Zee A. Quantum Field Theory in a Nutshell

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102 | II. Dirac and the Spinor

Furthermore, ψ

transforms as a spinor. Let’s check: Under a Lorentz transformation

ψ →e

−(i/4)ω

μν

ψ, complex conjugating we have ψ

∗

→ e

+(i/4)ω

μν

(σ

μν

)

∗

; hence ψ

→

+(i/4)ω

μν

(σ

μν

)

∗

−(i/4)ω

μν

. [Recall from (10) that σ

μν

is defined with an explicit

i.]

Note that C

= γ

=−C in both the Dirac and the Weyl bases.

Majorana neutrino

Since ψ

transforms as a spinor, Majorana

noted that Lorentz invariance allows not only

the Dirac equation i∂ψ = mψ but also the Majorana equation

i∂ψ = mψ

(31)

Complex conjugating this equation and multiplying by γ

, we have −γ

iγ

μ∗

∂

∗

m(−γ

)ψ , that is, i∂ψ

= mψ . Thus, −∂

ψ =i∂(i∂ψ) = i∂mψ

= m

ψ . As we antic-

ipated, m is indeed the mass, known as a Majorana mass, of the particle associated with

ψ.

The Majorana equation (31) can be obtained from the Lagrangian

L =

ψi∂ψ −

m(ψ

Cψ +

ψC

) (32)

upon varying

ψ .

Since ψ and ψ

carry opposite charge, the Majorana equation, unlike the Dirac equation,

can only be applied to electrically neutral fields. However, as ψ

is right handed if ψ is left

handed, the Majorana equation, again unlike the Dirac equation, preserves handedness.

Thus, the Majorana equation is almost tailor made for the neutrino.

From its conception the neutrino was assumed to be massless, but couple of years ago

experimentalists established that it has a small but nonvanishing mass. As of this writing, it

is not known whether the neutrino mass is Dirac or Majorana. We will see in chapter VII.7

that a Majorana mass for the neutrino arises naturally in the SO(10) grand unified theory.

Finally, there is the possibility that ψ =ψ

, in which case ψ is known as a Majorana

spinor.

Time reversal

Finally, we come to time reversal,

which as you probably know, is much more confusing

to discuss than parity and charge conjugation. In a famous 1932 paper Wigner showed that

Ettore Majorana had a brilliant but tragically short career. In his early thirties, he disappeared off the coast

of Sicily during a boat trip. The precise cause of his death remains a mystery. See F. Guerra and N. Robotti, Ettore

Majorana: Aspects of His Scientific and Academic Activity.

Upon recalling that C is antisymmetric, you may have worried that ψ

Cψ = C

αβ

vanishes. In future

chapters we will learn that ψ has to be treated as anticommuting “Grassmannian numbers.”

Incidentally, I do not feel that we completely understand the implications of time-reversal invariance. See

A. Zee, “Night thoughts on consciousness and time reversal,” in: Art and Symmetry in Experimental Physics: pp.

246–249 .

II.1. The Dirac Equation | 103

time reversal is represented by an antiunitary operator. Since this peculiar feature already

appears in nonrelativistic quantum physics, it is in some sense not the responsibility of a

book on relativistic quantum field theory to explain time reversal as an antiunitary operator.

Nevertheless, let me try to be as clear as possible. I adopt the approach of “letting the

physics, namely the equations, lead us.”

Take the Schr

odinger equation i(∂/∂t)(t) = H (t) (and for definiteness, think of

H =−(1/2m)∇

+ V(x), just simple one particle nonrelativistic quantum mechanics.)

We suppress the dependence of  on x. Consider the transformation t →t



=−t .We

want to find a 



) such that i(∂/∂t



)



) = H



). Write 



) = T (t), where T

is some operator to be determined (up to some arbitrary phase factor η). Plugging in, we

have i[∂/∂(−t)]T (t) =HT(t). Multiply by T

−1

, and we obtain T

−1

(−i)T (∂/∂t)(t) =

−1

HT(t). Since H does not involve time in any way, we want T

−1

H = HT

−1

. Then

−1

(−i)T (∂/∂t)(t) = H (t). We are forced to conclude, as Wigner was, that

−1

(−i)T =i (33)

Speaking colloquially, we can say that in quantum physics time goes with an i and so

flipping time means flipping i as well.

Let T = UK, where K complex conjugates everything to its right. Then T

−1

= KU

−1

and (33) holds if U

−1

iU = i , that is, if U

−1

is just an ordinary (unitary) operator that does

nothing to i. We will determine U as we go along. The presence of K makes T “antiunitary.”

