quantum field theory by peskin_chap03 solution

Solutions to Problems in Peskin and Schroeder,

An Introduction To Quantum Field Theory

Homer Reid

March 13, 2003

Chapter 3

Problem 3.1

Part a. The commutation relations are

[Jµν , Jρσ] = i(

gνρJµσ − gµρJνσ − gνσJµρ + gµσJνρ)

.

Then we have

[Ki,Kj] = [J0i, J0j ]

= i(

gi0J0j − g00J ij − gijJ00 + g0jJ i0)

We can replace g00 = 1, J00 = 0, and gi0 = gj0 = 0 since we know i and j arespace indices. Then

[Ki,Kj] = −iJ ij

= −εijkLk.

Next,

[Li,Kn] =1

2εijk[Jjk, J0n]

This is really just a fancy way of using the properties of the ε tensor to write

=1

2

{

[Jpq, J0n] − [Jqp, J0n]}

1

Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3 2

where p = 1 + ((i + 1) mod 3) and q = 1 + ((i + 2) mod 3). In the secondcommutator we can switch q and p and simultaneously flip the sign since J isantisymmetric to obtain

= [Jpq, J0n]

= i(

gq0Jpn − gp0Jqn − gqnJp0 + gpnJq0)

= i(

gpnJq0 − gqnJp0)

.

But only one of gpn and gqn can be nonzero, and whichever one is nonzero hasthe value −1 since p and q are space indices. So we get

[Li,Kn] =

{

−iJ i+2,0, (n = i+ 1)

+iJ i+1,0, (n = i+ 2)

(where all addition is to be carried out modulo 3).

= −iεinkJk0

= iεinkJ0k

= iεinkKk.

Finally, the last commutator is calculated through a calculation exactly anal-ogous to the one we just went through, and actually the answer is given to usin the problem, namely,

[Li, Lj ] = iεijkLk.

Next, we have

[J+,J−] =1

4

[

(L + iK), (L − iK)]

=1

4

{

[L,L] + i [K,L] − i [L,K] + [K,K]}

= 2i [K,L]

= 2i{

[K1, L1] i + [K2, L2] j + [K3, L3] k}

= 0

since [Ki, Lj ] 6= 0 only for i 6= j as we saw above. Next,

[

J i+, J

j+

]

=1

4

[

(Li + iKi), (Lj + iKj)]

=1

4

{

[Li, Lj] + i[Ki, Lj] + i[Li,Kj ] − [Ki,Kj ]}

=1

4

{

iεijkLk + iεijk + iεijkKk + iεijkLk}

=1

2εijkJj

+


so the J+ operators satisfy the angular momentum commutation relations amongthemselves. The demonstration that the same is true for the J− operators isjust too similar to what we just did to warrant repeating.

Part b. For the (12 , 0) representation of the Lorentz group we want to take

J+ =1

2(L + iK) =

1

2~σ, J− =

1

2(L − iK) = 0.

These are solved by taking L = 12~σ, K = − i

2~σ. Then the transformation law forthe (1

2 , 0) representation is

Φ( 1

2,0) → e−iωµνJµν

Φ( 1

2,0)

= e−i(~θ·L+~β·K)Φ( 1

2,0)

= e−i2

~θ·~σ− 1

2

~β·~σΦ( 1

2,0).

For the (0, 12 ) representations we want to take

J+ =1

2(L + iK) = 0, J− =

1

2(L − iK) =

1

2~σ

which is obtained by taking L = 12~σ, K = i

2~σ. Then for this object the trans-formation law is

Φ(0, 1

2) → e−

i2

~θ·~σ+ 1

2

~β·~σΦ(0, 1

2).

Comparing with Peskin and Schroeder equation 3.37, this tells us to identify

Φ( 1

2,0) = ψL, Φ(0, 1

2) = ψR.

