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Relativistic Quantum Field Theory
Section 6: Relation between relativistic equations and the Lorentz group
A Group G is a set with a rule for assigning to every (ordered) pair of elements a third
element, satisfying:
• If f, g ∈ G then h = f g ∈ G (closure)
• For f, g, h ∈ G, f (gh) = (f g)h (associative)
• There is an identity element i, such that for all f ∈ G, if = f i = f
• Every element f ∈ G has an inverse f −1, such that f f −1 = f −1f = i.
The Lorentz transformations are 4 × 4 matrices Λ that relate how coordinates in two inertial
reference frames are related x ν = Λνµxµ (some examples of Lorentz transformations are
given in the Revision Notes that have been handed out). Lorentz transformations form a
continuous group called SO(3,1). It has an infinite number of elements, since the parameters
of rotation and boost take on a continuum of values. A general Lorentz transformation has
6 real parameters (θ, ω), to which there correspond 6 generators
 0 0 0 0
0 0 0
0
0
0 0 0
0 0
0 0 
 0 0 0 −1 
 0 0 1 0 
J1 = −i 
J2 = −i
J3 = −i
 0 0 0 1 
 0 0 0 0 
 0 −1 0 0 






0 0 −1 0 


 0 1 0 0 

 0 0 0 0 
and
 0 1 0 0
0 0 1 0
0 0 0 1
1 0 0 0 
 0 0 0 0 
 0 0 0 0 
K1 = −i 
K2 = −i
K3 = −i
.
 0 0 0 0 
 1 0 0 0 
 0 0 0 0 






0 0 0 0 


 0 0 0 0 

 1 0 0 0 
The generators of rotation J are Hermitian and those of boost K are anti-Hermitian. In-
finitesimal transformations are given by, for example for a boost in the z-direction
Λ(δω3) = 1 + iK3δω3,
so that the corresponding finite boost is Λ(ω3) = exp(iK3ω3). A general Lorentz transfor-
mation (group element) is
Λ = exp (iJ · θ + iK · ω) .
(1)
The commutation relations of the generators can be calculated to give
[Ji, Jj] = i ijkJk
[Ki, Kj] = −i ijkJk
[Ji, Kj] = i ijkKk .
It is easier to understand the algebra by first defining
1
A ≡
(J + iK)
2
1
B ≡
(J − iK)
2
1

then,
[Ai, Aj] = i ijkAk
[Bi, Bj] = i ijkBk
[Ai, Bj] = 0 .
Observe that A and B both have the algebra of SU(2) (compare to commutation relations
for the Pauli matrices). Thus A and B each generate SU(2), so the Lorentz algebra is
SU (2) × SU (2). States transforming in a well-defined way will be labelled by two angular
momenta (j, j ), the first corresponding to A and second to B.
Representations of SU (2) × SU (2)
• (0, 0) is the simplest and corresponds to the scalar field case. The field transforms
trivially under a Lorentz transformation. Restricting to a linear equation, we seek
a equation which satisfies Dφ = 0 and in the primed frame retains the same form
D φ = 0, where φ(x) = φ (x = Λx). If assume a quadratic equation, then it implies
D = ∂µ∂µ + m2.
• ( 1 , 0) and (0, 1 ) - called the fundamental representation - the algebra of SU(2) for
2
2
angular momentum 1/2 is explicitly realized by the set of Pauli matrices, σ/2. A
general Lorentz transformation in terms of A and B can be written as
Λ = exp (iA · Θ1 + iB · Θ2) .
If this is compared to Eq. (1) it implies Θ1 = θ − iω and Θ2 = θ + iω.
Examining now the two cases, under ( 1 , 0) one state transforms as spin 1/2 and the
2
other as a scalar. This requires B = 0, which implies J = σ/2 and K = −iσ/2, where
θ and ω are both three component quantities. Thus a spinor φ1 transforms as
σ
φ1 → exp i
· (θ − iω) φ
2
1 ≡ M φ1.
(2)
Similarly for (0, 1 ) the spinor φ
2
2 associated with it transforms as
σ
φ2 → exp i
· (θ + iω) φ2 ≡ N φ2.
(3)
2
If we want to consider also improper Lorentz transformations, we can no longer consider
these 2-spinors separately. For example under parity, velocity v → −v, hence the
generator K changes sign, K → −K like a vector but J does not chance sign, J → J ,
like an axial vector. Thus under parity the representations get interchanged (j, 0) ↔
(0, j), thus φ1 ↔ φ2. To have a spinor in which a parity transformation can be done,
it requires a 4-spinor
φ1
ψ =
φ2
which under Lorentz transformation behaves as
φ
exp i σ · (θ − iω)
0
1
2
φ
→ 
 1
φ2

