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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:
S = 14
Zd4xFF
; with F = @A @A: (1)
(a) Maxwell's equations We now derive the equations of motion from the action.
Note that
@[email protected](@A)
= ;
= 0:
Then from the rst equality we get:
@
@(@A)
FF
= 4F:
Now substitute this into Euler-Lagrange equation, we have
0 = @
@[email protected](@A)
= @F (2)
This is sometimes called the \second pair" Maxwell's equations. The so-called \rst
pair" comes directly from the denition of F = @A @A, and reads
@F + @F + @F = 0: (3)
The familiar electric and magnetic eld strengths can be written as Ei = F 0i andijkBk = F ij , respectively. From this we deduce the Maxwell's equations in terms ofEi and Bi:
@iEi = 0; [email protected] @0Ei = 0; [email protected] = 0; @iBi = 0: (4)E-mail: [email protected]
1
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version)
(b) The energy-momentum tensor The energy-momentum tensor can be dened
to be the Nother current of the space-time translational symmetry. Under space-time
translation the vector A transforms as,
A = @A : (5)
Thus~T =
@[email protected](@A)
@A L = [email protected] + 14FF
: (6)
Obviously, this tensor is not symmetric. However, we can add an additional term @K
to ~T with K being antisymmetric to its rst two indices. It's easy to see that this
term does not aect the conservation of ~T . 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 can
also rewrite it in terms of Ei and Bi:
T 00 =1
2(EiEi +BiBi); T i0 = T 0i = ijkEjBk; etc. (8)
2 The complex scalar eld
The Lagrangian is given by:
L = @@m2: (9)
(a) The conjugate momenta of and :
= _; ~ [email protected]@ _
= _ = : (10)
The canonical commutation relations:
[(x); (y)] = [(x); (y)] = i(x y); (11)
The rest of commutators are all zero.
The Hamiltonian:
H =
Zd3x
_+ _ L = Z d3x +r r+m2: (12)
(b) Now we Fourier transform the eld as:
(x) =
Zd3p
(2)31p2Ep
ape
ipx + bypeipx
; (13)
thus:
(x) =Z
d3p
(2)31p2Ep
bpe
ipx + aypeipx
: (14)
2
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version)
Feed all these into the Hamiltonian:
H =
Zd3x
_ _+r r+m2
=
Zd3x
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
EpEq
aype
ipx bpeipx
aqeiqx byqeiqx
+ p q
aype
ipx bpeipx
aqeiqx byqeiqx
+m2
aype
ipx + bpeipx
aqeiqx + byqe
iqx
=
Zd3x
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
(EpEq + p q+m2)
aypaqe
i(pq)x + bpbyqei(pq)x
(EpEq + p qm2)
bqaqe
i(p+q)x + aypbyqe
i(p+q)x
=
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
(EpEq + p q+m2)
aypaqe
i(EpEq)t + bpbyqei(EpEq)t
(2)3(3)(p q)
(EpEq + p qm2)bqaqe
i(Ep+Eq)t + aypbyqe
i(Ep+Eq)t(2)3(3)(p+ q)
=
Zd3x
E2p + p2 +m2
2Ep
aypap + bpb
yp
=
Zd3xEp
aypap + b
ypbp + [bp; b
yp]: (15)
Note that the last term contributes an innite constant. It is normally explained as the
vacuum energy. We simply drop it:
H =
Zd3xEp
aypap + b
ypbp
: (16)
Where we have used the mass-shell condition: Ep =pm2 + p2. Hence we at once nd
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
Zd3x
_ _: (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
Zd3x
_ _
= i
Zd3x
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
bpe
ipx + aypeipx
@@t
aqe
iqx + byqeiqx
@@t
bpe
ipx + aypeipx
aqe
iqx + byqeiqx
=
Zd3x
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
Eq
bpe
ipx + aypeipx
aqe
iqx byqeiqx
Epbpe
ipx aypeipx
aqeiqx + byqe
iqx
=
Zd3x
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
(Eq Ep)
bpaqe
i(p+q)x aypbyqei(p+q)x
+ (Eq + Ep)aypaqe
i(pq)x bpbyqei(pq)x
=
Zd3p
(2)3p2Ep
d3q
(2)3p2Eq
(Eq Ep)
bpaqe
i(Ep+Eq)t aypbyqei(Ep+Eqt)(2)3(3)(p+ q)
+ (Eq + Ep)aypaqe
i(EpEq)t bpbyqei(EpEq)t(2)3(3)(p q)
=
Zd3p
(2)32Ep 2Ep(aypap bpbyp)
=
Zd3p
(2)3aypap bypbp
; (18)
where the last equal sign holds up to an innitely large constant term, as we did when
calculating the Hamiltonian in (b). Then the commutators follow straightforwardly:
[Q; ay] = ay; [Q; by] = by: (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 = @[email protected] m2yii; (20)where i 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 i ! Uijj with Uija 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 a, witha Pauli matrices. Then 4 independent Nother currents are associated, which are given
by
j = @[email protected](@i)
i @[email protected](@i )
i = (@i )(ii) (@i)(ii )
@(@i)ai @L
@(@i )ai =
i
2
h(@
i )ijj (@i)ijj
i: (21)
4
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 Nother charges are given by
Q = iZ
d3x_ii i _i
;
Qa = i2
Zd3x
_i (
a)ijj i (a)ij _j: (22)
Repeating the derivations above, we can also rewrite these charges in terms of creation
and annihilation operators, as
Q =
Zd3p
(2)3
ayipaip byipbip
;
Qa =1
2
Zd3p
(2)3
ayip
aijaip byipaijbip
: (23)
The generalization to n-component complex scalar is straightforward. In this case
we only need to replace the generators a=2 of SU(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] =
Zd3p
(2)3d3q
(2)3Eq
hayipaip byipbip
;ayjqajq + b
yjqbjq
i=
Zd3p
(2)3d3q
(2)3Eq
ayip[aip; a
yjq]ajq + a
yjq[a
yip; ajq]aip + (a! b)
=
Zd3p
(2)3d3q
(2)3Eq
ayipaiq ayiqaip + (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] =
Zd3p
(2)3d3q
(2)3
hayipt
aijajp byiptaijbjp
;aykqt
bk`a`q bykqtbk`b`q
i=
Zd3p
(2)3d3q
(2)3
ayipt
aijt
bj`a`q aykqtbk`ta`jajp + (a! b)
(2)3(3)(p q)
= ifabcZ
d3p
(2)3
ayipt
cijajp byiptcijbjp
= ifabcQc; (25)
and similarly, [Q;Q] = [Qa; Q] = 0.
3 The spacelike correlation function
We evaluate the correlation function of a scalar eld at two points:
D(x y) = h0j(x)(y)j0i; (26)with x y being spacelike. Since any spacelike interval x y can be transformed to aform such that x0 y0 = 0, thus we will simply take:
x0 y0 = 0; and jx yj2 = r2 > 0: (27)
5
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 2 (draft version)
Now:
D(x y) =Z
d3p
(2)31
2Epeip(xy) =
Zd3p
(2)31
2pm2 + p2
eip(xy)
=1
(2)3
Z 20
d'
Z 11
d cos
Z 10
dpp2
2pm2 + p2
eipr cos
=i
2(2)2r
Z 11
dppeiprpm2 + 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) = 142r
Z 1m
derp2 m2 =
m
42rK1(mr): (29)
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 ; J] = i(gJ gJ gnuJ + gJ): (1)
(a) Let us redene the generators as Li = 12 ijkJjk (All Latin indices denote spatial
components), where Li generate rotations, and Ki generate boosts. The commutators
of them can be deduced straightforwardly to be:
[Li; Lj ] = iijkLk; [Ki;Kj ] = iijkLk: (2)
If we further dene J i =12 (L
i iKi), then the commutators become
[J i; Jj] = i
ijkJk; [Ji+; J
j] = 0: (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 nite dimensional representations of the Lorentz
group by a pair (j+; j), where j = 0; 1=2; 1; 3=2; 2; are labels of irreducible repre-sentations of SU(2).
We study two specic cases.
1. ( 12 ; 0). Following the denition, we have Ji+ represented by
12
i and J i representedby 0. This implies
Li = (J i+ + Ji) =
12
i; Ki = i(J i+ J i) = i2i: (4)
Hence a eld under this representation transforms as:
! eiii=2ii=2 : (5)E-mail: [email protected]
1
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
2. ( 12 ; 0). In this case, Ji+ ! 0, J i ! 12 i. Then
Li = (J i+ + Ji) =
12
i; Ki = i(J i+ J i) = i2i: (6)Hence a eld under this representation transforms as:
! eiii=2+ii=2 : (7)
We see that a eld under the representation (12 ; 0) and (0;12 ) are precisely the left-handed
spinor L and right-handed spinor R, respectively.
(c) Let us consider the case of ( 12 ;12 ). To put the eld associated with this represen-
tation into a familiar form, we note that a left-handed spinor can also be rewritten as
row, which transforms under the Lorentz transformation as:
TL2 ! TL2
1 + i2
ii + 12 ii: (8)
Then the eld under the representation (12 ;12 ) can be written as a tensor with spinor
indices:
R TL
2 V = V 0 + V 3 V 1 iV 2V 1 + iV 2 V 0 V 3
!: (9)
In what follows we will prove that V is in fact a Lorentz vector.
A quantity V is called a Lorentz vector, if it satises the following transformation
law:
V ! V ; (10)where =
i2!(J) in its innitesimal form. We further note that:
(J) = i(
): (11)
and also, !ij = ijkk, !0i = !i0 = i, then the combination V = V ii + V 0
transforms according to
V ii !ij
i
2!mn(J
mn)ij
V ji +
i
2!0n(J
0n)i0 i
2!n0(J
n0)0i
V 0i
=ij i2 mnkk(i)(mi nj mj ni )
V ji +
ii(i)(ni )V 0i=V ii ijkV ijk + V 0ii;
V 0 ! V 0 + i
2!0n(J
0n)0i i2!n0(J
n0)0i
V i
= V 0 + ii(ini )V i = V 0 + iV i:
In total, we have
V !i ijkjk + iV i + (1 + ii)V 0: (12)
If we can reach the same conclusion by treating the combination V a matrix trans-
forming under the representation (12 ;12 ), then our original statement will be proved. In
fact:
V !1 i
2jj +
1
2jj
V
1 +
i
2jj +
1
2jj
=i +
i
2j [i; j ] +
1
2jfi; jg
V i + (1 + ii)V 0
=i ijkjk + iV i + (1 + ii)V 0; (13)
as expected. Hence we proved that V is a Lorentz vector.