Part c. Parametrizing the matrix as suggested, we have

Φ( 1

2, 12) =

(

V 0 + V 3 V 1 − iV 2

V 1 + iV 2 V 0 − V 3

)

= V µσµ

where σ0 =�, σi = the ith Pauli matrix. Then applying the transformation

described, we obtain

[

e−1

2(i~θ+~β)·~σ

]T

Φ( 1

2, 1

2)

[

e−1

2(i~θ−~β)·~σ

]

≈[

1 − 1

2(i~θ + ~β) · ~σ

]

V µσµ

[

1 − 1

2(i~θ − ~β) · ~σ

]

= V µσµ − 1

2(i~θ + ~β) · ~σV µσµ − 1

2(i~θ − ~β) · V µσµ~σ + O(θ2)

≈ V µσµ − 1

2V µi~θ · {~σ, σµ} −

1

2V µi~θ · [~σ, σµ]


But for the σ matrices we have the following properties:{

σ0, σ0}

= 2σ0

{

σi, σ0}

= 2σi

{

σi, σj}

= −2δij

[

σ0, σ0]

= 0[

σi, σ0]

= 0[

σi, σj]

= 2ieijkσk

so the above equation becomes[

e−1

2(i~θ+~β)·~σ

]T

Φ( 1

2, 1

2)

[

e−1

2(i~θ−~β)·~σ

]

=

[

V µ − i

2ωρσJ

ρσµνV

ν

]

σµ

= V ′µσµ

where we packed the transformation parameters ~θ and ~β into ωρσ according toωij = εijkθk and ω0i = βi and where Jρσ is just the matrix defined in Peskin& Schroeder equation (3.18). So this all just says that V µ transforms as a fourvector.

Problem 3.2

The Dirac equation for u(p) is

/pu(p) = mu(p). (1)

We will also need the corresponding equation for u(p). Taking the adjoint of(1),

u†(p)(/p)† = mu†(p).

On the LHS we insert a factor of γ0γ0 =�, and we multiply both sides by γ0

on the left to obtainu(p)γ0/p†γ0 = mu(p). (2)

We have

γ0γµ†γ0 =

(

0 11 0

) (

0 σµ

−σµ 0

)† (

0 11 0

)

= −(

0 11 0

) (

0 σµ

−σµ 0

) (

0 11 0

)

(since σi† = σi)

=

(

0 σµ

−σµ 0

)

= γµ.


So that was easy enough. Plugging into (2), we find the Dirac equation for u.

u(p)/p = mu(p). (3)

Having established that preliminary result, we next observe

iσµνqν = −1

2[γµ, γν ] (p′ν − pν)

= −1

2

[

γµγνp′ν − γµγνpν − γνγµp′ν + γνγµpν

]

. (4)

But from the anticommutation relation for the γ matrices we know

γµγν = −γνγµ + 2δµν .

Plugging this in above, we can rewrite (4) as

iσµνqν = −1

2

[

− 2/p′γµ − 2γµ/p+ 2δµνγµpν + 2δµνγµp′ν

]

(5)

= /p′γµ + γµ/p− γµ(p′ν + pν) (6)

Finally, we just sandwich (6) between u(p) and u(p′) and use (1) and (2) toobtain

u(p′)[

iσµν

]

u(p) = u(p′)[

/p′γµ + γµ/p− γµ(p′ν + pν)]

u(p)

= u(p′)[

[2m− (p′ + pν)]γµ]

u(p)

or

u(p′)γµu(p) = u(p′)

[

p′ + pν

2m+iσµνqν

2m

]

u(p)

as advertised.

Problem 3.3

(a) We have/k0uR0 = /k0/k1uL0 (7)

Let’s work out the commutator of /k0 and /k1 :

[/k0,/k1] = k0µγµk1νγ

ν − k1νγνk0µγ

µ

= k0µk1ν

[

γµγν − γνγµ]

= 2k0µk1ν

(

γµγν − gµν)


The first term vanishes because the γs anticommute, and the second term van-ishes because k0 · k1 = 0 :

= 0.

Hence we can interchange /k1 and /k0 in (7):

/k0/k1uL0 = /k1/k0uL0 (8)

But uL0 is the spinor for a massless fermion with momentum k0, so its Diracequation is just /k0uL0 = 0. Hence (8) vanishes.

Next,

/puL,R(p) =1√

2p · k0

/p/pUR0,L0 = 0

since /p/p = p2� = 0.

(b). We have

uL0 =

( √k0 · σξ√k0 · σξ

)

=√E

( √�+ σ3ξ√� − σ3ξ

)

=√

2E

(

1 00 0

)

ξ(

0 00 1

)

ξ

Since k0 is the momentum of a particle moving in the negative z direction, forthis to be a left-handed spinor requires that we take ξ =

(

10

)

, so

uLO =√

2E

1000

.