We check that this works for a spinless particle in a plane wave state (t) = e



x−Et)

Plugging in, we have 



) = T (t ) = UK(t) = U

∗

(t) = Ue

−i(



x−Et)

; since  has

only one component, U is just a phase factor

η that we can choose to be 1. Rewriting, we

have 



(t) = e

−i(



x+Et)

= e

i(−



x−Et)

. Indeed, 



describes a plane wave moving in the

opposite direction. Crucially, 



(t) ∝ e

−iEt

and thus has positive energy as it should. Note

that acting on a spinless particle T

= UKUK = UU

∗

=+1.

Next consider a spin

nonrelativistic electron. Acting with T on the spin-up state





we want to obtain the spin-down state





. Thus, we need a nontrivial matrix U = ησ

to flip the spin:





= U





= iη





Similarly, T acting on the spin-down state produces the spin-up state. Note that acting on

a spin

particle

= ησ

Kησ

K = ησ

∗

KK =−1

This is the origin of Kramer’s degeneracy: In a system with an odd number of electrons

in an electric field, no matter how complicated, each energy level is twofold degenerate.

The proof is very simple: Since the system is time reversal invariant,  and T have the

same energy. Suppose they actually represent the same state. Then T= e

iα

 , but then

It is a phase factor rather than an arbitrary complex number because we require that |



=||

104 | II. Dirac and the Spinor

 =T(T)= Te

iα

 =e

−iα

T= =− .So and Tmust represent two distinct

states.

All of this is beautiful stuff, which as I noted earlier you could and should have learned

in a decent course on quantum mechanics. My responsibility here is to show you how it

works for the Dirac equation. Multiplying (1) by γ

from the left, we have i(∂/∂t)ψ(t ) =

H ψ(t) with H =−iγ

∂

+ γ

m. Once again, we want i(∂/∂t



)ψ



) = Hψ



) with



) = T ψ(t) and T some operator to be determined. The discussion above carries

over if T

−1

HT = H , that is, KU

−1

HUK =H . Thus, we require KU

−1

UK = γ

and

−1

(iγ

)UK = iγ

. Multiplying by K on the left and on the right, we see that

we have to solve for a U such that U

−1

U = γ

0∗

and U

−1

U =−γ

i∗

. We now restrict

ourselves to the Dirac and Weyl bases, in both of which γ

is the only imaginary guy. Okay,

what flips γ

and γ

but not γ

and γ

? Well, U =ηγ

(with η an arbitrary phase factor)

works:



) = ηγ

Kψ(t) (34)

Since the γ

’s are the same in both the Dirac and the Weyl bases, in either we have from

(4)

U = η(σ

⊗ iτ

)(σ

⊗ iτ

) = ηiσ

⊗ 1

As we expect, acting on the 2-component spinors contained in ψ , the time reversal operator

T involves multiplying by iσ

. Note also that as in the nonrelativistic case T

ψ =−ψ.

It may not have escaped your notice that γ

appears in the parity operator (18), γ

charge conjugation (30), and γ

in time reversal (34). If we change a Dirac particle to its

antiparticle and flip spacetime, γ

appears.

CPT theorem

There exists a profound theorem stating that any local Lorentz invariant field theory must

be invariant under

CP T , the combined action of charge conjugation, parity, and time

reversal. The pedestrian proof consists simply of checking that any Lorentz invariant local

interaction you can write down [such as (25)], while it may break charge conjugation,

parity, or time reversal separately, respects CP T . The more fundamental proof involves

considerable formal machinery that I will not develop here. You are urged to read about

the phenomenological study of charge conjugation, parity, time reversal, and CP T , surely

one of the most fascinating chapters in the history of physics.

A rather pedantic point, but potentially confusing to some students, is that I distinguish carefully between the

action of charge conjugation C and the matrix C: Charge conjugation C involves taking the complex conjugate of

ψ and then scrambling the components with Cγ

. Similarly, I distinguish between the operation of time reversal

T and the matrix T .

See, e.g., J. J. Sakurai, Invariance Principles and Elementary Particles and E. D. Commins, Weak Interactions.

II.1. The Dirac Equation | 105

Two stories

I end this chapter with two of my favorite physics stories—one short and one long.

Paul Dirac was notoriously a man of few words. Dick Feynman told the story that when

he first met Dirac at a conference, Dirac said after a long silence, “I have an equation; do

you have one too?”