2
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
2 The Gordon identity
In this problem we derive the Gordon identity,
u(p0)u(p) = u(p0) p0 + p
2m+
i(p0 p)2m
u(p): (14)
Let us start from the right hand side:
RHS. =1
2mu(p0)
(p0 + p) + i(p0 p)
u(p)
=1
2mu(p0)
(p0 + p)
1
2[; ](p0 p)
u(p)
=1
2mu(p0)
12f; g(p0 + p)
1
2[; ](p0 p)
u(p)
=1
2mu(p0)
=p0 + =p
u(p) = u(p0)u(p) = LHS;
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 k0 and k1 be xed four-vectors satisfying k
20 = 0, k
21 = 1 and
k0 k1 = 0. With these two reference momenta, we dene the following spinors:
1. Let uL0 be left-handed spinor with momentum k0;
2. Let uR0 = =k1uL0;
3. For any lightlike momentum p (p2 = 0), dene:
uL(p) =1p
2p k0 =puR0; uR(p) =
1p2p k=puL0: (15)
(a) We show that =k0uR0 = 0 and =puL(p) = =puR(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,
=k0uR0 = =k0=k1uL0 = (2g )k0k1uL0 = 2k0 k1uL0 =k1=k0uL0 = 0; (16)
and, by denition,
=puL(p) =1p
2p k0 =p=puR0 =
1p2p k0
p2uR0 = 0: (17)
In the same way, we can show that =puR(p) = 0.
3
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
(b) Now we choose k0 = (E; 0; 0;E) and k1 = (0; 1; 0; 0). Then in the Weylrepresentation, we have:
=k0uL0 = 0 )
[email protected] 0 0 0
0 0 0 2E
2E 0 0 0
0 0 0 0
1CCCAuL0 = 0: (18)Thus uL0 can be chosen to be (0;
p2E; 0; 0)T , and:
uR0 = =k1uL0 =
[email protected] 0 0 1
0 0 1 0
0 1 0 01 0 0 0
1CCCAuL0 [email protected]
0
0
p2E0
1CCCA : (19)Let p = (p0; p1; p2; p3), then:
uL(p) =1p
2p k0 =puR0
=1p
2E(p0 + p3)
[email protected] 0 p0 + p3 p1 ip20 0 p1 + ip2 p0 p3
p0 p3 p1 + ip2 0 0p1 ip2 p0 + p3 0 0
1CCCAuR0
=1p
p0 + p3
[email protected](p0 + p3)(p1 + ip2)
0
0
1CCCA : (20)In the same way, we get:
uR(p) =1p
p0 + p3
0
p1 + ip2p0 + p3
1CCCA : (21)
(c) We construct explicitly the spinor product s(p; q) and t(p; q).
s(p; q) = uR(p)uL(q) =(p1 + ip2)(q0 + q3) (q1 + iq2)(p0 + p3)p
(p0 + p3)(q0 + q3); (22)
t(p; q) = uL(p)uR(q) =(q1 iq2)(p0 + p3) (p1 ip2)(q0 + q3)p
(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 js(p; q)j2:
js(p; q)j2 =p1(q0 + q3) q1(p0 + p3)
2+p2(q0 + q3) q2(p0 + p3)
2(p0 + p3)(q0 + q3)
=(p21 + p22)q0 + q3p0 + p3
+ (q21 + q22)
p0 + p3q0 + q3
2(p1q1 + p2q2)
=2(p0q0 p1q1 p2q2 p3q3) = 2p q: (24)Where we have used the lightlike properties p2 = q2 = 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 Solution to P&S, Chapter 3 (draft version)
4 Majorana fermions
(a) We at rst study a two-component massive spinor lying in (12 ; 0) representation,
transforming according to ! UL(). It satises the following equation of motion:
[email protected] im2 = 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 satised, we note that is invariant under the simultane-
ous transformations of its Lorentz indices and spinor indices. That is U()U(1) =
. This implies
UR()UL(
1) = ;
as can be easily seen in chiral basis. Then, the combination @ transforms as @ !
UR()@UL(
1). As a result, the rst term of the equation of motion transforms as
[email protected]! iUR()@UL(1)UL() = UR()[email protected]
: (26)
To show the full equation of motion is covariant, we also need to show that the second
term i2 transforms in the same way. To see this, we note that in the innitesimalform,
UL = 1 iii=2 ii=2; UR = 1 iii=2 + ii=2:Then, under an innitesimal Lorentz transformation, transforms as:
! (1 iii=2 ii=2); ) ! (1 + iii=2 ii=2)
) 2 ! 2(1 + ii()i=2 i()i=2) = (1 iii=2 + ii=2)2:That is to say, 2 is a right-handed spinor that transforms as 2 ! UR()2.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()@ im2 = 0:
Combining this and the original equation to eliminate , we get
(@2 +m2) = 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 can be derived
from the following action through the variation principle:
S =
Zd4x
yi @+ im
2(T2 y2)
: (28)
Firstly, let us check that this action is real, namely S = S. In fact,
S =Z
d4x
T i @ im
2(y2 T2)
5
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
The rst term can be rearranged as
T i @ = (T i @)T = (@y) i = yi @+ total derivative:Thus we see that S = S.
Now we vary the action with respect to y, that gives
0 =S
y= i @ im
2 22 = 0; (29)
which is exactly the Majorana equation.
(c) Let us rewrite the Dirac Lagrangian in terms of two-component spinors:
L = ([email protected] m)
=y1 iT2 2
0 11 0
! m [email protected]@ m
! 1
i22
!= iy1
@1 + iT2
@2 imT2
21 y122
= [email protected] + i
y2
@2 imT2
21 y122; (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 ! ei now be-comes 1 ! ei1, 2 ! ei2. The associated Nother current is
J = = y11 y22: (31)
To show its divergence @J vanishes, we make use of the equations of motion:
[email protected] im22 = 0;[email protected] im21 = 0;i(@
y1)
imT2 2 = 0;i(@
y2)
imT1 2 = 0:Then we have
@J = (@
y1)
1 + y1
@1 (@y2)2 [email protected]= m
T2
21 + y1
22 T1 22 y221= 0: (32)
In a similar way, one can also show that the Nother currents associated with the global
symmetries of Majorana elds have vanishing divergence.
(e) To quantize the Majorana theory, we introduce the canonical anticommutation
relation, a(x);
yb(y)
= ab
(3)(x y);and also expand the Majorana eld into modes. To motivate the mode expansion, we
note that the Majorana Langrangian can be obtained by replacing the spinor 2 in the
Dirac Lagrangian (30) with 1. Then, according to our experience in Dirac theory, it
can be found that
(x) =
Zd3p
(2)3
rp 2Ep
Xa
haaa(p)e
ipx + (i2)aaya(p)eipxi: (33)
6
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
Then with the canonical anticommutation relation above, we can nd the anticommu-
tators between annihilation and creation operators:
faa(p); ayb(q)g = ab(3)(p q); faa(p); ab(q)g = faya(p); ayb(q)g = 0: (34)On the other hand, the Hamiltonian of the theory can be obtained by Legendre trans-
forming the Lagrangian:
H =
Zd3x
L _
_ L=
Zd3x
iy r+ im
2
y2 T2: (35)
Then we can also represent the Hamiltonian H in terms of modes:
H =
Zd3x
Zd3pd3q
(2)6p2Ep2Eq
Xa;b
yaa
ya(p)e
ipx + Ta (i2)aa(p)e
ipx
(pp )y(q )pq bab(q)e
iqx (i2)bayb(q)eiqx
+im
2
yaa
ya(p)e
ipx + Ta (i2)aa(p)e
ipx
(pp )y2(pq )ba
yb(q)e
iqx + (i2)bab(q)eiqx
im2
Ta aa(p)e
ipx + ya(i2)aya(p)e
ipx
(pp )T2pq bab(q)e
iqx + (i2)bayb(q)eiqx
=
Zd3x
Zd3pd3q
(2)6p2Ep2Eq
Xa;b
aya(p)ab(q)
ya
(pp )y(q )pq
+im
2(pp )y2(pq )(i2) im
2(i2)(
pp )T2pq
be
i(pq)x
+ aya(p)ayb(q)
ya
(pp )y(q )pq (i2) + im
2(pp )y2(pq )
im2(i2)(
pp )T2pq (i2)
b e
i(p+q)x
+ aa(p)ab(q)Ta
(i2)(
pp )y(q )pq + im
2(i2)(
pp )y2(pq )(i2)
im2(pp )T2pq
be
i(p+q)x
+ aa(p)ayb(q)
Ta
(i2)(pp )y(q )pq (i2) + im
2(i2)(
pp )y2(pq )
im2(pp )T2pq (i2)
b e
i(pq)x
=
Zd3p
(2)32Ep
Xa;b
aya(p)ab(p)
ya
(pp )y(p )pp
+im
2(pp )y2(pp )(i2) im
2(i2)(
pp )T2pp
b
+ aya(p)ayb(p)ya
(pp )y(p )
p~p (i2) + im
2(pp )y2(
p~p )
im2(i2)(
pp )T2
p~p (i2)
b
+ aa(p)ab(p)Ta(i2)(
pp )y(p )
p~p + im
2(i2)(
pp )y2(
p~p )(i2)
7
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
im2(pp )T2
p~p
b
+ aa(p)ayb(p)
Ta
(i2)(pp )y(p )pp (i2) + im
2(i2)(
pp )y2(pp )
im2(pp )T2pp (i2)
b
=
Zd3p
(2)32Ep
Xa;b
1
2
E2p + jpj2 +m2
haya(p)ab(p)
yab aa(p)ayb(p)Ta b
i=
Zd3p
(2)3Ep2
Xa
haya(p)aa(p) aa(p)aya(p)
i=
Zd3p
(2)3EpXa
aya(p)aa(p): (36)
In the calculation above, each step goes as follows in turn: (1) Substituting the mode
expansion for into the Hamiltonian. (2) Collecting the terms into four groups, charac-
terized by aya, ayay, aa and aay. (3) Integrating over d3x to produce a delta function,with which one can further nish the integration over d3q. (4) Using the following
relations to simplify the spinor matrices:
(p )2 = (p )2 = E2p + jpj2; (p )(p ) = p2 = m2; p = 12 (p p ):
In this step, the ayay and aa terms vanish, while the aay and aya terms remain. (5)Using the normalization yab = ab to eliminate spinors. (6) Using the anticommutatorfaa(p); aya(p)g = (3)(0) to further simplify the expression. In this step we have throwaway a constant term 12 Ep(3)(0) in the integrand. The minus sign of this termindicates that the vacuum energy contributed by Majorana eld is negative. With these
steps done, we nd the desired result, as shown above.
5 Supersymmetry
(a) In this problem we briey study the Wess-Zumino model. Maybe it is the simplest
supersymmetric model in 4 dimensional spacetime. Firstly let us consider the massless
case, in which the Lagrangian is given by
L = @@+ [email protected]+ F F; (37)
where is a complex scalar eld, is a Weyl fermion, and F is a complex auxiliary
scalar eld. By auxiliary we mean a eld with no kinetic term in the Lagrangian and
thus it does not propagate, or equivalently, it has no particle excitation. However, in
the following, we will see that it is crucial to maintain the o-shell supersymmtry of the
theory.
