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**Unformatted text preview: **Solutions to Peskin & Schroeder
Chapter 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:
∫
1
S=−
d4 x Fµν F µν ,
with Fµν = ∂µ Aν − ∂ν Aµ .
(1)
4
(a) Maxwell’s equations
Note that We now derive the equations of motion from the action.
∂Fµν
= 0.
∂Aλ ∂Fµν
= δµλ δνκ − δνλ δµκ ,
∂(∂λ Aκ )
Then from the ﬁrst equality we get:
(
)
∂
Fµν F µν = 4F λκ .
∂(∂λ Aκ )
Now substitute this into Euler-Lagrange equation, we have
(
0 = ∂µ ∂L )
∂L
−
= −∂µ F µν
∂(∂µ Aν )
∂Aν (2) This is sometimes called the “second pair” Maxwell’s equations. The so-called “ﬁrst
pair” comes directly from the deﬁnition of Fµν = ∂µ Aν − ∂ν Aµ , and reads
∂λ Fµν + ∂µ Fνλ + ∂ν Fµλ = 0. (3) The familiar electric and magnetic ﬁeld strengths can be written as E i = −F 0i and
ϵijk B k = −F ij , respectively. From this we deduce the Maxwell’s equations in terms of
E i and B i :
∂ i E i = 0,
∗ E-mail: ϵijk ∂ j B k − ∂ 0 E i = 0, [email protected] 1 ϵijk ∂ j E k = 0, ∂ i B i = 0. (4) 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 N¨other current of the space-time translational symmetry. Under space-time
translation the vector Aµ transforms as,
δ µ Aν = ∂ µ Aν .
Thus
T˜µν = (5) ∂L
1
∂ ν Aλ − η µν L = −F µλ ∂ ν Aλ + η µν Fλκ F λκ .
∂(∂µ Aλ )
4 (6) Obviously, this tensor is not symmetric. However, we can add an additional term ∂λ K λµν
to T˜µν with K λµν being antisymmetric to its ﬁrst two indices. It’s easy to see that this
term does not aﬀect the conservation of T˜µν . Thus if we choose K λµν = F µλ Aν , then:
T µν = T˜µν + ∂λ K λµν = F µλ Fλ ν + 1 µν
η Fλκ F λκ .
4 (7) Now this tensor is symmetric. It is called the Belinfante tensor in literature. We can
also rewrite it in terms of E i and B i :
T 00 = 2 1 i i
(E E + B i B i ),
2 T i0 = T 0i = ϵijk E j B k , etc. (8) The complex scalar field
The Lagrangian is given by:
L = ∂µ ϕ∗ ∂ µ ϕ − m2 ϕ∗ ϕ. (a) (9) The conjugate momenta of ϕ and ϕ∗ :
π= ∂L
= ϕ˙ ∗ ,
∂ ϕ˙ π
˜= ∂L
= ϕ˙ = π ∗ .
∂ ϕ˙ ∗ (10) The canonical commutation relations:
[ϕ(x), π(y)] = [ϕ∗ (x), π ∗ (y)] = iδ(x − y),
The rest of commutators are all zero.
The Hamiltonian:
∫
∫
(
)
(
)
H = d3 x π ϕ˙ + π ∗ ϕ˙ ∗ − L = d3 x π ∗ π + ∇ϕ∗ · ∇ϕ + m2 ϕ∗ ϕ .
(b) Now we Fourier transform the ﬁeld ϕ as:
∫
)
1 (
d3 p
−ip·x
† ip·x
√
,
ϕ(x) =
a
e
+
b
e
p
p
(2π)3 2Ep thus:
ϕ∗ (x) = ∫ )
1 ( −ip·x
d3 p
√
bp e
+ a†p eip·x .