Next,

uR0 = /k1uL0

=

(

0 −σ1

σ1 0

)

uL0

=√

2E

0001


Next,

uL(p) =1√

p0 + p3/p

0001

=1√

p0 + p3

−p1 + ip2

p0 + p3

00

Finally,

uR(p) =1√

p0 + p3/p

1000

=1√

p0 + p3

00

p0 + p3

p1 + ip2

.

(c). Using the explicit forms we found above,

s(p1, p2) = uR(p1)uL(p2)

=1

√

(p01 + p3

1)(p02 + p3

2)

(

0 0 p01 + p3

1 p11 − ip2

1

)

(

0�

�0

)

−p12 + ip2

2

p02 + p3

2

00

=(p0

1 + p31)(−p1

2 + ip22) + (p1

1 − ip21)(p

02 + p3

2)√

(p01 + p3

1)(p02 + p3

2)

Separating into real and imaginary parts,

=

[

(p02 + p3

2)p11 − (p0

1 + p31)p

12

]

+ i[

(p01 + p3

1)p22 − (p0

2 + p32)p

21

]

√

(p01 + p3

1)(p02 + p3

2)

From this it is evident that s(p1, p2) = −s(p2, p1). The squared magnitude is

|s(p1, p2)|2 =

[

(p02 + p3

2)p11 − (p0

1 + p31)p

12

]2+

[

(p01 + p3

1)p22 − (p0

2 + p32)p

21

]2

(p01 + p3

1)(p02 + p3

2)

=(p0

1 + p31)

2(p122 + p22

2 ) + (p02 + p3

2)2(p12

1 + p221 ) − 2(p0

2 + p32)(p

01 + p3

1)(p11p

12 + p2

1p22)

(p01 + p3

1)(p02 + p3

2)

=p01 + p3

1

p02 + p3

2

(p122 + p22

2 ) +p02 + p3

2

p01 + p3

1

(p121 + p22

1 ) − 2(p11p

12 + p2

1p22). (9)


But since p1 is lightlike, we have

p121 + p22

1 = p021 − p32

1

and similarly for p2. Plugging these in to (9), we obtain

|s(p1, p2)|2 = −p01 + p3

1

p02 + p3

2

(p022 + p32

2 ) − p02 + p3

2

p01 + p3

1

(p021 + p32

1 ) − 2(p11p

12 + p2

1p22).

Now simply multiplying out and simplifying we obtain

= 2p01p

02 − 2p1

1p12 − 2p1

1p12 − 2p3

1p32

= 2p1 · p2.

Problem 3.4

Part a. We suppose we have a wavefunction χ that we know satisfies theequation

iσµ∂µχ(x) = imσ2χ∗(x). (10)

Now suppose a friend of ours in a different reference frame (related to ours bya Lorentz transformation Λ) wanted to do some physics in his reference frame,but he didn’t have values for the wavefunction χ in his frame, so he asked to usethe values of χ that we have tabulated in our frame. We are happy to let himuse our values, but we have to remind him to be careful. First of all, if he wantsto know the value of the wavefunction at a spacetime point x in his frame, hehas to ask us for the value of our wavefunction at the point Λ−1x in our frame.Second, once we look up the value of our wavefunction at that point and reportit to him, he then has to transform the result according to P & S equation 3.37to make sure the wavefunction has the right orientation in his reference frame.To remember all of this we can write the function χ′(x′), which gives the valueof the wavefunction in our friend’s frame evaluated at a point x′ in our friend’sframe, as follows:

χ′(x′) = e−i2(θ−iβ)·~σχ

(

Λ−1x′)

.

Let’s now see what our friend would obtain on applying the operator iσµ∂µ tohis wavefunction:

iσµ∂µχ′(x′) = iσµ∂µe

− i2(θ−iβ)·~σχ

(

Λ−1x′)

But ∂µ is just a scalar operator and passes right through the transformationmatrix:

= iσµe−i2(θ−iβ)·~σ(Λ−1)ν

µ |∂νχ|x=Λ−1x′ (11)

We now use the commutation identity

σµe−i2(~θ−i~β)·~σ = e−

i2(~θ−i~β)·~σΛµ

νσν


where Λ is the Lorentz transformation matrix corresponding to the parameters~θ, ~β. Using this in (11), we obtain

iσµ∂µχ′(x′) = i

[

e−i2(θ−iβ)·~σ

] [

(Λ−1)µρσ

ρ] [

(Λ−1)νµ

] ∣

∣

∣∂νχ∣

∣

∣

x=Λ−1x′

We can simplify the product of the matrices in the second and third squarebrackets with the identity

