Zee A. Quantum Field Theory in a Nutshell

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II.1 The Dirac Equation

Staring into a fire

According to a physics legend, apparently even true, Dirac was staring into a fire one

evening in 1928 when he realized that he wanted, for reasons that are no longer relevant,

a relativistic wave equation linear in spacetime derivatives ∂

≡ ∂/∂x

. At that time, the

Klein-Gordon equation (∂

+ m

)ϕ = 0, which describes a free particle of mass m and

quadratic in spacetime derivatives, was already well known. This is in fact the equation of

motion of the scalar field theory we studied earlier.

At first sight, what Dirac wanted does not make sense. The equation is supposed to

have the form “some linear combination of ∂

acting on some field ψ is equal to some

constant times the field.” Denote the linear combination by c

∂

. If the c

’s are four

ordinary numbers, then the four-vector c

defines some direction and the equation cannot

be Lorentz invariant.

Nevertheless, let us follow Dirac and write, using modern notation,

(iγ

∂

− m)ψ =0 (1)

At this point, the four quantities iγ

are just the coefficients of ∂

and m is just a constant.

We have already argued that γ

cannot simply be four numbers. Well, let us see what these

objects have to be in order for this equation to contain the correct physics.

Acting on (1) with (iγ

∂

+ m), Dirac obtained −(γ

∂

+ m

)ψ = 0. It is tradi-

tional to define, in addition to the commutator [A, B] = AB − BA familiar from quan-

tum mechanics, the anticommutator {A, B}=AB +BA. Since derivatives commute,

∂

{γ

, γ

}∂

∂

, and we have (

{γ

, γ

}∂

∂

+ m

)ψ = 0. In a moment of

inspiration Dirac realized that if

{γ

, γ

}=2η

μν

(2)

with η

μν

the Minkowski metric he would obtain (∂

)ψ =0, which describes a particle

of mass m, and thus (1) would also describe a particle of mass m.

94 | II. Dirac and the Spinor

Since η

μν

is a diagonal matrix with diagonal elements η

= 1 and η

=−1, (2) says

that (γ

)

= 1, (γ

)

=−1, and γ

=−γ

for μ = ν. This last statement, that the

coefficients γ

anticommute with each other, implies that they indeed cannot be ordinary

numbers. Dirac’s thought would make sense if we could find four such objects.

Clifford algebra

A set of objects γ

(clearly d of them in d-dimensional spacetime) satisfying the relation

(2) is said to form a Clifford algebra. I will develop the mathematics of Clifford algebra

later. Suffice it for you to check here that the following 4 by 4 matrices satisfy (2):



I 0

0 −I



= I ⊗ τ

(3)



0 σ

−σ



= σ

⊗ iτ

(4)

Here σ and τ denote the standard Pauli matrices. For historical reasons the four matrices

are known as gamma matrices—not a very imaginative name! (Our convention is such

that whether an index on a Pauli matrix is upper or lower has no significance. On the other

hand, we define γ

≡ η

μν

and it does matter whether the index on a gamma matrix is

upper or lower; it is to be treated just like the index on any Lorentz vector. This convention

is useful because then γ

∂

= γ

∂

The direct product notation is convenient for computation: For example, γ

= (σ

⊗

iτ

)(σ

⊗ iτ

) = (σ

⊗ i

) =−(σ

⊗ I) and thus {γ

, γ

}=−{σ

, σ

}⊗I =

−2δ

as desired.

You can convince yourself that the γ

’s cannot be smaller than 4 by 4 matrices. The

mathematics forces the Dirac spinor ψ to have 4 components! The physical content of

the Dirac equation (1) is most transparent if we transform to momentum space: we plug

ψ(x) =



p/(2π)

−ipx

ψ(p) into (1) and obtain

(γ

− m)ψ(p) = 0 (5)

Since (5) is Lorentz invariant, as we will show below, we can examine its physical content

in any frame, in particular the rest frame p

= (m,



0), in which it becomes

(γ

− 1)ψ = 0 (6)

As (γ

−1)

=−2(γ

−1) we recognize (γ

−1) as a projection operator up to a trivial nor-

malization. Indeed, using the explicit form in (3), we see that there is nothing mysterious

to Dirac’s equation: When written out, (6) reads



0 I



ψ =0

thus telling us that 2 of the 4 components in ψ are zero.

