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
[J µν , J ρσ ] = i g νρ J µ σ − g µρ J νσ − g νσ J µρ + g µσ J νρ .
Then we have
[K i , K j ] = [J 0i , J 0j ]
= i g i0 J 0j − g 00 J ij − g ij J 00 + g 0j J i0
We can replace g 00 = 1, J 00 = 0, and g i0 = g j0 = 0 since we know i and j are
space indices. Then
[K i , K j ] = −iJ ij
= −ijk Lk .
Next,
[Li , K n ] =
1 ijk jk 0n
[J , J ]
2
This is really just a fancy way of using the properties of the tensor to write
=
o
1 n pq 0n
[J , J ] − [J qp , J 0n ]
2
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 second
commutator we can switch q and p and simultaneously flip the sign since J is
antisymmetric to obtain
= [J pq , J 0n ]
= i g q0 J pn − g p0 J qn − g qn J p0 + g pn J q0
= i g pn J q0 − g qn J p0 .
But only one of g pn and g qn can be nonzero,
the value −1 since p and q are space indices.
(
−iJ i+2,0 ,
i
n
[L , K ] =
+iJ i+1,0 ,
and whichever one is nonzero has
So we get
(n = i + 1)
(n = i + 2)
(where all addition is to be carried out modulo 3).
= −iink J k0
= iink J 0k
= iink K k .
Finally, the last commutator is calculated through a calculation exactly analogous to the one we just went through, and actually the answer is given to us
in the problem, namely,
[Li , Lj ] = iijk Lk .
Next, we have
i
1h
(L + iK), (L − iK)
4
o
1n
=
[L, L] + i [K, L] − i [L, K] + [K, K]
4
= 2i [K, L]
n
o
= 2i [K 1 , L1 ] î + [K 2 , L2 ] ĵ + [K 3 , L3 ] k̂
[J+ , J− ] =
=0
since [K i , Lj ] 6= 0 only for i 6= j as we saw above. Next,
h
i
i 1h
j
i
J+
, J+
=
(Li + iK i ), (Lj + iK j )
4
1 i j
[L , L ] + i[K i , Lj ] + i[Li , K j ] − [K i , K j ]
=
4
1 ijk k
i L + iijk + iijk K k + iijk Lk
=
4
1 ijk j
= J+
2
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
3
so the J+ operators satisfy the angular momentum commutation relations among
themselves. The demonstration that the same is true for the J− operators is
just too similar to what we just did to warrant repeating.
Part b.
For the ( 21 , 0) representation of the Lorentz group we want to take
J+ =
1
1
(L + iK) = ~σ ,
2
2
1
(L − iK) = 0.
2
J− =
These are solved by taking L = 12 ~σ , K = − 2i ~σ . Then the transformation law for
the ( 21 , 0) representation is
Φ( 12 ,0)
→
µν
e−iωµν J Φ( 21 ,0)
~
~
~
1
=
e−i(θ·L+β·K) Φ( 21 ,0)
=
e− 2 θ·~σ− 2 β·~σ Φ( 12 ,0) .
i
~
For the (0, 21 ) representations we want to take
J+ =
1
(L + iK) = 0,
2
J− =
1
1
(L − iK) = ~σ
2
2
which is obtained by taking L = 21 ~σ , K = 2i ~σ . Then for this object the transformation law is
i~
1~
Φ(0, 12 ) → e− 2 θ·~σ+ 2 β·~σ Φ(0, 12 ) .
Comparing with Peskin and Schroeder equation 3.37, this tells us to identify
Φ( 12 ,0) = ψL ,
Part c.
Φ(0, 21 ) = ψR .
Parametrizing the matrix as suggested, we have
0
V + V 3 V 1 − iV 2
Φ( 21 , 12 ) =
