Applying Regge theory
Peter Landshoff
[email protected]
Much of the work I will describe was done in collaboration with Sandy Donnachie
History
• 1935: Yukawa predicted the existence of the pion — its
exchange generates the static strong interaction
• 1960s: Nearly everybody worked on the applications of Regge theory, which sums
the exchanges of many particles and generates the high-energy strong interaction
• The known particles not are enough — we need to include exchange of another
object, the pomeron
• 1970s: QCD is discovered — the BFKL equation uses perturbative QCD to
generate pomeron exchange as gluon exchange, but it makes total cross sections
rise with energy much faster than is observed
Basic beliefs
• When two protons collide, most of the cross section results from a long-range force
between them
• That force is quantum chromodynamics (QCD)
• Although QCD is weak at short range and so can be calculated by perturbation
theory, this is not the case at long range
• For long range the only theory we have is Regge Theory, but it has its limitations
• Regge Theory models the exchange of families of particles ρ, ω, f2 , a2 etc
• But to describe data it needs another exchange, called the “pomeron”
• Pomeron exchange is probably the exchange of a family of glueballs
Because of our inability to calculate, we have to inform the theory with information
from the data, and build models
Linear particle trajectories
Plot of spins of families of particles against their squared masses:
f ;a
(t) 6
[ 6℄ [ 6℄
[ 5℄
5
a4 ; f 4
4
!3 ; 3
3
f2 ; a2
2
; !
1
0
0
1
2
3
4
5
6
7
8
t (GeV2 )
4 degenerate familes of particles: α(t) ≈ 12 + 0.9t
The particles in square brackets are listed in the data tables, but there is some uncertainty about whether they exist.
The function α(t) is called a Regge trajectory.
Regge trajectories
10
α(t) 9
8
7
6
5
4
3
2
1
0
4
f, a2 families
0.70 + 0.80 t
α(t)
∆15/2
3
N13/2
∆11/2
2
N9/2
∆7/2
1
N5/2
baryon trajectories
-0.38+t
∆
p
0
1
2
3
4
5 6
t (GeV2 )
7
8
First daughter α1 (t) = α0 (t) − 1
rho, omega families
0.44 + 0.92 t
0
9
10
0
1
2
3
4
5
6
2
t (GeV )
Second daughter α2 (t) = α0 (t) − 2 etc
C = +1 and C = −1 trajectories not exactly degenerate
Why are trajectories almost straight lines?
Cannot be exactly so: meson trajectories become complex for t > 4m2π
and baryon trajectories for (mN + mπ )2
Im α(t) at a resonance is proportional to its width
7
Regge theory
Regge theory sums the exchanges of many particles.
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Define
s = (P1 + P2 )2 = squared CM energy
t = (P3 − P1 )2 = squared momentum transfer
At large s but |t| << s each trajectory α(t) contributes to the amplitude a power of s:
A(s, t) ∼ B(t) sα(t)−1
We know only a little about the function B(t) —- later
Total cross sections
Optical theorem:
σ TOT (s) = Im A(s, t = 0)
So each trajectory contributes a fixed power
α(0)−1
s
− 12
≈s
for ρ, ω, f2 , a2 trajectories
The contribution from the ρ trajectory sums the exchanges of ρ, ρ3 , ρ5 , . . .
80
pBARp total cross section
σ
(mb) 70
60
50
40
10
100
√
s (GeV)
Agrees with experiment only at rather low energies
1000
Total cross section data
Total cross sections rise gently at large s
So we need another trajectory αIP (t) with αIP (0) a little > 1
σ 110
(mb)
100
σ 30
(mb)
28
p̄p : 21.70s0.0808 + 98.39s−0.4525
pp : 21.70s0.0808 + 56.08s−0.4525
90
80
π − p : 13.63s0.0808 + 36.02s−0.4525
π + p : 13.63s0.0808 + 27.56s−0.4525
26
70
24
60
50
22
40
30
100
1000
√
s (GeV)
10
(mb)
0.0808
K p : 11.93s
+ 25.33s
K + p : 11.93s0.0808 + 7.58s−0.4525
−
√
s (GeV)
26
σ
(mb)
24
20
10000
0.2
−0.4525
p : 0:0677s0:0808 + 0:129s
0:4525
0.18
22
0.16
20
0.14
0.12
18
0.1
10
16
10
√
s (GeV)
100
ps (GeV)100
100
• We call this the pomeron trajectory (after the Russian physicist Pomeranchuk)
Coupling to nucleon
21.7
3
=
≈
coupling to pion
13.63
2
• The pomeron seems to couple to the separate quarks in a hadron – quark counting rule
• These were fits made in 1992 – best value for αIP (0) now is 1.096 rather than
1.0808
• The pomeron couples a bit more weakly to s quarks – much more so for c quarks
• Pomeranchuk theorem: σ pp /σ p̄p ∼ 1 at high energy. So the pomeron contributions
to AB and ĀB scattering are equal: pomeron exchange has C-parity +1
• f2 , a2 also have C = +1 and so contribute equally to AB and ĀB scattering. But
ρ, ω have C = −1 and so contribute to AB and ĀB scattering with opposite signs.
