Zee A. Quantum Field Theory in a Nutshell

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402 | VII. Grand Unification

sum counts for a power of N . In the large N limit (we can think of N as the inverse

temperature), we evaluate Z by steepest descent and minimize E, obtaining



(λ

) =



n=k

− λ

(15)

which in the continuum limit, as the poles in (15) merge into a cut, becomes V



(λ) =



dμ[ρ(μ)/(λ − μ)], where ρ(μ) is the unknown function we want to solve for and P

denotes principal value.

Defining as before G(z) =



dμ[ρ(μ)/(z − μ)] we see that our equation for ρ(μ) can be

written as Re G(λ + iε) =



(λ). In other words, G(z) is a real analytic function with cuts

along the real axis. We are given the real part of G(z) on the cut and are to solve for the

imaginary part. Br

ezin, Itzykson, Parisi, and Zuber have given an elegant solution of this

problem. Assume for simplicity that V(z)is an even polynomial and that there is only one

cut (see exercise VII.4.7). Invoke symmetry and, incorporating what we know, postulate

the form

G(z) =





(z) − P(z)



− a



with P(z) an unknown even polynomial. Remarkably, the requirement G(z) → 1/z for

large z completely determines P(z). Pedagogically, it is clearest to go to a specific example,

say V(z)=

+gz

. Since V



(z) is a cubic polynomial in z, P(z)has to be a quadratic

(even) polynomial in z. Taking the limit z →∞and requiring the coefficients of z

and

of z in G(z) to vanish and the coefficient of 1/z to be 1 gives us three equations for three

unknowns [namely a and the two unknowns in P(z)]. The density of eigenvalues is then

determined to be ρ(E) = (1/π )P (E)

√

− E

I think the lesson to take away here is that Feynman diagrams, in spite of their historical

importance in quantum electrodynamics and their usefulness in helping us visualize what

is going on, are vastly overrated. Surely, nobody imagines that QCD, even large N QCD,

will one day be solved by summing Feynman diagrams. What is needed is the analog of

the Dyson gas approach for large N QCD. Conversely, if a reader of this book manages to

calculate G(z) by summing planar diagrams (after all, the answer is known!), the insight

he or she gains might conceivably be useful in seeing how to deal with planar diagrams

in large N QCD.

Field theories in the large N limit

A number of field theories have also been solved in the large N expansion. I will tell you

about one example, the Gross-Neveu model, partly because it has some of the flavor of

QCD. The model is defined by

S(ψ) =



⎡

⎣



a=1

i∂ψ





a=1



⎤

⎦

(16)

Recall from chapter III.3 that this theory should be renormalizable in (1 +1)-dimensional

spacetime. For some finite N , say N = 3, this theory certainly appears no easier to solve

VII.4. Large N Expansion | 403

than any other fully interacting field theory. But as we will see, as N →∞we can extract

a lot of interesting physics.

Using the identity (A.14) we can rewrite the theory as

S(ψ, σ)=







a=1

(i∂ − σ)ψ

−



(17)

By introducing the scalar field σ(x)we have “undone” the four-fermion interaction. (Recall

that we used the same trick in chapter III.5.) You will note that the physics involved is

similar to that behind the introduction of the weak boson to generate the Fermi interaction.

Using what we learned in chapters II.5 and IV.3 we can immediately integrate out the

fermion fields to obtain an action written purely in terms of the σ field

S(σ) =−



− iN tr log(i∂ − σ) (18)

Note the factor of N in front of the tr log term coming from the integration over N fermion

fields. With the malice of forethought we, or rather Gross and Neveu, have introduced an

explicit factor of 1/N in the coupling strength in (16), so that the two terms in (18) both scale

as N. Thus, the path integral Z =



Dσe

iS(σ)

may be evaluated by the steepest descent or

stationary phase method in the large N limit. We simply extremize S(σ).

