomposed into two represen-
tations of SU(2) with generators Nj and Nj , respectively. Note that Nj and Nj are hermitian and
linearly independent. Nevertheless, there is an interesting relation between the two: Consider the
hermitian conjugate representation of the Lorentz group as related to the fundamental one by eq.
(7.5). The representation of the generators Nj , Nj is
NCj =
1
2(JCj − iKC
j ) =1
2(JFj + iKF
j ) = NFj ,
NCj =
1
2(JCj + iKC
j ) =1
2(JFj − iKF
j ) = NFj .
This implies that the role of Nj and Nj is interchanged in the hermitian conjugate representation.
Representations of SU(2) are characterized by spin n of half integer or integer value. Accordingly,
the representations of the Lorentz group can be classified as (2n+ 1, 2n+ 1). For example
(1, 1) = scalar or singlet,
(2, 1) = left-handed spinor,
(1, 2) = right-handed spinor,
(2, 2) = vector.
7.2 Pauli formalism
In the non-relativistic description of spin-1/2 particles due to Pauli the generators of rotation are
given by
Ji =1
2σi ,
where the hermitian Pauli matrices are given by
σ1 =
(0 1
1 0
), σ2 =
(0 −ii 0
), σ3 =
(1 0
0 −1
),
– 53 –
and fulfill the algebraic relation
σi · σj = δij12 + iεijkσk.
In other words, the Pauli matrices provide a mapping between the space of rotations SO(3) and
the space of unitary matrices SU(2). More concrete, a rotation
Λi j = δi j + δωi j
corresponds to
L(Λ) = 12 +i
4δωij εijk σk.
By exponentiating this one obtains the mapping. Note, however, that the group SU(2) covers SO(3)
twice in the sense that a rotation by 360 degree corresponds to L(Λ) = −12.
7.3 Dirac formalism
Left and right handed spinor representation We will construct the left and right handed
spinor representations of the Lorentz group by using that they agree with the Pauli representation
for normal (spatial) rotations. When acting on the left-handed representation (2,1), the generator
Nj vanishes. Since Jj = Nj + Nj and Kj = i(Nj − Nj) one has
Nj = Jj = −iKj =1
2σj , Nj = 0.
Using (7.3) and (7.4) this yields for the left handed spinor representation
(M jkL ) = εjklNl =
1
2εjkl σl
(M j0L ) = iNj = i
1
2σj .
(7.6)
Note that (M j0L ) receives a factor 1/v in Wick space so that it becomes − 1
2σj in Euclidean space.
As the name suggests, this representation acts in the space of left-handed spinors which are two-
components entities, for example
ψL =
((ψL)1
(ψL)2
).
We also use a notation with explicit indices (ψL)a with a = 1, 2. The infinitesimal transformation
in (7.1) reads with the matrices (7.6)
δ(ψL)a =i
2δωµν(Mµν
L ) ba (ψL)b. (7.7)
Similarly one finds for the right-handed spinor representation (1,2) using
Nj = 0, Nj = Jj = iKj =1
2σj ,
the relations
(M jkR ) = εjklNl =
1
2εjkl σl
(M j0R ) = −iNj = −i1
2σj .
(7.8)
In Wick space (M j0R ) receives an additional factor 1/v and becomes 1
2σj in the Euclidean limit.
The representation (7.8) acts in the space of right handed spinors, for example
ψR =
((ψR)1
(ψR)2
).
– 54 –
For right handed spinors we will also use a notation with an explicit index that has a dot in order
to distinguish it from a left-handed index, (ψR)a with a = 1, 2. The infinitesimal transformation in
(7.1) reads with the matrices in (7.8)
δ(ψR)a =i
2δωµν(Mµν
R )ab
(ψR)b.
