Zee A. Quantum Field Theory in a Nutshell

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112 | II. Dirac and the Spinor

It is OK: Noting that δ

(3)

( p) = [1/(2π)

]



i p x

, we see that δ

(3)

(



0) = [1/(2π)

]



(we encounter the same maneuver in exercise I.8.2) and so the last term contributes to H

=−





) (19)

(since in natural units we have  = 1 and hence h = 2π). We have an energy −

in each

unit-size phase-space cell (1/h

p in the sense of statistical mechanics, for each spin

and for the electron and positron separately (hence the factor of 2) . This infinite additive

term E

is precisely the analog of the zero point energy

ω of the harmonic oscillator you

encountered in your quantum mechanics course. But it comes in with a minus sign!

The sign is bizarre and peculiar! Each mode of the Dirac field contributes −

ω to

the vacuum energy. In contrast, each mode of a scalar field contributes

ω as we saw

in chapter I.8. This fact is of crucial importance in the development of supersymmetry,

which we will discuss in chapter VIII.4.

Fermion propagating through spacetime

In analogy with (I.8.14), the propagator for the electron is given by iS

αβ

(x) ≡

0|Tψ

(x)

(0) |0, where the argument of

ψ has been set to 0 by translation invariance.

As we will see, the anticommuting character of ψ requires us to define the time-ordered

product with a minus sign, namely

T ψ(x)

ψ(0) ≡ θ(x

)ψ(x)

ψ(0) − θ(−x

)

ψ(0)ψ(x) (20)

Referring to (10), we obtain for x

> 0,

iS(x) =0|ψ(x)

ψ(0) |0=



(2π)

/m)



u(p, s)¯u(p, s)e

−ipx



(2π)

/m)

p + m

−ipx

For x

< 0, we have to be a bit careful about the spinorial indices:

αβ

(x) =−0|

(0)ψ

(x) |0

=−



(2π)

/m)



¯v

(p, s)v

(p, s)e

−ipx

=−



(2π)

/m)

(

p − m

)

αβ

−ipx

using the identity (9).

Putting things together we obtain

iS(x) =



(2π)

/m)



θ(x

)

p + m

−ipx

− θ(−x

)

p − m

+ipx



(21)

We will now show that this fermion propagator can be written more elegantly as a

4-dimensional integral:

II.2. Quantizing the Dirac Field | 113

iS(x) =i



(2π)

−ip

p + m

− m

+ iε



(2π)

−ip

p − m + iε

(22)

To show that (22) is indeed equivalent to (21) we go through essentially the same steps as

after (I.8.14). In the complex p

plane the integrand has poles at p

=±



p

+ m

− iε 

±(E

−iε).Forx

> 0 the factor e

−ip

tells us to close the contour in the lower half-plane.

We go around the pole at +(E

− iε) clockwise and obtain

iS(x) =(−i)i



(2π)

−ip

p + m

producing the first term in (21). For x

< 0 we are now told to close the contour in the

upper half-plane and thus we go around the pole at −(E

− iε) anticlockwise. We obtain

iS(x) =i



(2π)

+iE

+i p

x

−2E

(−E

−p γ +m)

and flipping p we have

iS(x) =−



(2π)

−p γ −m) =−



(2π)

( p − m)

precisely the second term in (21) with the minus sign and all. Thus, we must define the

time-ordered product with the minus sign as in (20).

After all these steps, we see that in momentum space the fermion propagator has the

elegant form

iS(p) =

p − m + iε

(23)

This makes perfect sense: S(p) comes out to be the inverse of the Dirac operator p − m,

just as the scalar boson propagator D(k) = 1/(k

− m

+ iε) is the inverse of the Klein-

Gordon operator k

− m

Poetic but confusing metaphors

In closing this chapter let me ask you some rhetorical questions. Did I speak of an

electron going backward in time? Did I mumble something about a sea of negative energy

electrons? This metaphorical language, when used by brilliant minds, the likes of Dirac

and Feynman, was evocative and inspirational, but unfortunately confused generations

of physics students and physicists. The presentation given here is in the modern spirit,

which seeks to avoid these potentially confusing metaphors.