• Probably pomeron exchange corresponds to the exchange of glueballs
Miracle
σ
(mb)
50
p̄p : 21.70s0.0808 + 98.39s−0.4525
pp : 21.70s0.0808 + 56.08s−0.4525
0.15
0.14
gamma p
45
0.13
40
0.12
35
0.11
30
0.1
25
0.09
6
10
50
√
s (GeV)
0.08
4
6
8
10 12 14 16 18 20
The cross sections rise smoothly even from very low energies where only pions can
be produced. Thresholds for heavier particles, jets etc make no difference.
√
The pomeron term is there down to s = 6 GeV or lower
Unitarity
Unit probability that a given initial state results in some final state, and that any final
state came from some initial state.
2
*
i
X
X
• Optical theorem σ TOT (s) = Im A(s, t = 0)
• Froissart-Lukaszuk-Martin bound
σ TOT (s) <
π
2
log
(s/s0 )
2
mπ
for some unknown s0 — probably of the order of 1 GeV2 .
At LHC energy, this gives σ TOT < 4.3 barns
So the bound has little to do with physics!
• More restrictive is the bound on the elastic partial-wave amplitude
Im Aℓ (s) = |Aℓ (s)|2 + inelastic terms
so that |Aℓ (s)| < 1
• Note that these bounds do not apply to photon or lepton beams.
Impact parameter
In the CM frame
P3 = (E, p − 12 q)
P4 = (E, −p + 12 q)
P1 = (E, p + 12 q)
P2 = (E, −p − 21 q)
with (p + 12 q)2 = (p − 21 q)2 so that p.q = 0 and therefore q is in the two-dimensional
space perpendicular to p. Also t = −q2 .
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Write the amplitude as a 2-dimensional Fourier integral
∫
d2 b e−iq.b Ã(s, b2 )
∫
1
2
iq.b
2
Ã(s, b2 ) =
d
q
e
A(s,
−q
)
16π 2
A(s, −q2 ) = 4
b is called the “impact parameter”. Roughly speaking, it is the transverse distance
between the two scattering particles.
Eikonal representation
Define
χ(s, b) = − log(1 + 2iÃ/s)
so that
Ã(s, b2 ) = 12 is(1 − e−χ(s,b) )
∫
and
A(s, −q2 ) = 2is
d2 b e−iq.b (1 − e−χ(s,b) ).
It can be shown that the unitarity condition |Aℓ (s)| < 1 is just
Re χ(s, b) ≥ 0
so this is an easy way to satisfy unitarity when one makes models
Multiple exchanges
∫
A(s, −q2 ) = 2is
d2 b e−iq.b
(
)
χ2
χ3
(−χ)n
χ−
+
... −
... .
2!
3!
n!
If we choose χ(s, b) so that the first term represents the exchange of a single pomeron,
then the next term will be two-pomeron exchange, then three-pomeron etc.
+
+
+ ....
But this is only a model: we do not know how to calculate multiple exchanges properly.
Only single exchange is a simple power of s. So in our fits with s0.08 this is an
effective power, representing single plus mutliple exchanges.
This is usually good enough, but not always.
Multiple exchanges
√
This is data for elastic αα scattering at s = 126 GeV from the CERN Intersecting
Storage Rings:
100
d=dt
b
(m
2
GeV
)
10
1
0.06
0.1
0.14
t
j j
(GeV
0.18
2
)
The energy is high enough for pomeron exchange to dominate
The shape of the curve is similar to the intensity pattern for optical diffraction, so
pomeron exchange is often called diffractive scattering
At small t the dominant contribtion is from one pair of nucleons scattering
But the dip comes from interference with double exchange: two pairs of nucleons
scattering, calculated by Glauber formula which is just the eikonal model
Proton-proton scattering
d=dt
b
(m
GeV
1
2
)
0.1
0.01
0.001
62 GeV
0.0001
1e-05
23 GeV
1e-06
1e-07
0.5
1
1.5
2
2.5
3
t
j j
3.5
(GeV
4
2
4.5
5
)
Now the scattering is mainly between pairs of constituent quarks:
For pp scattering there are 3 × 3 = 9 pairings and for πp there are 2 × 3 = 6
Ratio of coefficients of s0.08 in fits to pp and πp total cross sections is 21.7/13.63 ≈ 3/2
Note:
• The forward peak gets steeper as the energy increases
• The dip is at larger |t| than in αα scattering, which makes the theory more complicated
The Regge formula
The theory of pomeron exchange uses the same mathematics as for optical diffraction: promote the orbital angular momentum ℓ into a complex variable and transform
the partial-wave series into an integral
After some quite subtle mathematics find that the exchange of particles associated
with a Regge trajectory α(t) contributes at high energy
A(s, t) = β(t) ξ(α(t)) (s/s0 )α(t)−1
for |t| << s and fixed s0 .