3
(2π)
2Ep 2 (11) (12) (13) (14) Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version) Feed all these into the Hamiltonian:
∫
(
)
H = d3 x ϕ˙ ∗ ϕ˙ + ∇ϕ∗ · ∇ϕ + m2 ϕ∗ ϕ
∫
∫
d3 p
d3 q
√
√
= d3 x
3
(2π) 2Ep (2π)3 2Eq
[
(
)(
)
× Ep Eq a†p eip·x − bp e−ip·x aq e−iq·x − b†q eiq·x
(
)(
)
+ p · q a†p eip·x − bp e−ip·x aq e−iq·x − b†q eiq·x
(
)(
)]
+ m2 a†p eip·x + bp e−ip·x aq e−iq·x + b†q eiq·x
∫
∫
d3 p
d3 q
√
√
= d3 x
3
(2π) 2Ep (2π)3 2Eq
[
(
)
× (Ep Eq + p · q + m2 ) a†p aq ei(p−q)·x + bp b†q e−i(p−q)·x
(
)]
−i(p+q)·x
† † i(p+q)·x
− (Ep Eq + p · q − m ) bq aq e
+ ap bq e
2 ∫
= d3 p
d3 q
√
√
(2π)3 2Ep (2π)3 2Eq
[
(
)
× (Ep Eq + p · q + m2 ) a†p aq ei(Ep −Eq )t + bp b†q e−i(Ep −Eq )t (2π)3 δ (3) (p − q)
(
)
−i(Ep +Eq )t
† † i(Ep +Eq )t
3 (3)
− (Ep Eq + p · q − m ) bq aq e
+ ap bq e
(2π) δ (p + q)
2 ∫ )
Ep2 + p2 + m2 ( †
ap ap + bp b†p
2Ep
∫
(
)
= d3 x Ep a†p ap + b†p bp + [bp , b†p ] .
= d3 x (15) Note that the last term contributes an inﬁnite constant. It is normally explained as the
vacuum energy. We simply drop it:
∫
(
)
H = d3 x Ep a†p ap + b†p bp .
(16)
Where we have used the mass-shell condition: Ep =
two sets of particles with the same mass m. √ m2 + p2 . Hence we at once ﬁnd (c) The theory is invariant under the global transformation: ϕ → eiθ ϕ, ϕ∗ → e−iθ ϕ∗ .
The corresponding conserved charge is:
∫
(
)
Q = i d3 x ϕ∗ ϕ˙ − ϕ˙ ∗ ϕ .
(17) 3 Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version) Rewrite this in terms of the creation and annihilation operators:
∫
(
)
Q = i d3 x ϕ∗ ϕ˙ − ϕ˙ ∗ ϕ
[(
∫
∫
) (
)
d3 p
d3 q
−ip·x
† ip·x ∂
−iq·x
† iq·x
√
√
= i d3 x
b
e
+
a
e
a
e
+
b
e
p
q
p
q
∂t
(2π)3 2Ep (2π)3 2Eq
) (
)
∂ ( −ip·x
−
bp e
+ a†p eip·x · aq e−iq·x + b†q eiq·x
∂t
[ (
∫
∫
)(
)
d3 p
d3 q
3
−ip·x
† ip·x
−iq·x
† iq·x
√
√
e
e
= d x
E
b
e
+
a
a
e
−
b
q
p
q
p
q
(2π)3 2Ep (2π)3 2Eq
(
)(
)]
− Ep bp e−ip·x − a†p eip·x aq e−iq·x + b†q eiq·x
[
∫
∫
(
)
d3 q
d3 p
−i(p+q)·x
† † i(p+q)·x
√
√
= d3 x
(E
−
E
)
b
a
e
b
e
−
a
q
p
p q
p q
(2π)3 2Ep (2π)3 2Eq
(
)
+ (Eq + Ep ) a†p aq ei(p−q)·x − bp b†q e−i(p−q)·x
∫
d3 p
d3 q
√
√
=
(2π)3 2Ep (2π)3 2Eq
[
(
)
× (Eq − Ep ) bp aq e−i(Ep +Eq )t − a†p b†q ei(Ep +Eq t) (2π)3 δ (3) (p + q)
(
)
+ (Eq + Ep ) a†p aq ei(Ep −Eq )t − bp b†q e−i(Ep −Eq )t (2π)3 δ (3) (p − q)
∫
d3 p
=
· 2Ep (a†p ap − bp b†p )
(2π)3 2Ep
∫
)
d3 p ( †
=
ap ap − b†p bp ,
(18)
3
(2π)
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 = ∂µ Φ†i ∂ µ Φi − m2 Φ†i Φi , (20) where Φi with i = 1, 2 is a two-component complex scalar. Then it is straightforward to
see that the Lagrangian is invariant under the U (2) transformation Φi → Uij Φj with Uij
a matrix in fundamental (self) representation of U (2) group. The U (2) group, locally
isomorphic to SU (2) × U (1), is generated by 4 independent generators 1 and 12 τ a , with
τ a Pauli matrices. Then 4 independent N¨other currents are associated, which are given
by
∂L
∂L
∆Φi −
∆Φ∗i = −(∂µ Φ∗i )(iΦi ) − (∂µ Φi )(−iΦ∗i )
∂(∂ µ Φi )
∂(∂ µ Φ∗i )
∂L
∂L
i[
a
∗
∗
a ∗
∆
Φ
−
(∂
Φ
)τ
Φ
−
(∂
Φ
)τ
Φ
jµa = −
∆
Φ
=
−
i
µ i ij j
µ i ij j .
