Solutions to Peskin & Schroeder

Chapter 15

Zhong-Zhi Xianyu∗

Institute of Modern Physics and Center for High Energy Physics,

Tsinghua University, Beijing, 100084

Draft version: March 12, 2013

1 Brute-force computations in SU(3)

(a) The dimension of SU(N) group is d = N2 − 1, when N = 3 we get d = 8.

(b) It’s easy to see that t1, t2, t3 generate a SU(2) subgroup of SU(3). Thus we have

f ijk = ǫijk for i, j, k = 1, 2, 3. Just take another example, let’s check [t6, t7]:

[t6, t7] = i(− 12 t

3 +√32 t8),

thus we get

f678 =√32 , f673 = − 1

2 .

Then what about f376?

[t3, t7] = i2 t

6 ⇒ f376 = 12 = −f673.

(c) C(F ) = 12 . Here F represents fundamental representation.

(d) C2(F ) =43 , d(F ) = 3, d(G) = 8, thus we see that d(F )C2(F ) = d(G)C(F ).

2 Adjoint representation of SU(2)

The structure constants for SU(2) is fabc = ǫabc, thus we can write down the repre-

sentation matrices for its generators directly from

(tbG)ac = ifabc = iǫabc.

More explicitly,

t1G =

0 0 0

0 0 −i

0 i 0

, t2G =

0 0 i

0 0 0

−i 0 0

, t3G =

0 −i 0

i 0 0

0 0 0

, (1)

∗E-mail: xianyuzhongzhi@gmail.com

1

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

Then,

C(G) = tr (t1Gt1G) = tr (t2Gt

2G) = tr (t3Gt

3G) = 2,

C2(G)I3 = t1Gt1G + t2Gt

2G + t3Gt

3G = 2I3 ⇒ C2(G) = 2.

Here I3 is the 3× 3 unit matrix.

3 Coulomb potential

(a) We calculate vacuum expectation value for Wilson loop UP (z, z), defined by

UP (z, z) = exp

[

− ie

∮

P

dxµAµ(x)

]

. (2)

By definition, we have

〈UP (z, z)〉 =

∫

DAµ exp

[

iS[Aµ]− ie

∮

P

dxµAµ(x)

]

, (3)

where

S[Aµ] =

∫

d4x[

− 14 FµνF

µν − 12ξ (∂

µAµ)2]

. (4)

Thus 〈UP (z, z)〉 is simply a Gaussian integral, and can be worked out directly, as

〈UP (z, z)〉 = exp

[

−1

2

(− ie

∮

P

dxµ)(

− ie

∮

P

dyν)∫

d4k

(2π)4−igµνk2 + iǫ

e−ik·(x−y)

]

(5)

Here we have set ξ → 0 to simplify the calculation. Working out the momentum integral,

we get

〈UP (z, z)〉 = exp

[

−e2

8π2

∮

P

dxµ∮

P

dyνgµν

(x − y)2

]

. (6)

The momentum integration goes as follows

∫d4k

(2π)4e−ik·(x−y)

k2 + iǫ= i

∫d4kE(2π)4

eikE·(x−y)

−k2E

=i

(2π)4

∫ 2π

0

dψ

∫ π

0

dφ sinφ

∫ π

0

dθ sin2 θ

∫ ∞

0

dkE k3E

eikE|x−y| cos θ

−k2E

= −i

4π3

∫ ∞

0

dkE kE

∫ π

0

dθ sin2 θeikE |x−y| cos θ

= −i

4π2

∫ ∞

0

dkE kEJ1(kE |x− y|

)

kE |x− y|= −

i

4π2(x− y)2. (7)

Where J1(x) is Bessel function and we use the fact that∫∞0

dxJ1(x) = 1.