ξ(α(t)) is a phase factor:
{
ξ(α(t)) =
e
1
− 2 iπα(t)
1
− 2 iπα(t)
−ie
C = +1 exchange
C = −1 exchange
The theory does not tell us what is the function β(t), except
It is real for t < 0
It has a pole for values of t corresponding to each particle on the trajectory
(unimportant when we are interested in t < 0)
Photon coupling to a nucleon
p
p/
The coupling to a quark is (quark charge) ×γ µ
To get the coupling to the nucleon we couple to each quark in turn and take account
of the nucleon wave function, giving
κ
F2 (t)iσ µν (p′ν − pν )
eF1 (t)γ +
2m
µ
F1 (0) = F2 (0) = 1
e is the sum of the quark charges, ie the charge of the nucleon
κ is its anomalous magnetic moment, ie the amount it differs from the Dirac-equation
value
If this is used to calculate scattering from photon exchange, the last term flips the
helicity of the nucleon
The photon is a mixture of isospin 0 and 1. The last term comes almost entirely from
the I = 1 part, because κp = 1.79 κn = −1.91 so that 12 (κp + κn ) ≈ 0
The F2 term flips the helicity of the nucleon
The proton form factors
Define
GE (t) = F1 (t) +
Data:
GM (t) ≈ µGE (t)
t
4m2p F2 (t)
GE (t) ≈
This gives
F1 (t) ≈
µp = 1 + κp
http://www.phy.anl.gov/theory/PHYTI09/
PHYTI09 fichiers/ArringtonQCD09.pdf
4m2p −2.79t
1
2
4mp −t (1−t/0.71)2
GM (t) = F1 (t) + F2 (t)
1
1−t/0.71)2
Pomeron coupling to nucleon (1974!)
σ
(mb)
0.0808
σ 110
(mb)
100
p̄n : 21.70s
+ 92.71s
50 pn : 21.70s0.0808 + 54.77s−0.4525
−0.4525
p̄p : 21.70s0.0808 + 98.39s−0.4525
pp : 21.70s0.0808 + 56.08s−0.4525
90
45
80
70
60
40
50
40
35
30
10
√
s (GeV)
100
100
1000
√
s (GeV)
10000
Coefficient of s0.0808 same as for pp and p̄p: the pomeron is isosinglet
Also C = +1: it couples equally to quarks and antiquarks
Big assumption: its coupling to quarks is just like an isoscalar photon, so to a hadron
it is nβF1 (t) where F1 (t) is its electromagnetic form factor and β is a constant.
n is the number of constituent (quarks + antiquarks) in the hadron: n = 3 for a
nucleon and 2 for a pion.
The absence of an F2 term means no helicity flip – fits data
[Note that the coefficients of the s−0.4525 term in pn are also not very different from
pp, so the ω couples to the nucleon more strongly than the ρ]
Pomeron exchange
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A(s, t) =
ū(p3 ) 3γ µ βIP F1 (t) u(p1 ) · ū(p4 ) 3γ µ βIP F1 (t) u(p2 )
s
e
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1
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(s/s0 )αIP (t)−1
P
Assume the pomeron trajectory is linear (like ρ, ω, f2 , a2 ):
αIP (t) = ϵIP + αI′P t
The choice of s0 is not critical, but s0 = 1/αI′P works well
βIP and ϵIP are known from σ TOT
The only free parameter is αI′P
[3βIP F1 (t)]4 ′ 2(ϵIP +α′IP t)
dσ
=
(αIP s)
dt
4π
Include also the ρ, ω, f2 , a2 trajectory
αR (t) = 1 − 0.4525 + 0.9t
Proton-proton scattering
√
′
Fix α from the very-low-t data at some energy, say s =53 Gev:
d=dt
mb GeV 2
(mb GeV 2 )
200
100
100
10
50
1
0
0.01
0.02
0.03
t
j j
(GeV2 )
0.04
0.05
0
0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
jtj
(GeV2 )
Determines α′ = 0.25 GeV−2
Then the formula works well out to larger t at the same energy
It also fits well to pp and pp̄ elastic scattering data at all other available energies.
Because F1 (t) is raised to the power 4 in the formula, this gives a good test that it is
the correct form factor, but why this should be so is not understood.
Note that the curves do not include photon exchange, which adds e2 /t to the amplitude and contributes significantly at very small t.