Exercises

II.2.1 Use Noether’s theorem to derive the conserved current J

ψγ

ψ . Calculate [Q, ψ], thus showing that

b and d

†

must carry the same charge.

II.2.2 Quantize the Dirac field in a box of volume of V and show that the vacuum energy E

is indeed

proportional to V . [Hint: The integral over momentum



p is replaced by a sum over discrete values

of the momentum.]

II.3 Lorentz Group and Weyl Spinors

The Lorentz algebra

In chapter II.1 we followed Dirac’s brilliantly idiosyncratic way of deriving his equation.

We develop here a more logical and mathematical theory of the Dirac spinor. A deeper

understanding of the Dirac spinor not only gives us a certain satisfaction, but is also

indispensable, as we will see later, in studying supersymmetry, one of the foundational

concepts of superstring theory; and of course, most of the fundamental particles such as

the electron and the quarks carry spin

and are described by spinor fields.

Let us begin by reminding ourselves how the rotation group works. The three generators

(i = 1, 2, 3 or x , y, z) of the rotation group satisfy the commutation relation

, J

] =i

ij k

(1)

When acting on the spacetime coordinates, written as a column vector

⎛

⎜

⎝

⎞

⎟

⎠

the generators of rotations are represented by the hermitean matrices

1 =

⎛

⎜

⎝

000 0

000−i

00i 0

⎞

⎟

⎠

(2)

with J

and J

obtained by cyclic permutations. You should verify by laboriously multi-

plying these three matrices that (1) is satisfied. Note that the signs of J

are fixed by the

commutation relation (1).

II.3. Lorentz Group and Spinors | 115

Now add the Lorentz boosts. A boost in the x ≡ x

direction transforms the spacetime

coordinates:



= (cosh ϕ) t + (sinh ϕ) x; x



= (sinh ϕ) t + (cosh ϕ) x (3)

or for infinitesimal ϕ



= t +ϕx; x



= x + ϕt (4)

In other words, the infinitesimal generator of a Lorentz boost in the x direction is repre-

sented by the hermitean matrix (x

≡ t as usual)

⎛

⎜

⎝

0100

1000

0000

⎞

⎟

⎠

(5)

Similarly,

⎛

⎜

⎝

0010

0000

1000

0000

⎞

⎟

⎠

(6)

I leave it to you to write down K

. Note that K

is defined to be antihermitean.

Check that [J

, K

] = i

ij k

. To see that this implies that the boost generators K

transform as a 3-vector



K under rotation, as you would expect, apply a rotation through

an infinitesimal angle θ around the 3-axis. Then (you might wish to review the material in

appendix B at this point) K

→e

iθJ

−iθJ

+iθ[J

, K

] +O(θ

) =K

+iθ (iK

) +

O(θ

) = cosθK

− sinθK

to the order indicated.

You are now about to do one of the most significant calculations in the history of

twentieth century physics. By brute force compute [K

, K

], evidently an antisymmetric

matrix. You will discover that it is equal to −iJ

. Two Lorentz boosts produce a rotation!

(You might recall from your course on electromagnetism that this mathematical fact is

responsible for the physics of the Thomas precession.)

Mathematically, the generators of the Lorentz group satisfy the following algebra [known

to the cognoscenti as SO(3, 1)]:

, J

] =i

ij k

(7)

, K

] =i

ij k

(8)

, K

] =−i

ij k

(9)

Note the all-important minus sign!

116 | II. Dirac and the Spinor

How do we study this algebra? The crucial observation is that the algebra falls apart into

two pieces if we form the combinations J

±i

≡

± iK

). You should check that

, J

] =i

ij k

(10)

−i

, J

−j

] =i

ij k

−k

(11)

and most remarkably

, J

−j

] =0 (12)

This last commutation relation tells us that J

and J

−

form two separate SU(2) algebras.

(For more, see appendix B.)