II.1. The Dirac Equation | 95

This makes perfect sense since we know that the electron has 2 physical degrees of

freedom, not 4. Viewed in this light, the mysterious Dirac equation is no more and no less

than a projection that gets rid of the unwanted degrees of freedom. Compare our discussion

of the equation of motion of a massive spin 1 particle (chapter I.5). There also, 1 of the 4

components of A

is projected out. Indeed, the Klein-Gordon equation (∂

)ϕ(x) = 0

just projects out those Fourier components ϕ(k) not satisfying the mass shell condition

= m

. Our discussion provides a unified view of the equations of motion in relativistic

physics: They just project out the unphysical components.

A convenient notation introduced by Feynman, a ≡ γ

for any 4-vector a

, is now

standard. The Dirac equation then reads (i∂ −m)ψ =0.

Cousins of the gamma matrices

Under a Lorentz transformation x

ν

= 

, the 4 components of the vector field A

transform like, well, a vector. How do the 4 components of ψ transform? Surely not in the

same way as A

since even under rotation ψ and A

transform quite differently: one as

spin

and the other as spin 1. Let us write ψ(x)→ψ



) ≡S()ψ(x) and try to determine

the 4 by 4 matrix S().

It is a good idea to first sort out (and name) the 16 linearly independent 4 by 4 matrices.

We already know five of them: the identity matrix and the γ

’s. The strategy is simply

to multiply the γ

’s together, thus generating more 4 by 4 matrices until we get all 16.

Since the square of a gamma matrix γ

is equal to ±1 and the γ

’s anticommute with

each other, we have to consider only γ

, γ

, and γ

with μ, ν , λ, and ρ all

different from one another. Thus, the only product of four gamma matrices that we have

to consider is

≡ iγ

(7)

This combination is so important that it has its own name! (The peculiar name comes

about because in some old-fashioned notation the time coordinate was called x

with a

corresponding γ

.) We have

= i(I ⊗ τ

)(σ

⊗ iτ

)(σ

⊗ iτ

)(σ

⊗ iτ

) = i

(I ⊗ τ

)(σ

⊗ τ

)

and so

= I ⊗ τ



0 I

I 0



(8)

With the factor of i included, γ

is manifestly hermitean. An important property is that γ

anticommutes with the γ

’s:

{γ

, γ

}=0 (9)

96 | II. Dirac and the Spinor

Continuing, we see that the products of three gamma matrices, all different, can be

written as γ

(e.g., γ

=−iγ

). Finally, using (2) we can write the product of

two gamma matrices as γ

= η

μν

− iσ

μν

, where

μν

≡

[γ

, γ

] (10)

There are 4

3/2 = 6 of these σ

μν

matrices.

Count them, we got all 16. The set of 16 matrices {1, γ

, σ

μν

, γ

} forms a

complete basis of the space of all 4 by 4 matrices, that is, any 4 by 4 matrix can be written

as a linear combination of these 16 matrices.

It is instructive to write out σ

μν

explicitly in the representation (3) and (4):

= i



0 σ



(11)

= ε

ij k



0 σ



(12)

We see that σ

are just the Pauli matrices doubly stacked, for example,



0 σ



Lorentz transformation

Recall from a course on quantum mechanics that a general rotation can be written as



with



J the 3 generators of rotation and



θ 3 rotation parameters. Recall also that the

Lorentz group contains boosts in addition to rotations, with



K denoting the 3 generators of

boosts. Recall from a course on electromagnetism that the 6 generators {



J ,



K} transform

under the Lorentz group as the components of an antisymmetric tensor just like the

electromagnetic field F

μν

and thus can be denoted by J

μν

. I will discuss these matters

in more detail in chapter II.3. For the moment, suffice it to note that with this notation

we can write a Lorentz transformation as  = e

−

μν

, with J

generating rotations,

generating boosts, and the antisymmetric tensor ω

μν

=−ω

νμ

with its 6 = 4

3/2

components corresponding to the 3 rotation and 3 boost parameters.

Given the preceding discussion and the fact that there are six matrices σ

μν

, we suspect

that up to an overall numerical factor the σ

μν

’s must represent the 6 generators J

μν

of the

Lorentz group acting on a spinor. In fact, our suspicion is confirmed by thinking about

what a rotation e

−

does. Referring to (12) we see that if J

is represented by

this would correspond exactly to how a spin

particle transforms in quantum mechanics.