V 1 + iV 2 V 0 − V 3
= V µ σµ
where σ0 = , σi = the ith Pauli matrix. Then applying the transformation
described, we obtain
h 1
h 1
iT
i
~ ~
~ ~
e− 2 (iθ+β)·~σ Φ( 21 , 12 ) e− 2 (iθ−β)·~σ
1 ~ ~
1 ~ ~
µ
≈ 1 − (iθ + β) · ~σ V σµ 1 − (iθ − β) · ~σ
2
2
1 ~ ~
1 ~ ~
µ
µ
= V σµ − (iθ + β) · ~σ V σµ − (iθ − β) · V µ σµ ~σ + O(θ2 )
2
2
1 µ ~
1 µ~
µ
≈ V σµ − V iθ · {~σ , σµ } − V iθ · [~σ , σµ ]
2
2
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
4
But for the σ matrices we have the following properties:
0 0
σ , σ = 2σ 0
i 0
σ , σ = 2σ i
i j
σ , σ = −2δ ij
0 0
σ ,σ = 0
i 0
σ ,σ = 0
i j
σ , σ = 2ieijk σ k
so the above equation becomes
h 1
iT
i
h 1
~ ~
~ ~
e− 2 (iθ+β)·~σ Φ( 21 , 12 ) e− 2 (iθ−β)·~σ
i
ρσ ν
µ
= V − ωρσ Jµν V σµ
2
= V 0µ σµ
~ into ωρσ according to
where we packed the transformation parameters ~θ and β
ijk k
i
ρσ
ωij = θ and ω0i = β and where J is just the matrix defined in Peskin
& Schroeder equation (3.18). So this all just says that V µ transforms as a four
vector.
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 γ 0 γ 0 = , and we multiply both sides by γ 0
on the left to obtain
u(p)γ 0 /
p† γ 0 = mu(p).
(2)
We have
γ 0 γ µ† γ 0 =
=−
† 0
σµ
0 1
−σ µ 0
1 0
µ
0 1
0
σ
0 1
1 0
−σ µ 0
1 0
1
0
0
−σ µ
σµ
0
0
1
(since σ i† = σ i )
=
= γ µ.
5
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
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
1
iσ µν qν = − [γ µ , γ ν ] (p0ν − pν )
2
i
1h
= − γ µ γ ν p0ν − γ µ γ ν pν − γ ν γ µ p0ν + γ ν γ µ pν .
2
(4)
But from the anticommutation relation for the γ matrices we know
γ µ γ ν = −γ ν γ µ + 2δ µν .
Plugging this in above, we can rewrite (4) as
i
1h
− 2p/0 γ µ − 2γ µ /
p + 2δ µν γ µ pν + 2δ µν γ µ p0ν
2
= p/0 γ µ + γ µ /
p − γ µ (p0ν + pν )
iσ µν qν = −
(5)
(6)
Finally, we just sandwich (6) between u(p) and u(p0 ) and use (1) and (2) to
obtain
h
h
i
i
u(p0 ) iσµν u(p) = u(p0 ) p/0 γ µ + γ µ /
p − γ µ (p0ν + pν ) u(p)
h
i
= u(p0 ) [2m − (p0 + pν )]γ µ u(p)
or
u(p0 )γ µ u(p) = u(p0 )
iσ µν qν
p0 + pν
u(p)
+
2m
2m
as advertised.
Problem 3.3
(a) We have
k/0 uR0 = k/0k/1 uL0
Let’s work out the commutator of k/0 and k/1 :
[k/0 ,k/1 ] = k0µ γ µ k1ν γ ν − k1ν γ ν k0µ γ µ
h
i
= k0µ k1ν γ µ γ ν − γ ν γ µ
= 2k0µ k1ν γ µ γ ν − g µν
(7)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
6
The first term vanishes because the γs anticommute, and the second term vanishes because k0 · k1 = 0 :
= 0.
Hence we can interchange k/1 and k/0 in (7):
k/0k/1 uL0 = k/1k/0 uL0
(8)
But uL0 is the spinor for a massless fermion with momentum k0 , so its Dirac
equation is just k/0 uL0 = 0. Hence (8) vanishes.
Next,
1
/puL,R (p) = √
/p/pUR0,L0 = 0
2p · k0
since /
p/
p = p2 = 0.
(b). We have
uL0
√
k0 · σξ
√
=
k0 · σξ
√
√
+ σ3 ξ
= E √
− σ3 ξ
 1 0
ξ
√ 
0 0 = 2E 

