Greiner W., Schrammm S., Stein E. Quantum Chromodynamics
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6.1 The Drell–Yan Process 409
problem is that there is no competing model to QCD at this time. One does not
therefore know how specific the predictions of QCD really are. There may be
other models that describe scaling violations, 3-jet events, and so on equally well.
Conversely many theoreticians consider the fact that there is only one theoretical
candidate for describing quark–quark interactions to be the strongest argument
for the correctness of QCD.
As a result of the analysis described here we conclude that Λ lies in a region
between 100 and 200 MeV. (Λ depends on the renormalization procedure. The
values given are for the
MS scheme; see Sect. 5.1.)
EXERCISE
6.1 The Drell–Yan Cross Section
Problem. Derive the elementary Drell–Yan cross section in (6.5).
Solution. According to the techniques given in Chap. 2, we get
dσ =
e
4
2M
2
d
3
q
1
(2π)
3
2E
d
3
q
2
(2π)
3
1
4
F
2
M
4
(2π)
4
δ
4
( p
1
+ p
2
−q
1
−q
2
). (1)
Here we have simply neglected the quark and gluon masses, thus getting the
flux factor to 1/M
2
.(M is the invariant mass of the two quarks.) Spin-averaging
yields a factor of 1/4, and F
2
is calculated from
F
2
=tr( p/
1
γ
µ
p/
2
γ
ν
) tr(q/
1
γ
µ
q/
2
γ
ν
)
=16( p
µ
1
p
ν
2
+ p
µ
2
p
ν
1
−g
µν
p
1
· p
2
)(q
1µ
q
2ν
+q
2µ
q
1ν
−g
µν
q
1
·q
2
)
=32(q
1
· p
1
q
2
· p
2
+q
1
· p
2
q
2
· p
1
). (2)
This is simplest in the center-of-momentum frame, where
dσ =
e
4
8M
2
d
3
q
1
(2π)
2
M
6
F
2
δ(2q
0
1
−2 p
0
1
). (3)
With p
1
= (E, 0, 0, E), p
2
= (E, 0, 0, −E), q
1
= (E, E sin θ cos φ, E sin θ
×sin φ, E cos θ) and q
2
=(E, −E sin θ cos φ, −E sin θ sin φ, −E cos θ),itfol-
lows that
F
2
=32
E
4
(1 −cos θ)
2
+E
4
(1 +cos θ)
2
=64E
4
(1 +cos
2
θ) , (4)
and therefore
dσ =
e
4
8M
2
d(cos θ) E
2
4πM
6
64E
4
(1 +cos
2
θ) (5)
410 6. Perturbative QCD II: The Drell–Yan Process and Small-x Physics
Exercise 6.1
σ =
e
4
8M
2
1
4π
8
3
=
4πα
2
3M
2
. (6)
This is the relation we used in (6.5).
EXERCISE
6.2 The One-Gluon Contribution to the Drell–Yan Cross Section
Problem. Calculate the total Drell–Yan cross section for one-gluon emission,
i.e., integrate (6.12) with bounds (6.17).
Solution. As s, u,andt are related by (introducing B for simplicity)
s +t +u = M
2
+m
2
G
= B , (1)
we have to choose which variables we want to treat as independent. We use t
and s, thus the integral becomes
8πα
s
αe
2
q
9s
2
t
max
t
min
dt
B −s
t
−1 +
t +s − B −s −B
B −s −t
+
2Bs
t(B −s −t)
−M
2
m
2
G
1
t
2
+
1
(B −s −t)
2
=
8πα
s
αe
2
q
9s
2
(B −s) ln t −2t +(s − B) ln(B −s −t)
+
2Bs
B −s
ln(t) −ln(B −s −t)
−M
2
m
2
G
−
1
t
+
1
B −s −t
t
max
t
min
.