Invariant symbols From the group-theoretic relation
(2, 1)× (2, 1) = (1, 1)A + (3, 1)S ,
it follows that there must be a Lorentz-singlet with two left-handed spinor indices and that it has
to be anti-symmetric. The corresponding invariant symbol can be taken as εab with components
ε21 = 1, ε12 = −1 and ε11 = ε22 = 0. Indeed one finds that
(MµνL ) c
a εcb + (MµνL ) c
b εac = 0. (7.9)
(This is essentialy due to σjσ2 + σ2σTj = 0 for j = 1, 2, 3.) It is natural to use εab and its inverse
εab to pull the indices a, b, c up and down. For clarity the non-vanishing components are
ε12 = −ε21 = ε21 = −ε12 = 1. (7.10)
The symbol δab is also invariant when spinors with upper left-handed indices have the Lorentz-
transformation behavior
δ(ψL)a = − i2δωµν(ψL)b(Mµν
L ) ab .
From eq. (7.9) it follows also that
(MµνL )ab = (Mµν
L )ba,
so that
εab(MµνL )ab = (Mµν
L ) aa = 0.
In a completely analogous way the relation
(1, 2)× (1, 2) = (1, 1)A + (1, 3)S
implies that there is a Lorentz singlet with two right-handed spinor indices. The corresponding
symbol can be taken as εab, with inverse εab, with components as in (7.10). This symbol is used
to lower and raise right-handed indices. Spinors with lower right handed index transform under
Lorentz-transformations as
δ(ψR)a = − i2δωµν(ψR)b(M
µνR )b a. (7.11)
Consider now an object with a left-handed and a right-handed index. It is in the representation
(2, 2) which should also contain the vector. There is therefore an invariant symbol which can be
chosen as
(σµ)aa = (12, ~σ),
and similarly
(σµ)aa = (12,−~σ).
It turns out that the matrices for infinitesimal Lorentz transformations can be written as
(MµνL ) b
a =i
4(σµσν − σν σµ) b
a ,
(MµνR )a
b=i
4(σµσν − σνσµ)a
b.
– 55 –
Some useful identities are
(σµ)aa(σµ)bb = −2 εabεab ,
(σµ)aa(σµ)bb = −2εabεab,
εabεab(σµ)aa(σν)bb = −2 ηµν ,
(σµ)aa = εabεab(σµ)bb,
(σµσν + σν σµ) ba = −2 ηµνδba,
Tr(σµσν) = Tr(σµσν) = −2 ηµν ,
σµσν σµ = 2 σν ,
σµσνσµ = 2σν .
Complex conjugation Note that the matrices (7.6) and (7.8) are hermitian conjugate of each
other, i. e.
(MµνL )† = Mµν
R , (MµνR )† = Mµν
L .
The hermitian conjugate of the Lorentz transformation (7.7) is given by
[δ(ψL)a]†
= − i2δω∗µν [(ψL)b]
† [(Mµν
L ) ba
]†︸ ︷︷ ︸=(Mµν
R )b a
. (7.12)
For δωµν ∈ R this is of the same form as eq. (7.11). In Minkowski space it is therefore consistent
to take ψ†L to be a right-handed spinor with lower dotted index, we write
[(ψL)a]†
= (ψ†L)a,
and in an analogous way one finds that it is consistent to write[(ψR)a
]†= (ψ†R)a.
So far we have considered Minkowski space only. In Euclidean space or for more general complex
δωµν the hermitian conjugation is more complicated. For complex δωµν eq. (7.12) constitutes
a transformation behavior that is not of any already discussed type. For a consistent analytic
continuation it is actually necessary to have all fields transforming such that the infinitesimal
transformation law involves only δωµν (and not δω∗µν). We define therefore new fields
(ψL)a, (ψR)a,
with transformation laws
δ(ψL)a = − i2δωµν(ψL)b(M
µνR )b a,
δ(ψR)a = − i2δωµν(ψR)b(Mµν
L ) ab .
Only in Minkowski space one may identify (ψL)a = (ψ†L)a and (ψR)a = (ψ†R)a.