From algebra to representation

This means that you can simply use what you have already learned about angular mo-

mentum in elementary quantum mechanics and the representation of SU(2) to deter-

mine all the representations of SO(3, 1). As you know, the representations of SU(2)

are labeled by j = 0,

,1,

...

. We can think of each representation as consisting of

(2j +1) objects ψ

with m =−j , −j + 1,

...

, j − 1, j that transform into each other

under SU(2). It follows immediately that the representations of SO(3, 1) are labeled by

, j

−

) with j

and j

−

each taking on the values 0,

,1,

, . . . . Each representation

consists of (2j

+ 1)(2j

−

+ 1) objects ψ

−

with m

=−j

, −j

+ 1,...,j

− 1, j

and m

−

=−j

−

, −j

−

+ 1,...,j

−

− 1, j

−

Thus, the representations of SO(3, 1) are (0, 0), (

,0), (0,

), (1, 0), (0, 1), (

), and

so on, in order of increasing dimension. We recognize the 1-dimensional representation

(0, 0) as clearly the trivial one, the Lorentz scalar. By counting dimensions, we expect

that the 4-dimensional representation (

) has to be the Lorentz vector, the defining

representation of the Lorentz group (see exercise II.3.1).

Spinor representations

What about the representation (

,0)? Let us write the two objects as ψ

with α = 1, 2.

Well, what does the notation (

,0) mean? It says that J

+ iK

) acting on ψ

represented by

while J

−i

− iK

) acting on ψ

is represented by 0. By adding

and subtracting we find that

(13)

and

(14)

II.3. Lorentz Group and Spinors | 117

where the equal sign means “represented by” in this context. (By convention we do not

distinguish between upper and lower indices on the 3-dimensional quantities J

, K

, and

.) Note again that K

is anti-hermitean.

Similarly, let us denote the two objects in (0,

) by the peculiar symbol ¯χ

˙α

. I should em-

phasize the trivial but potentially confusing point that unlike the bar used in chapter II.1,

the bar on ¯χ

˙α

is a typographical element: Think of the symbol ¯χ as a letter in the Hittite

alphabet if you like. Similarly, the symbol ˙α bears no relation to α: we do not obtain ˙α

by operating on α in any way. The rather strange notation is known informally as “dotted

and undotted” and more formally as the van der Waerden notation—a bit excessive for our

rather modest purposes at this point but I introduce it because it is the notation used in

supersymmetric physics and superstring theory. (Incidentally, Dirac allegedly said that he

wished he had invented the dotted and undotted notation.) Repeating the same steps as

above, you will find that on the representation (0,

) we have J

and iK

=−

The minus sign is crucial.

The 2-component spinors ψ

and ¯χ

˙α

are called Weyl spinors and furnish perfectly good

representations of the Lorentz group. Why then does the Dirac spinor have 4 components?

The reason is parity. Under parity, x →−x and p →−p, and thus



J →



J and



K →−



K ,

and so



↔



−

. In other words, under parity the representations (

,0) ↔ (0,

). There-

fore, to describe the electron we must use both of these 2-dimensional representations, or

in mathematical notation, the 4-dimensional reducible representation (

,0) ⊕ (0,

We thus stack two Weyl spinors together to form a Dirac spinor

 =



¯χ

˙α



(15)

The spinor (p) is of course a function of 4-momentum p [and by implication also ψ

(p)

and ¯χ

˙α

(p)] but we will suppress the p dependence for the time being. Referring to (13)

and (14) we see that acting on  the generators of rotation



J =



σ 0

σ



where once again the equality means “represented by,” and the generators of boost



K =



σ 0

0 −

σ



Note once again the all-important minus sign.