0
exp i σ · (θ + iω)
φ2
2

2

and under parity
φ
0 1
1
φ
→ 
 1
φ2
 1 0  φ2
We will relabel the 2-spinors at this stage to a notation that better reflects their relation
by parity transformation, thus
φ1 → φR φ2 → φL.
Consider a spinor of mass m at rest φR(p = 0) and a transformed one with momentum
p, φR(p), they are related by,
1
φR(p) = exp( σ · ω)φ
2
R(0)
γ + 1 1/2
γ − 1 1/2
=
+ σ · ˆ
p
φ
2
2
R(0)
E + m + σ · p
=
φ
2m(E + m)]1/2 R(0).
(4)
where we used cosh(ω/2) = [(γ +1)/2]1/2, sinh(ω/2) = [(γ −1)/2]1/2, and γ = E(p)/m.
Similarly
E + m − σ · p
φL(p) =
φ
2m(E + m)]1/2 L(0).
(5)
At rest, one cannot define spin as either left (L) of right (R) handed, so φR(0) = φL(0),
thus from Eqs. (4) and (5) it implies
 −m p0 +σ ·p  φR(p)

= 0 ,
(6)
p
φ
0 − σ · p
−m
 L(p)
which can be written as
(γµpµ − m)ψ(p) = 0 ,
(7)
where
φ
ψ(p) =
R(p)
,
φL(p)
0 −σ
0 1
γ =
and
γ0 =
.
(8)
σ
0
1 0
Eq. (7) is the Dirac equation for massive spin-1/2 particles in momentum space. In
particular Eq. (6) is in the Weyl representation. Noting that the matrices in Eq. (8)
satisfy {γµ, γν} = 2gµν, one has freedom to choose other representations such as the
Dirac representation given in Section 1 and primarily used in this course.
• ( 1 , 1 ) - this contains the four vector. To see this construct a general 2 index object
2
2
ψαα , in which the respective indices transform as (1/2, 0) and (0, 1/2), thus
ψββ = MβαNβ α ψαα ,
3

where M and N are the SU(2) transformations defined in Eqs. (2) and (3) respectively.
Thus ψαα is a four component quantity. Consider the expression
Aµ ≡ ψαα (σ2σµ)α α,
(9)
where σ0 = I and σi, i = 1, 2, 3, are the standard Pauli matrices. The above equation
is simply re-expressing these four independent components of ψαα in a different way.
Let us show that as implied by the notation Aµ indeed transforms like a Lorentz four-
vector. Consider for example the case of a rotation,
A µ = ψαα (σ2σµ)α α = MαβNα β ψββ (σ2σµ)α α
σ
σ

1 +
· (iθ + ω)
1 +
· (iθ − ω)
ψ
2
ββ (σ2σµ)α α
αβ
2
α β
σ
σ
= Aµ + iθ ·
αβ (σ2σµ)
α β (σ2σµ)
2
β α +
2
α β
ψββ
σ
σT
= Aµ + iθ · σ2σµ
+
σ2σµ
ψ
2
2
ββ .
(10)
β β
Now observe that σT = σ∗ and
σ2σ∗ = −σσ2
which leads from last line in Eq. (10) to
σ
σ
A µ = Aµ + iθ · σ2 σµ

σµ
ψ
2
2
ββ
β β
1
= Aµ + iθ · σ2 [σµ, σ]
ψ
2
ββ .
β β
For µ = 0 the commutator in the last line is zero, thus A 0 = A0 as expected for a
rotation. For the three spatial directions recall for Pauli matrices
σi, σk = 2i ikjσj
so that
A i = Ai − θk ikj (σ2σj)β βψββ = Ai − θk ikjAj,
where the last line follows from comparison with Eq. (9). Similarly one can show for
a Lorentz boost that Eq. (9) transforms like a Lorentz four-vector (tutorial problem).
A product of rotations and boosts can then generate any arbitrary Lorentz transfor-
mation.
If we now seek an equation which remains form invariant in any Lorentz frame and is
second order, the most general form would be
∂µ∂µAν + m2Aν + ξ∂ν(∂µAµ) = 0.
For m = 0 and ξ = 0, this is the Maxwell equations and for m = 0 this is the equation
for a massive vector field.
A. Berera, October 2008
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