Incidentally, we can see the judiciousness of the choice a = 1 in large N QCD in the

same way. Integrating out the quarks in (1) we get

S =−



tr F

μν

+ N tr log(i(∂ −iA) − m)

and thus the two terms both scale as N and can balance each other. The increase in the

number of degrees of freedom has to be offset by a weakening of the coupling.

To study the ground state behavior of the theory, we restrict our attention to field

configurations σ(x) that do not depend on x. (In other words, we are not expecting

translation symmetry to be spontaneously broken.) We can immediately take over the result

you got in exercise IV.3.3 and write the effective potential

V(σ)=

2g(μ)

4π



log

− 3



(19)

We have imposed the condition (1/N)[d

V (σ )/dσ

σ =μ

= 1/g(μ)

as the definition of

the mass scale dependent coupling g(μ) (compare IV.3.18). The statement that V(σ) is

independent of μ immediately gives

g(μ)

−

g(μ



)

log



(20)

As μ →∞, g(μ) → 0. Remarkably, this theory is asymptotically free, just like QCD. If we

want to, we can work backward to find the flow equation

dμ

g(μ) =−

2π

g(μ)

...

(21)

The theory in its different incarnations, (16), (17), and (18), enjoys a discrete Z

symme-

try under which ψ

→γ

and σ →−σ . As in chapter IV.3, this symmetry is dynamically

404 | VII. Grand Unification

broken by quantum fluctuations. The minimum of V(σ)occurs at σ

min

=μe

1−π/g(μ)

and

so according to (17) the fermions acquire a mass

= σ

min

= μe

1−π/g(μ)

(22)

Note that this highly nontrivial result can hardly be seen by staring at (16) and we have no

way of proving it for finite N . In the spirit of the large N approach, however, we expect that

the fermion mass might be given by m

=μe

1−π/g(μ)

+O(1/N

) so that (22) would be a

decent approximation even for say, N =3. Since m

is physically measurable, it better not

depend on μ. You can check that.

This theory also exhibits dimensional transmutation as described in the previous chap-

ter. We start out with a theory with a dimensionless coupling g and end up with a dimen-

sional fermion mass m

. Indeed, any other quantity with dimension of mass would have

to be equal to m

times a pure number.

Dynamically generated kinks

I discuss the existence of kinks and solitons in chapter V.6. You clearly understood that the

existence of such objects follows from general considerations of symmetry and topology,

rather than from detailed dynamics. Here we have a (1 + 1)-dimensional theory with a

discrete Z

symmetry, so we certainly expect a kink, namely a time independent configu-

ration σ(x) (henceforth x will denote only the spatial coordinate and will no longer label

a generic point in spacetime) such that σ(−∞) =−σ

min

and σ(+∞) = σ

min

. [Obviously,

there is also the antikink with σ(−∞) = σ

min

and σ(+∞) =−σ

min

At first sight, it would seem almost impossible to determine the precise shape of the

kink. In principle, we have to evaluate tr log[i∂ − σ(x)] for an arbitrary function σ(x)such

that σ(+∞) =−σ(−∞) (and as I explained in chapter IV.3, this involves finding all the

eigenvalues of the operator i∂ −σ(x), summing over the logarithm of the eigenvalues),

and then varying this functional of σ(x)to find the optimal shape of the kink.

Remarkably, the shape can actually be determined thanks to a clever observation.

analogy with the steps leading to (IV.3.24) we note that

tr log[i∂ −σ(x)] = tr log γ

[i∂ −σ(x)]γ

= tr log(−1)[i∂ + σ(x)]

and thus up to an irrelevant additive constant

tr log(i∂ −σ(x))=

tr log[i∂ −σ(x)][i∂ +σ(x)]

tr log



−∂

+ iγ



(x) − [σ(x)]



(23)

C. Callan, S. Coleman, D. Gross, and A. Zee, (unpublished). See D. J. Gross, “Applications of the Renormal-

ization Group to High-Energy Physics,” in: R. Balian and J. Zinn-Justin, eds., Methods in Field Theory, p. 247. By

the way, I recommend this book to students of field theory.

