4Spinors
III Quantum Field Theory
4.1 The Lorentz group and the Lorentz algebra
So far, we have only been working with scalar fields. These are pretty boring,
since when we change coordinates, the values of the field remain unchanged. One
can certainly imagine more exciting fields, like the electromagnetic potential.
Ignoring issues of choice of gauge, we can imagine the electromagnetic potential
as a 4vector
A
µ
at each point in spacetime. When we change coordinates by a
Lorentz transformation Λ, the field transforms by
A
µ
(x) 7→ Λ
µ
ν
A
ν
(Λ
−1
x).
Note that we had to put a Λ
−1
inside
A
ν
because the names of the points have
changed. Λ
−1
x
in the new coordinate system labels the same point as
x
in the
old coordinate system.
In general, we can consider vectorvalued fields that transform when we
change coordinates. If we were to construct such a field, then given any Lorentz
transformation Λ, we need to produce a corresponding transformation D(Λ) of
the field. If our field
φ
takes values in a vector space
V
(usually
R
n
), then this
D(Λ) should be a linear map V → V . The transformation rule is then
x 7→ Λx,
φ 7→ D(Λ)φ.
We want to be sure this assignment of
D
(Λ) behaves sensibly. For example, if we
apply the Lorentz transformation Λ =
1
, i.e. we do nothing, then
D
(
1
) should
not do anything as well. So we must have
D(1) = 1.
Now if we do Λ
1
and then Λ
2
, then our field will transform first by
D
(Λ
1
), then
D
(Λ
2
). For the universe to make sense, this had better be equal to
D
(Λ
2
Λ
1
).
So we require
D(Λ
1
)D(Λ
2
) = D(Λ
1
Λ
2
)
for any Λ
1
,
Λ
2
. Mathematically, we call this a representation of the Lorentz
group.
Definition
(Lorentz group)
.
The Lorentz group, denoted O(1
,
3), is the group
of all Lorentz transformations. Explicitly, it is given by
O(1, 3) = {Λ ∈ M
4×4
: Λ
T
ηΛ = η},
where
η = η
µν
=
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
is the Minkowski metric. Alternatively, O(1
,
3) is the group of all matrices Λ
such that
hΛx, Λyi = hx, yi,
for all
x, y ∈ R
1+3
, where
hx, yi
denotes the inner product given by the Minkowski
metric
hx, yi = x
T
ηy.
Definition
(Representation of the Lorentz group)
.
A representation of the
Lorentz group is a vector space
V
and a linear map
D
(Λ) :
V → V
for each
Λ ∈ O(1, 3) such that
D(1) = 1, D(Λ
1
)D(Λ
2
) = D(Λ
1
Λ
2
)
for any Λ
1
, Λ
2
∈ O(1, 3).
The space V is called the representation space.
So to construct interesting fields, we need to find interesting representations!
There is one representation that sends each Λ to the matrix Λ acting on
R
1+3
,
but that is really boring.
To find more representations, we will consider “infinitesimal” Lorentz trans
formations. We will find representations of these infinitesimal Lorentz trans
formations, and then attempt to patch them up to get a representation of the
Lorentz group. It turns out that we will fail to patch them up, but instead we
will end up with something more interesting!
We write our Lorentz transformation Λ as
Λ
µ
ν
= δ
µ
ν
+ εω
µ
ν
+ O(ε
2
).
The definition of a Lorentz transformation requires
Λ
µ
σ
Λ
ν
ρ
η
σρ
= η
µν
.
Putting it in, we have
(δ
µ
σ
+ εω
µ
σ
)(δ
ν
ρ
+ εω
ν
ρ
)η
ρσ
+ O(ε
2
) = η
µν
.
So we find that we need
ω
µν
+ ω
νµ
= 0.
So
ω
is antisymmetric. Thus we figured that an infinitesimal Lorentz transfor
mation is an antisymmetric matrix. This is known as the Lie algebra of the
Lorentz group.
Definition (Lorentz algebra). The Lorentz algebra is
o(1, 3) = {ω ∈ M
4×4
: ω
µν
+ ω
νµ
= 0}.
It is important that when we say that ω is antisymmetric, we mean exactly
ω
µν
+
ω
νµ
= 0. Usually, when we write out a matrix, we write out the entries of
ω
µ
ν
instead, and the matrix we see will not in general be antisymmetric, as we
will see in the examples below.
A 4
×
4 antisymmetric matrix has 6 free components. These 6 components
in fact correspond to three infinitesimal rotations and three infinitesimal boosts.
We introduce a basis given by the confusing expression
(M
ρσ
)
µ
ν
= η
ρµ
δ
σ
ν
− η
σµ
δ
ρ
ν
.