(2)
We now write
t
max/min
=−
1
2
s −B ∓
(s −B)
2
−4M
2
m
2
G
, (3)
σ
=
8πα
s
αe
2
q
9s
2
B
2
+s
2
B −s
ln
t
B −s −t
−2t +M
2
m
2
G
1
B −s −t
−
1
t
t
max
t
min
=
8πα
s
αe
2
q
9s
2
⎡
⎣
2
B
2
+s
2
B −s
ln
⎛
⎝
s −B −
(s −B)
2
−4M
2
m
2
G
s −B +
(s −B)
2
−4M
2
m
2
G
⎞
⎠
−2
(s −B)
2
−4M
2
m
2
G
+M
2
m
2
G
(−2)
2
(s −B)
2
−4M
2
m
2
G
(s −B)
2
−(s − B)
2
+4M
2
m
2
G
2
⎤
⎦
(4)
6.1 The Drell–Yan Process 411
and obtain the final expression
σ
=
8παα
s
e
2
q
9s
⎡
⎣
2
1 + B
2
/s
2
1 − B/s
ln
⎛
⎝
s −B +
(s −B)
2
−4M
2
m
2
G
s −B −
(s −B)
2
−4M
2
m
2
G
⎞
⎠
−4
6
(s −B)
2
−4M
2
m
2
G
s
2
⎤
⎦
, (5)
B = M
2
+m
2
G
.
Obviously for m
G
→0 the logarithmic divergence is recovered.
EXERCISE
6.3 The Drell–Yan Process as Decay of a Heavy Photon
Problem. Derive the relation between the cross sections for q +
q → γ
∗
→
µ
+
+µ
−
and q +q → γ
∗
:
dσ
dM
2
(q +q →µ
+
+µ
−
) = σ(q +q → γ
∗
)
α
3πM
2
1 −
4m
2
M
2
3/2
dM
2
.
Solution. The decay cross section for q +
q → γ
∗
→µ
+
+µ
−
is
dσ =
Q
2
q
e
4
M
4
1
2p
0
2p
0
W
µν
L
µν
(2π)
4−6
δ( p + p
−k −k
)
d
3
k d
3
k
4k
0
k
0
, (1)
where p, p
, k,andk
are the quark, antiquark, µ
−
,andµ
+
momenta. The virtual
photon has the four-momentum q with q
2
= M
2
. We now multiply the right-hand
side of (1) by
δ
4
(q −k −k
)
d
3
q
2q
0
dM
2
=δ
4
(q −k −k
) d
4
q = 1 , (2)
giving
dσ =
Q
2
q
e
4
M
4
1
16k
0
k
0
p
0
p
0
2q
0
(2π)
−2
W
µν
L
µν
×δ
4
( p + p
−q)δ
4
(q −k −k
)d
3
k d
3
k
d
3
q dM
2
. (3)
We perform the k and k
integrations
I
µν
=
d
3
k d
3
k
(2π)
−2
L
µν
δ
4
(q −k −k
) (4)
=
4
(2π)
2
d
3
k d
3
k
k
µ
k
ν
+k
ν
k
µ
−g
µν
(k ·k
+m
2
)
δ
4
(q −k −k
).
Exercise 6.2
412 6. Perturbative QCD II: The Drell–Yan Process and Small-x Physics
Exercise 6.3
We then evaluate this expression in the rest frame of the virtual photon, q =
(M, 0, 0, 0), k
0
= M/2:
I
µν
=
1
π
2
d
3
k d
3
k
−2k
µ
k
ν
+k
µ
q
ν
+q
µ
k
ν
−g
µν
(−k
2
+k ·q +m
2
)
δ(q
0
−2k
0
)δ
3
(k−k
)
=
1
π
2
d
3
k
2k
µ
k
0
δ
ν0
+2k
ν
k
0
δ
µ0
−2k
µ
k
ν
−g
µν
(k
0
q
0
)
δ(q
0
−2k
0
). (5)
Because the integral over just one component of k vanishes, this expression is
diagonal. For µ = ν = 0weget
I
00
=
1
π
2
4π
dkk
2
2(k
0
)
2
−2(k
0
)
2
δ
q
0
−2
"
k
2
+m
2
=0 . (6)
This is also required by current conservation:
q
ν
I
µν
=q
µ
I
µν
= 0 . (7)
For µ = ν = j we find that