= i[


] [

(Λ−1)µρσ

ρ] [

(Λ−1)νµ

] ∣

∣

∣∂νχ∣

∣

∣

x=Λ−1x′

=[


] [

iσν∣

∣

∣∂νχ

∣

∣

∣

x=Λ−1x′

]

But now we can just plug in from (10):

=[


]

[

imσ2χ(

Λ−1x′)]

= imσ2χ′(x′).

So our friend finds his wavefunction satisfies the same equation ours does.Next, we write equation (10) and its complex conjugate in the form:

σµ∂µχ = mσ2χ∗ (12)

σ∗µ∂µχ∗ = −mσ2χ. (13)

Multiplying the first of these by σ2/m, we solve for χ∗:

χ∗ =1

mσ2σµ∂µχ

and plugging into (13) yields

1

mσ∗µσ2σν∂µ∂νχ = −mσ2χ

Finally, we use the identity σ∗µσ2 = σ2σµ (P & S equation (3.38)) to write

1

mσ2σµσν∂ν∂µχ = −mσ2χ

or

(∂µ∂µ −m2)χ = 0.

Part b. The action density is

S =

∫

d4x

[

χ†iσµ∂µχ+im

2

(

χTσ2χ− χ†σ2χ∗)

]

.


For the first term in the action density, we take the complex conjugate and recallthe rule (αβ)∗ = β∗α∗ for Grassmann numbers:

{

χ†iσµ∂µχ}∗

={

χ∗1 [iσµ∂µχ]1

}∗+

{

χ∗2 [iσµ∂µχ]2

}∗

={

[−iσ∗µ∂µχ∗]1 χ1

}

+{

[−iσ∗µ∂µχ∗]2 χ2

}

= −iσ∗µ∂χ†χ (14)

The complex conjugate of the second term in the action density is

[

im

2

(


]∗

= − im2

(

− iχ1χ2 + iχ2χ1 + iχ∗1χ

∗2 − iχ∗

2χ∗1

)∗

= − im2

(

+ iχ∗2χ

∗1 − iχ∗

1χ∗2 − iχ2χ1 + iχ1χ2

)

= +

[

im

2

(


]∗

(15)

so this term is its own complex conjugate.Hence we have

S − S∗ =

∫

d4x[

χ†iσµ∂µχ+ iσ∗µ∂χ†χ]

=

∫

d4xiσ∗µ∂µ

[

χ†χ]

= 0

since this is the integral of a perfect differential over all space and we assumethe integrand vanishes at the surface.

Next we write the action density in the form

S = χ∗1 [iσ∂µχ]1 + χ∗

2 [iσ∂µχ]2 +im

2

{

χ1[σ2χ]1 + χ2[σ

2χ]2 − χ∗1[σ

2χ∗]1 − χ∗2[σ

2χ∗]2}

Varying with respect to χ∗1 and χ∗

2 we obtain

0 =∂S∂χ∗

1

= i [σµ∂µχ]1 −im

2[σ2χ

∗]1

0 =∂S∂χ∗

2

= i [σµ∂µχ]2 −im

2[σ2χ

∗]2

which are just the 1 and 2 components of (10).

Part c. The Dirac Lagrangian density is

L = iψγµ∂µψ −mψψ

= i(

ψ∗TL ψ∗T

R

)

(

0�

�0

)(

0 σµ

σµ 0

)

∂µ

(

ψL

ψR

)

−m(

ψ∗TL ψ∗T

R

)

(

0�

�0

) (

ψL

ψR

)

= iψ∗TL σµ∂µψL + iψ∗T

R σµ∂µψR −mψ∗TL ψL −mψ∗T

R ψR


In terms of χ1 and χ2 we have

L = iχ∗T1 σµ∂µχ1 + i(iσ2χ∗

2)∗Tσµ∂µ(iσ2χ∗

2) −mχ∗T1 χ1 −mχ∗T

2 χ2.