VII.4. Large N Expansion | 405

Since γ

has eigenvalues ±i, this is equal to



tr log{−∂

+ σ



(x) − [σ(x)]

}+tr log{−∂

− σ



(x) − [σ(x)]

}



but these two terms are equal by parity (space reflection) and hence

tr log[i∂ −σ(x)] = tr log{−∂

− σ



(x) − [σ(x)]

}

Referring to (18) we see that S(σ) is the sum of two terms, a term quadratic in σ(x)

and a term that depends only on the combination σ



(x) + [σ(x)]

. But we know that σ

min

minimizes S(σ). Thus, the soliton is given by the solution of the ordinary differential

equation



(x) + [σ(x)]

= σ

min

(24)

namely σ(x) = σ

min

tanh σ

min

x. The soliton would be observed as an object of size

1/σ

min

= 1/m

. I leave it to you to show that its mass is given by

(25)

Precisely as theorized in the last chapter, the ratio m

comes out to be a pure number,

N/π, as it must.

By an even more clever method that I do not have space to describe, Dashen, Hasslacher,

and Neveu were able to study time dependent configurations of σ and determine the mass

spectrum of this model.

Exercises

VII.4.1 Since the number of gluons only differs by one, it is generally argued that it does not make any difference

whether we choose to study the U(N) theory or the SU (N ) theory. Discuss how the gluon propagator in

a U(N) theory differs from the gluon propagator in an SU(N) theory and decide which one is easier.

VII.4.2 As a challenge, solve large N QCD in (1 + 1)-dimensional spacetime. [Hint: The key is that in (1 + 1)-

dimensional spacetime with a suitable gauge choice we can integrate out the gauge potential A

.] For

help, see ’t Hooft, Under the Spell of the Gauge Principle, p. 443.

VII.4.3 Show that if we had chosen to calculate G(z) ≡(1/N) tr(1/z − ϕ), we would have to connect the two

open ends of the quark propagator. We see that figures VII.4.5b and d lead to the same diagram. Complete

the calculation of G(z) in this way.

VII.4.4 Suppose the random matrix ϕ is real symmetric rather than hermitean. Show that the Feynman rules

are more complicated. Calculate the density of eigenvalues. [Hint: The double-line propagator can twist.]

VII.4.5 For hermitean random matrices ϕ, calculate

(z, w) ≡



z − ϕ

w − ϕ



−



z − ϕ



w − ϕ



for V(ϕ)=

using Feynman diagrams. [Note that this is a much simpler object to study than the

object we need to study in order to learn about localization (see exercise VI.6.1).] Show that by taking

suitable imaginary parts we can extract the correlation of the density of eigenvalues with itself. For help,

see E. Br

ezin and A. Zee, Phys. Rev. E51: p. 5442, 1995.

406 | VII. Grand Unification

VII.4.6 Use the Faddeev-Popov method to calculate J in the Dyson gas approach.

VII.4.7 For V(ϕ)=

+gϕ

, determine ρ(E).Form

sufficiently negative (the double well potential again)

we expect the density of eigenvalues to split into two pieces. This is evident from the Dyson gas picture.

Find the critical value m

.Form

the assumption of G(z) having only one cut used in the text fails.

Show how to calculate ρ(E) in this regime.

VII.4.8 Calculate the mass of the soliton (25).

VII.5 Grand Unification

Crying out for unification

A gauge theory is specified by a group and the representations the matter fields belong

to. Let us go back to chapter VII.2 and make a catalogue for the SU(3) ⊗ SU(2) ⊗ U(1)

theory. For example, the left handed up and down quarks are in a doublet





with

hypercharge

Y =

. Let us denote this by (3, 2,

)

, with the three numbers indicating

how these fields transform under SU(3) ⊗ SU(2) ⊗ U(1). Similarly, the right handed up

quark is (3, 1,

)

. The leptons are (1, 2, −

)

and (1, 1, −1)

, where the “1” in the first

entry indicates that these fields do not participate in the strong interaction. Writing it all

down, we see that the quarks and leptons of each family are placed in

(3, 2,

)

, (3, 1,

)

, (3, 1, −

)

, (1, 2, −

)

, and (1, 1, −1)

(1)

This motley collection of representations practically cries out for further unification.