Enrico Fermi did not usually take notes, but during the 1948 Pocono conference (see

chapter I.7) he took voluminous notes during Julian Schwinger’s lecture. When he got

back to Chicago, he assembled a group consisting of two professors, Edward Teller and

Gregory Wentzel, and four graduate students, Geoff Chew, Murph Goldberger, Marshall

Rosenbluth, and Chen-Ning Yang (all to become major figures later). The group met in

Fermi’s office several times a week, a couple of hours each time, to try to figure out what

Schwinger had done. After 6 weeks, everyone was exhausted. Then someone asked, “Didn’t

Feynman also speak?” The three professors, who had attended the conference, said yes.

But when pressed, not Fermi, nor Teller, nor Wentzel could recall what Feynman had said.

All they remembered was his strange notation: p with a funny slash through it.

Exercises

II.1.1 Show that the following bilinears in the spinor field

ψψ,

ψγ

ψ ,

ψσ

μν

ψ,

ψγ

ψ , and

ψγ

ψ trans-

form under the Lorentz group and parity as a scalar, a vector, a tensor, a pseudovector or axial vector, and

a pseudoscalar, respectively. [Hint: For example,

ψγ

ψ →

ψ[1 + (i/4)ωσ ]γ

[1 −(i/4)ωσ ]ψ under

an infinitesimal Lorentz transformation and →

ψγ

ψ under parity. Work out these transforma-

tion laws and show that they define an axial vector.]

II.1.2 Write all the bilinears in the preceding exercise in terms of ψ

and ψ

II.1.3 Solve ( p − m)ψ(p) = 0 explicitly (by rotational invariance it suffices to solve it for p along the 3rd

direction, say). Verify that indeed χ is much smaller than φ for a slowly moving electron. What happens

for a fast moving electron?

II.1.4 Exploiting the fact that χ is much smaller than φ for a slowly moving electron, find the approximate

equation satisfied by φ.

II.1.5 For a relativistic electron moving along the z-axis, perform a rotation around the z-axis. In other words,

study the effect of e

−(i/4)ωσ

on ψ(p) and verify the assertion in the text regarding ψ

and ψ

II.1.6 Solve the massless Dirac equation.

II.1.7 Show explicitly that (25) violates parity.

II.1.8 The defining equation for C evidently fixes C only up to an overall constant. Show that this constant is

fixed by requiring (ψ

)

= ψ .

C. N. Yang, Lecture at the Schwinger Memorial Session of the American Physical Society meeting in

Washington D. C., 1995.

106 | II. Dirac and the Spinor

II.1.9 Show that the charge conjugate of a left handed field is right handed and vice versa.

II.1.10 Show that ψCψ is a Lorentz scalar.

II.1.11 Work out the Dirac equation in (1 + 1)-dimensional spacetime.

II.1.12 Work out the Dirac equation in (2 + 1)-dimensional spacetime. Show that the apparently innocuous

mass term violates parity and time reversal. [Hint: The three γ

’s are just the three Pauli matrices with

appropriate factors of i.]

II.2 Quantizing the Dirac Field

Anticommutation

We will use the canonical formalism of chapter I.8 to quantize the Dirac field.

Long and careful study of atomic spectroscopy revealed that the wave function of two

electrons had to be antisymmetric upon exchange of their quantum numbers. It follows

that we cannot put two electrons into the same energy level so that they will have the

same quantum numbers. In 1928 Jordan and Wigner showed how this requirement of an

antisymmetric wave function can be formalized by having the creation and annihilation

operators for electrons satisfy anticommutation rather than commutation relations as in

(I.8.12).

Let us start out with a state with no electron |0 and denote by b

†

the operator creating

an electron with the quantum numbers α. In other words, the state b

†

|0 is the state

with an electron having the quantum numbers α. Now suppose we want to have another

electron with the quantum numbers β , so we construct the state b

†

|0. For this to be

antisymmetric upon interchanging α and β we must have

†

, b

†

}≡b

†

+ b

†

= 0 (1)

Upon hermitean conjugation, we have {b

, b

}=0. In particular, b

†

= 0, so that we

cannot create two electrons with the same quantum numbers.

To this anticommutation relation we add

, b

†

}=δ

αβ

(2)

One way of arguing for this is to say that we would like the number operator to be

N =



†

, just as in the bosonic case. Show with one line of algebra that [AB , C] =

A[B , C] + [A, C]B or [AB , C] = A{B , C}−{A, C}B. (A heuristic way of remembering

the minus sign in the anticommuting case is that we have to move C past B in order for

C to do its anticommuting with A.) For the desired number operator to work we need

[



†

, b

†

] =+b

†

(so that as usual N |0=0, and Nb

†

|0=b

†

|0) and so we have (2).