The supersymmetry transformation in its innitesimal form is given by:
= iT2; (38a) = F + (@)
2; (38b)
F = [email protected]; (38c)
where is a 2-component Grassmann variable. Now let us show that the Lagrangian
is invariant (up to a total divergence) under this supersymmetric transformation. This
8
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
can be checked term by term, as follows:
(@@) = i
@
([email protected]) =F y + [email protected]
= iF [email protected]+ [email protected](@
) iT2(@@)
+ [email protected] + [email protected]@
= iF [email protected]+ [email protected](@
) iT2(@2)
+ [email protected] + [email protected];
(F F ) = i(@y)F iF [email protected];
where we have used @@ = @2. Now summing the three terms above, we get:
L = [email protected]@+ yF + T2
@
i; (39)
which is indeed a total derivative.
(b) Now let us add the mass term in to the original massless Lagrangian:
L = mF + 12 imT2+ c.c. (40)Let us show that this mass term is also invariant under the supersymmetry transforma-
tion, up to a total derivative:
(L) = imT2F [email protected]+ 12 im[TF + y(2)T ()[email protected]]2+ 12 im
T2[F + (@)2] + c.c.
= 12 imF (T2 T2) [email protected] 12 im(@)y+ 12 im(@)T ()T + c.c.
= 12 imF (T2 T2) [email protected](y)+ 12 im(@)[
y+ T ()T ] + c.c
= [email protected](y) + c.c (41)
where we have used the following relations:
(2)T = 2; 2()T2 = ; T2 = T2; y = T ()T :
Now let us write down the Lagrangian with the mass term:
L = @@+ [email protected]+ F F +mF + 12 im
T2+ c.c.: (42)
Varying the Lagrangian with respect to F , we get the corresponding equation of motion:
F = m: (43)
Substitute this algebraic equation back into the Lagrangian to eliminate the eld F , we
get
L = @@m2+ [email protected]+ 12imT2+ c.c.
: (44)
Thus we see that the scalar eld and the spinor eld have the same mass.
9
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
(c) We can also include interactions into this model. Generally, we can write a La-
grangian with nontrivial interactions containing elds i, i and Fi (i = 1; ; n), as
L = @i @i + yi [email protected] + F i Fi +Fi
@W []
@i+
i
2
@2W []
2j + c.c.
; (45)
where W [] is an arbitrary function of i.
To see this Lagrangian is supersymmetry invariant, we only need to check the inter-
actions terms in the square bracket:
Fi
@W []
@i+
i
2
@2W []
2j + c.c.
= iy(@i) @W
@i+ Fi
@2W
@[email protected](iT2j) + i
2
@3W
@[email protected]@k(iT2k)Ti 2j
+i
2
@2W
hTFi +
y(2)T ()[email protected] +
Ti
2Fj +
@j2i
+ c.c.:
The term proportional to @[email protected] vanishes. To see this, we note that the partial deriva-
tives with respect to i are commutable, hence @[email protected]@[email protected] is totally symmetric
on i; j; k. However, we also have the following identity:
(T2k)(Ti
2j) + (T2i)(
Tj
2k) + (T2j)(
Tk
2i) = 0; (46)
which can be directly checked by brute force. Then it can be easily seen that the
@[email protected] term vanishes indeed. On the other hand, the terms containing F also sum
to zero, which is also straightforward to justify. Hence the terms left now are
iy(@i) @[email protected]
@[email protected](2)T ()T (@i)2j
@W
@i
+ iyi
@2W
@[email protected]@j i @
2W
@[email protected](@i)j
@W
@i
; (47)
which is a total derivative. Thus we conclude that the Lagrangian (45) is supersymmet-
rically invariant up to a total derivative.
Let us end up with a explicit example, in which we choose n = 1 and W [] = g3=3.
Then the Lagrangian (45) becomes
L = @@+ [email protected]+ F F +gF2 + iT2+ c.c.
: (48)
We can eliminate F by solving it from its eld equation,
F + g()2 = 0: (49)
Substituting this back into the Lagrangian, we get
L = @@+ [email protected] g2()2 + ig(T2 y2): (50)
This is a Lagrangian of massless complex scalar and a Weyl spinor, with 4 and Yukawa
interactions. The eld equations can be easily got from by variations.
10
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
6 Fierz transformations
In this problem, we derive the generalized Fierz transformation, with which one can
express (u1Au2)(u3
Bu4) as a linear combination of (u1Cu4)(u3
Du2), where A is
any normalized Dirac matrices from the following set:1; ; = i2 [
; ]; 5; 5 = i0123 :(a) The Dirac matrices A are normalized according to
tr (AB) = 4AB : (51)
For instance, the unit element 1 is already normalized, since tr (1 1) = 4. For Diracmatrices containing one , we calculate the trace in Weyl representation without loss
of generality. Then the representation of
=
0
0
!
gives tr () = 2 tr () (no sum on ). For = 0, we have tr (00) = 2 tr (122) =4, and for = i = 1; 2; 3, we have tr (ii) = 2 tr (ii) = 2 tr (122) = 4 (no sumon i). Thus the normalized gamma matrices are 0 and ii.
In the same way, we can work out the rest of the normalized Dirac matrices, as:
tr (0i0i) = 2 tr (ii) = 4; (no sum on i)tr (ijij) = 2 tr (kk) = 4; (no sum on i; j; k)
tr (55) = 4;
tr (5050) = 4; tr (5i5i) = 4:
Thus the 16 normalized elements are:1; 0; ii; i0i; ij ; 5; i50; 5i
: (52)
(b) Now we derive the desired Fierz identity, which can be written as:
(u1Au2)(u3
Bu4) =XC;D
CABCD(u1Cu4)(u3
Du2): (53)
Left-multiplying the equality by (u2Fu3)(u4
Eu1), we get:
(u2Fu3)(u4
Eu1)(u1Au2)(u3
Bu4) =XCD
CABCD tr (EC) tr (FD): (54)
The left hand side:
(u2Fu3)(u4
Eu1)(u1Au2)(u3
Bu4) = u4EAFBu4 = tr (
EAFB);
the right hand side:XC;D
CABCD tr (EC) tr (FD) =
XC;D
CABCD4EC4FD = 16CABEF ;
thus we conclude:
CABCD =116 tr (
CADB): (55)
11
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
(c) Now we derive two Fierz identities as particular cases of the results above. The
rst one is:
(u1u2)(u3u4) =XC;D
tr (CD)
16(u1
Cu4)(u3Du2): (56)
The traces on the right hand side do not vanish only when C = D, thus we get:
(u1u2)(u3u4) =XC
14 (u1
Cu4)(u3Cu2)
= 14
h(u1u4)(u3u2) + (u1
u4)(u3u2) +12 (u1
u4)(u3u2)
(u15u4)(u35u2) + (u15u4)(u35u2)i: (57)
The second example is:
(u1u2)(u3u4) =
XC;D
tr (CD)
16(u1
Cu4)(u3Du2): (58)
Again, the traces vanish if C 6= C / D with C a commuting number, whichimplies that C = D. That is,
(u1u2)(u3u4) =
XC
tr (CC)
16(u1
Cu4)(u3Cu2)
= 14
h4(u1u4)(u3u2) 2(u1u4)(u3u2)
2(u15u4)(u35u2) 4(u15u4)(u35u2)i: (59)
We note that the normalization of Dirac matrices has been properly taken into account
by raising or lowering of Lorentz indices.
7 The discrete symmetries P , C and T
(a) In this problem, we will work out the C, P and T transformations of the bilinear , with = i2 [
; ]. Firstly,
P (t;x) (t;x)P = i2 (t;x)0[; ]0 (t;x):
With the relations 0[0; i]0 = [0; i] and 0[i; j ]0 = [i; j ], we get:
P (t;x) (t;x)P =
( (t;x)0i (t;x); (t;x)ij (t;x): (60)
Secondly,
T (t;x) (t;x)T = i2 (t;x)(13)[; ](13) (t;x):Note that gamma matrices keep invariant under transposition, except 2, which changes
the sign. Thus we have:
T (t;x) (t;x)T =
( (t;x)0i (t;x); (t;x)ij (t;x): (61)
Thirdly,
C (t;x) (t;x)C = i2 (i02 )T(i 02)T = 02()T 02 :Note that 0 and 2 are symmetric while 1 and 3 are antisymmetric, we have
C (t;x) (t;x)C = (t;x) (t;x): (62)
12
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
(b) Now we work out the C, P and T transformation properties of a scalar eld .
Our starting point is
PapP = ap; TapT = ap; CapC = bp:
Then, for a complex scalar eld
(x) =
Zd3k
(2)31p2k0
hake
ikx + bykeikxi; (63)
we have
P(t;x)P =
Zd3k
(2)31p2k0
hakei(k
0tkx) + bykei(k0tkx)
i= (t;x): (64a)
T(t;x)T =
Zd3k
(2)31p2k0
hakei(k
0tkx) + bykei(k0tkx)
i= (t;x): (64b)
C(t;x)C =
Zd3k
(2)31p2k0
hbke
i(k0tkx) + aykei(k0tkx)
i= (t;x): (64c)
As a consequence, we can deduce the C, P , and T transformation properties of the
current J = [email protected] (@), as follows:
PJ(t;x)P = (1)s()i(t;x)@(t;x) @(t;x)(t;x)= (1)s()J(t;x); (65a)
where s() is the label for space-time indices that equals to 0 when = 0 and 1 when
= 1; 2; 3. In the similar way, we have
TJ(t;x)T = (1)s()J(t;x); (65b)CJ(t;x)C = J(t;x): (65c)
One should be careful when playing with T | it is antihermitian rather than hermitian,
and anticommutes, rather than commutes, withp1.
(c) Any Lorentz-scalar hermitian local operator O(x) constructed from (x) and (x)can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and
contains either (x) or (x) only. Thus to prove that O(x) is an operator of CPT = +1,it is enough to show that all Lorentz-tensor hermitian operators constructed from either
(x) or (x) have correct CPT value. For operators constructed from (x), this has been
done as listed in Table on Page 71 of Peskin & Schroeder; and for operators constructed
from (x), we note that all such operators can be decomposed further into a product
(including Lorentz inner product) of operators of the form
(@1 @my)(@1 @n) + c.c
together with the metric tensor . But it is easy to show that any operator of this
form has the correct CPT value, namely, has the same CPT value as a Lorentz tensor
of rank (m+n). Therefore we conclude that any Lorentz-scalar hermitian local operator
constructed from and has CPT = +1.
13
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 3 (draft version)
8 Bound states
(a) A positronium bound state with orbital angular momentum L and total spin S
can be build by linear superposition of an electron state and a positron state, with the
spatial wave function L(k) as the amplitude. Symbolically we have
jL; Si Xk
L(k)ay(k; s)by(k; s0)j0i:
Then, apply the space-inversion operator P , we get
P jL; Si =Xk
L(k)abay(k; s)by(k; s0)j0i = (1)LabXk
L(k)ay(k; s)by(k; s0)j0i:
(66)
Note that b = a, we conclude that P jL; Si = ()L+1jL; Si. Similarly,
CjL; Si =Xk
L(k)by(k; s)ay(k; s0)j0i = (1)L+S
Xk
L(k)by(k; s0)ay(k; s)j0i:
(67)
That is, CjL; Si = (1)L+S jL; Si. Then its easy to nd the P and C eigenvalues ofvarious states, listed as follows:
SL 1S 3S 1P 3P 1D 3D
P + + C + + +
(b) We know that a photon has parity eigenvalue 1 and C-eigenvalue 1. Thus wesee that the decay into 2 photons are allowed for 1S state but forbidden for 3S state
due to C-violation. That is, 3S has to decay into at least 3 photons.