i
∂(∂ µ Φi )
∂(∂ µ Φ∗i )
2
jµ = − 4 (21) Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version) The overall sign is chosen such that the particle carry positive charge, as will be seen in
the following. Then the corresponding N¨other charges are given by
∫
)
( ∗
˙ i Φi − Φ∗i Φ
˙i ,
Q = − i d3 x Φ
∫
[
i
˙j .
Qa = −
d3 x Φ˙ ∗i (τ a )ij Φj − Φ∗i (τ a )ij Φ
(22)
2
Repeating the derivations above, we can also rewrite these charges in terms of creation
and annihilation operators, as
∫
)
d3 p ( †
†
Q=
a
a
−
b
b
,
ip
ip
ip
ip
(2π)3
∫
)
1
d3 p ( † a
† a
Qa =
a
τ
a
−
b
τ
b
.
(23)
ip
ip
ij
ij
ip
ip
2
(2π)3
The generalization to n-component complex scalar is straightforward. In this case
we only need to replace the generators τ a /2 of SU (2) group to the generators ta in the
fundamental representation with commutation relation [ta , tb ] = if abc tc .
Then we are ready to calculate the commutators among all these N¨other 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
∫
[(
) ( †
)]
d3 p d3 q
†
†
†
[Q, H] =
E
a
a
−
b
b
,
a
a
+
b
b
q
ip ip
ip ip
jq jq
jq jq
(2π)3 (2π)3
∫
)
(
d3 p d3 q
Eq a†ip [aip , a†jq ]ajq + a†jq [a†ip , ajq ]aip + (a → b)
=
3
3
(2π) (2π)
∫
(
)
d3 p d3 q
=
Eq a†ip aiq − a†iq aip + (a → b) (2π)3 δ (3) (p − q)
3
3
(2π) (2π)
= 0. (24) Similar calculation gives [Qa , H] = 0. Then we consider the commutation among internal
U (N ) charges:
∫
) ( † b
)]
d3 p d3 q [( † a
† a
† b
[Qa , Qb ] =
a
t
a
−
b
t
b
,
a
t
a
−
b
t
b
jp
jp
ℓq
ℓq
ip ij
ip ij
kq kℓ
kq kℓ
(2π)3 (2π)3
∫
)
(
3
3
d p d q
† a b
† b a
=
a
t
a
−
a
t
t
a
+
(a
→
b)
(2π)3 δ (3) (p − q)
t
ℓq
jp
kq kℓ ℓj
(2π)3 (2π)3 ip ij jℓ
∫
)
d3 p ( † c
† c
= if abc
a
t
a
−
b
t
b
jp
jp
ip ij
(2π)3 ip ij
= if abc Qc , (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|ϕ(x)ϕ(y)|0⟩, (26) with x − y being spacelike. Since any spacelike interval x − y can be transformed to a
form such that x0 − y 0 = 0, thus we will simply take:
x0 − y 0 = 0, |x − y|2 = r2 > 0. and
5 (27) Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version) Now:
∫ ∫
d3 p 1 −ip·(x−y)
d3 p
1
√
D(x − y) =
e
=
eip·(x−y)
3
3
(2π) 2Ep
(2π) 2 m2 + p2
∫ 2π
∫ 1
∫ ∞
1
p2
√
=
eipr cos θ
dφ
d
cos
θ
dp
(2π)3 0
2 m2 + p2
−1
0
∫ ∞
−i
peipr
=
dp √
2
2(2π) r −∞
m2 + p 2 (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
∫ ∞
1
ρe−ρr
m
√
D(x − y) =
dρ
K (mr).