Parity forces us to have a 4-component spinor but we know on the other hand that the

electron has only two physical degrees of freedom. Let us go to the rest frame. We must

project out two of the components contained in (p

) with the rest momentum p

≡

(m,



0). With the benefit of hindsight, we write the projection operator as P =

(1 − γ

You are probably guessing from the notation that γ

will turn out to be one of the gamma

matrices, but at this point, logically γ

is just some 4 by 4 matrix. The condition P

= P

implies that (γ

)

=1 so that the eigenvalues of γ

are ±1. Since ψ

↔¯χ

˙α

under parity we

naturally guess that ψ

and ¯χ

˙α

correspond to the left and right handed fields of chapter II.1.

118 | II. Dirac and the Spinor

We cannot simply use the projection to set for example ¯χ

˙α

to 0. Parity means that we should

treat ψ

and ¯χ

˙α

on the same footing. We choose





or more explicitly

⎛

⎜

⎝

0010

0001

1000

0100

⎞

⎟

⎠

(Different choices of γ

correspond to the different basis choices discussed in chapter II.1.)

In other words, in the rest frame ψ

−¯χ

˙α

= 0. The projection to two degrees of freedom

can be written as

(γ

− 1)(p

) = 0 (16)

Indeed, we recognize this as just the Weyl basis introduced in chapter II.1.

The Dirac equation

We have derived the Dirac equation, a bit in disguise!

Since our derivation is based on a step-by-step study of the spinor representation of the

Lorentz group, we know how to obtain the equation satisfied by (p) for any p: We simply

boost. Writing (p) =e

−i ϕ



(p

), we have (e

−i ϕ



i ϕ



−1)(p) =0. Introducing the

notation γ

/m ≡ e

−i ϕ



i ϕ



, we obtain the Dirac equation

(γ

− m)(p) = 0 (17)

You can work out the details as an exercise.

The derivation here represents the deep group theoretic way of looking at the Dirac

equation: It is a projection boosted into an arbitrary frame.

Note that this is an example of the power of symmetry, which pervades modern physics

and this book: Our knowledge of how the electron field transforms under the rotation

group, namely that it has spin

, allows us to know how it transforms under the Lorentz

group. Symmetry rules!

In appendix E we will develop the dotted and undotted notation further for later use in

the chapter on supersymmetry.

In light of your deeper group theoretic understanding it is a good idea to reread chap-

ter II.1 and compare it with this chapter.

II.3. Lorentz Group and Spinors | 119

Exercises

II.3.1 Show by explicit computation that (

) is indeed the Lorentz vector.

II.3.2 Work out how the six objects contained in the (1, 0) and (0, 1) transform under the Lorentz group.

Recall from your course on electromagnetism how the electric and magnetic fields



E and



B transform.

Conclude that the electromagnetic field in fact transforms as (1, 0) ⊕ (0, 1). Show that it is parity that

once again forces us to use a reducible representation.

II.3.3 Show that

−i ϕ



i ϕ





0 e

−ϕ σ

ϕ σ



and

ϕ σ

= cosh ϕ +σ

ˆϕ sinh ϕ

with the unit vector ˆϕ ≡ϕ/ϕ. Identifying p = m ˆϕ sinh ϕ, derive the Dirac equation. Show that



0 σ

−σ



II.3.4 Show that a spin

particle can be described by a vector-spinor 

αμ

, namely a Dirac spinor carrying a

Lorentz index. Find the corresponding equations of motion, known as the Rarita-Schwinger equations.

[Hint: The object 

αμ

has 16 components, which we need to cut down to 2

+ 1 = 4 components.]

II.4 Spin-Statistics Connection

There is no one fact in the physical world which has

a greater impact on the way things are, than the Pauli

exclusion principle.

Degrees of intellectual incompleteness

In a course on nonrelativistic quantum mechanics you learned about the Pauli exclusion

principle

and its later generalization stating that particles with half integer spins, such as

electrons, obey Fermi-Dirac statistics and want to stay apart, while in contrast particles with

integer spins, such as photons or pairs of electrons, obey Bose-Einstein statistics and love

to stick together. From the microscopic structure of atoms to the macroscopic structure

of neutron stars, a dazzling wealth of physical phenomena would be incomprehensible

without this spin-statistics rule. Many elements of condensed matter physics, for instance,

band structure, Fermi liquid theory, superfluidity, superconductivity, quantum Hall effect,

and so on and so forth, are consequences of this rule.