More precisely, separate the 4 components of the Dirac spinor into 2 sets of 2 components:

ψ =





(13)

II.1. The Dirac Equation | 97

From (12) we see that under a rotation around the 3rd axis, φ → e

−iω

φ and χ →

−iω

χ . It is gratifying to see that φ and χ transform like 2-component Pauli spinors.

We have thus figured out that a Lorentz transformation  acting on ψ is represented

by S() = e

−(i/4)ω

μν

, and so, acting on ψ , the generators J

μν

are indeed represented

μν

. Therefore we would expect that if ψ(x) satisfies the Dirac equation (1) then



) ≡ S()ψ(x) would satisfy the Dirac equation in the primed frame,

(iγ

∂



− m)ψ



) = 0 (14)

where ∂



≡ ∂/∂x

μ

. To show this, calculate [σ

μν

, γ

] = 2i(γ

νλ

− γ

μλ

) and hence

for ω infinitesimal Sγ

−1

= γ

− (i/4)ω

μν

[σ

μν

, γ

] = γ

+ γ

. Building up a finite

Lorentz transformation by compounding infinitesimal transformations (just as in the

standard discussion of the rotation group in quantum mechanics), we have Sγ

−1



Dirac bilinears

The Clifford algebra tells us that (γ

)

=+1 and (γ

)

=−1; hence the necessity for the i

in (4). One consequence of the i is that γ

is hermitean while γ

is antihermitean, a fact

conveniently expressed as

(γ

)

†

= γ

(15)

Thus, contrary to what you might think, the bilinear ψ

†

ψ is not hermitean; rather,

ψγ

ψ is hermitean with

ψ ≡ψ

†

. The necessity for introducing

ψ in addition to ψ

†

relativistic physics is traced back to the (+, −, −, −) signature of the Minkowski metric.

It follows that (σ

μν

)

†

=γ

μν

. Hence, S()

†

=γ

(i/4)ω

μν

, (which incidentally,

clearly shows that S is not unitary, a fact we knew since σ

is not hermitean), and so



) = ψ(x)

†

S()

†

ψ(x)e

+(i/4)ω

μν

. (16)

We have



)ψ



) =

ψ(x)e

+(i/4)ω

μν

−(i/4)ω

μν

ψ(x) =

ψ(x)ψ(x)

You are probably used to writing ψ

†

ψ in nonrelativistic physics. In relativistic physics you

have to get used to writing

ψψ.Itis

ψψ, not ψ

†

ψ , that transforms as a Lorentz scalar.

There are obviously 16 Dirac bilinears

ψψ that we can form, corresponding to the 16

linearly independent ’s. You can now work out how various fermion bilinears transform

(exercise II.1.1). The notation is rather nice: Various objects transform the way it looks like

they should transform. We simply look at the Lorentz indices they carry. Thus,

ψ(x)γ

ψ(x)

transforms as a Lorentz vector.

98 | II. Dirac and the Spinor

Parity

An important discrete symmetry in physics is that of parity or reflection in a mirror

→ x

μ

= (x

, −x) (17)

Multiply the Dirac equation (1) by γ

: γ

(iγ

∂

− m)ψ(x) = 0 = (iγ

∂



− m)γ

ψ(x),

where ∂



≡ ∂/∂x

μ

. Thus,



) ≡ ηγ

ψ(x) (18)

satisfies the Dirac equation in the space-reflected world (where η is an arbitrary phase that

we can set to 1).

Note, for example,



)ψ



) =

ψ(x)ψ(x) but



)γ



) =

ψ(x)γ

ψ(x)=

−

ψ(x)γ

ψ(x). Under a Lorentz transformation

ψ(x)γ

ψ(x) and

ψ(x)ψ(x) transform in

the same way but under space reflection they transform in an opposite way; in other words,

while

ψ(x)ψ(x) transforms as a scalar,

ψ(x)γ

ψ(x) transforms as a pseudoscalar.

You are now ready to do the all-important exercises in this chapter.

The Dirac Lagrangian

An interesting question: What Lagrangian would give Dirac’s equation? The answer is

L =

ψ(i∂ −m)ψ (19)

Since ψ is complex we can vary ψ and

ψ independently to obtain the Euler-Lagrange

equation of motion. Thus, ∂

(δL/δ∂

ψ) − δL/δψ =0 gives ∂

ψγ

) +m

ψ =0, which

upon hermitean conjugation and multiplication by γ

gives the Dirac equation (1). The

other variational equation ∂

(δL/δ∂

ψ) −δL/δ

ψ = 0 gives the Dirac equation even more

directly. (If you are disturbed by the asymmetric treatment of ψ and

ψ , you can always

integrate by parts in the action, have ∂

act on

ψ in the Lagrangian, and then average the

two forms of the Lagrangian. The action S =



xL treats ψ and

ψ symmetrically.)