14
Solutions to Peskin & Schroeder
Chapter 4
Zhong-Zhi Xianyu
Institute of Modern Physics and Center for High Energy Physics,
Tsinghua University, Beijing, 100084
Draft version: December 9, 2013
1 Scalar field with a classical source
In this problem we consider the theory with the following Hamiltonian:
H = H0
d3 j(t,x)(x), (1)
where H0 is the Hamiltonian for free Klein-Gordon field , and j is a classical source.
(a) Let us calculate the probability that the source creates no particles. Obviously,
the corresponding amplitude is given by the inner product of the needed in-state and
the out-state. In our case, both in- and out-state are vacuum state. Thus:
P (0) =out0|0in
2 = limt(1i)
0|ei2Ht|02
=0|T exp{ i
d4xHint
}|02 =
0|T exp{i
d4x j(x)I (x)}|02. (2)
(b) Now we expand this probability P (0) to j2. The amplitude reads:
0|T exp{i
d4x j(x)I (x)
}|0 =1 1
2
d4xd4y j(x)0|TI(x)I(y)|0j(y) +O(j4)
=1 12
d4xd4y j(x)j(y)
d3p
(2)31
2Ep+O(j4)
=1 12
d3p
(2)31
2Ep|j(p)|2 +O(j4). (3)
Thus:
P (0) = |1 12+O(j4)|2 = 1 +O(j4), (4)
where:
d3p
(2)31
2Ep|j(p)|2. (5)
E-mail: [email protected]
1
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
(c) We can calculate the probability P (0) exactly, to perform this, we calculate the
j2n term of the expansion:
i2n
(2n)!
d4x1 d4x2n j(x1) j(x2n)0|T(x1) (x2n)|0
=i2n(2n 1)(2n 3) 3 1
(2n)!
d4x1 d4x2n j(x1) j(x2n)
d3p1 d3pn
(2)3n1
2nEp1 Epneip1(x1x2) eipn(x2n1x2n)
=(1)n2nn!
( d3p(2)3
|j(p)|22Ep
)n=
(/2)2n!
. (6)
Thus:
P (0) =( n=0
(/2)nn!
)2= e. (7)
(d) Now we calculate the probability that the source creates one particle with mo-
mentum k. This time, we have:
P (k) =k|T exp{i
d4x j(x)I (x)
}|02 (8)
Expanding the amplitude to the first order in j, we get:
P (k) =k|0+ i
d4x j(x)
d3p
(2)3eipx2Ep
k|ap|0+O(j2)2
=i
d3p
(2)3j(p)2Ep
2Ep(2)
3(p k)2 = |j(k)|2 +O(j3). (9)
If we go on to work out all the terms, we will get:
P (k) =
n
i(2n+ 1)(2n+ 1)(2n 1) 3 1(2n+ 1)!
jn+1(k)2 = |j(k)|2e|j(k)|. (10)
(e) To calculate the probability that the source creates n particles, we write down the
relevant amplitude:
d3k1 d3kn
(2)3n2nEk1 Ekn
k1 kn|T exp{i
d4x j(x)I (x)
}|0. (11)
Expanding this amplitude in terms of j, we find that the first nonvanishing term is
the one of nth order in j. Repeat the similar calculations above, we can find that the
amplitude is:
in
n!
d3k1 d3kn
(2)3n2nEk1 Ekn
d4x1 d4xn j1 jnk1 kn|1 n|0+O(jn+2)
=in
n!
d3k1 d3kn jn(k)
(2)3n2nEk1 Ekn
n=0
(1)n2nn!
( d3p(2)3
|j(p)|22Ep
)n(12)
Then we see the probability is given by:
P (n) =n
n!e, (13)
which is a Poisson distribution.
2
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
(f) Its quite easy to check that the Poisson distribution P (n) satisfies the following
identities: n
P (n) = 1. (14)
N =n
nP (n) = . (15)
The first one is almost trivial, and the second one can be obtained by acting dd on
both sides of the first identity. If we apply dd again on the second identity, we get:
(N N)2 = N2 N2 = . (16)
2 Decay of a scalar particle
This problem is based on the following Lagrangian:
L = 12()
2 12M22 +
1
2()
2 12m22 . (17)
When M > 2m, a particle can decay into two particles. We want to calculate the
lifetime of the particle to lowest order in .
According to the eqn (4.86) on P.107, the decay rate is given by:d =
1
2M
d3p1d
3p2(2)6
1
4Ep1Ep2
M((0) (p1)(p2))2(2)4(4)(p p1 p2).(18)
To lowest order in , the amplitude M is given by:iM = 2i. (19)
The delta function in our case reads:
(4)(p p1 p2) = (M Ep1 Ep2)(3)(p1 + p2), (20)thus:
=1
2 2
2
M
d3p1d
3p2(2)6
1
4Ep1Ep2(2)4(M Ep1 Ep2)(3)(p1 + p2), (21)
where an additional factor of 1/2 takes account of two identical s in final state. Fur-
thermore, there are two mass-shell constraints:
m2 + p2i = E2pi
i = 1, 2. (22)
Hence:
=2
M
d3p1(2)3
1
4E2p1(2)(M 2Ep1) =
2
8M
(1 4m
2
M2
)1/2. (23)
Then the lifetime of is:
= 1 =8M
2
(1 4m
2
M2
)1/2. (24)
3 Linear sigma model
In this problem, we study the linear sigma model, provided by the following La-
grangian:
L = 12 ii 12 m2ii 14 (ii)2. (25)Where is a N -component scalar.
3
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
(a) We firstly compute the following differential cross sections to the leading order in
:
(12 12), (11 22), (11 11).Since the masses of all incoming and outgoing particles are identical, the cross section
is simply given by ( dd
)COM
=|M|2642s
, (26)
where s is the square of COM energy, and M is the scattering amplitude. With thehelp of Feynman rules, its quite easy to get
M(12 12) =M(11 22) = 2i;M(11 11) = 6i. (27)
Immediately, we get
(12 12) = M(11 22) = 2
162s;
(11 11) = 92
162s. (28)
(b) Now we study the symmetry broken case, that is, m2 = 2 < 0. Then, the scalarmultiplet can be parameterized as
= (1, , N1, + v)T , (29)
where v is the VEV of ||, and equals to2/ at tree level.
Substitute this into the Lagrangian, we get
L = 12 (k)2 + 12 ()2 12 (22)2 3
kk
4 4 2 2(kk) 4 (kk)2. (30)
Then its easy to read the Feynman rules from this expression:
k=
i
k2 22 ; (31a)
k=
iij
k2; (31b)
= 6iv; (31c)
i j
= 2ivij ; (31d)
= 6i; (31e)
4
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
i j
= 2iij ; (31f)
i j
k
= 2i(ijk + ikj + ijk). (31g)
(c) With the Feynman rules derived in (b), we can compute the amplitude
M[i(p1)j(p2) k(p3)(p4)],as:
M = (2iv)2[ is 22
ijk +i
t 22 ikj +
i
u 22 ijk
]
2i(ijk + ikj + ijk), (32)
where s, t, u are Mandelstam variables (See Section 5.4). Then, at the threshold pi = 0,
we have s = t = u = 0, and M vanishes.On the other hand, if N = 2, then there is only one component in , thus the
amplitude reduces to
M = 2i[ 22s 22 +
22
t 22 +22
u 22 + 3]
= 2i[ s+ t+ u
22+O(p4)
]. (33)
In the second line we perform the Taylor expansion on s, t and u, which are of order
O(p2). Note that s+ t+ u = 4m2 = 0, thus we see that O(p2) terms are also canceledout.
(d) We minimize the potential with a small symmetry breaking term:
V = 2ii + 4 (ii)2 aN , (34)
which yields the following equation that determines the VEV:
( 2 + ii)i = aiN . (35)Thus, up to linear order in a, the VEV i = (0, , 0, v) is
v =
2
+
a
22. (36)
Now we repeat the derivation in (b) with this new VEV, and write the Lagrangian in
terms of new field variable i and , as
L = 12 (k)2 + 12 ()2 12a
kk 12 (22)2
v3 vkk 14 4 2 2(kk) 4 (kk)2. (37)
5
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
The ij k amplitude is still given by
M = (2iv)2[ is 22
ijk +i
t 22 ikj +
i
u 22 ijk
]
2i(ijk + ikj + ijk). (38)
However this amplitude does not vanishes at the threshold. Since the vertices 6=
exactly even at tree level, and also s, t and u are not exactly zero in this case due to
nonzero mass of i. Both deviations are proportional to a, thus we conclude that the
amplitude M is also proportional to a.
4 Rutherford scattering
The Rutherford scattering is the scattering of an election by the coulomb field of a
nucleus. In this problem, we calculate the cross section by treating the electromagnetic
field as a given classical potential A(x). Then the interaction Hamiltonian is:
HI =
d3x eA. (39)
(a) We first calculate the T -matrix to lowest order. In fact:
outp|pin =p|T exp(i
d4xHI)|p = p|p ie
d4xA(x)p||p+O(e2)
=p|p ie
d4xA(x)u(p)u(p)ei(p
p)x +O(e2)
=(2)4(4)(p p) ieu(p)u(p)A(p p) +O(e2) (40)
But on the other hand,
outp|pin = p|S|p = p|p+ p|iT |p. (41)
Thus to the first order of e, we get:
p|iT |p = ieu(p)u(p)A(p p). (42)
(b) Now we calculate the cross section d in terms of the matrix elements iM.The incident wave packet | is defined to be:
| =
d3k
(2)3eibk2Ek
(k)|k, (43)
where b is the impact parameter.