Quantum statistics, one of the most subtle concepts in physics, rests on the fact that in

the quantum world, all elementary particles and hence all atoms, are absolutely identical

to, and thus indistinguishable from, one other.

It should be recognized as a triumph of

quantum field theory that it is able to explain absolute identity and indistinguishability

easily and naturally. Every electron in the universe is an excitation in one and the same

electron field ψ . Otherwise, one might be able to imagine that the electrons we now

I. Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem,p.21.

While a student in Cambridge, E. C. Stoner came to within a hair of stating the exclusion principle.

Pauli himself in his famous paper (Zeit. f. Physik 31: 765, 1925) only claimed to “summarize and generalize

Stoner’s idea.” However, later in his Nobel Prize lecture Pauli was characteristically ungenerous toward Stoner’s

contribution. A detailed and fascinating history of the spin and statistics connection may be found in Duck and

Sudarshan, op. cit.

Early in life, I read in one of George Gamow’s popular physics books that he could not explain quantum

statistics—all he could manage for Fermi statistics was an analogy, invoking Greta Garbo’s famous remark “I

vont to be alone.”—and that one would have to go to school to learn about it. Perhaps this spurs me, later in life,

to write popular physics books also. See A. Zee, Einstein’s Universe,p.x.

II.4. Spin-Statistics Connection | 121

know came off an assembly line somewhere in the early universe and could all be slightly

different owing to some negligence in the manufacturing process.

While the spin-statistics rule has such a profound impact in quantum mechanics, its

explanation had to wait for the development of relativistic quantum field theory. Imagine

a civilization that for some reason developed quantum mechanics but has yet to discover

special relativity. Physicists in this civilization eventually realize that they have to invent

some rule to account for the phenomena mentioned above, none of which involves motion

fast compared to the speed of light. Physics would have been intellectually unsatisfying

and incomplete.

One interesting criterion in comparing different areas of physics is their degree of

intellectual incompleteness.

Certainly, in physics we often accept a rule that cannot be explained until we move to the

next level. For instance, in much of physics, we take as a given the fact that the charge of the

proton and the charge of electron are exactly equal and opposite. Quantum electrodynamics

by itself is not capable of explaining this striking fact either. This fact, charge quantization,

can only be deduced by embedding quantum electrodynamics into a larger structure, such

as a grand unified theory, as we will see in chapter VII.6. (In chapter IV.4 we will learn that

the existence of magnetic monopoles implies charge quantization, but monopoles do not

exist in pure quantum electrodynamics.)

Thus, the explanation of the spin-statistics connection, by Fierz and by Pauli in the late

1930s, and by L

uders and Zumino and by Burgoyne in the late 1950s, ranks as one of the

great triumphs of relativistic quantum field theory. I do not have the space to give a general

and rigorous proof

here. I will merely sketch what goes terribly wrong if we violate the

spin-statistics connection.

The price of perversity

A basic quantum principle states that if two observables commute then they are simul-

taneously diagonalizable and hence observable. A basic relativistic principle states that if

two spacetime points are spacelike with respect to each other then no signal can propagate

between them, and hence the measurement of an observable at one of the points cannot

influence the measurement of another observable at the other point.

Consider the charge density J

= i(ϕ

†

∂

ϕ − ∂

†

ϕ) in a charged scalar field theory.

According to the two fundamental principles just enunciated, J

(x , t =0) and J

(y , t =0)

should commute for x =y. In calculating the commutator of J

(x , t = 0) with J

(y , t = 0),

we simply use the fact that ϕ(x , t = 0) and ∂

ϕ(x, t = 0) commute with ϕ(y , t = 0) and

∂

ϕ(y, t =0 ), so we just move the field at x steadily past the field at y. The commutator

vanishes almost trivially.

See I. Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem, and R. F. Streater and A. S.

Wightman, PCT, Spin Statistics, and All That.