I
jj
=
1
π
2
d
3
k
−2(k
j
)
2
+
M
2
2
δ
q
0
−2
"
k
2
+m
2
. (8)
We substitute k
j
k
j
→1/3 k
2
and get
I
jj
=
4
π
dkk
2
−
2
3
k
2
+2k
2
δ
M −2
"
k
2
+m
2
=
4
π
4
3
M
2
4
−m
2
M
2
4
−m
2
M
2
2
=
2
3π
M
4
4
1 −
4m
2
M
2
3/2
. (9)
As I
µν
must be proportional to g
µν
−q
µ
q
ν
/q
2
, owing to current conservation,
it is easy to obtain (note that g
µν
I
µν
= I
00
−3I
jj
)
I
µν
=
−g
µν
+
q
µ
q
ν
q
2
M
4
6π
1 −
4m
2
M
2
3/2
. (10)
Substituting this result back into (3) gives
dσ =
Q
2
q
e
4
M
4
1
4M
2
p
0
p
0
2q
0
W
µν
δ
4
( p + p
−q)
×
−g
µν
+
q
µ
q
ν
q
2
M
4
6π
1 −
4m
2
M
2
3/2
d
3
q dM
2
. (11)
6.1 The Drell–Yan Process 413
To bring this into the form of dσ(q +
¯
q → γ
∗
), we substitute
−g
µν
+
q
µ
q
ν
q
2
=
ε
ε
µ
ε
∗ν
(12)
and get
dσ =
e
2
6πM
2
Q
2
q
e
2
W
µν
ε
ε
µ
ε
∗ν
(2π)
4−3
δ
4
( p + p
−q)
d
3
q
4p
0
p
0
·2q
0
× dM
2
1
2π
1 −
4m
2
M
2
3/2
=
α
3πM
2
1 −
4m
2
M
2
3/2
σ(q +
¯
q → γ
∗
) dM
2
. (13)
This completes our proof.
EXERCISE
6.4 Heavy Photon Decay Into Quark, Antiquark, and Gluon
Problem. Derive (6.48).
Solution. The differential decay rate for a one-body decay in its rest frame is
simply
dΓ =
1
2M
|M
1
+M
2
|
2
d
3
p
1
(2π)
3
2E
1
d
3
p
2
(2π)
3
2E
2
d
3
p
3
(2π)
3
2E
3
×(2π)
4
δ
4
(k − p
1
− p
2
− p
3
) (1)
with
M
2
=
¯
u(p
2
, s
2
)
−igγ
ν
λ
a
2
i( p/
2
+ p/
3
)
( p
2
+ p
3
)
2
×(−i ee
q
γ
µ
)v(p
1
, s
1
)
˜
ε
µ
(k)ε
ν
( p
3
) (2)
and
M
1
=−
¯
u(p
2
, s
2
)(−i ee
q
γ
µ
)
(−i)( p/
1
+ p/
3
)
( p
1
+ p
3
)
2
×
−igγ
ν
λ
a
2
v( p
1
, s
1
)
˜
ε
µ
(k)ε
ν
( p
3
). (3)
Exercise 6.3
414 6. Perturbative QCD II: The Drell–Yan Process and Small-x Physics
Exercise 6.4
The factor 1/2M in (1) is just the photon field normalization. In the photon
rest system all four-products are rather simple for zero masses:
2 p
1
· p
3
= ( p
1
+ p
3
)
2
= (k
γ
− p
2
)
2
= k
2
γ
−2 k
γ
· p
2
= M
2
−2M
2
x
2
= M
2
(1 −2x
2
),
2 p
2
· p
3
= M
2
(1 −2x
1
),
2 p
1
· p
2
= M
2
(1 −2x
3
), (4)
where x
i
is the energy fraction carried by the parton i. Due to (4), |M
1
+M
2
|
2
depends only on x
1
, x
2
,andx
3
=1 −x
1
−x
2
, such that the phase-space integral
is very easy.
d
3
p
1
(2π)
3
2E
1
d
3
p
2
(2π)
3
2E
2
d
3
p
3
(2π)
3
2E
3
(2π)
4
δ
4
(k − p
1
− p
2
− p
3
) ···
=
d
3
p
1
d
3
p
2
(2π)
5
2
3
E
1
E
2
E
3
δ
(
M −E
1
−E
2
−E
3
)
··· .