Shrinkage of the forward peak
Because the formula contains the factor
′
2α′IP t
(αIP s)
= exp(−2αI′P log(αI′P s)|t|)
the contribution of pomeron exchange to the forward peak in dσ/dt becomes steeper
as the energy increases:
dσ/dt
(mb GeV−2 )
100
1800 GeV
53 GeV
10
1
0
0.05
0.1
0.15
0.2
2
|t| (GeV )
Note the discrepancy between the data from the two Tevatron experiments.
Pion form factor
e
q
γ*
F (t) 1
0.9
0.8
π
p
0.7
0.6
0.5
0.4
0.3
0
0.2
0.4
t
0.6
t
j j
Measure ep → epπ at t = 0
The exchanged π is almost on shell
So gives pion form factor Fπ (q 2 )
(Chew-Low theory)
The curve is
1
Fπ (t) =
1 − t/m20
m20 = 0.5 GeV2 ≈ m2ρ
0.8
(GeV2 )
1
Pion-proton elastic scattering
dσ
[3βIP F1 (t)]4 ′ 2(ϵIP +α′IP t)
=
(αIP s)
dt
4π
pp
[2βIP Fπ (t)]2 [3βIP F1 (t)]2 ′ 2(ϵIP +α′IP t)
dσ
=
(αIP s)
dt
4π
πp
d=dt
10
(mb GeV 2 )
1
0.1
0
0.1
0.2
0.3
t
j j
Pomeron exchange dominates already at
to the nucleon but not to the pion
√
0.4
0.5
0.6
0.7
(GeV2 )
s = 19.4 GeV, because ω couples strongly
Proton-proton elastic scattering at large t
For |t| greater than about 3 Gev2 , the data are consistent with being energy-independent
and fit well to a simple power of t:
dσ/dt = 0.09 t−8
dσ/dt
(mb GeV−2 )
"27.4"
"30.5"
"44.6"
"52.8"
"62.1"
"fit"
1e-06
1e-07
1e-08
1e-09
1e-10
1e-11
5
10
2
|t| (GeV )
This behaviour is what is calculated from triple-gluon exchange
Why this simple mechanism, with no higher-order perturbative QCD corrections?
Note that triple-gluon exchange is C = −1
amplitudes are opposite in sign
— its contributions to the pp and p̄p
Double pomeron exchange
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What we know about double exchange:
αIP (t) = 1 + ϵIP + αI′P (t)
Phase
e
αIP IP (t) = 1 + 2ϵIP + 12 αI′P (t)
1
− 2 iπαIP (t)
−e
1
− 2 iπαIP IP (t)
So IP IP has less steep t-dependence than IP and is opposite in sign at small t
But:
Not simple power sαIP IP (t) – there are also unknown log factors
Unknown factor FIP IP (t), both t-dependence and magnitude
∫
A(s, −q ) = 2is
2
2
d be
−iq.b
(
)
χ2
χ−
+ ...
2!
If choose first term to be single exchange IP , second term has the right structure to
be IP IP , but is wrong in detail
Effective power
The IP IP term gives a negative contribution to the total cross sections
200
180
Blue line 17.55 s0.110 IP
Red crosses IP + IP IP
Red line 18.23 s0.096 effective power
Black points TOTEM data
2013 fit
160
Donnachie and Landshoff
140
120
100
80
60
40
10
100
1000
10000
Current fit to pp and p̄p total cross
sections with IP, IP IP, ρ, ω, f2 , a2
120
d
dt
100
IP
1
0
0
1
80
IP IP
60
40
t
j j
100
1000
10000
Dips in proton-proton scattering
√
Really deep dip at s =31 GeV
Both real and imaginary parts of amplitude
almost vanish at same t value
1
− 2 iπαIP (t)
IP and IP IP have different phases e
100
30.54 GeV pp
1
0.01
1
− 2 iπαIP IP (t)
0.0001
1e-06
0
1
2
3
4
5
and −e
Adjust IP IP to cancel imaginary parts
Need another term to cancel real parts: perhaps ggg
But ggg is C = −1 so then no dip in p̄p scattering
dσ/dt
(mb GeV−2 )
0.0001
pp and pBARp at 53 GeV
1e-05
1.1
1.2
1.3
1.4 1.5
|t| (GeV2 )
1.6
1.7
100
100
10
23 GeV pp
10
1
1
0.1
0.1
0.01
0.01
0.001
0.001
0.0001
0.0001
1e-05
1e-05
1e-06
1e-06
1e-07
1e-07
0
1
2
3
4
5
100
30.54 GeV pp
0
1
2
3
4
5
3
4
5
100
53 GeV pp
10
1
62 GeV pp
1
0.1
0.01
0.01
0.0001
0.001
0.0001
1e-06
1e-05
1e-06
1e-08
1e-07
0
2
4
6
8
10
0
1
2
1000
1000
7000 GeV pp
100
10
8000 GeV pp
1
0.1
500
0.01
0.001
0.0001
1e-05
300
0
0.5
1
1.5
2
1000
2.5
0.002
1000
800
800
600
600
400
31 GeV pp
0.006
400
200
0.01
0.014
0.018
31 GeV ppbar
200
0
0.005
0.01
0.015
0.02
0
0.005
0.01
0.015
0.02
100
100
53 GeV ppbar
600 GeV ppbar
1
1
0.01
0.01
0.0001
0.0001
1e-06
1e-06
0
1
2
3
4
5
0
1
2
3
4
5
1000
100
10
Details: A Donnachie and P V Landshoff
Physics Letters B727 (2013) 500
http://arxiv.org/abs/1309.1292
1800 Gev ppbar
1
0.1
0.01
0.001
Challenge: get a better fit to all the data!