(b) Now taking a narrow rectangular Wilson loop P with width R in x1 direction

(0 < x1 < 1) and length T in x0 direction (0 < x0 < T ) and evaluate 〈UP 〉. When

the integral over dx and dy go independently over the loop, divergence will occur as

|x− y|2 → 0. But what we want to show is the dependence of 〈UP 〉 on the geometry of

the loop, namely the width R and length T , which should be divergence free. Therefore,

when T ≫ R, the integral in Wilson loop is mainly contributed by time direction and

can be expressed as

〈UP (z, z)〉 ≃ exp

[

−2e2

8π2

∫ T

0

dx0∫ 0

T

dy01

(x0 − y0)2 −R2 − iǫ

]

, (8)

2

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

and we have add a small imaginary part to the denominator for the reason that will be

clear. Carry out the integration, we find

∫ T

0

dx0∫ 0

T

dy01

(x0 − y0)2 −R2 − iǫ

T≫R−−−→

2T

Rarctanh

( T

R+ iǫ

)

= −i

π

T

R.

Therefore,

〈UP 〉 = exp( ie2

4πR· T)

= e−iV (R)T , (9)

which gives the familiar result V (R) = −e2/4πR.

(c) For the Wilson loop of a non-Abelian gauge group, we have

UP (z, z) = tr

{

P exp[

− ig

∮

P

dxµAaµ(x)t

ar

]}

, (10)

where tar is the matrices of the group generators in representation r. We expand this

expression to the order of g2,

UP (z, z) = tr (1)− g2∮

P

dxµ∮

P

dyνAaµ(x)A

bν (y) tr (t

ar t

br) +O(g3)

= tr (1)

[

1− g2C2(r)

∮

P

dxµ∮

P

dyνAaµ(x)A

bν(y)

]

+O(g3). (11)

Compared with the Abelian case, we see that to order g2, the non-Abelian result is

given by making the replacement e2 → g2C2(r). Therefore we conclude that V (R) =

−g2C2(r)/4πR in non-Abelian case.

4 Scalar propagator in a gauge theory

In this problem we study very briefly the heat kernel representation of Green func-

tions/propagator of a scalar field living within a gauge field background.

(a) To begin with, we consider the simplest case, in which the background gauge

field vanishes. Then we can represent the Green function DF (x, y) of the Klein-Gordon

equation, defined to be

(∂2 +m2)DF (x, y) = −iδ(4)(x− y) (12)

with proper boundary conditions, by the following integral over the heat kernel function

D(x, y, T ):

DF (x, y) =

∫ ∞

0

dT D(x, y, T ). (13)

The heat kernel satisfies the following “Schrodinger equation”:

[

i∂

∂T− (∂2 +m2)

]

D(x, y, T ) = iδ(T )δ(4)(x− y). (14)

The solution to this equation can be represented by

D(x, y, T ) = 〈x|e−iHT |y〉 =

∫d4k

(2π)4d4k′

(2π)4〈x|k〉〈k|e−iHT |k′〉〈k′|y〉

=

∫d4k

(2π)4d4k′

(2π)4e−i(−k2+m2)T e−ik·x+ik′·y(2π)4δ(4)(k − k′)

3

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

=

∫d4k

(2π)4ei(k

2−m2)T e−ik·(x−y), (15)

with H = ∂2 +m2. Integrating this result over T , with the +iǫ prescription, we recover

the Feynman propagator for a scalar field:∫ ∞

0

dT D(x, y, T ) =

∫d4k

(2π)4e−ik·(x−y)

∫ ∞

0

dT ei(k2−m2+iǫ)T

=

∫d4k

(2π)4ie−ik·(x−y)

k2 −m2 + iǫ. (16)

(b) Now let us turn on a background Abelian gauge field Aµ(x). The corresponding

“Schrodinger equation” then becomes[

i∂

∂T−((∂µ − ieAµ(x)

)2+m2

)]

D(x, y, T ) = iδ(T )δ(4)(x− y), (17)

the solution of which, 〈x|e−iHT |y〉, can also be expressed as a path integral,

〈x|e−iHT |y〉 = limN→∞

∫ N∏

i=1

(

dxi⟨xi∣∣ exp

{

− i∆t[(∂µ − ieAµ(x)

)2+m2

]}∣∣xi−1

⟩)