Using E
1
= x
1
M , E
2
= x
2
M and E
3
=(1 −x
1
−x
2
)M we can simplify the
formula above. In the δ function we express
E
2
3
= p
2
3
=
(
k− p
1
− p
2
)
2
=
(
p
1
+ p
2
)
2
and get
d
3
p
1
(2π)
3
2E
1
d
3
p
2
(2π)
3
2E
2
d
3
p
3
(2π)
3
2E
3
(2π)
4
δ
4
(k − p
1
− p
2
− p
3
) ···
=
d
3
p
1
d
3
p
2
2
5
π
5
2
3
x
1
x
2
(1 −x
1
−x
2
)M
3
×δ
M −x
1
M −x
2
M −
( p
1
− p
2
)
2
···
=
4πE
2
1
dE
1
2πE
2
2
dE
2
dcos(θ
12
)
2
8
π
5
x
1
x
2
(1 −x
1
−x
2
)M
3
×δ
M(1 −x
1
−x
2
) −M
"
2x
1
x
2
cos(θ
12
)
···
=
x
2
1
M
3
x
2
2
M
3
dx
1
dx
2
2
5
π
3
x
1
x
2
(1 −x
1
−x
2
)M
3
√
2x
1
x
2
cos(θ
12
)
Mx
1
x
2
√
2x
1
x
2
cos(θ
12
)=1−x
1
−x
2
···
=
M
2
dx
1
dx
2
2
5
π
3
··· (5)
such that the decay rate reads
dΓ =
M
64π
3
dx
1
dx
2
|M
1
+M
2
|
2
. (6)
6.1 The Drell–Yan Process 415
We now make a special choice for the polarization vectors of the gluon. The
reason for this will become clear later:
ε
ε
µ
ε
∗
µ
=−g
µµ
+
n
µ
p
3µ
+n
µ
p
3µ
n · p
3
=Σ
µµ
, (7)
˜
"
˜
ε
ν
˜
ε
∗
ν
=−g
νν
. (8)
The choice (7) is equivalent to choosing the axial gauge n · A = 0. n
µ
is an arbi-
trary four-vector with n
2
= 0andn ·ε = 0, while for the photon we choose the
covariant Feynman gauge.
S
11
=
spins
M
1
M
∗
1
=−g
2
e
2
e
2
q
tr
#
λ
a
λ
a
4
&
Σ
µµ
×tr
#
p/
2
γ
ν
p/
1
+ p/
3
( p
1
+ p
3
)
2
γ
µ
p/
1
γ
µ
p/
1
+ p/
3
( p
1
+ p
3
)
2
γ
ν
&
, (9)
S
12
=+
spins
(M
1
M
∗
2
+M
2
M
∗
1
)
= 4 g
2
e
2
e
2
q
Σ
µµ
tr
#
p/
2
γ
ν
p/
1
+ p/
3
( p
1
+ p
3
)
2
γ
µ
p/
1
γ
ν
p/
2
+ p/
3
( p
2
+ p
3
)
2
γ
µ
&
+tr
#
p/
2
γ
µ
p/
2
+ p/
3
( p
2
+ p
3
)
2
γ
ν
p/
1
γ
µ
p/
1
+ p/
3
( p
1
+ p
3
)
2
γ
ν
&
, (10)
S
22
=
spins
M
2
M
∗
2
=−4 g
2
e
2
e
2
q
Σ
µµ
tr
#
p/
2
γ
µ
p/
2
+ p/
3
( p
2
+ p
3
)
2
γ
ν
p/
1
γ
ν
p/
2
+ p/
3
( p
2
+ p
3
)
2
γ
µ
&
. (11)
We shall use the gauge
n
µ
=
k
γ
−
p
2
2x
2
µ
, (12)
n
2
= M
2
−2
x
2
M
2
2x
2
+0 =0 . (13)
This choice will result in a particularly simple result and will thus be justified
