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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

4

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

4