0 0
ξ
0 1




Since k0 is the momentum of a particle moving in the negative
z direction, for
this to be a left-handed spinor requires that we take ξ = 10 , so
 
1
√  0 

uLO = 2E 
 0 .
0
Next,
uR0 = k/1 uL0
0 −σ 1
uL0
=
σ1
0
 
0
√  0 

= 2E 
 0 
1
7
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
Next,


0
 0 
1
uL (p) = √
/p  
p0 + p3  0 
1

−p1 + ip2
 p0 + p3
1

=√
0
p0 + p3 
0




Finally,

1
 0 
1

/p 
uR (p) = √
p0 + p3  0 
0

0

1
0

=√

p
+
p3
p0 + p3
0
p1 + ip2



.

(c). Using the explicit forms we found above,
s(p1 , p2 ) = uR (p1 )uL (p2 )
1
= p 0
0
(p1 + p31 )(p02 + p32 )
=
0 p01 + p31
p11 − ip21
(p01 + p31 )(−p12 + ip22 ) + (p11 − ip21 )(p02 + p32 )
p
(p01 + p31 )(p02 + p32 )
0
0
Separating into real and imaginary parts,
0
(p2 + p32 )p11 − (p01 + p31 )p12 + i (p01 + p31 )p22 − (p02 + p32 )p21
p
=
(p01 + p31 )(p02 + p32 )