The probability that a scattered electron will be found within an infinitesimal element
d3p centered at p is:
P =d3p
(2)31
2Ep
outp|in2
=d3p
(2)31
2Ep
d3kd3k
(2)62Ek2Ek
(k)(k)(outp|kin
)(outp|kin
)eib(kk
)
=d3p
(2)31
2Ep
d3kd3k
(2)62Ek2Ek
(k)(k)(p|iT |k
)(p|iT |k
)eib(kk
). (44)
6
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
In the last equality we have excluded the trivial scattering part from the S-matrix. Note
that:
p|iT |p = iM(2)(Ep Ep), (45)we have:
P =d3p
(2)31
2Ep
d3kd3k
(2)62Ek2Ek
(k)(k)|iM|2(2)2(EpEk)(EpEk)eib(kk)
(46)
The cross section d is given by:
d =
d2b P (b), (47)
thus the integration over b gives a delta function:
d2b eib(kk
) = (2)2(2)(k k). (48)
The other two delta functions in the integrand can be modified as follows:
(Ek Ek) = Ekk
(k k) =1
v(k k), (49)
where we have used |v| = v = v. Taking all these delta functions into account, we get:
d =d3p
(2)31
2Ep
d3k
(2)32Ek
1
v(k)(k)|iM|2(2)(Ep Ek). (50)
Since the momentum of the wave packet should be localized around its central value,
we can pull out the quantities involving energy Ek outside the integral:
d =d3p
(2)31
2Ep
1
2Ek
1
v(2)|M|2(Ep Ek)
d3k
(2)3(k)(k). (51)
Recall the normalization of the wave packet:
d3k
(2)3(k)(k) = 1, (52)
then:
d =d3p
(2)31
2Ep
1
2Ek
1
v|M(k p)|2(2)(Ep Ek). (53)
We can further integrate over |p| to get the differential cross section d/d:
d
d=
dp p2
(2)31
2Ep
1
2Ek
1
v|M(k p)|2(2)(Ep Ek)
=
dp p2
(2)31
2Ep
1
2Ek
1
v|M(k p)|2(2) Ek
k(p k)
=1
(4)2|M(k, )|2. (54)
In the last line we work out the integral by virtue of delta function, which constrains the
outgoing momentum |p| = |k| but leave the angle between p and k arbitrary. Thusthe amplitude M(k, ) is a function of momentum |k| and angle .
7
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 4 (draft version)
(c) We work directly for the relativistic case. Firstly the Coulomb potential A0 =
Ze/4r in momentum space is
A0(q) =Ze
|q|2 . (55)
This can be easily worked out by Fourier transformation, with a regulator emr in-serted:
A0(q,m)
d3x eipxemrZe
4r=
Ze
|q|2 +m2 . (56)
This is simply Yukawa potential, and Coulomb potential is a limiting case when m 0.The amplitude is given by
iM(k, ) = ieu(p)A(q)u(p) with q = p k. (57)
Then we have the module square of amplitude with initial spin averaged and final spin
summed (See 5.1 of Peskin & Schroeder for details), as:12
spin
|iM(k, )|2 = 12 e2A(q)A (q)spin
u(p)u(k)u(k)u(p)
= 12 e2A(q)A (q) tr
[(/p+m)
(/k +m)]
=2e2[2(p A)(k A) + (m2 (k p))A2]. (58)
Note that
A0(q) =Ze
|p k|2 =Ze
4|k|2 sin2(/2) , (59)
thus
12
spin
|iM(k, )|2 = Z2e4
(1 v2 sin2 2
)4|k|4v2 sin4(/2) , (60)
andd
d=
Z22(1 v2 sin2 2
)4|k|2v2 sin4(/2) (61)
In non-relativistic case, this formula reduces to
d
d=
Z22
4m2v4 sin4(/2)(62)
8
Solutions to Peskin & Schroeder
Chapter 5
Zhong-Zhi Xianyu
Institute of Modern Physics and Center for High Energy Physics,
Tsinghua University, Beijing, 100084
Draft version: November 8, 2012
1 Coulomb scattering
In this problem we continue our study of the Coulomb scattering in Problem 4.4.
Here we consider the relativistic case. Let's rst recall some main points considered
before. The Coulomb potential A0 = Ze=4r in momentum space is
A0(q) =Ze
jqj2 : (1)
Then the scattering amplitude is given by
iM(k; ) = ieu(p) ~A(q)u(p) with q = p k: (2)
Then we can derive the squared amplitude with initial spin averaged and nal spin
summed, as:
1
2
Xspin
jiM(k; )j2 = 12e2 ~A(q) ~A(q)
Xspin
u(p)u(k)u(k)u(p)
=1
2e2 ~A(q) ~A(q) tr
h
(=p+m)
(=k +m)i
= 2e2h2(p ~A)(k ~A) + m2 (k p) ~A2i: (3)
Note that~A0(q) =
Ze
jp kj2 =Ze
4jkj2 sin2(=2) ; (4)
thus1
2
Xspin
jiM(k; )j2 = Z2e41 v2 sin2 2
4jkj4v2 sin4(=2) ; (5)
Now, from the result of Problem 4.4(b), we know that
d
d
=
1
(4)2
12
Xspin
jM(k; )j2=
Z221 v2 sin2 2
4jkj2v2 sin4(=2) (6)
E-mail: [email protected]
1
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
% k1 k2 -
- p1 p2 %
e Z
e Z
Figure 1: The scattering of an electron by a charged heavy particle Z . All initial momentago inward and all nal momenta go outward.
This is the formula for relativistic electron scatted by Coulomb potential, and is called
Mott formula.
Now we give an alternative derivation of the Mott formula, by considering the cross
section of eZ ! eZ . When the mass of goes to innity and the charge of istaken to be Ze, this cross section will reduces to Mott formula. The relevant amplitude
is shown in Figure 1, which reads
iM = Z(ie)2u(p1)u(k1) it
U(p2)U(k2); (7)
where u is the spinor for electron and U is the spinor for muon, t = (k1 p1)2 is oneof three Mandelstam variables. Then the squared amplitude with initial spin averaged
and nal spin summed is
14
Xspin
jiMj2 = Z2e4
t2trh
(=k1 +m)
(=p1 +m)itrh
(=k2 +M)(=p2 +M)
i=
Z2e4
t2
h16m2M2 8M2(k1 p1) + 8(k1 p2)(k2 p1)
8m2(k2 p2) + 8(k1 k2)(p1 p2)i: (8)
Note that the cross section is given by dd
CM
=1
2Ee2Ejvk1 vk2 jjp1j
(2)24ECM
14
XjMj2
: (9)
When the mass of goes to innity, we have E ' ECM 'M , vk2 ' 0, and jp1j ' jk1j.Then the expression above can be simplied to d
d
CM
=1
16(2)2M2
14
XjMj2
: (10)
When M ! 1, only terms proportional to M2 are relevant in jMj2. To evaluate thissquared amplitude further, we assign each momentum a specic value in CM frame:
k1 = (E; 0; 0; k); p1 ' (E; sin ; 0; k cos );k2 ' (M; 0; 0;k); p2 ' (M;k sin ; 0;k cos ); (11)
then t = (k1 p1)2 = 4k2 sin2 2 , and
14
XjiMj2 = Z
2e4(1 v2 sin2 2 )k2v2 sin2 2
M2 +O(M): (12)
Substituting this into the cross section, and sendingM !1, we reach the Mott formulaagain:
d
d
=
Z221 v2 sin2 2
4jkj2v2 sin4(=2) (13)
2
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
k1
p2
k2
p1
e e+
e e+
k1
p2p1
k2e e+
e e+
Figure 2: Bhabha scattering at tree level. All initial momenta go inward and all nal momenta
go outward.
2 Bhabha scattering
The Bhabha scattering is the process e+e ! e+e. At the tree level, it consistsof two diagrams, as shown in Figure 2. The minus sign before the t-channel diagram
comes from the exchange of two fermion eld operators when contracting with in and
out states. In fact, the s- and t-channel diagrams correspond to the following two ways
of contraction, respectively:
hp1p2j =A =A jk1k2i; hp1p2j =A =A jk1k2i: (14)
In the high energy limit, we can omit the mass of electrons, then the amplitude for the
whole scattering process is:
iM = (ie)2v(k2)
u(k1)is
u(p1)v(p2) u(p1)u(k1) itv(k2)v(p2)
: (15)
Where we have used the Mandelstam variables s, t and u. They are dened as:
s = (k1 + k2)2; t = (p1 k1)2; u = (p2 k1)2: (16)
In the massless case, k21 = k22 = p
21 = p
22 = 0, thus we have:
s = 2k1 k2 = 2p1 p2; t = 2p1 k1 = 2p2 k2; u = 2p2 k1 = 2p1 k2: (17)
We want to get the unpolarized cross section, thus we must average the ingoing spins
and sum over outgoing spins. That is:
1
4
Xspin
jMj2 = e4
4s2
Xv(k2)u(k1)u(p1)v(p2)2+
e4
4t2
Xu(p1)u(k1)v(k2)v(p2)2 e
4
4st
Xhv(p2)u(p1)u(k1)
v(k2)u(p1)u(k1)v(k2)v(p2) + c.c.
i=
e4
4s2tr (=k1
=k2) tr (=p2=p1) +
e4
4t2tr (=k1
=p1
) tr (=p2=k2)
e4
4st
htr (=k1
=k2=p2=p1) + c.c.
i=
2e4(u2 + t2)
s2+
2e4(u2 + s2)
t2+
4e4u2
st
= 2e4t2
s2+
s2
t2+ u2
1s+
1
t
2: (18)
3
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
In the center-of-mass frame, we have k01 = k02 k0, and k1 = k2, thus the total
energy E2CM = (k01 + k
02)2 = 4k2 = s. According to the formula for the cross section in
the four identical particles' case (Eq.4.85): dd
CM
=1
642ECM
14
XjMj2
; (19)
thus dd
CM
=2
2s
t2
s2+
s2
t2+ u2
1s+
1
t
2; (20)
where = e2=4 is the ne structure constant. We integrate this over the angle ' to
get: dd cos
CM
=2
s
t2
s2+
s2
t2+ u2
1s+
1
t
2: (21)
3 The spinor products (2)
In this problem we continue our study of spinor product method in last chapter. The
formulae needed in the following are:
uL(p) =1p
2p k0 =puR0; uR(p) =
1p2p k0 =
puL0: (22)
s(p1; p2) = uR(p1)uL(p2); t(p1; p2) = uL(p1)uR(p2): (23)
For detailed explanation for these relations, see Problem 3.3
(a) Firstly, let us prove the following relation:
js(p1; p2)j2 = 2p1 p2: (24)
We make use of the another two relations,
uL0uL0 =1 5
2=k0; uR0uR0
1 + 5
2=k0: (25)
which are direct consequences of the familiar spin-sum formulaP
u0u0 = =k0. We now
generalize this to:
uL(p)uL(p) =1 5
2=p; uR(p)uR(p) =
1 + 5
2=p: (26)
We prove the rst one:
uL(p)uL(p) =1
2p k0 =puR0uR0=p =1
2p k0 =p1 + 5
2=k0=p
=1
2p k01 5
2=p=k0=p =
1
2p k01 5
2(2p k =k0=p)=p
=1 5
2=p 1
2p k01 5
2=k0p
2 =1 5
2=p: (27)
The last equality holds because p is lightlike. Then we get:
js(p1; p2)j2 =juR(p1)uL(p2)j2 = truL(p2)uL(p2)uR(p1)uR(p1)
=1
4tr(1 5)=p2(1 5)=p1
= 2p1 p2: (28)
4
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
(b) Now we prove the relation:
tr (12 n) = tr (n 21); (29)
where i = 0; 1; 2; 3; 5.