Who would have constructed the universe by throwing this strange looking list down?

What we would like to have is a larger gauge group G containing SU(3) ⊗ SU(2) ⊗

U(1), such that this laundry list of representations is unified into (ideally) one great big

representation. The gauge bosons in G [but not in SU(3) ⊗ SU(2) ⊗U(1) of course] would

couple the representations in (1) to each other.

Before we start searching for G, note that since gauge transformations commute with

the Lorentz group, these desired gauge transformations cannot change left handed fields

to right handed fields. So let us change all the fields in (1) to left handed fields. Recall from

exercise II.1.9 that charge conjugation changes left handed fields to right handed fields

and vice versa. Thus, instead of (1) we can write

(3, 2,

), (3

∗

,1,−

), (3

∗

,1,

), (1, 2, −

), and (1, 1, 1) (2)

We now omit the subscripts L and R: everybody is left handed.

408 | VII. Grand Unification

A perfect fit

The smallest group that contains SU(3) ⊗ SU(2) ⊗ U(1) is SU(5). (If you are shaky

about group theory, study appendix B now.) Recall that SU(5) has 5

− 1 = 24 generators.

Explicitly, the generators are represented by 5 by 5 hermitean traceless matrices acting on

five objects we denote by ψ

with μ =1, 2, . . . , 5. [These five objects form the fundamental

or defining representation of SU(5).]

It is now obvious how we can fit SU(3) and SU(2) into SU(5). Of the 24 matrices that

generate SU(5), eight have the form



A 0



and three the form



0 B



, where A represents

3 by 3 hermitean traceless matrices (of which there are 3

− 1 = 8, the so-called Gell-

Mann matrices) and B represents 2 by 2 hermitean traceless matrices (of which there

are 2

− 1 = 3, namely the Pauli matrices). Clearly, the former generate an SU(3) and the

latter an SU(2). Furthermore, the 5 by 5 hermitean traceless matrix

Y =

⎛

⎜

⎝

−

0000

0 −

000

00−

000

0000

⎞

⎟

⎠

(3)

generates a U(1). Without being coy about it, we have already called this matrix the

hypercharge

Y .

In other words, if we separate the index μ ={α, i} with α = 1, 2, 3 and i = 4, 5, then

the SU(3) acts on the index α and the SU(2) acts on the index i . Thus, the three objects

transform as a 3-dimensional representation under SU(3) and hence could be a 3

ora3

∗

. Let us choose ψ

as transforming as 3; we will see shortly that this is the right

choice with Y/2 given as in (3). The three objects ψ

do not transform under SU(2)

and hence each of them belongs to the singlet 1 representation. Furthermore, they carry

hypercharge −

as we can read off from (3). To sum up, ψ

transform as (3, 1, −

)

under SU(3) ⊗ SU(2) ⊗U(1). On the other hand, the two objects ψ

transform as 1 under

SU(3) and 2 under SU(2), and carry hypercharge

; thus they transform as (1, 2,

).In

other words, we embed SU(3) ⊗ SU(2) ⊗U(1) into SU(5) by specifying how the defining

representation of SU(5) decomposes into representations of SU(3) ⊗ SU(2) ⊗ U(1)

5 →(3, 1, −

) ⊕ (1, 2,

) (4)

Taking the conjugate we see that

∗

→ (3

∗

,1,

) ⊕ (1, 2, −

) (5)

Inspecting (2), we see that (3

∗

,1,

) and (1, 2, −

) appear on the list. We are on the right

track! The fields in these two representations fit snugly into 5

∗

This accounts for five of the fields contained in (2); we still have the ten fields

(3, 2,

), (3

∗

,1,−

), and (1, 1, 1) (6)

VII.5. Grand Unification | 409

Consider the next representation of SU(5) in order of size, namely the antisymmetric

tensor representation ψ

μν

. Its dimension is (5 ×4)/2 =10, precisely the number we want,

if only the quantum numbers under SU(3) ⊗ SU(2) ⊗ U(1) work out!