,

(18)

where we have identify x = xN , y = x0, and ∆t = T/N . Then,

〈xi|e−i∆t[(∂µ−ieAµ(x))

2+m2]|xi−1〉

=

∫d4ki(2π)4

〈xi|e−i∆t[∂2−ieAµ(x)∂

µ+m2]|ki〉〈ki|e−i∆t[−ie∂µAµ(x)−e2A2(x)]|xi−1〉

=

∫d4ki(2π)4

〈xi|e−i∆t[−k2

i+eAµ(xi)kµ

i+m2]|ki〉〈ki|e

−i∆t[ekµ

iAµ(xi−1)−e2A2(xi−1)]|xi−1〉

=

∫d4ki(2π)4

e−i∆t[−k2

i+eki·(A(xi)+A(xi−1))−e2A2(xi−1)+m2−iǫ]e−iki·(xi−xi−1)

= C exp

[

−i∆t

4

( xi − xi−1

∆t+ eA(xi) + eA(xi−1)

)2

− i∆t(m2 − e2A2(xi−1))

]

⇒ C exp

[

−i∆t

4

( dx

dt

)2

− i∆teA(x) ·dx

dt− i∆tm2

]

. (19)

In the last line we take the continuum limit, and C is an irrelevant normalization con-

stant. Then we get

D(x, y, T ) =

∫

Dx exp

[

− i

∫ T

0

dt

(( dx

dt

)2

+m2

)

− ie

∫ T

0

dx(t) · A(x(t))

]

(20)

5 Casimir operator computations

(a) In the language of angular momentum theory, we can take common eigenfunctions

of J2 =∑

a TaT a and Jz = T 3 to be the representation basis. Then the representation

matrix for T 3 is diagonal:

t3j = diag (−j,−j + 1, · · · , j − 1, j).

Thus

tr (t3j t3j ) =

j∑

m=−j

m2 = 13 j(j + 1)(2j + 1).

4

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

Then we have

tr (t3rt3r) =

∑

i

tr (t3jit3ji) =

13

∑

i

ji(ji + 1)(2ji + 1) = C(r),

which implies that

3C(r) =∑

i

ji(ji + 1)(2ji + 1). (21)

(b) Let the SU(2) subgroup be spanned by T 1, T 1 and T 3. Then in fundamental

representation, the representation matrices for SU(2) subgroup of SU(N) can be taken

as

tiN =

(

τi/2 02×(N−2)

0(N−2)×2 0(N−2)×(N−2)

)

. (22)

Where τi (i = 1, 2, 3) are Pauli matrices. We see that the representation matrices for

SU(2) decomposes into a doublet and (N − 2) singlet. Then it’s easy to find that

C(N) = 13

(12 (

12 + 1)(2 · 1

2 + 1))= 1

2 , (23)

by formula in (a).

In adjoint representation, the representationmatrices (ti)ab = ifaib (a, b = 1, · · · , N2−

1, i = 1, 2, 3). Thus we need to know some information about structure constants. Here

we give a handwaving illustration by analyzing the structure of fundamental representa-

tion matrices a little bit more. Note that there’re three types of representation matrices,

listed as follows. For convenience, let’s call them tA, tB and tC :

tA =

(

A2×2 02×(N−2)

0(N−2)×2 0(N−2)×(N−2)

)

. (24)

tB =

(

A2×2 B2×(N−2)

B†(N−2)×2 0(N−2)×(N−2)

)

. (25)

tC =

(

− 12 tr (C)I2×2 02×(N−2)

0(N−2)×2 C(N−2)×(N−2)

)

. (26)

In which, tA is just the representation matrices for SU(2) subgroup. Thus we see that

there are 3 tA, 2(N − 2) tB and (N − 2)2 tC in total. It’s also obvious that there is no

way to generate a tA from commutators between two tC or between a tB and tC , the

only way to generate tA are commutators between two tA or between to tB. Then, tAcommutators correspond to the triplet representation os SU(2) subgroup, and 2(N−2)-

tB commutators correspond to the doublet representation of SU(2). In this way we

see that adjoint representation matrices for SU(2) subgroup decompose into 1 triplet,

2(N − 1) doublets and (N − 2)2 singlets.