To make things easier, let us perform the proof in Weyl representation, without loss
of generality. Then it's easy to check that
()T =
(
; = 0; 2; 5;
; = 1; 3: (30)
Then, we dene M = 13, and it can be easily shown that M1M = ()T , andM1M = 1. Then we have:
tr (12 n) = tr (M11MM12M M1nM)= tr
(1)T
2)T (n)T = tr (n 21)T
= tr (n 21): (31)
With this formula in hand, we can derive the equality
uL(p1)uL(p2) = uR(p2)
uR(p1) (32)
as follows
LHS = CuR0=p1=p2uR0 = C tr (=p1
=p2)
= C tr (=p2=p1) = CuL0=p2
=p1uL0 = RHS;
in which C 2p(p1 k0)(p2 k0)1.(c) The way of proving the Fierz identity
uL(p1)uL(p2)[]ab = 2
uL(p2)uL(p1) + uR(p1)uR(p2)
ab
(33)
has been indicated in the book. The right hand side of this identity, as a Dirac matrix,
which we denoted by M , can be written as a linear combination of 16 matrices listed
in Problem 3.6. In addition, it is easy to check directly that M = M5. Thus Mmust have the form
M = 1 5
2
V
+ 1 + 5
2
W
:
Each of the coecients V and W can be determined by projecting out the other one
with the aid of trace technology, that is,
V =1
2tr
1 5
2
M= uL(p1)
uL(p2); (34)
W =1
2tr
1 + 5
2
M= uR(p2)
uR(p1) = uL(p1)uL(p2): (35)
The last equality follows from (32). Substituting V and W back, we nally get the
left hand side of the Fierz identity, which nishes the proof.
5
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
(d) The amplitude for the process at leading order in is given by
iM = (ie2)uR(k2)uR(k1) isvR(p1)vR(p2): (36)
To make use of the Fierz identity, we multiply (33), with the momenta variables changed
to p1 ! k1 and p2 ! k2, byvR(p1)
aand
vR(p2)
b, and also take account of (32),
which leads to
uR(k2)uR(k1)vR(p1)vR(p2)
= 2vR(p1)uL(k2)uL(k1)vR(p2) + vR(p1)uR(k1)uR(k2)vR(p2)
= 2s(p1; k2)t(k1; p2): (37)
Then,
jiMj2 = 4e4
s2js(p1; k2)j2jt(k1; p2)j2 = 16e
4
s2(p1 k2)(k1 p2) = e4(1 + cos )2; (38)
andd
d
(e+Le
R ! +LR) =
jiMj2642Ecm
=2
4Ecm(1 + cos )2: (39)
It is straightforward to work out the dierential cross section for other polarized pro-
cesses in similar ways. For instance,
d
d
(e+Le
R ! +RL ) =
e4jt(p1; k1)j2js(k2; p2)j2642Ecm
=2
4Ecm(1 cos )2: (40)
(e) Now we recalculate the Bhabha scattering studied in Problem 5.2, by evaluating
all the polarized amplitudes. For instance,
iM(e+LeR ! e+LeR)
= (ie)2uR(k2)
uR(k1)isvR(p1)vR(p2)
uR(p1)uR(k1) itvR(k2)vR(p2)
= 2ie2
s(p1; k2)t(k1; p2)
s s(k2; p1)t(k1; p2)
t
: (41)
Similarly,
iM(e+LeR ! e+ReL ) = 2ie2t(p1; k1)s(k2; p2)
s; (42)
iM(e+ReL ! e+LeR) = 2ie2s(p1; k1)t(k2; p2)
s; (43)
iM(e+ReL ! e+ReL ) = 2ie2t(p1; k2)s(k1; p2)
s t(k2; p1)s(k1; p2)
t
; (44)
iM(e+ReR ! e+ReR) = 2ie2t(k2; k1)s(p1; p2)
t; (45)
iM(e+LeL ! e+LeL ) = 2ie2s(k2; k1)t(p1; p2)
t: (46)
Squaring the amplitudes and including the kinematic factors, we nd the polarized
dierential cross sections as
d
d
(e+Le
R ! e+LeR) =
d
d
(e+Re
L ! e+ReL ) =
2u2
2s
1s+
1
t
2; (47)
6
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
d
d
(e+Le
R ! e+ReL ) =
d
d
(e+Re
L ! e+LeR) =
2
2s
t2
s2; (48)
d
d
(e+Re
R ! e+ReR) =
d
d
(e+Le
L ! e+LeL ) =
2
2s
s2
t2: (49)
Therefore we recover the result obtained in Problem 5.2:
d
d
(e+e ! e+e) =
2
2s
t2
s2+
s2
t2+ u2
1s+
1
t
2: (50)
4 Positronium lifetimes
In this problem we study the decay of positronium (Ps) in its S and P states. To
begin with, we recall the formalism developed in the Peskin & Schroeder that treats
the problem of bound states with nonrelativistic quantum mechanics. The positronium
state jPsi, as a bound state of an electron-positron pair, can be represented in terms ofelectron and positron's state vectors, as
jPsi =p2MP
Zd3k
(2)3 (k)Cab
1p2m
jea (k)i1p2m
je+b (k)i; (51)
where m is the electron's mass, MP is the mass of the positronium, which can be taken
to be 2m as a good approximation, a and b are spin labels, the coecient Cab depends
on the spin conguration of jPsi, and (k) is the momentum space wave function forthe positronium in nonrelativistic quantum mechanics. In real space, we have
100(r) =
r(mr)
3
exp(mrr); (52)
21i(r) =
r(mr=2)
5
xi exp(mrr=2): (53)
where mr = m=2 is the reduced mass. Then the amplitude of the decay process Ps! 2can be represented in terms of the amplitude for the process e+e ! 2 as
M(Ps! 2) = 1pm
Zd3k
(2)3 (k)CabcMea (k)e+b (k)! 2: (54)
We put a hat on the amplitude of e+e ! 2. In the following, a hatted amplitudealways refer to this process.
(a) In this part we study the decay of the S-state positronium. As stated above, we
have to know the amplitude of the process e+e ! 2, which is illustrated in Figure 3with the B replaced with , and is given by
icM = (ie)2(p1)(p2) v(k2)
i(=k1 =p1 +m)(k1 p1)2 m2
+ i(=k1 =p2 +m)(k1 p2)2 m2
u(k1); (55)
where the spinors can be written in terms of two-component spinors and 0 in thechiral representation as
u(k1) =
pk1 pk1
; v(k2) =
pk2 0
pk2 0: (56)
7
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
We also write as:
=
0
0
!;
where = (1; i) and = (1;i) with i the three Pauli matrices. Then theamplitude can be brought into the following form,
icM = ie2(p1)(p2)0y t(k1 p1)2 m2 + u
(k1 p2)2 m2; (57)
with
t =p
k2 pk1
pk2
pk1
m
+p
k2 pk1
pk2
pk1
(k1 p1);
u =p
k2 pk1
pk2
pk1
m
+p
k2 pk1
pk2
pk1
(k1 p2):
In the rest of the part (a), we take the nonrelativistic limit, with the momenta chosen
to be
k1 = k2 = (m; 0; 0; 0); p
1 = (m; 0; 0;m); p
2 = (m; 0; 0;m): (58)
Accordingly, we can assign the polarization vectors for nal photons to be
(p1) =1p2(0; 1;i; 0); (p2) = 1p2 (0;1;i; 0): (59)
Now substituting the momenta (58) into (57), noticing thatpki =
pki =
pm (i =
1; 2), and (k1 p1)2 = (k1 p2)2 = m2, and also using the trick that one can freelymake the substitution ! since the temporal component of the polarizationvectors always vanishes, we get a much more simplied expression,
icM = ie2(p1)(p2)0y3 3: (60)The positronium can lie in spin-0 (singlet) state or spin-1 (triplit) state. In the former
case, we specify the polarizations of nal photons in all possible ways, and also make
the substitution 0y ! 1p2(See (5.49) of Peskin & Schroeder), which leads to
icMs++ = icMs = i2p2e2; icMs+ = icMs+ = 0; (61)where the subscripts denote nal photons' polarizations and s means singlet. We show
the mid-step for calculating iMs++ as an example:
icMs++ = ie22 tr h(1 + i2)3(1 + i2) (1 + i2)3(1 + i2)0yi = i2p2e2:In the same way, we can calculate the case of triplet initial state. This time, we make
the substitution 0y ! n =p2, with n = (x^ iy^)=p2 or n = z^, corresponding tothree independent polarizations. But it is straightforward to show that the amplitudes
with these initial polarizations all vanish, which is consistent with our earlier results by
using symmetry arguments in Problem 3.8.
8
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
Therefore it is enough to consider the singlet state only. The amplitude for the decay
of a positronium in its 1S0 state into 2 then follows directly from (54), as
M(1S0 ! 2) = (x = 0)pm
cMs; (62)where (x = 0) =
p(m=2)3= according to (52). Then the squared amplitude with
nal photons' polarizations summed isXspin
M(1S0 ! 2)2 = j (0)j22m
jMs++j2 + jMsj2
= 165m2: (63)
Finally we nd the decay width of the process Ps(1S0)! 2, to be
(1S0 ! 2) = 12
1
4m
Zd3p1d
3p2(2)62E12E2
PM(1S0 ! 2)2(2)4(4)(pPs p1 p2)=
1
2
1
4m
Zd3p1
(2)34m2PM(1S0 ! 2)2(2)(m E1)
=1
25m; (64)
where an additional factor of 1=2 follows from the fact that the two photons in the nal
state are identical particles.
(b) To study the decay of P state (l = 1) positronium, we should keep one power of
3-momenta of initial electron and positron. Thus we set the momenta of initial and nal
particles, and also the polarization vectors of the latter, in ee+ ! 2, to be
k1 = (E; 0; 0; k); k2 = (E; 0; 0;k);
p1 = (E;E sin ; 0; E cos ); p2 = (E;E sin ; 0;E cos );
(p1) =1p2(0; cos ;i; sin ); (p2) = 1p2 (0; cos ;i; sin ): (65)
Here we have the approximate expression up to linear order in k:pk1 =
pk2 =
pm k
2pm3 +O(k2);p
k2 =pk1 =
pm+
k
2pm3 +O(k2);
1
(k1 p1)2 m2 = 1
2m2 k cos
2m3+O(k2);
1
(k1 p2)2 m2 = 1
2m2+
k cos
2m3+O(k2):
Consequently,
t = 2m2(1s +
3c) mk3 + 3 + 23+O(k2);
u = 2m2(1s + 3c) mk3 + 3 + 23
+O(k2);
where we use the shorthand notation s = sin and c = cos . We can use these
expansion to nd the terms in the amplitude icM of linear order in k, to beicMO(k) = ie2(p1)(p2) k2m0yh 2c(1s + 3c) 2c(1s + 3c)
9
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
+3 + 3 + 23
+3 + 3 + 23
i;
(66)
Feeding in the polarization vectors of photons, and also make the substitution 0y !n =p2 or 1=p2 for triplet and singlet positronium, respectively, as done in last part,we get
icM##jO(k) = 0; icM##jO(k) = i2s(1 + c)e2k=m;icM"#+#" jO(k) = i2p2e2k=m; icM"#+#" jO(k) = i2p2s2e2k=m;icM""jO(k) = 0; icM""jO(k) = i2s(1 + c)e2k=m;icM"##" jO(k) = 0; icM"##" jO(k) = 0: (67)
The vanishing results in the last line indicate that S = 0 state of P -wave positronium
cannot decay to two photons.