−p12 + ip22
 p02 + p32 




0
0

From this it is evident that s(p1 , p2 ) = −s(p2 , p1 ). The squared magnitude is
0
2 2
(p2 + p32 )p11 − (p01 + p31 )p12 + (p01 + p31 )p22 − (p02 + p32 )p21
2
|s(p1 , p2 )| =
(p01 + p31 )(p02 + p32 )
22
0
3 2 12
22
0
3
0
3
1 1
2 2
(p0 + p31 )2 (p12
2 + p2 ) + (p2 + p2 ) (p1 + p1 ) − 2(p2 + p2 )(p1 + p1 )(p1 p2 + p1 p2 )
= 1
0
3
0
3
(p1 + p1 )(p2 + p2 )
0
3
0
3
p2 + p2 12
p + p1 12
1 1
2 2
(p + p22
(p + p22
(9)
= 10
2 )+ 0
1 ) − 2(p1 p2 + p1 p2 ).
p2 + p32 2
p1 + p31 1
8
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
But since p1 is lightlike, we have
22
02
32
p12
1 + p1 = p1 − p1
and similarly for p2 . Plugging these in to (9), we obtain
|s(p1 , p2 )|2 = −
p02 + p32 02
p01 + p31 02
32
1 1
2 2
(p
+
p
)
−
(p + p32
2
2
1 ) − 2(p1 p2 + p1 p2 ).
p02 + p32
p01 + p31 1
Now simply multiplying out and simplifying we obtain
= 2p01 p02 − 2p11 p12 − 2p11 p12 − 2p31 p32
= 2p1 · p2 .
Problem 3.4
Part a. We suppose we have a wavefunction χ that we know satisfies the
equation
iσ µ ∂µ χ(x) = imσ 2 χ∗ (x).
(10)
Now suppose a friend of ours in a different reference frame (related to ours by
a 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 use
the values of χ that we have tabulated in our frame. We are happy to let him
use our values, but we have to remind him to be careful. First of all, if he wants
to know the value of the wavefunction at a spacetime point x in his frame, he
has to ask us for the value of our wavefunction at the point Λ−1 x in our frame.
Second, once we look up the value of our wavefunction at that point and report
it to him, he then has to transform the result according to P & S equation 3.37
to make sure the wavefunction has the right orientation in his reference frame.
To remember all of this we can write the function χ0 (x0 ), which gives the value
of the wavefunction in our friend’s frame evaluated at a point x0 in our friend’s
frame, as follows:
i
χ0 (x0 ) = e− 2 (θ−iβ)·~σ χ Λ−1 x0 .
Let’s now see what our friend would obtain on applying the operator iσ µ ∂µ to
his wavefunction:
i
iσ µ ∂µ χ0 (x0 ) = iσ µ ∂µ e− 2 (θ−iβ)·~σ χ Λ−1 x0
But ∂µ is just a scalar operator and passes right through the transformation
matrix:
i
= iσ µ e− 2 (θ−iβ)·~σ (Λ−1 )νµ |∂ν χ|x=Λ−1 x0
We now use the commutation identity
i
~
~
i
~
~
σ µ e− 2 (θ−iβ)·~σ = e− 2 (θ−iβ)·~σ Λµν σ ν
(11)
9
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
where Λ is the Lorentz transformation matrix corresponding to the parameters
~
~ Using this in (11), we obtain
θ, β.
i
ih
ih
h i
iσ µ ∂µ χ0 (x0 ) = i e− 2 (θ−iβ)·~σ (Λ−1 )µρ σ ρ (Λ−1 )νµ ∂ν χ
x=Λ−1 x0