(29)
=
2r 1
2
2
4π 2 r m
4π
ρ −m 6 Solutions to Peskin & Schroeder
Chapter 3
Zhong-Zhi Xianyu∗
Institute of Modern Physics and Center for High Energy Physics,
Tsinghua University, Beijing, 100084 Draft version: November 8, 2012 1 Lorentz group The Lorentz group can be generated by its generators via exponential mappings.
The generators satisfy the following commutation relation:
[J µν , J ρσ ] = i(g νρ J µσ − g µρ J νσ − g nuσ J µρ + g µσ J νρ ). (1) (a) Let us redeﬁne the generators as Li = 12 ϵijk J jk (All Latin indices denote spatial
components), where Li generate rotations, and K i generate boosts. The commutators
of them can be deduced straightforwardly to be:
[K i , K j ] = −iϵijk Lk . [Li , Lj ] = iϵijk Lk ,
i
If we further deﬁne J±
= 1
2 (2) (Li ± iK i ), then the commutators become
j
i
[J+
, J− = 0. j
k
i = iϵijk J±
,
[J±
, J± (3) Thus we see that the algebra of the Lorentz group is a direct sum of two identical algebra
su(2).
(b) It follows that we can classify the ﬁnite dimensional representations of the Lorentz
group by a pair (j+ , j− ), where j± = 0, 1/2, 1, 3/2, 2, · · · are labels of irreducible representations of SU (2).
We study two speciﬁc cases.
i
1. ( 12 , 0). Following the deﬁnition, we have J+
represented by
by 0. This implies
i
i
Li = (J+
+ J−
)= 1
2 1
2 i
σ i and J−
represented i
i
K i = −i(J+
− J−
) = − 2i σ i . σi , (4) Hence a ﬁeld ψ under this representation transforms as:
ψ → e−iθ
∗ E-mail: [email protected] 1 i σ i /2−η i σ i /2 ψ. (5) Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version) i
i
2. ( 12 , 0). In this case, J+
→ 0, J−
→
i
i
Li = (J+
+ J−
)= 1
2 1
2 σ i . Then
i
i
K i = −i(J+
− J−
)= σi , i i
2σ . (6) Hence a ﬁeld ψ under this representation transforms as:
ψ → e−iθ i σ i /2+η i σ i /2 ψ. (7) We see that a ﬁeld under the representation ( 12 , 0) and (0, 12 ) are precisely the left-handed
spinor ψL and right-handed spinor ψR , respectively.
(c) Let us consider the case of ( 21 , 12 ). To put the ﬁeld associated with this representation into a familiar form, we note that a left-handed spinor can also be rewritten as
row, which transforms under the Lorentz transformation as:
(
)
T 2
T 2
ψL
σ → ψL
σ 1 + 2i θi σ i + 12 η i σ i .
(8)
Then the ﬁeld under the representation ( 12 , 12 ) can be written as a tensor with spinor
indices:
)
(
V 0 + V 3 V 1 − iV 2
T 2
µ
ψR ψL σ ≡ V σ
¯µ =
.
(9)
V 1 + iV 2 V 0 − V 3
In what follows we will prove that V µ is in fact a Lorentz vector.
A quantity V µ is called a Lorentz vector, if it satisﬁes the following transformation
law:
V µ → Λµν V ν ,
(10)
where Λµν = δνµ − 2i ωρσ (J ρσ )µ ν in its inﬁnitesimal form. We further note that:
(J ρσ )µν = i(δµρ δνσ − δνρ δµσ ). (11) and also, ωij = ϵijk θk , ω0i = −ωi0 = η i , then the combination V µ σ
¯µ = V i σ i + V 0
transforms according to
(
)
(
)
i
i
i
V i σ i → δji − ωmn (J mn )i j V j σ i + − ω0n (J 0n )i 0 − ωn0 (J n0 )0 i V 0 σ i
2
2)
2
(
(
)
i
k
m n
m n
j i
i
i
= δj − 2 ϵmnk θ (−i)(δi δj − δj δi ) V σ + − iη (−i)(−δin ) V 0 σ i
=V i σ i − ϵijk V i θj σ k + V 0 η i σ i ,
)
i
i
ω0n (J 0n )0i − ωn0 (J n0 )0i V i
2
2
(
) i
0
i
n
0
= V + − iη (iδi ) V = V + η i V i . V0 → V0+ ( − In total, we have
(
)
V µσ
¯µ → σ i − ϵijk θj σ k + η i V i + (1 + η i σ i )V 0 . (12) If we can reach the same conclusion by treating the combination V µ σ
¯µ a matrix transforming under the representation ( 12 , 12 ), then our original statement will be proved. In
fact:
)
(
)
(
i
i
1
1
V µσ
¯ µ → 1 − θ j σ j + η j σ j V µ σµ 1 + θ j σ j + η j σ j
2
2
2
2
(
i j i j
1 j i j ) i
i
= σ + θ [σ , σ ] + η {σ , σ } V + (1 + η i σ i )V 0
2
2
( i
) i
ijk j k
i
= σ − ϵ θ σ + η V + (1 + η i σ i )V 0 ,
(13)
as expected. Hence we proved that V µ is a Lorentz vector.