Connection between Lorentz group and SL(2,C) Consider the Lorentz transformation of
a left-handed spinor
(δψL)a =i
2δωµν(Mµν
L ) ba (ψL)b .
Decompose the infinitesimal Lorentz transformation into a boost
δω0j = −δωj0 = δϕj
– 56 –
and rotation
δωjk = εjkmδθm
and use
M j0 = −M0j =i
2σj
M jk =1
2εjklσl
so that
(δψL)a =i
2
(− iδϕj + δθj
)(σj)
ba (ψL)b .
In other words, the representation is of the form
U(Λ) = exp(1
2(ϕj + iθj)σj
).
Because there are real and imaginary parts, this is a general 2× 2 matrix, although with vanishing
determinant. We have established a map between Lorentz transformations SO(3, 1) and the group
SL(2,C) of linear transformations by complex 2 × 2 matrices with uni determinant. The Lorentz
transformations of left-handed spinors can be represented as elements of SL(2,C).
Exercise: Show that for a given matrix B in SL(2,C) the associated Lorentz transformation
matrix Λµν is
Λµν =1
2tr(σµ ·B · σν ·B†) .
Weyl fermions We now have everything to understand massless relativistic fermions. The La-
grange density for Weyl fermions is
L = i(ψL)a(σµ)ab∂µ(ψL)b .
Exercise: Check invariance under rotations and Lorentz boosts.
To obtain the equation of motion we vary with respect to ψL
δ
δ(ψL)a(x)S =
δ
δ(ψL)a(x)
∫d4xL
= i(σµ)ab∂µ(ψL)b(x)!= 0 .
This is the Weyl equation for a left-handed fermion. In Fourier representation
ψL(x) ∼ eiEt+ip·x
one has
(E 12 + p · σ)ψL = 0 .
Multiplying with E 12 − p · σ from the left gives(E212 − pipjσi, σj
)ψL = 0 .
With σi, σj = δij 12 the dispersion relation is therefore
E2 − p2 = 0
which describes massless particles, indeed.
– 57 –
Helicity Particle spin was defined for massive particles in the rest frame and we could choose
one axis, say the z-axis for the label of states. For massless particles this does not work, they
have no rest frame. One defines spin with respect to the momentum axis and defines helicity to be
determined by the operator
h =σ · pE
=σ · p|p|
.
For the left-handed fermions
hψL =σ · p|p|
ψL = (−1)ψL
so they have helicity −1. Analogously for right-handed fermions one sees they are massless particles
with helicity 1. These are right-handed Weyl fermions.
Exercise: Consider the Lagrangian
L = i(ψR)a(σµ)ab∂µ(ψR)b
and show it describes massless particles with helicity 1.
Transformations of fields So far we have discussed how the ”internal” indices of a field trans-
form under Lorentz transformations. However, a field depends on a space-time position xµ which
also transforms. This is already the case for a scalar field
φ(x) 7→ φ′(x) = φ(Λ−1x) .
(A maximum at xµ is moved to a maximum at Λµνxν .) In infinitesimal form
(Λ−1)µν = δµν − δωµν
and thus
φ(x) 7→ φ′(x) = φ(x)− xνδωµν∂µφ(x) .
This can be written as
φ′(x) = (1 +i
2δωµνMµν)
with generator
Mµν = −i(xµ∂ν − xν∂µ) .
Indeed, these generators form a representation of the Lorentz algebra
[Mµν ,Mρσ] = i(ηµρMνσ − ηµσMνρ − ηνρMµσ + ηνσMµρ) .
For higher representation fields, the complete generator contains Mµν and the generator of ”inter-
nal” transformations, for example
(ψL)a(x) 7→ (ψ′L)a(x) = (δ ba +
i
2δωµν(Mµν) b
a )(ψL)b(x)
with
(Mµν) ba := (Mµν
L ) ba +Mµνδ b
a .