108 | II. Dirac and the Spinor

The Dirac field

Let us now turn to the free Dirac Lagrangian

L =

ψ(i∂ −m)ψ (3)

The momentum conjugate to ψ is π

= δL/δ∂

= iψ

†

. We anticipate that the correct

canonical procedure requires imposing the anticommutation relation:

{ψ

(x, t), ψ

†

(



0, t)}=δ

(3)

(x)δ

αβ

(4)

We will derive this below.

The Dirac field satisfies

(i∂ − m)ψ = 0 (5)

Plugging in plane waves u(p, s)e

−ipx

and v(p, s)e

ipx

for ψ , we have

(p − m)u(p, s) =0 (6)

and

(p + m)v(p, s) = 0 (7)

The index s =±1 reminds us that each of these two equations has two solutions, spin

up and spin down. Evidently, under a Lorentz transformation the two spinors u and v

transform in the same way as ψ . Thus, if we define ¯u ≡ u

†

and ¯v ≡ v

†

, then ¯uu and

¯vv are Lorentz scalars.

This subject is full of “peculiar” signs and so I will proceed very carefully and show you

how every sign makes sense.

First, since (6) and (7) are linear we have to fix the normalization of u and v. Since

¯u(p, s)u(p, s) and ¯v(p, s)v(p, s) are Lorentz scalars, the normalization condition we

impose on them in the rest frame will hold in any frame.

Our strategy is to do things in the rest frame using a particular basis and then invoke

Lorentz invariance and basis independence. In the rest frame, (6) and (7) reduce to

(γ

− 1)u = 0 and (γ

+ 1)v = 0. In particular, in the Dirac basis γ



I 0

0 −I



, so the

two independent spinors u (labeled by spin s =±1) have the form

⎛

⎜

⎝

⎞

⎟

⎠

and

⎛

⎜

⎝

⎞

⎟

⎠

II.2. Quantizing the Dirac Field | 109

while the two independent spinors v have the form

⎛

⎜

⎝

⎞

⎟

⎠

and

⎛

⎜

⎝

⎞

⎟

⎠

The normalization conditions we have implicitly chosen are then ¯u(p, s)u(p, s) = 1 and

¯v(p, s)v(p, s) =−1. Note the minus sign thrust upon us. Clearly, we also have the orthog-

onality condition ¯uv = 0 and ¯vu = 0. Lorentz invariance and basis independence then tell

us that these four relations hold in general.

Furthermore, in the rest frame



(p, s)¯u

(p, s) =



I 0



αβ

(γ

+ 1)

αβ

and



(p, s)¯v

(p, s) =



0 −I



αβ

(γ

− 1)

αβ

Thus, in general



(p, s)¯u

(p, s) =



p + m



αβ

(8)

and



(p, s)¯v

(p, s) =



p − m



αβ

(9)

Another way of deriving (8) is to note that the left hand side isa4by4matrix (it is like a

column vector multiplied by a row vector on the right) and so must be a linear combination

of the sixteen 4 by 4 matrices we listed in chapter II.1. Argue that γ

and γ

are ruled out

by parity and that σ

μν

is ruled out by Lorentz invariance and the fact that only one Lorentz

vector, namely p

, is available. Hence the right hand side must be a linear combination of

p and m. Fix the relative coefficient by acting with p − m from the left. The normalization

is fixed by setting α = β and summing over α. Similarly for (9). In particular, setting α = β

and summing over α, we recover ¯v(p, s)v(p, s) =−1.

We are now ready to promote ψ(x) to an operator. In analogy with (I.8.11) we expand

the field in plane waves

(x) =



(2π)

/m)



[b(p, s)u

(p, s)e

−ipx

+ d

†

(p, s)v

(p, s)e

ipx

] (10)

(Here E

= p



p

+ m

and px = p

.) The normalization factor (E

/m)

slightly different from that in (I.8.11) for reasons we will see. Otherwise, the rationale

The notation is standard. See e.g., J. A. Bjorken and S. D. Drell, Relativistic Quantum Mechanics.

110 | II. Dirac and the Spinor

for (10) is essentially the same as in (I.8.11). We integrate over momentum p, sum over

spin s, expand in plane waves, and give names to the coefficients in the expansion. Because

ψ is complex, we have, similar to the complex scalar field in chapter I.8, a b operator and

a d

†

operator.

Just as in chapter I.8, the operators b and d

†

must carry the same charge. Thus, if b

annihilates an electron with charge e =−|e|, d

†

must remove charge e; that is, it creates

a positron with charge −e =|e|.