0.0001
1e-05
0
0.5
1
1.5
2
2.5
100
TOTEM pp 8 TeV
March 2015
10
0
0.05
0.1
t (GeV^2)
0.15
0.2
A C = −1 term like ggg which survives at high energy is called an odderon.
There have been many attempts to see signs of an odderon at t = 0.
A sensitive test is thought to be in
ReA(s, t) ρ=
ImA(s, t) t=0
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
pp
0
pbarp
0
100
Not there?
1000
10000
100
1000
10000
Glueball trajectory??
Our fit has
αIP (t) = 1.1 + 0.165t
If this linear form extends to positive t it should pass through the mass of a
J P C = 2++ glueball
Experimental situation with glueballs obscure
Lattice calculations:
Bali et al (1993)
2.27 ± 0.1 GeV
Morningstar et al (1999)
2.4 ± 0.12 GeV
Chen et al (2006)
2.39 ± 0.12 GeV
2.6
Gregory et al (2012)
2620 ± 0.05 GeV 2.4
2.2
2
1.8
1.6
1.4
1.2
1
0
1
2
3
4
5
6
7
8
Summary so far
Effective-power approach describes small-t elastic scattering and total cross sections:
pp p̄p πp Kp γp
But for larger t need to consider
IP + IP IP + ggg + . . . ?
Easy to fit subsets of data, but universal fit more challenging
100
53 GeV pp and ppBAR
1
0.01
0.0001
1e-06
1e-08
0
2
4
6
8
10
Diffraction dissociation
AB → AX with particle A losing a very small fraction ξ of its momentum
t
A p
p
1
IP
B
p
ξ p1
Can think of the lower part of the diagram as the
pomeron IP scattering on particle B
A
1
X
2
MX
= (p1 + p2 − p′1 )2 ∼ ξs
2
If p is the momentum of one of the particles of system X, define its rapidity
Y =
1
2
log
E + pL
E − pL
(which is invariant under longitudinal Lorentz-frame boosts)
If ξ is small, the rapidity of particle p′1 is approximately
√
′2
2
log( s/m′1T )
m′2
1T = p1T + m
while the rapidity of the fastest particle in the system X is approximately 0
So there is a large rapidity gap
Mueller’s generalised optical theorem
Sum over systems of hadrons X
t
A p
p
1
IP
ξ p1
d2 σ
2
= DIP/a (t, ξ) σ IP b (MX
, t)
dt dξ
A
1
DIP/a (t, ξ) is the ”flux” of pomerons emitted by A:
B
p
2
X
2
9β
DIP/a (t, ξ) = IP2 (F1 (t))2 ξ 1−2αIP (t)
4π
2
σ IP b (MX
, t) is the cross section for a pomeron of squared mass t scattering on
particle B
Generalised optical theorem: it is calculated from the imaginary part of the forward
IP B elastic scattering amplitude
Triple-reggeon vertex
At large MX the amplitude IP B → IP B should be dominated by pomeron exchange:
t
t
a
a
A
A
d σ
dt dξ
2
c
B
Upper two pomerons: squared 4-momentum t
Lower pomeron: zero 4-momentum
We know all the factors, except the triple-pomeron
vertex VIP IP IP (t) in the middle
Complications: unless ξ is very small, also need ρ, ω, f2 , a2 instead of the top two
2
pomerons, and unless MX
= ξs is very large the same for the lower reggeon
So need
IP IP
IP IP
f2 IP
IP f2
f2 IP
ωIP
...
IP
f2
IP
IP
f2
ω
each with its unknown triple vertex in the middle
Usually, for simplicity, only terms of the form aac are considered:
FaA (t)FaA (t)FcB (0)Vcaa (t) ξ αc (0)−2αa (t) (αc′ s)αc (0)−1
Non-pomeron terms are often important!