7.4 Pioncare group
Poincare transformations consist of Lorentz transformations plus translations
xµ 7→ Λµνxν − bµ .
It is clear that these transformations form a group.
– 58 –
Exercise: Show the Poincare group indeed is a group with the composition law
(Λ2, b2) (Λ1, b1) = (Λ2Λ1,Λ2b1 + b2) .
As transformations of fields, translations are generated by the momentum operator
Pµ := −i∂µ .
For example
φ(x) 7→ φ′(x) = φ(Λ−1(x+ b))
≈ φ(xµ − δωµνxν + bµ)
= (1 +i
2δωµνMµν + ibµPµ)φ(x) .
One finds easily
[Pµ, Pν ] = 0 (7.13)
and[Mµν , Pρ] = i(δ µ
ρ P ν − δνρPµ)
[Mµν , P ρ] = i(ηµρP ν − ηνρPµ)(7.14)
which together with
[Mµν ,Mρσ] = i(ηµρMνσ − ηµσMνρ − ηνρMµσ + ηνσMµρ)
forms the Poincare algebra. The commutator (7.13) tells that the different components of the
energy-momentum operator can be diagonalized simultaneously, while (7.14) says that P ρ trans-
forms as a vector under Lorentz transformations.
Particles as representations One can understand particles as representations of the Poincare
algebra. Energy and momentum are the eigenvalues of
Pµ := −i∂µ
and the spin tells information about Mµν . One Casimir operator is
P 2 := PµPµ
which obviously commutes with Mµν and Pµ.
P 2 |p〉 = −m2 |p〉
gives the particle mass. The other Casimir follows from the Pauli-Lubanski vector
Wσ = −1
2εµνρσMµνP ρ .
In the particles rest frame P ρ gives pρ∗ = (m, 0, 0, 0) and
W0 = 0
Wj =1
2εjmnMmn = mJj
with spin operator Jj . The second Casimir of the Poincare algebra is WµWµ and
1
m2WµW
µ |p, j〉 = J2 |p, j〉 = j(j + 1) |p, j〉
so the states are characterized by the labels p and j. The commutation relations
[Wµ,Wν ] = iεµνρσWρPσ
reduce to the SO(3) algebra for the rest frame case: Pσ pσ∗ = (m, 0, 0, 0).
– 59 –
The little group For massive particles, spin can be characterized in terms of the rotation group
SO(3) in the particles rest frame. For massless particles this is more complicated and one must
pick another frame, for example where pµ = (p, 0, 0, p). The little group consists of transformations
that leave this invariant. This leads to a characterization in terms of helicity.
– 60 –
8 Conformal group
We are interested in transformations which describe a change of scale but preserves the angle
between line segments. These will lead to the conformal algebra which generates the conformal
group. An example of a conformal map is the Mercator projection
Consider transformations of the form
xµ 7→ xµ + ε ξµ(x) =: x′µ
for some infinitesimal ε. For a general metric tensor gµν the transformation is given by
gµν 7→ gµν∂xρ
∂x′µ∂xσ
∂x′ν=: g′µν
and any transformations such that
g′µν = Ω2(x)gµν
is a conformal transformation. These form a group called the conformal group. Expanding
Ω2(x) = 1 + εκ(x) +O(ε2) (8.1)
yields the conformal Killing equation
gµσ∂ρξµ + gρν∂σξ
ν + ξλ∂λgρσ + κgρσ = 0 (8.2)
up to first order in ε and ξµ is called the conformal Killing vector.
Exercise: Show that the expansion (8.1) implies the conformal Killing equation (8.2).
In the special case of the Minkowski metric gµν = ηµν the conformal Killing equation (8.2)
reduces to
∂σξρ + ∂ρξσ + κηρσ = 0 .
