7Quantum chromodynamics (QCD)
III The Standard Model
7.4 Deep inelastic scattering
In this chapter, we are going to take an electron, accelerate it to really high
speeds, and then smash it into a proton.
If we do this at low energies, then the proton appears pointlike. This is
Rutherford and Mott scattering we know and love from A-levels Physics. If we
increase the energy a bit, then the wavelength of the electron decreases, and
now the scattering would be sensitive to charge distributions within the proton.
But this is still elastic scattering. After the interactions, the proton remains a
proton and the electron remains an electron.
What we are interested in is inelastic scattering. At very high energies, what
tends to happen is that the proton breaks up into a lot of hadrons
X
. We can
depict this interaction as follows:
H
e
−
p
e
−
p
0
γ
X
θ
This led to the idea that hadrons are made up of partons. When we first
studied this, we thought these partons are weakly interacting, but nowadays, we
know this is due to asymptotic freedom.
Let’s try to understand this scattering. The final state
X
can be very
complicated in general, and we have no interest in this part. We are mostly
interested in the difference in momentum,
q = p − p
0
,
as well as the scattering angle
θ
. We will denote the mass of the initial hadron
H by M, and we shall treat the electron as being massless.
It is conventional to define
Q
2
≡ −q
2
= 2p ·p
0
= 2EE
0
(1 −cos θ) ≥ 0,
where
E
=
p
0
and
E
0
=
p
00
, since we assumed electrons are massless. We also let
ν = p
H
· q.
It is an easy manipulation to show that p
2
X
= (p
H
+ q)
2
≥ M
2
. This implies
Q
2
≤ 2ν.
For simplicity, we are going to consider the scattering in the rest frame of the
hadron. In this case, we simply have
ν = M(E − E
0
)
We can again compute the amplitude, which is confusingly also called M:
M = (ie)
2
¯u
e
(p
0
)γ
µ
u
e
(p)
−
ig
µν
q
2
hX|J
ν
h
|H(p
H
)i
.
Then we can write down the differential cross-section
dσ =
1
4ME|v
e
− v
H
|
d
3
p
0
(2π)
3
2p
00
X
X,p
X
(2π)
4
δ
(4)
(q + p
H
− p
X
)
1
2
X
spins
|M|
2
.
Note that in the rest frame of the hadron, we simply have |v
e
− v
H
| = 1.
We can’t actually compute this non-perturbatively. So we again have to
parametrize this. We can write
1
2
X
spins
|M|
2
=
e
4
2q
4
L
µν
hH(p
H
)|J
µ
h
|XihX|J
ν
h
|H(p
H
)i,
where
L
µν
= Tr
/
pγ
µ
/
p
0
γ
ν
) = 4(p
µ
p
0
ν
− g
µν
p ·p
0
+ p
ν
p
0
µ
.
We define another tensor
W
µν
H
=
1
4π
X
X
(2π)
4
δ
(4)
(p + p
H
− o
X
) hH|J
µ
h
|XihX|J
ν
H
|Hi.
Note that this
P
X
should also include the sum over the initial state spins. Then
we have
E
0
dσ
d
3
p
0
=
1
8ME(2π)
3
4π
e
4
2q
4
L
µν
W
µν
H
.
We now use our constraints on
W
µν
H
such as Lorentz covariance, current conser-
vation and parity, and argue that W
µν
H
can be written in the form
W
µν
H
=
−g
µν
+
q
µ
q
ν
q
2
W
1
(ν, Q
2
)
+
p
µ
H
−
p
H
· q
q
2
q
µ
p
ν
H
−
p
H
· q
q
2
q
ν
× W
2
(ν, Q
2
).
Now, using
q
µ
L
µν
= q
ν
L
µν
= 0,
we have
L
µν
W
µν
H
= 4(−2p ·p
0
+ 4p ·p
0
)W
1
+ 4(2p ·p
H
p
0
· p
H
− p
2
H
p ·p
0
)
= 4Q
2
W
1
+ 2M
2
(4EE
0
− Q
2
)W
2
.
