Peskin_Chap02

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Solutions to Peskin & SchroederChapter 2

Zhong-Zhi Xianyu

Institute of Modern Physics and Center for High Energy Physics,

Tsinghua University, Beijing, 100084

Draft version: November 8, 2012

1 Classical electromagnetism

In this problem we do some simple calculation on classical electrodynamics. The

action without source term is given by:

S= 1

4

d4x FF

, with F=A A. (1)

(a) Maxwells equations We now derive the equations of motion from the action.

Note that

F(A)

=

,

FA

= 0.

Then from the first equality we get:

(A)

FF

= 4F.

Now substitute this into Euler-Lagrange equation, we have

0 =

L(A)

L

A=F

(2)

This is sometimes called the second pair Maxwells equations. The so-called firstpair comes directly from the definition ofF=A A, and reads

F+ F +F = 0. (3)

The familiar electric and magnetic field strengths can be written as Ei = F0i and

ijkBk =Fij , respectively. From this we deduce the Maxwells equations in terms of

Ei and B i:

iEi = 0, ijkjBk 0Ei = 0, ijkjEk = 0, iBi = 0. (4)

E-mail: [email protected]

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Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version)

(b) The energy-momentum tensor The energy-momentum tensor can be defined

to be the Nother current of the space-time translational symmetry. Under space-time

translation the vectorA transforms as,

A =A. (5)

Thus

T = L

(A)A

L= FA+ 1

4FF

. (6)

Obviously, this tensor is not symmetric. However, we can add an additional termK

to T with K being antisymmetric to its first two indices. Its easy to see that this

term does not affect the conservation ofT. Thus if we choose K =FA, then:

T = T +K =FF +

1

4FF

. (7)

Now this tensor is symmetric. It is called the Belinfante tensor in literature. We canalso rewrite it in terms ofEi and B i:

T00 = 1

2(EiEi +BiBi), Ti0 =T0i =ijkEjBk, etc. (8)

2 The complex scalar field

The Lagrangian is given by:

L= m2. (9)

(a) The conjugate momenta of and

:

= L

= , =

L

= = . (10)

The canonical commutation relations:

[(x), (y)] = [(x), (y)] =i(x y), (11)

The rest of commutators are all zero.

The Hamiltonian:

H=

d3x

+ L=

d3x + +m2. (12)

(b) Now we Fourier transform the field as:

(x) =

d3p

(2)312Ep

ape

ipx +bp

eipx

, (13)

thus:

(x) =

d3p

(2)312Ep

bpe

ipx +ap

eipx

. (14)

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Feed all these into the Hamiltonian:

H=

d3x

+ +m2

=

d3x

d3

p(2)3

2Ep

d3

q(2)3

2Eq

EpEq

ap

eipx bpeipx

aqe

iqx bq

eiqx

+ p q

ap

eipx bpeipx

aqe

iqx bq

eiqx

+m2

ap

eipx +bpeipx

aqe

iqx +bq

eiqx

=

d3x

d3p

(2)3

2Ep

d3q

(2)3

2Eq

(EpEq+ p q +m2)apaqei(pq)x +bpbqei(pq)x (EpEq+ p q m

2)

bqaqei(p+q)x +a

pbq

ei(p+q)x

=

d3p

(2)3

2Ep

d3q

(2)3

2Eq

(EpEq+ p q +m

2)

ap

aqei(EpEq)t +bpb

q

ei(EpEq)t

(2)3(3)(p q)

(EpEq+ p q m2)

bqaqei(Ep+Eq)t +a

pbq

ei(Ep+Eq)t

(2)3(3)(p + q)

=

d3x

E2p

+ p2 +m2

2Ep ap

ap+bpbp

=

d3x Ep

ap

ap+bp

bp+ [bp, bp

]

. (15)

Note that the last term contributes an infinite constant. It is normally explained as the

vacuum energy. We simply drop it:

H=

d3x Ep

ap

ap+bp

bp

. (16)

Where we have used the mass-shell condition: Ep=

m2 + p2. Hence we at once find

two sets of particles with the same mass m.