For the case κ = 0 the solutions are of the form
ξµ(x) = aµ + ωµνxν
with an antisymmetric ωµν = −ωνµ. For the more interesting case of κ 6= 0 we consider
xµ 7→ λxµ (8.3)
where λ = 1 + εc with a real number c ∈ R.
– 61 –
Exercise: Show that for the transformation (8.3) the conformal Killing vector and κ are given by
ξµ = cxµ , κ = −2c .
Now turning to inversions
xµ 7→ xµ
x2
we have
dxµ =δµλx
2 − 2xλxµ
x4dxλ
and thus
ds2 = ηµνdxµdxν 7→ 1
x4ηµνdx
µdxν
is a conformal transformation. Still we want to find conformal transformations which can be ex-
pressed in an infinitesimal form. Therefore consider
xµinversion7→ xµ
x2
translation7→ xµ
x2+ aµ
inversion7→xµ
x2 + aµ
ηρσ
(xρ
x2 + aρ)(
xσ
x2 + aσ)
infinitesimal≈ xµ + aλ(ηµλx2 − 2xµxλ) +O(a2)
and thus
ξµ = aλ(ηµλx2 − 2xµxλ)
is a conformal Killing vector. Taking the derivative of the conformal Killing equation yields
∂ρ∂ρξσ + ∂σ∂ρξρ + ∂σκ = 0
and with κ = −2∂µξ
µ
d where d = ηµνηµν is the dimension of space-time we arrive at
d∂µ∂µξσ = (2− d)∂σ∂µξ
µ .
For d = 2 the equation ∂µ∂µξσ = 0 has infinitely many solutions but for d > 2 ξσ can be at most
quadratic in xµ. The generators of the conformal algebra are
D = −ixµ∂µPµ = −i∂µKµ = −i(ηµνx2 − 2xµxν)∂ν
Mµν = −i(xµ∂ν − xν∂µ)
where we already know the commutation relations of Pµ and Mµν . We notice that
[D, (xα1 ...xαd)] = d(xα1 ...xαd)
[D, (∂α1 ...∂αd)] = −(∂α1 ...∂αd)
and the commutation relations read
[D,Pµ] = iPµ
[D,Mµν ] = 0
[Kµ,Kν ] = 0
[Mµν ,Kρ] = i(ηµρKν − ηνρKµ)
[Kµ, P ν ] = 2i(Mµν + ηµνD) .
(8.4)
Exercise: Explicitly derive the commutation relations (8.4).
– 62 –
9 Non-Abelian gauge theories
A gauge theory has a local symmetry as opposed to a global symmetry. The fields are invariant
under
ψa(x)→ U ba (x) ψb(x) = exp
[iωA(x) TA(x)
] b
aψb(x)
where U ba (x) depends on space and time. This is possible with the help of gauge fields, like for
example the photon field Aµ(x) and its non-Abelian generalisation. Let us concentrate on the group
theoretic aspects. The gauge group of the Standard Model is
SU(3)× SU(2)× U(1) .
The fermion fields and the Higgs boson scalar field can be classified into representations of the
corresponding algebras. With respect to the strong interaction group SU(3)colour we need the
singlet 1
triplet 3
anti-triplet 3∗
representations. With respect to the weak interaction group SU(2) we need
singlets 1
doublets 2 .
Recall that SU(2) is pseudo-real so there is no 2∗. Finally with respect to the hypercharge group
U(1)Y we will classify fields by their charge as generalisations of electric charge q. The charge turn
out to be
0 , ±1
6, ±1
3, ±1
2,
2
3, ±1 .