Slow and fast electrons

Given a set of gamma matrices it is straightforward to solve the Dirac equation

( p − m)ψ(p) = 0 (20)

for ψ(p): It is a simple matrix equation (see exercise II.1.3).

Rotations consist of all linear transformations x

→R

such that det R =+1. Those transformations with

det R =−1 are composed of parity followed by a rotation. In (3 +1)-dimensional spacetime, parity can be defined

as reversing one of the spatial coordinates or all three spatial coordinates. The two operations are related by a

rotation. Note that in odd dimensional spacetime, parity is not the same as space inversion, in which all spatial

coordinates are reversed (see exercise II.1.12).

II.1. The Dirac Equation | 99

Note that if somebody uses the gamma matrices γ

, you are free to use instead γ

μ

−1

W with W any 4 by 4 matrix with an inverse. Obviously, γ

μ

also satisfy the Clifford

algebra. This freedom of choice corresponds to a simple change of basis. Physics cannot

depend on the choice of basis, but which basis is the most convenient depends on the

physics.

For example, suppose we want to study a slowly moving electron. Let us use the basis

defined by (3) and (4), and the 2-component decomposition of ψ (13). Since (6) tells us

that χ(p) = 0 for an electron at rest, we expect χ(p) to be much smaller than φ(p) for a

slowly moving electron.

In contrast, for momentum much larger than the mass, we can approximate (20) by

pψ(p) = 0. Multiplying on the left by γ

, we see that if ψ(p) is a solution then γ

ψ(p)

is also a solution since γ

anticommutes with γ

. Since (γ

)

= 1, we can form two

projection operators P

≡

(1 − γ

) and P

≡

(1 + γ

), satisfying P

= P

, P

= P

and P

=0. It is extremely useful to introduce the two combinations ψ

(1 −γ

)ψ

and ψ

(1 + γ

)ψ . Note that γ

=−ψ

and γ

=+ψ

. Physically, a relativistic

electron has two degrees of freedom known as helicities: it can spin either clockwise or

anticlockwise around the direction of motion. I leave it to you as an exercise to show that

and ψ

correspond to precisely these two possibilities. The subscripts L and R indicate

left and right handed. Thus, for fast moving electrons, a basis known as the Weyl basis,

designed so that γ

, rather than γ

, is diagonal, is more convenient. Instead of (3), we

choose



0 I

I 0



= I ⊗ τ

(21)

We keep γ

as in (4). This defines the Weyl basis. We now calculate

≡ iγ

= i(I ⊗ τ

)(σ

⊗ i

) =−(I ⊗ τ

) =



−I 0

0 I



(22)

which is indeed diagonal as desired. The decomposition into left and right handed fields is

of course defined regardless of what basis we feel like using, but in the Weyl basis we have

the nice feature that ψ

has two upper components and ψ

has two lower components.

The spinors ψ

and ψ

are known as Weyl spinors.

Note that in going from the Dirac to the Weyl basis γ

and γ

trade places (up to a sign):

Dirac : γ

diagonal; Weyl : γ

diagonal. (23)

Physics dictates which basis to use: We prefer to have γ

diagonal when we deal with

slowly moving spin

particles, while we prefer to have γ

diagonal when we deal with fast

moving spin

particles .

I note in passing that if we define σ

≡ (I , σ) and ¯σ

≡ (I , −σ) we can write



0 σ

¯σ



more compactly in the Weyl basis. (We develop this further in appendix E.)

100 | II. Dirac and the Spinor

Chirality or handedness

Regardless of whether a Dirac field ψ(x)is massive or massless, it is enormously useful to

decompose ψ into left and right handed fields ψ(x) =ψ

(x) + ψ

(x) ≡

(1 −γ

)ψ(x) +

(1 + γ

)ψ(x). As an exercise, show that you can write the Dirac Lagrangian as

L =

ψ(i∂ −m)ψ =

i∂ψ

− m(

) (24)

The kinetic energy connects left to left and right to right, while the mass term connects

left to right and right to left.

The transformation ψ → e

iθ

ψ leaves the Lagrangian L invariant. Applying Noether’s

theorem, we obtain the conserved current associated with this symmetry J

ψγ

ψ .