only a posteriori. From
n
µ
Σ
µµ
=n
µ
Σ
µµ
=0andp
3µ
Σ
µµ
= 0 (14)
we find that
p
1
+ p
2
+ p
3
−
p
2
2x
2
µ
Σ
µµ
=
p
1
− p
2
1 −2x
2
2x
2
µ
Σ
µµ
= 0 . (15)
Exercise 6.4
416 6. Perturbative QCD II: The Drell–Yan Process and Small-x Physics
Exercise 6.4
This allows us to substitute all momentum vectors occurring in the trace accord-
ing to
p
3
µ
,
p
3
µ
→0 ,
p
1
µ
→
p
2
µ
1 −2x
2
2x
2
,
p
1
µ
→
p
2
µ
1 −2x
2
2x
2
, (16)
which simplifies the calculation substantially. We start with S
12
and introduce
the notation c = 4g
2
e
2
e
2
q
:
S
11
=
2cΣ
µµ
M
4
(1 −2x
2
)
2
tr
$
p/
2
( p/
1
+ p/
3
)γ
µ
p/
1
γ
µ
( p/
1
+ p/
3
)
%
=
2cΣ
µµ
M
4
(1 −2x
2
)
2
M
2
(1 −2x
3
+1 −2x
1
)tr
$
γ
µ
p/
1
γ
µ
( p/
1
+ p/
3
)
%
−2 p
1
· p
3
tr
$
p/
2
γ
µ
p/
1
γ
µ
%
,
=
2cΣ
µµ
M
2
(1 −2x
2
)
2
2(1−x
3
−x
1
) 4
p
1µ
( p
1
+ p
3
)
µ
+ p
1µ
( p
1
+ p
3
)
µ
−g
µµ
p
1
· p
3
−
2 p
1
· p
3
M
2
4
p
2µ
p
1µ
+ p
2µ
p
1µ
−g
µµ
p
2
· p
1
.
(17)
With (14) this reads
S
11
=
2cΣ
µµ
M
2
(1 −2x
2
)
2
16x
2
p
1µ
p
1µ
−8(1 −2x
2
) p
2µ
p
1µ
−4x
2
(1 −2x
2
)M
2
g
µµ
+2(1 −2x
2
)M
2
(1 −2x
3
)g
µµ
. (18)
With (16) and
Σ
µµ
g
µµ
=−4 +2 =−2 (19)
this gives
S
11
=
2c
1 −2x
2
(−2)(−4x
2
+2 −4x
3
)
=−
4c
1 −2x
2
(−2+4x
1
)
=8c
1 −2x
1
1 −2x
2
. (20)
Analogously S
22
gives
S
22
=
2cΣ
µµ
M
4
(1 −2x
1
)
2
tr
p/
2
γ
µ
( p/
2
+ p/
3
) p/
1
( p/
2
+ p/
3
)γ
µ
, (21)
6.1 The Drell–Yan Process 417
which is the same as S
11
with µ and µ
interchanged and p
1
interchanged
with p
2
. Thus the analogy to (18) now reads
S
22
=
2c
M
2
(1 −2x
1
)
2
16 x
1
p
2µ
p
2µ
−8(1 −2x
1
) p
1µ
p
2µ
Σ
µµ
+8x
1
(1 −2x
1
)M
2
−4(1 −2x
1
)M
2
(1 −2x
3
)
. (22)
Again using (16) this simplifies substantially to
S
22
=
2c
M
2
(1 −2x
1
)
2
#
8 p
2µ
p
2µ
2x
1
−
(1 −2x
2
)(1 −2x
1
)
2x
2
+4M
2
(1 −2x
1
)(1 −2x
2
)
&
. (23)
With
Σ
µµ
p
2µ
p
2µ
=2
n · p
2
2 p
3
· p
2
2 n · p
3
=2
M
2
x
2
M
2
(1 −2x
1
)
M
2
(1 −2x
2
)
1−2x
3
2x
2
= M
2
4x
2
2
(1 −2x
1