2 Notes by Zhong-Zhi Xianyu 2 Solution to P&S, Chapter 3 (draft version) The Gordon identity
In this problem we derive the Gordon identity,
u
¯(p′ )γ µ u(p) = u
¯(p′ ) ( p′µ + pµ
2m + iσ µν (p′ν − pν ) )
u(p).
2m (14) Let us start from the right hand side:
)
(
1
u
¯(p′ ) (p′µ + pµ ) + iσ µν (p′ν − pν ) u(p)
2m
)
(
1
1
=
u
¯(p′ ) η µν (p′ν + pν ) − [γ µ , γ ν ](p′ν − pν ) u(p)
2m
2
(1
)
1
1
=
u
¯(p′ )
{γ µ , γ ν }(p′ν + pν ) − [γ µ , γ ν ](p′ν − pν ) u(p)
2m
2
2
(
)
1
′ µ
µ
=
u
¯(p′ ) p
¯(p′ )γ µ u(p) = LHS,
/γ +γ p
/ u(p) = u
2m RHS. = where we have used the commutator and anti-commutators of gamma matrices, as well
as the Dirac equation. 3 The spinor products In this problem, together with the Problems 5.3 and 5.6, we will develop a formalism
that can be used to calculating scattering amplitudes involving massless fermions or
vector particles. This method can profoundly simplify the calculations, especially in the
calculations of QCD. Here we will derive the basic fact that the spinor products can be
treated as the square root of the inner product of lightlike Lorentz vectors. Then, in
Problem 5.3 and 5.6, this relation will be put in use in calculating the amplitudes with
external spinors and external photons, respectively.
To begin with, let k0µ and k1µ be ﬁxed four-vectors satisfying k02 = 0, k12 = −1 and
k0 · k1 = 0. With these two reference momenta, we deﬁne the following spinors:
1. Let uL0 be left-handed spinor with momentum k0 ;
2. Let uR0 = k/1 uL0 ;
3. For any lightlike momentum p (p2 = 0), deﬁne:
uL (p) = √ 1
p
/uR0 ,
2p · k0 uR (p) = √ 1
p
/uL0 .
2p · k (15) (a) We show that k/0 uR0 = 0 and p
/uL (p) = p
/uR (p) = 0 for any lightlike p. That is,
uR0 is a massless spinor with momentum k0 , and uL (p), uR (p) are massless spinors with
momentum p. This is quite straightforward,
k/0 uR0 = k/0 k/1 uL0 = (2g µν − γ ν γ µ )k0µ k1ν uL0 = 2k0 · k1 uL0 − k/1 k/0 uL0 = 0, (16) and, by deﬁnition,
p
/uL (p) = √ 1
1
p2 uR0 = 0.
p
/p
/uR0 = √
2p · k0
2p · k0 In the same way, we can show that p
/uR (p) = 0. 3 (17) Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version) (b) Now we choose k0µ = (E, 0, 0, −E) and k1µ = (0, 1, 0, 0). Then in the Weyl
representation, we have: 0 0 0 0 0 0 0 2E k/0 uL0 = 0 ⇒ (18) uL0 = 0.
2E 0 0 0 0 0 0 0
√
Thus uL0 can be chosen to be (0, 2E, 0, 0)T , and: 0
0
0 0 1 0 0
0 1 0 (19)
uR0 = k/1 uL0 = uL0 = √ .