We rewrite the second term:

i(iσ2χ∗2)

∗Tσµ∂µ(iσ2χ∗2) = i(χT

2 σ2σµσ2∂µχ

∗2)

Now, σ2σµσ2 = +σµ if µ = 2 and −σµ otherwise; this is the same as -σµ∗, sowe obtain

= −i(χT2 σ

µ∗∂µχ∗2)

= (full derivative) + iχ∗T2 σµ∗∂µχ2

Then the Lagrangian density is (ignoring terms that vanish on integration)

L = iχ∗T1 σµ∂µχ1 + iχ∗T

2 σµ∗∂µχ2 −mχ∗T1 χ1 −mχ∗T

2 χ2.

Part d. Taking the divergence of the first current, we find

∂µJµ = ∂µχ

†σµχ+ χ†σµ∂µχ. (16)

The Majorana equation is

iσµ∂µχ = imσ2χ∗

and its adjoint isi∂µχ

†σµ = −imχTσ2.

Plugging into (16),

∂µJµ = −mχTσ2χ+mχ†σ2χ.

Turning now to the second current we are given, we recall from the previouspart that the terms in the Lagrangian density containing χ1 and χ2 are both ofthe same form, so we expect to get the same Majorana equation for both fields,and hence each part of the current gives a term of the form we found for thedivergence of the first current, i.e.

∂µJµ = m(χ†

1σ2χ1 − χT

1 σ2χ1) −m(χ†

2σ2χ2 − χT

2 σ2χ2). (17)

Problem 3.5

Part a. The Lagrangian density is

L = T1 + T2 + T3

T1 = ∂µφ∗∂µφ

T2 = χ†iσµ∂µχ

T3 = F ∗F.


We add to the fields the quantities

δφ = −iεTσ2χ

= ε2χ1 − ε1χ2

δχ = εF − σµσ2ε∗∂µφ

(=⇒ δχ† = ε†F ∗ − εTσ2σµ∂µφ∗)

δF = −iε†σµ∂µχ

= −iε∗1 [σµ∂µχ]1 − iε∗2 [σµ∂µχ]2 .

Since ε and χ are Grassmann, we have to be careful about taking the complexconjugate of a term in which they both appear. For example,

δφ∗ =[

− iεTσ2χ]∗

=

[

− i(

ε1 ε2

)

(

0 −ii 0

)

(

χ1 χ2

)

]∗

=[

ε1χ2 − ε2χ1

]∗

=[

χ∗2ε

∗1 − χ∗

1ε∗2

]

= −[

ε∗1χ∗2 − ε∗2χ

∗1

]

= −[

ε∗1χ∗2 − ε∗2χ

∗1

]

= −[

− iε∗Tσ2χ∗T]

= −[

i∗ε∗Tσ2∗χ∗T]

.

Hmmm. Well, I guess we don’t have to be too careful. We just have to put ina minus sign in addition to taking the normal complex conjugate of a term inwhich two Grassmann fields appear.

To first order in ε, the various terms in the Lagrangian transform are aug-mented as follows:

δT1 = −iεTσ2∂µφ∗∂µχ (≡ τ1)

− iε∗Tσ2∂µχ∗∂µφ (≡ τ2)

δT2 = iχ†σµ∂µFε (≡ τ3)

− iχ†σµσνσ2ε∗∂µ∂νφ (≡ τ4)

+ iF ∗ε†σµ∂µχ (≡ τ5)

− iεTσ2σνσµ∂νφ∗∂µχ (≡ τ6)

δT3 = −iF ∗ε†σµ∂µχ (≡ τ7)

− iF εTσ∗µ∂µχ∗ (≡ τ8)

We proceed to argue that these terms sum to zero.


1. τ1 and τ6 These guys sum to zero because σνσµ∂∗ν∂χ = ∂µχ∗∂µχ, whichfollows immediately from the anticommutation relations of the σ matrices.

2. τ2 and τ4 Again we use the anticommutation relations of the σ matrices toobtain

τ4 = −iχ†σ2ε∗∂µ∂µφ

Then, summing in τ2, we have

τ2 + τ4 = −ε∗Tσ2∂µχ∗∂µχ− iχ†σ2ε∗∂µ∂

µφ

= −∂µ

[

ε∗Tσ2χ∂µφ]

which is the divergence of a four-vector, and hence does not affect theLagrangian dynamics.