We now want to examine what happens as we take the energy
E → ∞
. In
this case, for a generic collision, we have
Q
2
→ ∞
, which necessarily implies
ν → ∞
. To understand how this behaves, it is helpful to introduce dimensionless
quantities
x =
Q
2
2ν
, y =
ν
p
H
· p
,
known as the Bjorken
x
and the inelasticity respectively. We can interpret
y
as the fractional energy loss of the electron. Then it is not difficult to see that
0
≤ x, y ≤
1. So these are indeed bounded quantities. In the rest frame of the
hadron, we further have
y =
ν
ME
=
E − E
0
E
.
This allows us to write L
µν
W
µν
H
as
L
µν
W
µν
H
≈ 8EM
xyW
1
+
(1 −y)
y
νW
2
,
where we dropped the −2M
2
Q
2
W
2
term, which is of lower order.
To understand the cross section, we need to simplify d
3
p
0
. We integrate out
the angular φ coordinate to obtain
d
3
p
0
7→ 2πE
02
d(cos θ) dE
0
.
We also note that by definition of Q, x, y, we have
dx = d
Q
2
2ν
= −2EE
0
d cos θ + (···) dE
0
dy = −
dE
0
E
.
Since (dE
0
)
2
= 0, the d
3
p
0
part becomes
d
3
p
0
7→ πE
0
dQ
2
dy = 2πE
0
ν dx dy.
Then we have
dσ
dx dy
=
1
8(2π)
2
1
EM
e
4
q
4
2πν · 8EM
xyW
1
+
(1 −y)
y
νW
2
=
8πα
2
ME
Q
4
xy
2
F
1
+ (1 −y)F
2
,
where
F
1
≡ W
2
, F
2
≡ νW
2
.
By varying
x
and
y
in experiments, we can figure out the values of
F
1
and
F
2
.
Moreover, we expect if we do other sorts of experiments that also involve hadrons,
then the same
F
1
and
F
2
will appear. So if we do other sorts of experiments and
measure the same
F
1
and
F
2
, we can increase our confidence that our theory is
correct.
Without doing more experiments, can we try to figure out something about
F
1
and
F
2
? We make a simplifying assumption that the electron interacts with
only a single constituent of the hadron:
H
e
−
p
e
−
p
0
q
k
k + q
X
0
θ
We further suppose that the EM interaction is unaffected by strong inter-
actions. This is known as factorization. This leading order model we have
constructed is known as the parton model. This was the model used before we
believed in QCD. Nowadays, since we do have QCD, we know these “partons” are
actually quarks, and we can use QCD to make some more accurate predictions.
We let f range over all partons. Then we can break up the sum
P
X
as
X
X
=
X
X
0
X
f
1
(2π)
3
Z
d
4
k Θ(k
0
)δ(k
2
)
X
spins
,
where we put δ(k
2
) because we assume that partons are massless.
To save time (and avoid unpleasantness), we are not going to go through the
details of the calculations. The result is that
W
µν
H
=
X
f
Z
d
4
k Tr
W
µν
f
Γ
H,f
(p
H
, k) +
¯
W
µν
f
¯
Γ
H,f
(p
H
, k)
,
where
W
µν
f
=
¯
W
µν
f
=
1
2
Q
2
f
γ
µ
(
/
k +
/
q)γ
ν
δ((k + q)
2
)
Γ
H,f
(p
H
, k)
βα
=
X
X
0
δ
(4)
(p
H
− k − p
0
X
) hH(p
H
)| ¯q
f,α
|X
0
ihX
0
|q
f,β
|H(p
H
)i,
where α, β are spinor indices.
In the deep inelastic scattering limit, putting everything together, we find
F
1
(x, Q
2
) ∼
1
2
X
f
Q
2
f
[q
f
(x) + ¯q
f
(x)]
F
2
(x, Q
2
) ∼ 2xF
1
,
for some functions
q
f
(
x
),
¯q
f
(
x
). These functions are known as the parton
distribution functions (PDF ’s). They are roughly the distribution of partons
with the longitudinal momentum function.
The very simple relation between
F
2
and
F
1
is called the Callan–Gross
relation, which suggests the partons are spin
1
2
. This relation between
F
2
and
F
1
is certainly something we can test in experiments, and indeed they happen.
We also see the Bjorken scaling phenomenon —
F
1
and
F
2
are independent of
Q
2
. This boils down to the fact that we are scattering with point-like particles.