We can simplify the product of the matrices in the second and third square
brackets with the identity
h i
ih
ih
i
= i e− 2 (θ−iβ)·~σ (Λ−1 )µρ σ ρ (Λ−1 )νµ ∂ν χ
x=Λ−1 x0
i
h i
ih
− 2 (θ−iβ)·~
ν
σ
= e
iσ ∂ν χ
x=Λ−1 x0
But now we can just plug in from (10):
h i
i
= e− 2 (θ−iβ)·~σ imσ 2 χ Λ−1 x0
= imσ 2 χ0 (x0 ).
So our friend finds his wavefunction satisfies the same equation ours does.
Next, we write equation (10) and its complex conjugate in the form:
σ µ ∂µ χ = mσ 2 χ∗
∗µ
∗
2
σ ∂µ χ = −mσ χ.
(12)
(13)
Multiplying the first of these by σ 2 /m, we solve for χ∗ :
χ∗ =
1 2 µ
σ σ ∂µ χ
m
and plugging into (13) yields
1 ∗µ 2 ν
σ σ σ ∂µ ∂ν χ = −mσ 2 χ
m
Finally, we use the identity σ ∗µ σ 2 = σ 2 σ µ (P & S equation (3.38)) to write
1 2 µ ν
σ σ σ ∂ν ∂µ χ = −mσ 2 χ
m
or
(∂µ ∂ µ − m2 )χ = 0.
Part b.
The action density is
Z
im T 2
S=
d4 x χ† iσ µ ∂µ χ +
χ σ χ − χ† σ 2 χ∗ .
2
10
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
For the first term in the action density, we take the complex conjugate and recall
the rule (αβ)∗ = β ∗ α∗ for Grassmann numbers:
∗ ∗ ∗
† µ
χ iσ ∂µ χ = χ∗1 [iσ µ ∂µ χ]1 + χ∗2 [iσ µ ∂µ χ]2
= [−iσ∗µ ∂µ χ∗ ]1 χ1 + [−iσ ∗µ ∂µ χ∗ ]2 χ2
= −iσ ∗µ ∂χ† χ
(14)
The complex conjugate of the second term in the action density is
∗
∗
im T 2
im − iχ1 χ2 + iχ2 χ1 + iχ∗1 χ∗2 − iχ∗2 χ∗1
χ σ χ − χ† σ 2 χ∗
=−
2
2
im + iχ∗2 χ∗1 − iχ∗1 χ∗2 − iχ2 χ1 + iχ1 χ2
=−
2
∗
im T 2
=+
χ σ χ − χ† σ 2 χ∗
(15)
2
so this term is its own complex conjugate.
Hence we have
Z
∗
S−S =
d4 x χ† iσ µ ∂µ χ + iσ ∗µ ∂χ† χ
Z
=
d4 xiσ ∗µ ∂µ χ† χ
=0
since this is the integral of a perfect differential over all space and we assume
the integrand vanishes at the surface.
Next we write the action density in the form
S = χ∗1 [iσ∂µ χ]1 + χ∗2 [iσ∂µ χ]2 +
im χ1 [σ 2 χ]1 + χ2 [σ 2 χ]2 − χ∗1 [σ 2 χ∗ ]1 − χ∗2 [σ 2 χ∗ ]2
2
Varying with respect to χ∗1 and χ∗2 we obtain
∂S
= i [σ µ ∂µ χ]1 −
∂χ∗1
∂S
= i [σ µ ∂µ χ]2 −
0=
∂χ∗2
0=
im
[σ2 χ∗ ]1
2
im
[σ2 χ∗ ]2
2
which are just the 1 and 2 components of (10).
Part c.
The Dirac Lagrangian density is
L = iψγ µ ∂µ ψ − mψψ
0
∗T
∗T
= i ψL
ψR
=
∗T µ
iψL
σ ∂µ ψL
+
0
∗T µ
iψR
σ ∂µ ψR
0
σµ
−
σµ
0
∂µ
∗T
mψL
ψL
−
ψL
ψR
∗T
− m ψL
∗T
mψR
ψR
∗T
ψR
0
0
ψL
ψR
11
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
In terms of χ1 and χ2 we have
µ
2 ∗ ∗T µ
2 ∗
∗T
∗T
L = iχ∗T
1 σ ∂µ χ1 + i(iσ χ2 ) σ ∂µ (iσ χ2 ) − mχ1 χ1 − mχ2 χ2 .
We rewrite the second term:
i(iσ 2 χ∗2 )∗T σ µ ∂µ (iσ 2 χ∗2 ) = i(χT2 σ 2 σ µ σ 2 ∂µ χ∗2 )
Now, σ 2 σ µ σ 2 = +σ µ if µ = 2 and −σ µ otherwise; this is the same as -σ µ∗ , so
we obtain
= −i(χT2 σ µ∗ ∂µ χ∗2 )
µ∗
= (full derivative) + iχ∗T
2 σ ∂µ χ2
Then the Lagrangian density is (ignoring terms that vanish on integration)