Diffraction dissociation data problems
∫ ∫ ξmax
Diff
σ (s) = dt ξmin dξ d2 σ/dtdξ
25
σ Diff
(mb)
ξmax = 0.05
20
Sensitive to ξmax , and more so to ξmin
15
Uppermost three points: ALICE at LHC
The curve rises as s0.08
10
Issues with the data – often unclear experiments
are measuring the same thing
5
0
100
1000
√
10000
s (GeV)
dσ
dt
UA4
10
Data at
546 GeV M 2 > 2 ξ < 0.05
0.1
0
0.2
s = 546 GeV
The line is CDF’s parametrisation of its data for
d2 σ/dtdξ integrated over ξ
CDF
1
√
0.4
|t|
0.6
0.8
1
Diffraction dissociation data fits
Data for d2 σ/dtdξ
s=2880 -t=0.11 (top) to 0.3 (bottom)
200
100
s=309 -t=0.036, 0.075, 0.131, 0.197
80
150
60
100
40
50
20
0
0.02
0.04
0.06
0.08
0.1
0.02
0.04
0.06
0.08
0.1
1000
d2 σ
dtdξ
1800 GeV t = −0.05
IP IP IP does not dominate until
ξ is extremely small
100
IRIRIP
IP IP IP
10
0.02
0.04
0.06
0.08
ξ
0.1
Double diffraction dissociation
Both initial particles lose a very small fraction of their momentum
t1
1
A
A
1’
1’
1’
ξ 1 p1
1
1
X
ξ 2 p2
2
B
B
t2
2
2
2’
2’
2’
d4 σ
2
= DIP/A (t1 , ξ1 )DIP/B (t2 , ξ2 )σ IP IP (MX
, t1 , t 2 )
dt1 dξ1 dt2 dξ2
2
) to be dominated by IP exchange
If MX is large enough for σIP IP (MX
d4 σ(s)
d2 σ(s) d2 σ(s) Tot
=
σ (s)
dt1 dξ1 dt2 dξ2
dt1 dξ1 dt2 dξ2 pp
Exclusive central production
1
1’
a
H
b
2
5 independent variables, eg
s = (p1 + p2 )2 s1 = (p′1 + pH )2 s2 = (pH + p′2 )2
t1 = (p′1 − p1 )2 t2 = (p′2 − p2 )2
In most events s1 , s2 are large and t1 , t2 small
2
s1 ∼ s ξ2 s2 ∼ s ξ1 ξ1 ξ2 s ∼ MH
2’
Energy dependence of amplitude: (αa′ sξ2 )αa (t1 ) (αb′ sξ1 )αb (t2 )
Square and apply
∫ 1
dξ1 dξ2 δ(ξ1 ξ2 − m2H /s)
2 /s
MH
σ(IRIP ) and σ(IP IP ) increase with energy, σ(IRIR) ∼ 1/s
Hope that this is a good mechanism to produce glueballs
ALICE has detected f0 (980) and f2 (1270) production in pp → ppππ double-gap
events
Inclusive central production
H
pp →HX
H
H
AGK cancellation
(Abramovskii, Gribov, Kancheli)
Simplest example: Drell-Yan pp → ℓ+ ℓ− X
Add initial-state interactions, and cross interactions between initial and final states
They all cancel in the inclusive cross section dσ/dq 2
But they do change the final state
Real photons
ρ
ρ
00
11
11111
00000
00
11
11111
0000000
11
00000
11111
11111
00000
Simplest form of vector dominance:
dσ
4π dσ 0
(t) = αEM 2
(ρ p → ρ0 p : t)
dt
γρ dt
γρ is ρ-photon coupling got from ρ → e+ e−
Assume IP coupling to ρ same as to π and include f2 , a2 exchange – need a fudge
factor 0.84 (ρ width? higher resonances?)