− 2E 0 −1 0 0
−1 0 0 0
0
Let pµ = (p0 , p1 , p2 , p3 ), then:
uL (p) = √ 1
p
/uR0
2p · k0 0 1
0 = √ 2E(p0 + p3 ) p0 − p3
−p1 − ip2 −(p0 + p3 )
−(p + ip )
1 1
2 =√ .
p0 + p3 0
0
In the same way, we get: (c) 0
0
−p1 + ip2
p0 + p3 p0 + p3
p1 + ip2
0
0 p1 − ip2
p0 − p3 uR0
0 0 (20) 0 1
0 uR (p) = √ .
p0 + p3 −p1 + ip2 p0 + p3 (21) We construct explicitly the spinor product s(p, q) and t(p, q).
s(p, q) = u
¯R (p)uL (q) = (p1 + ip2 )(q0 + q3 ) − (q1 + iq2 )(p0 + p3 )
√
;
(p0 + p3 )(q0 + q3 ) (22) t(p, q) = u
¯L (p)uR (q) = (q1 − iq2 )(p0 + p3 ) − (p1 − ip2 )(q0 + q3 )
√
.
(p0 + p3 )(q0 + q3 ) (23) It can be easily seen that s(p, q) = −s(q, p) and t(p, q) = (s(q, p))∗ .
Now we calculate the quantity |s(p, q)|2 :
(
)2 (
)2
p1 (q0 + q3 ) − q1 (p0 + p3 ) + p2 (q0 + q3 ) − q2 (p0 + p3 )
2
|s(p, q)| =
(p0 + p3 )(q0 + q3 )
q
+
q
p0 + p3
0
3
=(p21 + p22 )
+ (q12 + q22 )
− 2(p1 q1 + p2 q2 )
p0 + p3
q0 + q3
=2(p0 q0 − p1 q1 − p2 q2 − p3 q3 ) = 2p · q. (24) Where we have used the lightlike properties p2 = q 2 = 0. Thus we see that the spinor
product can be regarded as the square root of the 4-vector dot product for lightlike
vectors.
4 Notes by Zhong-Zhi Xianyu 4 Solution to P&S, Chapter 3 (draft version) Majorana fermions (a) We at ﬁrst study a two-component massive spinor χ lying in ( 12 , 0) representation,
transforming according to χ → UL (Λ)χ. It satisﬁes the following equation of motion:
i¯
σ µ ∂µ χ − imσ 2 χ∗ = 0. (25) To show this equation is indeed an admissible equation, we need to justify: 1) It is
relativistically covariant; 2) It is consistent with the mass-shell condition (namely the
Klein-Gordon equation).
To show the condition 1) is satisﬁed, we note that γ µ is invariant under the simultaneous transformations of its Lorentz indices and spinor indices. That is Λµ ν U (Λ)γ ν U (Λ−1 ) =
γ µ . This implies
Λµ ν UR (Λ)¯
σ ν UL (Λ−1 ) = σ
¯µ,
as can be easily seen in chiral basis. Then, the combination σ
¯ µ ∂µ transforms as σ
¯ µ ∂µ →
UR (Λ)¯
σ µ ∂µ UL (Λ−1 ). As a result, the ﬁrst term of the equation of motion transforms as
[ µ
i¯
σ µ ∂µ χ → iUR (Λ)¯
σ µ ∂µ UL (Λ−1 )UL (Λ)χ = UR (Λ) i¯
σ ∂µ χ . (26) To show the full equation of motion is covariant, we also need to show that the second
term iσ 2 χ∗ transforms in the same way. To see this, we note that in the inﬁnitesimal
form,
UL = 1 − iθi σ i /2 − η i σ i /2,
UR = 1 − iθi σ i /2 + η i σ i /2.
Then, under an inﬁnitesimal Lorentz transformation, χ transforms as:
χ → (1 − iθi σ i /2 − η i σ i /2)χ,
⇒ ⇒ χ∗ → (1 + iθi σ i /2 − η i σ i /2)χ∗ σ 2 χ∗ → σ 2 (1 + iθi (σ ∗ )i /2 − η i (σ ∗ )i /2)χ∗ = (1 − iθi σ i /2 + η i σ i /2)σ 2 χ∗ . That is to say, σ 2 χ∗ is a right-handed spinor that transforms as σ 2 χ∗ → UR (Λ)σ 2 χ∗ .