µ
∗T µ∗
∗T
∗T
L = iχ∗T
1 σ ∂µ χ1 + iχ2 σ ∂µ χ2 − mχ1 χ1 − mχ2 χ2 .
Part d.
Taking the divergence of the first current, we find
∂µ J µ = ∂µ χ† σ µ χ + χ† σ µ ∂µ χ.
(16)
The Majorana equation is
iσ µ ∂µ χ = imσ 2 χ∗
and its adjoint is
Plugging into (16),
i∂µ χ† σ µ = −imχT σ 2 .
∂µ J µ = −mχT σ 2 χ + mχ† σ 2 χ.
Turning now to the second current we are given, we recall from the previous
part that the terms in the Lagrangian density containing χ1 and χ2 are both of
the 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 the
divergence of the first current, i.e.
∂µ J µ = m(χ†1 σ 2 χ1 − χT1 σ 2 χ1 ) − m(χ†2 σ 2 χ2 − χT2 σ 2 χ2 ).
Problem 3.5
Part a.
The Lagrangian density is
L = T1 + T2 + T3
T1 = ∂µ φ∗ ∂ µ φ
T2 = χ† iσ µ ∂µ χ
T3 = F ∗ F.
(17)
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
12
We add to the fields the quantities
δφ = −iT σ 2 χ
= 2 χ1 − 1 χ2
δχ = F − σ µ σ 2 ∗ ∂µ φ
δχ† = † F ∗ − T σ 2 σ µ ∂µ φ∗ )
(=⇒
δF = −i† σ µ ∂µ χ
= −i∗1 [σ µ ∂µ χ]1 − i∗2 [σ µ ∂µ χ]2 .
Since and χ are Grassmann, we have to be careful about taking the complex
conjugate of a term in which they both appear. For example,
h
i∗
δφ∗ = − iT σ 2 χ
0 −i ∗
= − i 1 2
χ1 χ2
i 0
i∗
h
= 1 χ2 − 2 χ1
h
i
= χ∗2 ∗1 − χ∗1 ∗2
i
h
= − ∗1 χ∗2 − ∗2 χ∗1
h
i
= − ∗1 χ∗2 − ∗2 χ∗1
i
h
= − − i∗T σ 2 χ∗T
h
i
= − i∗ ∗T σ 2∗ χ∗T .
Hmmm. Well, I guess we don’t have to be too careful. We just have to put in
a minus sign in addition to taking the normal complex conjugate of a term in
which two Grassmann fields appear.
To first order in , the various terms in the Lagrangian transform are augmented as follows:
δT1 = −iT σ 2 ∂µ φ∗ ∂ µ χ
(≡ τ1 )
− i∗T σ 2 ∂µ χ∗ ∂ µ φ
† µ
δT2 = iχ σ ∂µ F † µ ν
2 ∗
− iχ σ σ σ ∂µ ∂ν φ
∗ † µ
+ iF σ ∂µ χ
T
2 ν
µ
∗
− i σ σ σ ∂ν φ ∂µ χ
∗ † µ
δT3 = −iF σ ∂µ χ
T
∗µ
− iF σ ∂µ χ
∗
We proceed to argue that these terms sum to zero.
(≡ τ2 )
(≡ τ3 )
(≡ τ4 )
(≡ τ5 )
(≡ τ6 )
(≡ τ7 )
(≡ τ8 )
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
13
1. τ1 and τ6 These guys sum to zero because σ ν σ µ ∂ν∗ ∂χ = ∂ µ χ∗ ∂ µ χ, which
follows immediately from the anticommutation relations of the σ matrices.
2. τ2 and τ4 Again we use the anticommutation relations of the σ matrices to
obtain
τ4 = −iχ† σ 2 ∗ ∂µ ∂ µ φ
Then, summing in τ2 , we have
τ2 + τ4 = −∗T σ 2 ∂µ χ∗ ∂ µ χ − iχ† σ 2 ∗ ∂µ ∂ µ φ
h
i
= −∂µ ∗T σ 2 χ∂ µ φ
which is the divergence of a four-vector, and hence does not affect the
Lagrangian dynamics.
3. τ3 and τ8 These guys are
h
i
iχ† σ µ ∂µ F − iF ∗ σ ∗µ ∂µ χ∗ = −i∂µ F χ† σ ∗µ 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 = ∂µ φ∗ ∂ µ φ+χ† iσ µ ∂µ χ+F ∗ F +mφF +mφ∗ F ∗ +
im T 2
χ σ χ − χ† σ 2 χ∗ . (18)
2