(c) The theory is invariant under the global transformation: ei, ei.The corresponding conserved charge is:

Q= i

d3x

. (17)

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Rewrite this in terms of the creation and annihilation operators:

Q= i

d3x

= i

d3x

d3

p(2)3

2Ep

d3

q(2)3

2Eq

bpeipx +apeipx

t

aqeiqx +bqeiqx

t

bpe

ipx +ap

eipx

aqeiqx +b

qeiqx

=

d3x

d3p

(2)3

2Ep

d3q

(2)3

2Eq

Eq

bpe

ipx +ap

eipx

aqeiqx b

qeiqx

Ep

bpe

ipx ap

eipx

aqeiqx +b

qeiqx

=

d3x

d3p

(2)3

2Ep

d3q

(2)3

2Eq

(Eq Ep)

bpaqe

i(p+q)x ap

bq

ei(p+q)x

+ (Eq+Ep)

apaqei(pq)x bpbqei(pq)x

=

d3p

(2)3

2Ep

d3q

(2)3

2Eq

(Eq Ep)

bpaqe

i(Ep+Eq)t ap

bq

ei(Ep+Eqt)

(2)3(3)(p + q)

+ (Eq+Ep)

ap

aqei(EpEq)t bpb

q

ei(EpEq)t

(2)3(3)(p q)

=

d3p

(2)32Ep 2Ep(a

p

ap bpbp

)

= d3p

(2)3a

p

ap b

p

bp, (18)where the last equal sign holds up to an infinitely large constant term, as we did when

calculating the Hamiltonian in (b). Then the commutators follow straightforwardly:

[Q, a] = a, [Q, b] =b. (19)

We see that the particle a carries one unit of positive charge, and b carries one unit of

negative charge.

(d) Now we consider the case with two complex scalars of same mass. In this case the

Lagrangian is given by

L= ii m2ii, (20)

where iwithi = 1, 2 is a two-component complex scalar. Then it is straightforward to

see that the Lagrangian is invariant under theU(2) transformation i Uijj withUija matrix in fundamental (self) representation ofU(2) group. TheU(2) group, locally

isomorphic toS U(2)U(1), is generated by 4 independent generators 1 and 12

a, with

a Pauli matrices. Then 4 independent Nother currents are associated, which are given

by

j = L

(i)i

L

(i )i =(

i )(ii) (i)(i

i )

ja= L

(i)ai

L

(i )ai =

i

2(i )ijj (i)ijj. (21)

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The overall sign is chosen such that the particle carry positive charge, as will be seen in

the following. Then the corresponding Nother charges are given by

Q= i d3x

i i

i i,

Qa = i

2

d3x

i (

a)ijj i (

a)ijj

. (22)

Repeating the derivations above, we can also rewrite these charges in terms of creation

and annihilation operators, as

Q=

d3p

(2)3

aipaip b

ipbip

,

Qa =1

2

d3p

(2)3

aip

aijaip b

ip

aijbip

. (23)

The generalization to n-component complex scalar is straightforward. In this case

we only need to replace the generators a/2 ofS U(2) group to the generators ta in the

fundamental representation with commutation relation [ta, tb] = ifabctc.

Then we are ready to calculate the commutators among all these Nother charges and

the Hamiltonian. Firstly we show that all charges of the U(N) group commute with the

Hamiltonian. For the U(1) generator, we have

[Q, H] =

d3p

(2)3d3q

(2)3Eq

aipaip b

ipbip

,

ajqajq+bjqbjq

=

d3p

(2)3d3q

(2)3Eq

aip[aip, a

jq]ajq+a

jq[a

ip, ajq]aip + (a b)

=

d3p

(2)3d3q

(2)3Eq

aipaiq a

iqaip + (a b)

(2)3(3)(p q)

= 0. (24)

Similar calculation gives [Qa, H] = 0. Then we consider the commutation among internal

U(N) charges:

[Qa, Qb] =

d3p

(2)3d3q

(2)3

aipt

aijajp b

ipt

aijbjp

,

akqtbkaq b

kqt

bkbq

=

d3p

(2)3d3q

(2)3

aipt

aijt

bjaq a

kqt

bkt

ajajp + (a b)

(2)3(3)(p q)

= ifabc

d3p

(2)3

aipt

cijajp b

ipt

cijbjp

= if

abc

Q

c

, (25)and similarly, [Q, Q] = [Qa, Q] = 0.

3 The spacelike correlation function

We evaluate the correlation function of a scalar field at two points:

D(x y) =0|(x)(y)|0, (26)

with x y being spacelike. Since any spacelike interval x y can be transformed to a

form such that x0 y0 = 0, thus we will simply take:

x0 y0 = 0, and |x y|2 =r2 >0. (27)

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Now:

D(x y) =

d3p

(2)31

2Epeip(xy) =

d3p

(2)31

2m2 +p2

eip(xy)

= 1(2)3

2

0

d 1

1

dcos

0

dp p2

2

m2 +p2eipr cos

= i

2(2)2r

dp peipr

m2 +p2(28)

Now we make the path deformation on p-complex plane, as is shown in Figure 2.3 in

Peskin & Schroeder. Then the integral becomes

D(x y) = 1

42r

m

d er

2 m2=

m

42rK1(mr). (29)

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