Thus we see the the two terms in the equation of motion transform in the same way
under the Lorentz transformation. In other words, this equation is Lorentz covariant.
To show the condition 2) also holds, we take the complex conjugation of the equation:
−i(¯
σ ∗ )µ ∂µ χ∗ − imσ 2 χ = 0.
Combining this and the original equation to eliminate χ∗ , we get
(∂ 2 + m2 )χ = 0, (27) which has the same form with the Klein-Gordon equation.
(b) Now we show that the equation of motion above for the spinor χ can be derived
from the following action through the variation principle:
[
∫
im T 2
† 2 ∗
4
†
(χ σ χ − χ σ χ ) .
(28)
S = d x χ i¯
σ · ∂χ +
2
Firstly, let us check that this action is real, namely S ∗ = S. In fact,
[
∫
im † 2 ∗
S ∗ = d4 x χT i¯
σ ∗ · ∂χ∗ −
(χ σ χ − χT σ 2 χ)
2
5 Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version) The ﬁrst term can be rearranged as
χT i¯
σ ∗ · ∂χ∗ = −(χT i¯
σ ∗ · ∂χ∗ )T = −(∂χ† ) · i¯
σ χ = χ† i¯
σ · ∂χ + total derivative.
Thus we see that S ∗ = S.
Now we vary the action with respect to χ† , that gives
0= δS
im
= i¯
σ · ∂χ −
· 2σ 2 χ∗ = 0,
δχ†
2 (29) which is exactly the Majorana equation.
(c) Let us rewrite the Dirac Lagrangian in terms of two-component spinors:
¯ ∂/ − m)ψ
L = ψ(i
(
)(
)(
)
(
) 0 1
µ
−m
iσ
∂
χ
µ
1
†
= χ1 −iχT2 σ 2
i¯
σ µ ∂µ −m
iσ 2 χ∗2
1 0
(
)
= iχ†1 σ
¯ µ ∂µ χ1 + iχT2 σ
¯ µ∗ ∂µ χ∗2 − im χT2 σ 2 χ1 − χ†1 σ 2 χ∗2
)
(
= iχ†1 σ
¯ µ ∂µ χ1 + iχ†2 σ
¯ µ ∂µ χ2 − im χT2 σ 2 χ1 − χ†1 σ 2 χ∗2 , (30) where the equality should be understood to hold up to a total derivative term.
(d) The familiar global U (1) symmetry of the Dirac Lagrangian ψ → eiα ψ now becomes χ1 → eiα χ1 , χ2 → e−iα χ2 . The associated N¨other current is
† µ
µ
¯ µ ψ = χ† σ
J µ = ψγ
¯ χ2 .
1 ¯ χ1 − χ2 σ (31) To show its divergence ∂µ J µ vanishes, we make use of the equations of motion:
i¯
σ µ ∂µ χ1 − imσ 2 χ∗2 = 0,
i¯
σ µ ∂µ χ2 − imσ 2 χ∗1 = 0,
i(∂µ χ†1 )¯
σ µ − imχT2 σ 2 = 0,
i(∂µ χ†2 )¯
σ µ − imχT1 σ 2 = 0.
Then we have
¯ µ ∂µ χ1 − (∂µ χ†2 )¯
σ µ χ2 − χ†2 σ
¯ µ ∂µ χ2
σ µ χ1 + χ†1 σ
∂µ J µ = (∂µ χ†1 )¯
)
( T 2
= m χ2 σ χ1 + χ†1 σ 2 χ∗2 − χT1 σ 2 χ2 − χ†2 σ 2 χ∗1 = 0. (32) In a similar way, one can also show that the N¨other currents associated with the global
symmetries of Majorana ﬁelds have vanishing divergence.
(e) To quantize the Majorana theory, we introduce the canonical anticommutation
relation,
{
}
χa (x), χ†b (y) = δab δ (3) (x − y),
and also expand the Majorana ﬁeld χ into modes. To motivate the mode expansion, we
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- Fall '05
- Magnetism, Energy, Quantum Field Theory, Trigraph, spinor, Dirac equation, Zhong-Zhi Xianyu