The Euler-Lagrange equations for F and F ∗ are
∂L
=0
∂F
∂L
=0
∂F ∗
=⇒
F ∗ = −mφ
=⇒
F = −mφ∗ .
Substituting in for F and F ∗ in (18),
im T 2
χ σ χ − χ† σ 2 χ∗
2
im T 2
∗ µ
† µ
2 ∗
† 2 ∗
= ∂µ φ ∂ φ + χ iσ ∂µ χ − m φ φ +
χ σ χ−χ σ χ
2
L = ∂µ φ∗ ∂ µ φ + χ† iσ µ ∂µ χ + m2 φ∗ φ − m2 φφ∗ − m2 φ∗ φ +
which is a Lagrangian with mass terms for both φ and χ with the same mass
m.
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
Part c.
L=
14
The Lagrangian density is
∂µ φ∗i ∂ µ φi
+
χ†i iσ µ ∂µ χi
+
Fi∗ Fi
∂W [φ]
i ∂ 2 W [φ] T 2
+ Fi
+
χ σ χj + c.c.
∂φi
2 ∂φi ∂φj i
With the choice W = gφ3 /3, we have
L = ∂µ φ∗ ∂ µ φ + χ† iσ µ ∂µ χ + F ∗ F + gF φ2 + igφχT σ 2 χ + gF ∗ φ∗2 − igφ∗ χ† σ 2 χ∗ .
The Euler-Lagrange equations for F and F ∗ are
F ∗ = −gφ2 ,
F = −gφ∗2 .
Plugging these in,
L = ∂µ φ∗ ∂ µ φ + χ† iσ µ ∂µ χ − g 2 φ∗2 φ2 + igφχT σ 2 χ − igφ∗ χ† σ 2 χ∗ .
The equation of motion for φ∗ is
∂µ
∂L
∂L
=
∂(∂µ φ∗ )
∂φ∗
or
∂µ ∂ µ φ = −2gφ∗ φ2 − igχ† σ 2 χ∗
To get the χ equation we write out the χ− dependent terms in L explicitly in
terms of the components of χ :
∂ − ∂
−∂1 + i∂2
χ1
0
3
∗
∗
L = i χ1 χ2
+ gφ(χ1 χ2 − χ2 χ1 ) − gφ∗ (χ∗1 χ∗2 + χ∗2 χ∗1 )
−∂1 − i∂2
∂0 + ∂3
χ2
= iχ∗1 (∂0 − ∂3 )χ1 + iχ∗2 (−∂1 − i∂2 )χ1 + 2gφχ1 χ2 + terms independent of χ1
The the Euler-Lagrange equation is
∂µ
∂L
∂L
=
∂(∂µ χ1 )
∂χ1
or
i(∂0 − ∂3 )χ∗1 + i(−∂1 − i∂2 )χ∗2 = 2gφχ2
This looks like the 1 component of
i∂µ χ† σ µ = 2igφσ2 χ.
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3
15
Problem 3.6
(a). We are given the normalization for the scalar and vector elements of the
set, namely
scalar:
γ 0 , iγ i
vector:
For the tensor elements we have
tr [γ µ , γ ν ][γ µ , γ ν ] = (γ µ γ ν − γ ν γ µ ) (γ µ γ ν − γ ν γ µ )
= µνµν − νµµν − µννµ + νµνµ
where we employed an obvious shorthand. Since µ 6= ν (otherwise the whole
thing vanishes) we can use the relation µν = −νµ to bubble the µs past the νs
to make each term look like the first:
= 4µνµν.
The trace of this is (P& S equation 5.5)
tr = 16(g µν g µν − g µµ g νν + g µν g νµ )
= −16g µµ g νν
(
−16,
µ, ν both spacelike
=
+16,
µ, ν one spacelike and one timelike
so the correctly normalized tensor elements of the {Γ} set are
1
+ [γ 0 , γ 1 ],
4
tensor:
1
− [γ 1 , γ 3 ],
4
For the pseudovectors we have
1
+ [γ 0 , γ 2 ],
4
1 2 3
− [γ , γ ],
4
tr γ µ γ 5 γ µ γ 5 = tr γ µ2 γ 5
= tr g µµ
(
4,
=
−4,
1
+ [γ 0 , γ 3 ],
4
1
− [γ 1 , γ 2 ].
4
2
µ=0
µ = 1, 2, 3
so the correctly normalized pseudovector elements are
pseudovector :
γ 0 γ 5 , iγ i γ 5
Finally, for the pseudoscalar element we already have trγ 5 γ 5 = 4, so the
correctly normalized pseudoscalar element of the {Γ} set is
pseudoscalar :
γ5.