d=dt jt=0 160
(b GeV 2 )
140
100
120
dσ/dt
(µb GeV−2 )
100
80
10
60
40
10
100
p
s (GeV)
1
0
0.1
0.2
0.3
0.4
2
|t| (GeV )
dσ/dt at t = 0
√
and at s = 94 GeV
0.5
Summary on inelastic diffractive processes
• There are severe problems with the lack of good data
• Single effective-power exchanges should be tried first
• But there are indications that for exclusive production of heavy systems, such as
Higgs or a pair of high-pT jets, initial and final state interactions are important –
people talk about rapidity gap survival probabilities
Models of the pomeron
There are many:
dipole,
111
111
000
000
111
111
000
000
11111111111
00000000000
111
111
000
000
111
111
000
000
111
111
000
000
11111
00000
11111111111
00000000000
11111
00000
111
111
000
000
111
111
000
000
111
111
000
000
111
111
000
000
0
1
0
1
11111111111
00000000000
111
111
000
000
0
1
0
1
111
111
000
000
0
1
0
1
111
111
000
000
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
111
111
000
000
0
1
0
1
111
000
000
0
1
0111
1
111
111
000
000
111
111
000
000
111
111
000
000
11111
00000
11111111111
00000000000
11111
00000
111
111
000
000
111
111
000
000
111
111
000
000
111
111
000
000
11111111111
00000000000
111
111
000
000
111
111
000
000
111
111
000
000
geometrical,
dual,
string,
color glass,
etc
Simplest QCD model: two-gluon exchange
At t = 0 the amplitude is proportional to
∫0
2
2 2
dk
[α
D(k
)]
S
−∞
where D(k 2 ) is the gluon propagator
The perturbative propagator D(k 2 ) = 1/k 2 would make the integral diverge
Introduce a length scale a: the largest distance the gluon can propagate in the
vacuum before confinement pulls it back
Simplest: D(k 2 ) = a2 /(1 + a2 k 2 )
Then it turns out that if a is much less than the nucleon radius both gluons want to
couple to the same quark in each nucleon – in line with the quark-counting rule for
total cross sections
(This is the Landshoff-Nachtmann model)
BFKL pomeron
(Balitsky, Fadin, Kuraev, Lipatov)
Sums generalised perturbative gluon ladder graphs
Leading-order calculation: effective power sϵ
ϵ = 12(αs /π) log 2 ≈ 0.4
But DISASTER: next-order calculation makes ϵ negative
0.4 would have been good!
Inelastic lepton scattering
k
γ*
Q = −q
2
q
X
2
(
W 2 = (p+q)2 = Q2
p
µ
ν
q
p
q
p
ν = p.q
Q2
x=
2ν
p.q
y=
p.k
)
1
− 1 +m2
x
Square and sum over X
Need imaginary part of virtual-Compton amplitude
(
qµ qν )
2
g +
F
(x,
Q
)
1
2
Q
1( µ
ν µ )( ν
ν ν)
+
p + 2q
p + 2 q F2 (x, Q2 )
ν
Q
Q
µν
Does Regge theory apply when W 2 >> Q2 , ie at small x ?
Deep inelastic total cross sections
We already have seen that it applies for the real γp total cross section
2
4π
α
EM
2 ) 2
σ γp (W ) =
F
(x,
Q
2
2
Q
Q =0
Adopt same definition for nonzero Q2
σ
(mb)
0.4
0.1
0.35
effective power of W^2 or 1/x
against Q^2
0.3
x<0.005
0.25
0.01
0.2
0.15
0.1
0.05
0.001
10
100
W (GeV)
0
0.1
1
10
Q2 from 0.25 to 90 GeV2 —- power increases from a bit less than 0.1 to about 0.35
Regge approach
F2 (x, Q2 ) = F0 (Q2 )x−e0 + F1 (Q2 )x−e1 at small x ???
e1 = 0.096 "soft pomeron"
e0 between 0.35 and 0.45 "hard pomeron"
1
1
e_0=0.44
e_0=0.36
soft
soft
0.1
0.1
hard
0.01
0.01
hard
x<0.001
0.001
x<0.001
0.001
1
10
1
10
Q^2
Q^2
Fit all x < 0.001 data with
2
F0 (Q ) = A0
(
1
Q2 )1+e0 (
Q 2 ) 2 e0
1+
2
a0 + Q
a0
Include also f2 , a2 exchange term e2 = −0.34
2
F1 (Q ) = A1
(
Q2 )1+e1
a1 + Q2
100
Q2 F2 (x, Q2 )
(GeV4 )
10
Q2 = 0.1 to 45 GeV2
1
0.1
0.01
1e-6
Fit data with x < 0.001
1e-5
e0 = 0.41
1e-4
x
e1 = 0.08
Very accurate data, though with some irregularities