– 63 –
Moreover the fermions transform as Weyl spinors under the Lorentz group, either left- or right-
handed. There are the following fields(νLeL
):
neutrino
electron, left-handed ,
(1, 2, −1
2
)(νL eL
):
anti-neutrino
anti-electron, right-handed ,
(1, 2,
1
2
)eR : electron , right-handed ,
(1, 1, −1
)eR : anti-electron , left-handed ,
(1, 1, 1
)(uLdL
):
up-quark
down-quark, left-handed ,
(3, 2,
1
6
)(uL dL
):
anti-up-quark
anti-down-quark, right-handed ,
(3∗, 2, −1
6
)uR : up-quark , right-handed ,
(3, 1,
2
3
)uR : anti-up-quark , left-handed ,
(3∗, 1, −2
3
)dR : down-quark , right-handed ,
(3, 1, −1
3
)dR : anti-down-quark , left-handed ,
(3∗, 1,
1
3
)φ : Higgs-doublet , scalar ,
(1, 2,
1
2
)where the last expression determines the particles local symmetries:
(SU(3)colour, SU(2), Y
). The
fields have several indices corresponding to the different groups, for example(uR)am
where a ∈ 1, 2 is the Lorentz spinor index and m ∈ r, g, b is the SU(3)colour index.
– 64 –
10 Grand unification
We now discuss a proposed extension of the Standard Model which leads to a unification of the
gauge groups into SU(5). Note that the SU(3) and SU(2) generators naturally fit into SU(5)
generators and similar for the spinors(
3× 3
)SU(3) (
2× 2
)SU(2)
ψ1
ψ2
ψ3
ψ4
ψ5
.
There are 52 − 1 = 24 generators of SU(5) corresponding to the hermitian traceless 5× 5 matrices.
Out of them eight generate SU(3) while three generate SU(2). Moreover there is one hermitian
traceless matrix
1
2Y =
− 1
3
− 13
− 13
12
12
.
That generates a U(1) subgroup which actually gives U(1)Y . The remaining generators correspond
to additional gauge bosons not present in the Standard Model so they are supposedly very heavy
or confined. We find the embedding
SU(5)→ SU(3)× SU(2)× U(1) .
Now let us consider representations. Take the fundamental representation of SU(5) the spinor ψm.
It decomposes
5 =
(3, 1, −1
3
)⊕(
1, 2,1
2
)in a natural way. The conjugate decomposes
5∗ =
(3∗, 1,
1
3
)⊕(
1, 2, −1
2
).
Indeed this could be the representation for
dR ,(νL eL
)dR ,
(νLeL
).
So what about the other representations? The next smallest representation is the anti-symmetric
tensor ψmn with dimension ten. We still need(3, 2,
1
6
),
(3∗, 1, −2
3
),
(1, 1, 1
)and the corresponding anti-fields. These are ten fields indeed. Now ψmn decomposes into irreducible
representations (3, 1, −1
3
)⊗A
(3, 1, −1
3
)=
(3∗, 1, −2
3
)uR(
3, 1, −1
3
)⊗A
(1, 2,
1
2
)=
(3, 2,
1
6
) (uLdL
)(
1, 2,1
2
)⊗A
(1, 2,
1
2
)=
(1, 1, 1
)eR .
– 65 –
Note that we have used here non-trivial relations discussed before such as for SU(3)
3⊗ 3 = 3∗A ⊕ 6S
or for SU(2)
2⊗ 2 = 1A ⊕ 3S .
The U(1) charges are simply added. Indeed things work out! The fermion fields of a single generation
in the Standard Model can be organised into the SU(5) representations
5∗ : dR,
(νLeL
)and
10 : uR, eL,
(uLdL
)as well as corresponding anti-fields. The scalar Higgs field could be part of a 5 scalar representation
but the corresponding field with quantum numbers(3, 1, −2
3
)is not present in the Standard Model and must be very heavy or otherwise suppressed. The
hypothetical SU(5) gauge bosons that are neither SU(2) nor SU(3) bosons could in principle
induce transitionsd→ e+
u→ u
and thus u+ d→ u+ e+ causinguud→ uu+ e+
p→ π0 + e+ .
The proton could therefore in principle decay. This has not been observed so the transition rate
must be very small.
– 66 –
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