(c) Now we prove that the state
jB(k)i =p2MP
Zd3p
(2)3 i(p)a
yp+k=2
ibyp+k=2j0i (68)
is a properly normalized state for the P -wave positronium. In fact,
hB(k)jB(k)i = 2MPZ
d3p0
(2)3d3p
(2)3 j (p
0) i(p)
h0jbp0+k=2jyap0+k=2ayp+k=2ibyp0+k=2j0i
= 2MP
Zd3p0
(2)3d3p
(2)3 j (p
0) i(p)
h0jbp0+k=2jyibyp0+k=2j0i(2)3(3)(p0 p)
= 2MP
Zd3p
(2)3 j (p) i(p)h0jbp+k=2jyibyp+k=2j0i
= 2MP
Zd3p
(2)3 j (p) i(p)h0j tr (jyi)j0i(2)3(3)(0)
= 2MP (2)3(3)(0); (69)
which is precisely the needed normalization of a state. In this calculation we have used
the anticommutation relations of creation and annihilation operators, as well as the
normalization of the wave function and the matrices.
(d) Now we evaluate the partial decay rate of the S = 1 P -wave positronium of
denite J into two photons. The states for the positronium is presented in (c), with the
matrices chosen as
=
8>>>>>>>>>>>:
1p6i; J = 0;
1
2ijknjk; J = 1;
1p3hijj ; J = 2;
(70)
and the wave function given by (53).
10
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
Firstly, consider the J = 0 state, in which case we have
iM(3P0 ! ) = 1pm
Zd3k
(2)3 i(k)
1p6iabicMea (k)e+b (k)! ; (71)
where ; = + or are labels of photons' polarizations and a; b =" or # are spinorindices. For amplitude icM, we only need the terms linear in k, as listed in (67). Let usrewrite this as
icMea (k)e+b (k)! = F ab;iki:In the same way, the wave function can also be put into the form of i(x) = x
if(r),
with r = jxj. Then the integration above can be carried out to be
iM(3P0 ! ) = ip6m
iabFab;j
@xj i(x)
x=0
=ip6m
iabFab;if(0): (72)
On the other hand, we have chose the direction of k to be in the x3-axis, then F ab;1 =
F ab;2 = 0 as a consequence. Therefore,
iM(3P0 ! ) = ip6m
f(0)F "";3 F ##;3
=
r7
24m sin : (73)
Square these amplitudes, sum over the photons' polarizations, and nish the phase space
integration in the same way as what we did in (a), we nally get the partial decay rate
of the J = 0 P -wave positronium into two photons to be
(3P0) =1
5767m: (74)
The positronium in 3P1 state, namely the case J = 1, cannot decay into two photons
by the conservation of the angular momentum, since the total angular momentum of
two physical photons cannot be 1. Therefore let us turn to the case of J = 2. In this
case we should average over the initial polarizations of the positronium, which can be
represented by the symmetric and traceless polarization tensors hijn , with n = 1; 2; ; 5the labeled of 5 independent polarizations. Let us choose these tensors to be
hij1 =1p2(i2j3 + i3j2); hij2 =
1p2(i1j3 + i3j1);
hij3 =1p2(i1j2 + i2j1); hij4 =
1p2(i1j1 i2j2);
hij5 =1p2(i1j1 i3j3): (75)
Then the decay amplitude for a specic polarization of J = 2 Ps can be represented as
iMn(3P2 ! ) = 1pm
Zd3k
(2)3 i(k)
1p3hijn
jabicMea (k)e+b (k)!
=1p3m
hijn jabF
ab;if(0): (76)
Now substituting all stus in, we nd the nonvanishing components of the decay ampli-
tude to be
iM2(3P2 ! ) =r
7
48im;
11
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
iM2(3P2 ! ) =r
7
48im sin2 ;
iM5(3P2 ! ) = 2r
7
48im sin2 : (77)
Squaring these amplitudes, summing over photon's polarizations and averaging the ini-
tial polarization of the positronium (by dividing the squared and summed amplitude by
5), we get
1
5
Xspin
Mn(3P2 ! 2)2 = 7m2120
(1 + sin2 + 4 sin4 ): (78)
Finally, we nish the phase space integration and get the partial decay rate of 3P2positronium into 2 photons to be
(3P2) =19
192007m: (79)
5 Physics of a massive vector boson
In this problem, the mass of electron is always set to zero.
(a) We rstly compute the cross section (e+e ! B) and the decay rate (B !e+e). For the cross section, the squared amplitude can be easily found to be
1
4
Xspin
jiMj2 = 14
Xspin
ig(i) v(p0)u(p)2 = 2g2(p p0): (80)Note that we have set the mass of electrons to be zero. Then the cross section can be
deduced from (4.79). Let's take the initial momenta to be
p = 12 (E; 0; 0; E); p0 = 12 (E; 0; 0;E); (81)
with E being the center-of-mass energy. Then it's easy to get
=g2
4E(2)(MB E) = g
2
4E(2)2MB(M
2B s) = g2(M2B s); (82)
where s = E2.
To deduce the decay rate, we should average polarizations of massive vector B instead
of two electrons. Thus the squared amplitude in this case reads
1
3
Xspin
jiMj2 = 83g2(p p0): (83)
The decay rate can be found from (4.86):
=1
2MB
Zd3p
(2)3d3p0
231
2Ep
1
2Ep0
13
PjMj2(2)4(4)(pB p p0)=
1
2MB
Zd3p
(2)31
4E2p
163 g
2E2p
(2)(MB 2Ep)
=4
(2)22MB
Zdp 43 g
2E2p12 (
12 MB Ep) =
g2MB12
: (84)
We see the cross section and the decay rate satisfy the following relation, as expected:
(e+e ! B) = 122
MB(B ! e+e)(sM2): (85)
12
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
k1
p2p1
k2e e+
B
+
k1
p1
k2
p2
e e+
B
Figure 3: The tree diagrams of the process ee+ ! +B. All initial momenta go inward andall nal momenta go outward.
(b) Now we calculate the cross section (ee+ ! +B) in COM frame. The relateddiagrams are shown in Figure 3. The amplitude reads:
iM = (ie)(ig)(p1)e(p2)v(k2)h
i
=k1 =p1
+
i
=k1 =p2
iu(k1); (86)
where is the polarization of photon while e is the polarization for B. Now we square
this amplitude,
1
4
Xspin
jiMj2 = 14e2g2gg tr
(=k1 =p1)t
+
(=p1 =k2)
u
=k1
(=k1 =p1)
t+
(=p1 =k2)u
=k2
= 8e2g2
(k1 p1)(k2 p1)
t2+
(k1 p1)(k2 p1)u2
+2(k1 k2)(k1 k2 k1 p1 k2 p1)
tu
= 2e2g2
u
t+
t
u+
2s(s+ t+ u)
tu
= 2e2g2
u
t+
t
u+
2sM2Btu
(87)
Then the cross section can be evaluated as dd
CM
=1
2Ek12Ek2 jvk1 vk2 jjp1j
(2)24ECM
14
PjMj2=
e2g2
322s
1 M
2B
s
ut+
t
u+
2sM2Btu
: (88)
We can also write this dierential cross section in terms of squared COM energy s and
scattering angle . To do this, we note that
s = E2CM; t = (M2B E2CM) sin2 2 ; u = (M2B E2CM) cos2 2 : (89)
Then we have dd
CM
=e2g2(1M2B=s)162s sin2
1 + cos2 +
4sM2B(sM2B)2
; (90)
and dd cos
CM
=g2(1M2B=s)
2s sin2
1 + cos2 +
4sM2B(sM2B)2
: (91)
13
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
(c) The dierential cross obtained in (b) diverges when ! 0 or ! . Now let usstudy the former case, namely ! 0.
If we cut of the integral from 2c ' m2e=s, then we have:Zc
dd cos
CM
sin d ' g2(1M2B=s)
2s
2 +
4sM2B(sM2B)2
Z 1m2e=s dt1 t2
' g2(1M2B=s)
4s
2 +
4sM2B(sM2B)2
log
sm2e
=
g2
2
1 +M4B=s2
sM2Blog
sm2e
(92)
Now we calculate the following expression:Z 10
dx f(x)(e+e ! B)ECM=(1x)s
=
Z 10
dx
2
1 + (1 x)2x
log sm2e
g2
M2B (1 x)s
=
g2
2
1 +M4B=s2
sM2Blog
sm2e
(93)
6 The spinor products (3)
This problem generalize the spinor product formalism to the processes involving
external photons.
(a) Firstly we can represent photon's polarization vectors in terms of spinors of denite
helicity. Let the momentum of the photon be k, and p be a lightlike momentum such
that p k 6= 0. Then, the polarization vector (k) of the photon can be taken to be
+(k) =1p4p k uR(k)
uR(p); (k) =
1p4p k uL(k)
uL(p); (94)
where the spinors uL;R(k) have been introduced in Problems 3.3 and 5.3. Now we use
this choice to calculate the polarization sum:
+(k)+ (k) +
(k)
(k)
=1
4p khuR(k)
uR(p)uR(p)uR(k) + uL(k)
uL(p)uL(p)uL(k)
i=
1
4p k tr=p
=k= g + p
k + pk
p k : (95)
When dotted into an amplitude with external photon, the second term of the result
vanishes. This justies the denitions above for photon's polarization vectors.
(b) Now we apply the formalism to the process e+e ! 2 in the massless limit. Therelevant diagrams are similar to those in Figure 3, except that one should replace the
label `B' by `'. To simplify expressions, we introduce the standard shorthand notations
as follows:
pi = uR(p); p] = uL(p); hp = uL(p); [p = uR(p): (96)
14
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 5 (draft version)
Then the spin products become s(p1; p2) = [p1p2] and t(p1; p2) = hp1p2i. Variousexpressions get simplied with this notation. For example, the Fierz identity (37) now
reads [k2k1i[p1p2i = 2[p1k2]hk1p2i. Similarly, we also have hk1k2]hp1p2] =
2hk1p1i[p2k2].Now we write down the expression for tree amplitude of e+Re
L ! RL. For illustra-
tion, we still keep the original expression as well as all explicit mid-steps. The auxiliary
lightlike momenta used in the polarization vectors are arbitrarily chosen such that the
calculation can be mostly simplied.
iM(e+ReL ! LR)
= (ie)2(p1)+(p2)uL(k2)
i
=k1 =p1
+
i
=k1 =p1
uL(k1)
= ie2 hk2p1][k1p2i4p(k2 p1)(k1 p2)
hk2(=k1 =p1)k1]t
+hk2(=k1 =p2)k1]
u
= ie2 hk2p1][k1p2i
2u
hk2k1]hk1k1] hk2p1]hp1k1]t
+hk2k1]hk1k1] hk2p2]hp2k1]
u
=2ie2u
hk1k2i[p1k1]hk2p2i[k1k1] hk2p1i[k1p1]hk2p2i[k1p1]t
+hk2k2i[k1p1]hk1p2i[k1k1] hk2k2i[p2p1]hp2p2i[k1k1]
u
= 2ie2
hk2p1i[k1p1]hk2p2i[k1p1]tu
; (97)
where we have used the spin sum identity =p = p]hp+ pi[p in the third equality, and alsothe Fierz transformations. Note that all spinor products like hppi and [pp], or hpkiand [pk] vanish. Square this amplitude, we getiM(e+ReL ! LR)2 = 4e4 tu : (98)In the same way, we calculate other polarized amplitudes:
iM(e+ReL ! RL)
= ie2 [k1p1ihk2p2]4p(k1 p1)(k2 p2)
[k2
(=k1 =p1)k1it
+[k2
(=k1 =p2)k1iu
= 2ie2
hk2p1i[k1p2]hk2p2i[k1p2]tu
(99)
Note that we have used a dierent set of auxiliary momenta in photons' polarizations.