)
(1 −2x
2
)(1 −2x
3
)
, (24)
2 n · p
3
= 2 p
1
· p
3
−(1 −2x
2
)
2 p
2
· p
3
2x
2
= M
2
(1 −2x
2
)
1 −
1 −2x
1
2x
2
= M
2
1 −2x
2
2x
2
2x
2
+2x
1
−1
= M
2
(1 −2x
2
)(1 −2x
3
)
2x
2
, (25)
we obtain
S
22
=
2c
(1 −2x
1
)
2
#
8
4x
2
2
(1 −2x
1
)
(1 −2x
2
)(1 −2x
3
)
2x
1
−
(1 −2x
2
)(1 −2x
1
)
2x
2
+4(1 −2x
1
)(1 −2x
2
)
&
=
8c
(1 −2x
1
)(1 −2x
2
)(1 −2x
3
)
16x
1
x
2
2
−4(1 −2x
2
)(1 −2x
1
)x
2
+(1 −2x
2
)
2
(1 −2x
3
)
=
8c
(1 −2x
1
)(1 −2x
2
)(1 −2x
3
)
16x
2
2
−16x
3
2
−16x
3
x
2
2
+4x
2
(1 −2x
2
)(1 −2x
2
−2x
3
) +(1 −2x
2
)
2
(1 −2x
3
)
Exercise 6.4
418 6. Perturbative QCD II: The Drell–Yan Process and Small-x Physics
Exercise 6.4
=
8c
(1 −2x
1
)(1 −2x
2
)(1 −2x
3
)
8x
2
2
−16x
3
x
2
2
+4x
2
(1 −2x
2
)(1 −2x
3
)
+(1 −2x
2
)
2
(1 −2x
3
)
=
8c
(1 −2x
1
)(1 −2x
2
)
8x
2
2
+4x
2
−8x
2
2
+(1 −2x
2
)
2
=
8c
(1 −2x
1
)(1 −2x
2
)
1 +4x
2
2
=8c
1 +4x
2
2
(1 −2x
1
)(1 −2x
2
)
. (26)
Finally we calculate S
12
.
S
12
=−
2c Σ
µµ
M
4
(1 −2x
1
)(1 −2x
2
)
2tr
p/
1
p/
2
γ
µ
p/
2
+ p/
3
p/
1
+ p/
3
γ
µ
=−
4c Σ
µµ
M
4
(1 −2x
1
)(1 −2x
2
)
tr
p/
1
p/
2
γ
µ
p/
2
p/
1
+ p/
3
p/
1
+ p/
2
p/
3
γ
µ
=−
4c Σ
µµ
M
4
(1 −2x
1
)(1 −2x
2
)
2
p
2
µ
tr
$
p/
1
p/
2
p/
1
γ
µ
%
+2
p
1
µ
tr
$
p/
1
p/
2
γ
µ
p/
3
%
+2
p
2
µ
tr
$
p/
1
p/
2
p/
3
γ
µ
%
=−
4c Σ
µµ
M
4
(1 −2x
1
)(1 −2x
2
)
4
4 p
2µ
p
1
· p
2
p
1µ
+2 p
1µ
p
1
· p
3
p
2µ
+2 p
1µ
p
1µ
p
2
· p
3
+2 p
2µ
p
1µ
p
2
· p
3
−2 p
2µ
p
1
· p
3
p
2µ
=−
16c Σ
µµ
M
2
(1 −2x
1
)(1 −2x
2
)
p
2µ
p
2µ
2
(1 −2x
2
)(1 −2x
3
)
2x
2
+
(1 −2x
2
)
2
2x
2
−(1 −2x
1
)
(1 −2x
2
)
2
(2x
2
)
2
+(1 −2x
1
)
1 −2x
2
2x
2
−(1 −2x
2
)
=−
16c
(1 −2x
1
)(1 −2x
2
)
4x
2
2
(1 −2x
1
)
(1 −2x
2
)(1 −2x
3
)
1 −2x
2
4x
2
2
×
4x
2
(1 −2x
3
) +2x
2
(1 −2x
2
) −(1 −2x
1
)(1 −2x
2
)
+2x
2
(1 −2x
1
) −4x
2
2
=−
16c
(1 −2x
2
)(1 −2x
3
)
×
4x
2
(1 −2x
3
) +(1 −2x
2
)(1 −2x
3
) +2x
2
(1 −2x
1
−2x
2
)
=−
16c
(1 −2x
2
)(1 −2x
3
)
(1 −2x
3
)
4x
2
+1 −2x
2
−2x
2
=−
16c
1 −2x
2
. (27)