A word on notation: in (10) b(p, s), d

†

(p, s), u(p, s), and v(p, s) are written as functions

of the 4-momentum p but strictly speaking they are functions of p only, with p

always

understood to be +



p

+ m

Thus let b

†

(p, s) and b(p, s) be the creation and annihilation operators for an electron

of momentum p and spin s. Our introductory discussion indicates that we should impose

{b(p, s), b

†



, s



)}=δ

(3)

( p −p



)δ



(11)

{b(p, s), b(p



, s



)}=0 (12)

†

(p, s), b

†



, s



)}=0 (13)

There is a corresponding set of relations for d

†

(p, s) and d(p, s) the creation and annihi-

lation operators for a positron, for instance,

{d(p, s), d

†



, s



)}=δ

(3)

( p −p



)δ



(14)

We now have to show that we indeed obtain (4). Write

ψ(0) =





(2π)



/m)





†



, s



) ¯u(p



, s



) + d(p



, s



) ¯v(p



, s



)]

Nothing to do but to plow ahead:

{ψ(x ,0),

ψ(0)}



(2π)

/m)



[u(p, s)¯u(p, s)e

i p

x

+ v(p, s)¯v(p, s)e

−i p

x

]

if we take b and b

†

to anticommute with d and d

†

. Using (8) and (9) we obtain

{ψ(x ,0),

ψ(0)}=



(2π)

(2E

)

[(p + m)e

i p

x

+ (p − m)e

−i p

x

]



(2π)

(2E

)

−i p

x

= γ

(3)

(x)

which is just (4) slightly disguised.

Similarly, writing schematically, we have {ψ , ψ }=0 and {ψ

†

, ψ

†

}=0.

We are of course free to normalize the spinors u and v however we like. One alternative

normalization is to define u and v as the u and v given here multiplied by (2m)

, thus

changing (8) and (9) to



u ¯u =p + m and



v ¯v =p − m. Multiplying the numerator and

denominator in (10) by (2m)

, we see that the normalization factor (E

/m)

is changed to

(2E

)

[thus making it the same as the normalization factor for the scalar field in (I.8.11)].

II.2. Quantizing the Dirac Field | 111

This alternative normalization (let us call it “any mass normalization”) is particularly

convenient when we deal with massless spin-

particles: we could set m = 0 everywhere

without ever encountering m in the denominator as in (8) and (9).

The advantage of the normalization used here (let us call it “rest normalization”) is that

the spinors assume simple forms in the rest frame, as we have just seen. This would prove

to be advantageous when we calculate the magnetic moment of the electron in chapter

III.6, for example. Of course, multiplying and dividing here and there by (2m)

is a trivial

operation, and there is not much sense in arguing over the relative advantages of one

normalization over another.

In chapter II.6 we will calculate electron scattering at energies high compared to the

mass m, so that effectively we could set m = 0. Actually, even then, “rest normalization”

has the slight advantage of providing a (rather weak) check on the calculation. We set m =0

everywhere we can, such as in the numerator of (8) and (9), but not where we can’t, such as

in the denominator. Then m must cancel out in physical quantities such as the differential

scattering cross section.

Energy of the vacuum

An important exercise at this point is to calculate the Hamiltonian starting with the

Hamiltonian density

H = π

∂ψ

∂t

− L =

ψ(iγ



∂ + m)ψ (15)

Inserting (10) into this expression and integrating, we have

H =



xH =



ψ(iγ



∂ + m)ψ =



ψiγ

∂ψ

∂t

(16)

which works out to be

H =





†

(p, s)b(p, s) − d(p, s)d

†

(p, s)] (17)

We can see the all important minus sign in (17) schematically: In (16)

ψ gives a factor

∼ (b

†

+ d), while ∂/∂t acting on ψ brings down a relative minus sign giving ∼ (b − d

†

thus giving us ∼ (b

†

+d)(b −d

†

) ∼b

†

b −dd

†

(orthogonality between spinors vu = 0 kills

the cross terms).

To bring the second term in (17) into the right order, we anticommute −d(p, s)d

†

(p, s)

= d

†

(p, s)d(p, s) − δ

(3)

(



0) so that

H =





†

(p, s)b(p, s) + d

†

(p, s)d(p, s)]

− δ

(3)

(







(18)

The first two terms tell us that each electron and each positron of momentum p and spin

s has exactly the same energy E

, as it should. But what about the last term? That δ

(3)

(



should fill us with fear and loathing.