Since we know that 5 →(3, 1, −

) ⊕(1, 2,

), we simply (again, see appendix B!) have to

work out the antisymmetric product of (3, 1, −

) ⊕ (1, 2,

) with itself, namely the direct

sum of (where ⊗

denotes the antisymmetric product)

(3, 1, −

) ⊗

(3, 1, −

) = (3

∗

,1,−

) (7)

(3, 1, −

) ⊗

(1, 2,

) = (3, 2, −

) = (3, 2,

) (8)

and

(1, 2,

) ⊗

(1, 2,

) = (1, 1, 1 ) (9)

[I will walk you through (7): In SU(3) 3 ⊗

3 = 3

∗

(remember ε

ij k

from appendix B?), in

SU(2) 1 ⊗

1 = 1, and in U(1) the hypercharges simply add −

−

=−

Lo and behold, these SU(3) ⊗ SU(2) ⊗U(1) representations form exactly the collection

of representations in (6). In other words,

10 → (3, 2,

) ⊕ (3

∗

,1,−

) ⊕ (1, 1, 1) (10)

The known quark and lepton fields in a given family fit perfectly into the 5

∗

and 10

representations of SU(5)!

I have just described the SU(5) grand unified theory of Georgi and Glashow. In spite of

the fact that the theory has not been directly verified by experiment, it is extremely difficult

for me and for many other physicists not to believe that SU(5) is at least structurally correct,

in view of the perfect group theoretic fit.

It is often convenient to display the contents of the representation 5

∗

and 10, using the

names given to the various fields historically. We write 5

∗

as a column vector





⎛

⎜

⎝

⎞

⎟

⎠

(11)

and the 10 as an antisymmetric matrix

μν

={ψ

αβ

, ψ

αi

, ψ

}

⎛

⎜

⎝

0 ¯u −¯udu

−¯u 0 ¯udu

¯u −¯u 0 du

−d −d −d 0 ¯e

−u −u −u −¯e 0

⎞

⎟

⎠

(12)

(I suppressed the color indices.)

410 | VII. Grand Unification

Deepening our understanding of physics

Aside from its esthetic appeal, grand unification deepens our understanding of physics

enormously.

1. Ever wondered why electric charge is quantized? Why don’t we see particles with

charge equal to

√

π times the electron’s charge? In quantum electrodynamics, you could

perfectly well write down

L =

ψ[i( ∂ −iA) − m]ψ +



[i( ∂ − i

√

π A) − m



]ψ



...

(13)

In contrast, in grand unified theory A

couples to a generator of the grand unifying

gauge group, and you know that the generators of any group such as SU(N ) (that is

not given by the direct product of U(1) with other groups) are forced by the nontrivial

commutation relations [T

, T

] = if

abc

to assume quantized values. For example, the

eigenvalues of T

in SU(2), which depend on the representation of course, must be

multiples of

. Within SU(3) ×SU(2) × U(1), we cannot understand charge quantization:

The generator of U(1) is not quantized. But upon grand unification into SU(5) [or more

generally any group without U(1) factors] electric charge is quantized.

The result here is deeply connected to Dirac’s remark (chapter IV.4) that electric charge is

quantized if the magnetic monopole exists. We know from chapter V.7 that spontaneously

broken nonabelian gauge theories such as the SU(5) theory contain the monopole.