Projecting into left and right handed fields we see that they transform the same way:

→ e

iθ

and ψ

→ e

iθ

If m = 0, L enjoys an additional symmetry, known as a chiral symmetry, under which

ψ →e

iφγ

ψ . Noether’s theorem tells us that the axial current J

5μ

≡

ψγ

ψ is conserved.

The left and right handed fields transform in opposite ways: ψ

→ e

−iφ

and ψ

→

iφ

. These points are particularly obvious when L is written in terms of ψ

and ψ

,as

in (24).

In 1956 Lee and Yang proposed that the weak interaction does not preserve parity. It was

eventually realized (with these four words I brush over a beautiful chapter in the history

of particle physics; I urge you to read about it!) that the weak interaction Lagrangian has

the generic form

L = G

(25)

where ψ

1,2 , 3,4

denotes four Dirac fields and G the Fermi coupling constant. This La-

grangian clearly violates parity: Under a spatial reflection, left handed fields are trans-

formed into right handed fields and vice versa.

Incidentally, henceforth when I say a Lagrangian has a certain form, I will usually

indicate only one or more of the relevant terms in the Lagrangian, as in (25). The other

terms in the Lagrangian, such as

(i∂ − m

)ψ

, are understood. If the term is not

hermitean, then it is understood that we also add its hermitean conjugate.

Interactions

As we saw in (25) given the classification of bilinears in the spinor field you worked out

in an exercise it is easy to introduce interactions. As another example, we can couple

a scalar field ϕ to the Dirac field by adding the term gϕ

ψψ (with g some coupling

constant) to the Lagrangian L =

ψ(i∂ − m)ψ (and of course also adding the Lagrangian

for ϕ). Similarly, we can couple a vector field A

by adding the term eA

ψγ

ψ .We

note that in this case we can introduce the covariant derivative D

= ∂

− ieA

and write

II.1. The Dirac Equation | 101

L =

ψ(i∂ −m)ψ +eA

ψγ

ψ =

ψ(iγ

−m)ψ . Thus, the Lagrangian for a Dirac field

interacting with a vector field of mass μ reads

L =

ψ(iγ

− m)ψ −

μν

−

(26)

If the mass μ vanishes, this is the Lagrangian for quantum electrodynamics. Varying with

respect to

ψ, we obtain the Dirac equation in the presence of an electromagnetic field:

[iγ

(∂

− ieA

) − m]ψ =0 (27)

Charge conjugation and antimatter

With coupling to the electromagnetic field, we have the concept of charge and hence of

charge conjugation. Let us try to flip the charge e. Take the complex conjugate of (27):

[−iγ

μ∗

(∂

+ieA

) −m]ψ

∗

=0. Complex conjugating (2) we see that the −γ

μ∗

also satisfy

the Clifford algebra and thus must be the γ

matrices expressed in a different basis, that

is, there exists a matrix Cγ

(the notation with an explicit factor of γ

is standard; see

below) such that −γ

μ∗

= (Cγ

)

−1

(Cγ

). Plugging in, we find that

[iγ

(∂

+ ieA

) − m]ψ

= 0 (28)

where we have defined ψ

≡ Cγ

∗

. Thus, if ψ is the field of the electron, then ψ

is the

field of a particle with a charge opposite to that of the electron but with the same mass,

namely the positron.

The discovery of antimatter was one of the most momentous in twentieth-century

physics. We will discuss antimatter in more detail in the next chapter.

It may be instructive to look at the specific form of the charge conjugation matrix C.We

can write the defining equation for C as Cγ

μ∗

−1

=−γ

. Complex conjugating the

equation (γ

)

†

= γ

, we obtain (γ

)

= γ

μ∗

if γ

is real. Thus,

(γ

)

=−C

−1

C (29)

which explains why C is defined with a γ

attached.

In both the Dirac and the Weyl bases γ

is the only imaginary gamma matrix. Then the

defining equation for C just says that Cγ

commutes with γ

but anticommutes with the

other three γ matrices. So evidently C = γ

[up to an arbitrary phase not fixed by (29)]

and indeed γ

μ∗

= γ

. Note that we have the simple (and satisfying) relation

= γ

∗

(30)

You can easily convince yourself (exercise II.1.9) that the charge conjugate of a left

handed field is right handed and vice versa. As we will see later, this fact turns out to

be crucial in the construction of grand unified theory. Experimentally, it is known that the

neutrino is left handed. Thus, we can now predict that the antineutrino is right handed.