Then we can calculate C(G), again, by using formula in (a), as:

C(G) = 13

[1(1 + 1)(2 · 1 + 1) + 2(N − 2) · 1

2 (12 + 1)(2 · 1

2 + 1)]= N. (27)

(c) Let U ∈ SU(N) be N × N unitary matrix, S be a symmetric N × N matrix,

and A be an antisymmetric N × N matrix. Then we can use S and A to build two

representations for SU(N) respectively, as

S → USUT , A→ UAUT .

5

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

It’s easy to verify that they are indeed representations. Let’s denote these two rep-

resentation by s and a. It’s also obvious to see that the dimensions of s and a are

d(s) = N(N + 1)/2 and d(a) = N(N − 1)/2 respectively.

Accordingly, the generator T a acts on S and A as:

S → T aS + S(T a)T , A→ T aA+A(T a)T . (28)

To get C2(s) and C2(a), we can make use of the formula

d(r)C2(r) = d(G)C(r). (29)

Thus we need to calculate C(r) and C(a). By formula in (a), we can take an generator

in SU(2) subgroup to simplify the calculation. Let’s take

t3N = 12 diag(1,−1, 0, · · · , 0),

Then we have:

S =

S11 · · · S1n

.... . .

...

Sn1 · · · Snn

→ t3NS + S(t3N )T =1

2

2S11 0 S13 · · · S1n

0 2S22 S23 · · · S2n

S31 S32 0 · · · 0...

......

. . ....

Sn1 Sn2 0 · · · 0

A =

0 A12 · · · A1n

A21 0. . .

......

. . .. . . An−1,n

An1 · · · An,n−1 0

→ t3NA+A(t3N )T =

1

2

0 0 A13 · · · A1n

0 0 A23 · · · A2n

A31 A32 0 · · · 0...

......

. . ....

An1 An2 0 · · · 0

Thus we see that the representation matrices for T 3, in both s representation and a

representation, are diagonal. They are:

t3s = diag(1, 0, 12 · · · , 1

2︸ ︷︷ ︸

N−2

, 1, 12 , · · · ,

12

︸ ︷︷ ︸

N−2

, 0, · · · , 0︸ ︷︷ ︸

(N−2)(N−1)/2

); (30)

t3a = diag(0, 12 , · · · ,

12

︸ ︷︷ ︸

2(N−2)

, 0, · · · , 0︸ ︷︷ ︸

(N−2)(N−3)/2

). (31)

Here we have rearrange the upper triangular elements of S and A by line.

Then we get

C(s) = tr (t3s)2 = 1

2 (N + 2); (32)

C(a) = tr (t3a)2 = 1

2 (N − 2). (33)

Then,

C2(s) =d(G)C(s)

d(s)=

(N2 − 1)(N + 2)/2

N(N + 1)/2=

(N − 1)(N + 2)

N; (34)

C2(a) =d(G)C(a)

d(a)=

(N2 − 1)(N − 2)/2

N(N − 1)/2=

(N + 1)(N − 2)

N. (35)

6

Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 15 (draft version)

At last let’s check the formula implied by (15.100) and (15.101):

(C2(r1) + C2(r2)

)d(r1)d(r2) =

∑

C2(ri)d(ri), (36)

in which the tensor product representation r1 × r2 decomposes into a direct sum of

irreducible representations ri. In our case, the direct sum of representation s and a is

equivalent to the tensor product representation of two copies of N . That is,

N ×N ∼= s+ a.

Thus, we have,

(C2(N) + C2(N)

)d(N)d(N) =

[ N2 − 1

2N+N2 − 1

2N

]

N2 = N(N2 − 1);

and

C2(s)d(s) + C2(a)d(a) =[C(s) + C(a)

]d(G) = N(N2 − 1).

Thus formula (36) indeed holds in our case.

7