After evaluating the rest two nonvanishing amplitudes, we get the squared polarized
amplitudes, as follows:M(e+ReL ! LR)2 = M(e+LeR ! RL)2 = 4e4 tu ; (100)M(e+LeR ! LR)2 = M(e+LeR ! LR)2 = 4e4 ut : (101)Then the dierential cross section follows straightforwardly,
d
d cos =
1
16s
1
4
Xspin
jiMj2=
22
s
tu+
u
t
; (102)
which is in accordance with (5.107) of Peskin & Schroeder.
15
Solutions to Peskin & Schroeder
Chapter 6
Zhong-Zhi Xianyu
Institute of Modern Physics and Center for High Energy Physics,
Tsinghua University, Beijing, 100084
Draft version: November 8, 2012
1 Rosenbluth formula
In this problem we derive the dierential cross section for the electron-proton scatter-
ing in the lab frame, assuming that the scattering energy is much higher than electron's
mass, and taking account of the form factors of the proton. The result is known as
Rosenbluth formula. The relevant diagram is shown in Figure 1. Let us rstly work out
k1 k2
p1 p2
e p
e p
Figure 1: The electron-proton scattering. The blob denotes form factors that includes the
eect of strong interaction. All initial momenta go inward and all nal momenta go outward.
the kinematics. In the lab frame, the momenta can be parameterized as
k1 = (E; 0; 0; E); p1 = (E0; E0 sin ; 0; E0 cos ); k2 = (M; 0; 0; 0); (1)
and p2 can be found by momentum conservation, k1 + k2 = p1 + p2. With the on-shell
condition p22 =M2, we nd that
E0 =ME
M + 2E sin2 2: (2)
We also use q = k1 p1 to denote the momentum transfer and t = q2 its square. Notethat we have set the electron mass to zero.
Now we write down the amplitudeM.
iM = (ie)2 U(p2)
F1(q
2) +iq2M
F2(q2)
U(k2)
itu(p1)u(k1); (3)
E-mail: [email protected]
1
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 6 (draft version)
where U is the spinor for the proton and u is for the electron, M is the mass of the
proton. At this stage, we convert this expression into a more convenient form by means
of the Gordon identity (see Problem 3.2):
iM = (ie)2 U(p2)
(F1 + F2) (p2 + k2)
2MF2
U(k2)
itu(p1)u(k1); (4)
Now, the modular-squared amplitude with initial spins averaged and nal spins
summed, is
1
4
XjMj2 = e
4
4q4tr
(F1 + F2) (p2 + k2)
2MF2
(=k2 +M)
(F1 + F2) (p2 + k2)
2MF2
(=p2 +M)
tr
=k1=p1
=
4e4M2
q4
2E2 + 2E02 + q2
(F1 + F2)
2
2F1F2 + F
22
1 + q
2
4M2
(E + E0)2 + q2
1 q24M2
: (5)
There are two terms in the square bracket in the last expression. We rewrite the rst
factor in the second term as
2F1F2 + F22
1 + q
2
4M2
= (F1 + F2)
2 F 21 + q2
4M2F22 ;
and combine the (F1 + F2)2 part into the rst term, which leads to
1
4
XjMj2 = 4e
4M2
q4
q4
2M2 (F1 + F2)2 + 4
F 21 q
2
4M2F22
EE0 cos2 2
;
where we have used the following two relations which can be easily justied:
E0 E = q22m (6)q2 = 4E0E sin2 2 : (7)
Now we can put the squared amplitude into its nal form:
1
4
XjMj2 = 16e
4E2M3
q4M + 2E sin2 2
F 21 q
2
4M2F22
cos2 2
q2
2M2(F1 + F2)
2 sin2 2
; (8)
On the other hand, we can derive the A+B ! 1+ 2 dierential cross section in thelab frame as
dL =1
2EA2EB jvA vB jZ
d3p1d3p2
(2)62E12E2jMj2(2)4(4)(p1 + p2 pA pB): (9)
In our case, EA = E, EB =M , E1 = E0, and jvA vB j ' 1, thus
dL =1
4EM
Zd3p1d
3p2(2)62E12E2
jMj2(2)4(4)(p1 + p2 pA pB)
=1
4EM
ZE02dE0d cos d'(2)32E02E2
jMj2(2)E0 + E2(E0) E M=
1
4EM
ZE02dE0d cos d'(2)22E02E2
jMj21 +
E0 E cos E2(E0)
1
E0 ME
M + 2E sin2 2
2
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 6 (draft version)
=1
4EM
Zd cos
8jMj2 E
0
M + 2E sin2 2
where we use the notation E2 = E2(E0) to emphasize that E2 is a function of E0. That
is,
E2 =pM2 + E2 + E02 2E0E cos :
Then, dd cos
L=
1
32M + 2E sin2 2
2 jMj2 (10)So nally we get the dierential cross section, the Rosenbluth formula, as d
d cos
L=
2
2E21 + 2EM sin
2 2
sin4 2
F 21 q
2
4M2F22
cos2 2
q2
2M2(F1 + F2)
2 sin2 2
: (11)
2 Equivalent photon approximation
In this problem we study the scattering of a very high energy electron from a target
in the forward scattering limit. The relevant matrix element is
M = (ie)u(p0)u(p) igq2
cM(q): (12)(a) First, the spinor product in the expression above can be expanded as
u(p0)u(p) = A q +B q + C 1 +D 2 : (13)
Now, using the fact that qu(p0)u(p) = 0, we have
0 = Aq2 +Bq ~q ' 4AEE0 sin2 2 +Bq ~q ) B 2:
(b) It is easy to nd that
1 = N(0; p0 cos p; 0;p0 sin ); 2 = (0; 0; 1; 0);
where N = (E2 + E02 2EE0 cos )1=2 is the normalization constant. Then, for theright-handed electron with spinor u+(p) =
p2E(0; 0; 1; 0)T and left-handed electron
with u(p) =p2E(0; 1; 0; 0)T , it is straightforward to show that
u+(p0) =
p2E0(0; 0; cos 2 ; sin
2 )
T ; u(p0) =p2E0( sin 2 ; cos 2 ; 0; 0); (14)
and,
u(p0) 1u(p) ' pEE0
E + E0
jE E0j; (15)
u(p0) 2u(p) ' ipEE0; (16)
u(p0) 1u(p) = u(p0) 2u(p) = 0: (17)
That is to say, we have
C = pEE0
E + E0
jE E0j; D = ipEE0: (18)
3
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 6 (draft version)
(c) The squared amplitude is given by
jMj2 = e2
(q2)2cM(q)cM(q)C1 +D2 C1 +D2 (19)
Averaging and summing over the initial and nal spins of the electron respectively, we
get
1
2
XjMj2 = e
2
2(q2)2cM(q)cM(q)jC+j2 + jCj21 1 + jD+j2 + jDj22 2
+C+D
+ + CD
1
2 +
C+D+ + C
D
2
1
=
e2
(q2)2cM(q)cM(q)EE02 E + E0
E E021
1 +
2 2
(20)
Then the cross section readsZd =
1
2E2Mt
Zd3p0
(2)32E0d3pt
(23)2Et
1
2
XjMj2
(2)4(4)
Ppi
=e2
2E2Mt
Zd3p0
(2)32E0EE02
3(q2)2
E + E0E E0
2+ 1
Z
d3pt(23)2Et
jcM(q)j2(2)4(4)Ppi= 1
2E2Mt
2
Zdxh1 +
2 xx
2i Z 0
d2 sin
4(1 cos )2
Z
d3pt(23)2Et
jcM(q)j2(2)4(4)Ppi: (21)where we have used the trick described in the nal project of Part I (radiation of gluon
jets) to separate the contractions of Lorentz indices, and x (E E0)=E. Now letus focus on the integral over the scattering angle in the last expression, which is
contributed from the following factor:Z 0
d2 sin
4(1 cos )4 Z0
d
which is logarithmically divergent as ! 0.
(d) We reintroduce the mass of the electron into the denominator to cut o the diver-
gence, namely, let q2 = 2(EE0 pp0 cos ) + 2m2. Then we can expand q2, treatingm2 and as small quantities, as
q2 ' (1 x)E22 x2
1 xm2:
Then the polar angle integration near = 0 becomesZ0
d 32 +
x2
(1 x)2m2
E2
2 1
2log
E2
m2: (22)
(e) Combining the results above, the cross section can be expressed asZd = 1
2E2Mt
2
Zdxh1 +
2 xx
2i Z0
d 32 +
x2
(1 x)2m2
E2
24
Notes by Zhong-Zhi Xianyu Solution to P&S, Chapter 6 (draft version)
Z
d3pt(23)2Et
jcM(q)j2(2)4(4)Ppi=
1
2E2Mt
2
Zdx
1 + (1 x)2x2
logE2
m2
Z
d3pt(23)2Et
jcM(q)j2(2)4(4)Ppi: (23)3 Exotic contributions to g 2(a) The 1-loop vertex correction from Higgs boson is
u(p0)u(p) = ip
2
2 Z ddk(2)d
i
(k p)2 m2hu(p0)
i
=k + =q m i
=k mu(p)
=i2
2
Z 10
dx
Z 1x0
dy
Zddk0
(2)d2u(p0)Nu(p)(k02 )3 ; (24)
with
N = (=k + =q +m)(=k +m);
k0 = k xp+ yq; = (1 x)m2 + xm2h x(1 x)p2 y(1 y)q2 + 2xyp q:
To put this correction into the following form:
= F1(q) +iq2m
F2(q); (25)
we rst rewrite N as
N = A +B(p0 + p) + C(p0 p);
where term proportional to (p0 p) can be thrown away by Ward identity q(q) = 0.This can be done by gamma matrix calculations, leading to the following result:
N =h
2d 1
k02 + (3 + 2x x2)m2 + (y xy y2)q2
i
+ (x2 1)m(p0 + p):
Then, use Gordon identity, we nd
N =h
2d 1
k02 + (x+ 1)2m2 + (y y2 xy)q2
i
+
i
2m 2m2(1 x2):
Comparing this with (25), we see that
F2(q = 0) = 2i2m2
Z 10
dx
Z 1x0
dy
Zd4k