2. Ever wondered why the proton charge is exactly equal and opposite to the electron

charge? This important fact allows us to construct the universe as we know it. Atoms must

be electrically neutral to some fantastic degree of accuracy for standard cosmology to work;

otherwise, electrostatic forces between macroscopic matter would tear the universe apart.

This remarkable fact is nicely incorporated into SU(5). It is fun to see how it goes.

Evaluating tr Q = 0 over the 5

∗

implies that 3Q

=−Q

−

. I have used the fact that the

strong interaction commutes with electromagnetism and hence quarks with different color

have the same charge. Now let us calculate the proton charge Q

= 2Q

+ Q

= 2(Q

+ 1) + Q

= 3Q

+ 2 = Q

−

+ 2 (14)

If Q

−

=−1, then Q

=−Q

−

, as is indeed the case!

3. Recall that in electroweak theory we defined tan θ = g

, with the coupling of the

gauge bosons g

+ g

(Y/2). Since the normalization of A

and B

is fixed by

their respective kinetic energy term, the relative strength of g

and g

is determined by

the normalization of Y/2 relative to T

. Let us evaluate tr T

and tr(Y/2)

on the defining

representation 5 : tr T

= (

)

+ (

)

and tr(Y/2)

= (

)

3 + (

)

2 =

Thus, T

and



3/5(Y /2) are normalized equally. So the correct grand unified combina-

tion is A

+ B



3/5(Y /2), and therefore tan θ =g



3/5or

sin

θ =

(15)

VII.5. Grand Unification | 411

at the grand unification scale. To compare with the experimental value of sin

θ we would

have to study how the couplings g

and g

flow under the renormalization group down to

low energies. We will postpone this discussion until the next chapter.

Freedom from anomaly

Recall from chapter VII.2 that the key to proving renormalizability of nonabelian gauge

theory is the ability to pass freely between the unitary gauge and the R

gauge. The

crucial ingredient is gauge invariance and the resulting Ward-Takahashi identities (see

chapter II.7).

Suddenly you start to worry. What about the chiral anomaly? The existence of the

anomaly means that some Ward-Takahashi identities fail to hold. For our theories to make

sense, they had better be free from anomalies. I remarked in chapter IV.7 that the historical

name “anomaly” makes it sound like some kind of sickness. Well, in a way, it is.

We should have already checked the SU(3) ⊗ SU(2) ⊗ U(1) theory for anomalies, but

we didn’t. I will let you do it as an exercise. Here I will show that the SU(5) theory is healthy.

If the SU(5) theory is anomaly-free, then a fortiori so is the SU(3) ⊗SU(2) ⊗ U(1) theory.

In chapter IV.7 I computed the anomaly in an abelian theory but as I remarked there

clearly all we have to do to generalize to a nonabelian theory is to insert a generator T

of the gauge group at each vertex of the triangle diagram in figure IV.7.1. Summing over

the various fermions running around the loop, we see that the anomaly is proportional

to A

abc

(R) ≡ tr(T

, T

}), where R denotes the representation to which the fermions

belong. We have to sum A

abc

(R) over all the representations in the theory, remembering

to associate opposite signs to left handed and right handed fermion fields. (It may be

helpful to remind yourself of remark 3 in chapter IV.7 and exercise IV.7.6.)

We are now ready to give the SU(5) theory a health check. First, all fermion fields in (2)

are left handed. Second, convince yourself (simply imagine calculating A

abc

for all possible

abc) that it suffices to set T

, T

, and T

all equal to

T ≡

⎛

⎜

⎝

20000

02000

002 0 0

000−30

000 0−3

⎞

⎟

⎠

a multiple of the hypercharge. Let us now evaluate tr T

on the 5

∗

representation,

tr T

∗

= 3(−2)

+ 2(+3)

= 30 (16)

and on the 10,

tr T

= 3(+4)

+ 6(−1)

+ (−6)

=−30 (17)

An apparent miracle! The anomaly cancels.