3. τ3 and τ8 These guys are

iχ†σµε∂µF − iF ε∗σ∗µ∂µχ∗ = −i∂µ

[

Fχ†σ∗µε]

again the divergence of a four-vector, and hence irrelevant.

4. τ5 and τ7 These guys just flat out cancel.

Part b. The composite Lagrangian is

L = ∂µφ∗∂µφ+χ†iσµ∂µχ+F ∗F+mφF+mφ∗F ∗+

im

2

[

χTσ2χ− χ†σ2χ∗]

. (18)

The Euler-Lagrange equations for F and F ∗ are

∂L∂F

= 0 =⇒ F ∗ = −mφ∂L∂F ∗

= 0 =⇒ F = −mφ∗.

Substituting in for F and F ∗ in (18),

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ+m2φ∗φ−m2φφ∗ −m2φ∗φ+

im

2

[


= ∂µφ∗∂µφ+ χ†iσµ∂µχ−m2φ∗φ+

im

2

[


which is a Lagrangian with mass terms for both φ and χ with the same massm.


Part c. The Lagrangian density is

L = ∂µφ∗i ∂

µφi + χ†i iσ

µ∂µχi + F ∗i Fi +

(

Fi

∂W [φ]

∂φi

+i

2

∂2W [φ]

∂φi∂φj

χTi σ

2χj + c.c.

)

With the choice W = gφ3/3, we have

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ+ F ∗F + gFφ2 + igφχTσ2χ+ gF ∗φ∗2 − igφ∗χ†σ2χ∗.

The Euler-Lagrange equations for F and F ∗ are

F ∗ = −gφ2, F = −gφ∗2.

Plugging these in,

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ− g2φ∗2φ2 + igφχTσ2χ− igφ∗χ†σ2χ∗.

The equation of motion for φ∗ is

∂µ

∂L∂(∂µφ∗)

=∂L∂φ∗

or

∂µ∂µφ = −2gφ∗φ2 − igχ†σ2χ∗

To get the χ equation we write out the χ− dependent terms in L explicitly interms of the components of χ :

L = i(

χ∗1 χ∗

2

)

(

∂0 − ∂3 −∂1 + i∂2

−∂1 − i∂2 ∂0 + ∂3

) (

χ1

χ2

)

+ gφ(χ1χ2 − χ2χ1) − gφ∗(χ∗1χ

∗2 + χ∗

2χ∗1)

= iχ∗1(∂0 − ∂3)χ1 + iχ∗

2(−∂1 − i∂2)χ1 + 2gφχ1χ2 + terms independent of χ1

The the Euler-Lagrange equation is

∂µ

∂L∂(∂µχ1)

=∂L∂χ1

or

i(∂0 − ∂3)χ∗1 + i(−∂1 − i∂2)χ

∗2 = 2gφχ2

This looks like the 1 component of

i∂µχ†σµ = 2igφσ2χ.


Problem 3.6

(a). We are given the normalization for the scalar and vector elements of theset, namely

scalar:�

vector: γ0, iγi

For the tensor elements we have

tr(

[γµ, γν ][γµ, γν ])

= (γµγν − γνγµ) (γµγν − γνγµ)

= µνµν − νµµν − µννµ+ νµνµ

where we employed an obvious shorthand. Since µ 6= ν (otherwise the wholething vanishes) we can use the relation µν = −νµ to bubble the µs past the νsto make each term look like the first:

= 4µνµν.

The trace of this is (P& S equation 5.5)

tr = 16(gµνgµν − gµµgνν + gµνgνµ)

= −16gµµgνν

=

{

−16, µ, ν both spacelike

+16, µ, ν one spacelike and one timelike

so the correctly normalized tensor elements of the {Γ} set are

tensor:+

1

4[γ0, γ1], +

1

4[γ0, γ2], +

1

4[γ0, γ3],

− 1

4[γ1, γ3], −1

4[γ2, γ3], − 1

4[γ1, γ2].

For the pseudovectors we have

tr γµγ5γµγ5 = tr γµ2γ52

= tr gµµ�

=

{

4, µ = 0

−4, µ = 1, 2, 3

so the correctly normalized pseudovector elements are

pseudovector : γ0γ5, iγiγ5

Finally, for the pseudoscalar element we already have trγ5γ5 = 4, so thecorrectly normalized pseudoscalar element of the {Γ} set is

pseudoscalar : γ5.

quantum field theory by peskin_chap03 solution

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