Here the
ρ
and
σ
count which basis vector (i.e. matrix) we are talking about,
and
µ, ν
are the rows and columns. Usually, we will just refer to the matrix as
M
ρσ
and not mention the indices µ, ν to avoid confusion.
Each basis element will have exactly one pair of nonzero entries. For example,
we have
(M
01
)
µ
ν
=
0 1 0 0
1 0 0 0
0 0 0 0
0 0 0 0
, (M
12
)
µ
ν
=
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
.
These generate a boost in the
x
1
direction and a rotation in the
x
1

x
2
plane
respectively, as we will later see.
By definition of a basis, we can write all matrices
ω
in the Lorentz algebra
as a linear combination of these:
ω =
1
2
Ω
ρσ
M
ρσ
.
Note that here
M
ρσ
and
M
σρ
will be the negative of each other, so
{M
ρσ
}
doesn’t
really form a basis. By symmetry, we will often choose Ω so that Ω
ρσ
=
−
Ω
σρ
as well.
Now we can talk about representations of the Lorentz algebra. In the case of
representations of the Lorentz group, we required that representations respect
multiplication. In the case of the Lorentz algebra, the right thing to ask to be
preserved is the commutator (cf. III Symmetries, Fields and Particles).
Definition
(Representation of Lorentz algebra)
.
A representation of the Lorentz
algebra is a collection of matrices that satisfy the same commutation relations as
the Lorentz algebra.
Formally, this is given by a vector space V and a linear map R(ω) : V → V
for each ω ∈ o(1, 3) such that
R(aω + bω
0
) = aR(ω) + bR(ω
0
), R([ω, ω
0
]) = [R(ω), R(ω
0
)]
for all ω, ω
0
∈ o(1, 3) and a, b ∈ R.
It turns out finding representations of the Lorentz algebra isn’t that bad. We
first note the following magic formula about the basis vectors of the Lorentz
algebra:
Proposition.
[M
ρσ
, M
τν
] = η
στ
M
ρν
− η
ρτ
M
σν
+ η
ρν
M
στ
− η
σν
M
ρτ
.
This can be proven, painfully, by writing out the matrices.
It turns out this is the only thing we need to satisfy:
Fact.
Given any vector space
V
and collection of linear maps
L
ρσ
:
V → V
,
they give a representation of the Lorentz algebra if and only if
[L
ρσ
, L
τν
] = η
στ
L
ρν
− η
ρτ
L
σν
+ η
ρν
L
στ
− η
σν
L
ρτ
.
Suppose we have already found a representation of the Lorentz algebra. How
can we find a representation of the Lorentz group? We will need to make use of
the following fact:
Fact.
Let Λ be a Lorentz transformation that preserves orientation and does not
reverse time. Then we can write Λ =
exp
(
ω
) for some
ω ∈ o
(1
,
3). In coordinates,
we can find some Ω
ρσ
such that
Λ = exp
1
2
Ω
ρσ
M
ρσ
. (∗)
Thus if we have found some representation
R
(
M
ρσ
), we can try to define a
representation of the Lorentz group by
R(Λ) = exp(R(ω)) = exp
1
2
Ω
ρσ
R(M
ρσ
)
. (†)
Now we see two potential problems with our approach. The first is that we can
only lift a representation of the Lorentz algebra to the parity and timepreserving
Lorentz transformations, and not all of them. Even worse, it might not even be
welldefined for these nice Lorentz transformations. We just know that we can
find some Ω
ρσ
such that (
∗
) holds. In general, there can be many such Ω
ρσ
, and
they might not give the same answer when we evaluate (†)!
Definition
(Restricted Lorentz group)
.
The restricted Lorentz group consists
of the elements in the Lorentz group that preserve orientation and direction of
time.
Example. Recall that we had
(M
12
)
µ
ν
=
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
We pick
Ω
12
= −Ω
21
= −φ
3
,
with all other entries zero. Then we have
1
2
Ω
ρσ
M
ρσ
=
0 0 0 0
0 0 φ
3
0
0 −φ
3
0 0
0 0 0 0
So we know
Λ = exp
1
2
Ω
ρσ
M
ρσ
=
1 0 0 0
0 cos φ
3
sin φ
3
0
0 −sin φ
3
cos φ
3
0
0 0 0 1
,
which is a rotation in the
x
1

x
2
plane by
φ
3
. It is clear that
φ
3
is not uniquely
defined by Λ. Indeed,
φ
3
+ 2
nπ
for any
n
would give the same element of the
Lorentz group.
More generally, given any rotation parameters
φ
= (
φ
1
, φ
2
, φ
3
), we obtain a
rotation by setting
Ω
ij
= −ε
ijk
φ
k
.