1e-3
Charm cross section
Q^2 F_2^c
100
HERA charm NLO e0=0.41
Q^2=2.5 to 650
10
1
0.1
1e-05
0.0001
0.001
0.01
x
The lines are 2/5 of the hard pomeron
contribution to the complete F2
σ
(µb)
Q2 = 0
10
2
=
5
1
0.1
10
100
W (GeV)
4
9
4
9
+
1
9
+
4
9
+
4
9
e2c
= 2
eu + e2d + e2s + e2c
The contribution from soft pomeron exchange
is negligible
J/ψ production
Omega experiment (1977):
p + Cu → J/ψ + X
= 0.15 ± 0.08
p̄ + Cu → J/ψ + X
σ(p̄ p → J/ψ X = 12 ± 5 nb
So the valence quarks of the beam couple to J/ψ: |cc̄⟩ mixes with |q q̄ >
at 10−4 level
12
valence contribution
10
8.7 GeV
nb
8
6
4
2
0
0
0.5
1
1.5
Y
2
2.5
J/ψ production
1000
dσ/dt
(nb)
1000
ZEUS muon data
0.08
σ TOT
(nb)
γ p → J/ψ p
0.28
100
interf
0.48
100
hard
0.68
soft
10
50
100
150
W (GeV)
A mixture of hard and soft pomeron contributions
100
W (GeV)
1000
Open points:
ZEUS, Q2 = 3.1, 6.8, 16.0 GeV2
γ ∗ p → J/ψ p
100
dσ/dt
(nb)
Solid points:
H1, Q2 = 3.2, 7.0, 22.4 GeV2
10
1
W=90 GeV
Q2 = 3.1, 6.6, 16 GeV2
100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
2
t (GeV )
Proton vertex: form factor F1 (t)
σ TOT
(nb)
10
J/ψ vertex:
1
1
2
1.0 + t/(MJ/ψ
+ Q2 ) 1 + Q2 /B
Additional (1 + Kt) for hardpom
B = 66.6 GeV2
K = 2.14 GeV−2
γ ∗ p → J/ψ p
50
100
W (GeV)
150
200
Perturbative QCD
Perturbative QCD and Regge theory have to live together
Singlet DGLAP evolution eqation:
∫ 1
∂
2
2
2
u(x,
Q
)
=
dz
P(z,
α
(Q
))
u(x/z,
Q
)
Q2
s
∂Q2
(∑ x
)
2
2
i (qi (x, Q ) + q̄i (x, Q ))
u(x, Q2 ) =
g(x, Q2 )
xF2 (x, Q2 ) = 94 u(x, Q2 ) + 19 d(x, Q2 ) + 19 s(x, Q2 ) + 49 c(x, Q2 )
Mellin transform with respect to x:
∫
1
2
u(N, Q ) =
dx x
N −1
0
∫
2
u(x, Q )
1
2
dz z N P(z, αs (Q2 ))
P(N, αs (Q )) =
0
Then
Q2
∂
2
2
2
u(N,
Q
)
=
P(N,
α
(Q
))
u(N,
Q
).
s
2
∂Q
Problem
If u(x, Q2 ) ∼ f (Q2 )x−ϵ at small x, then
f (Q2 )
u(N, Q ) ∼
N −ϵ
2
and
Q2
∂
2
2
2
f
(Q
)
=
P(N
=
ϵ,
α
(Q
))
f
(Q
).
s
2
∂Q
For small z
Pgg (z, αs ) ∼ (6αs /2π)(1/z)
Pgg (N, αs ) ∼ (6αs /2π)(1/N )
Pgg (N, αs ) is known to be finite at N=0
At small N expansion in powers of αs is illegal
Compare
√
2
α
α
s
s
N 2 + αs −N =
−
...
3
2N 8N
The expansion is certainly not OK for the soft pomeron N = 0.096,
but may be OK for the hard pomeron N ≈ 0.4
Procedure
1 Choose some Q2 . Calculate hardpom coefficent function fq (Q2 ) from the fit.
2 Fix the gluon coefficient function fg (x, Q2 ) by requiring that the
perturbative calculation of
γ ∗ + g → c c̄
at is equal to 0.4 fq (Q2 )
Use running mass
m(Q2 ) = m0c (αs (Q2 )/αs (m20c ))C C = 12/(33 − 2nf )
and αs (Q2 + 4m2 (Q2 )), g(x, Q2 + 4m2 (Q2 )).
According to PDG
m0c = 1.275 ± 0.025GeV m0b = 4.18 ± 0.02GeV
Require αs (MZ2 ) ≈ 0.1184 giving Λ5 = 230 MeV at NLO. Continuity at Q2 = m20b gives
Λ4 = 330 MeV.
3 Apply DGLAP evolution to calculate fq (Q2 ) and fg (x, Q2 ) at all Q2 .
4 Fit fg (x, Q2 ) to some simple function of Q2
5 Calculate F2c (x, Q2 ) and compare with data.
1
10
gluon e0=0.41 NLO
quark e0=0.41 NLO
B (Q2/b))^e0 (1+Q2/b)^(-e0/2)
0.1
1
0.01
0.1
10
100
B=0.75 b=19.0
1000
10
Q^2
100
100
Q^2
HERA charm NLO e0=0.41
Q^2=2.5 to 650
charm photoproduction
Q^2 F_2^c
10
10
1
1
0.1
1e-05
0.1
0.0001
0.001
x
0.01
10
100
W (GeV)
(use mass m0c and αs (4m20c )
Summary on electroproduction
• Small-x data show need for hard pomeron 0.35 < ϵ0 < 0.45
• Charm couples only to the hard pomeron. Why??
¯ component: both pomerons couple to it, but problems
• J/ψ contains a small ūu.dd
fitting LHC data
• DGLAP evolution can be applied only to the hard pomeron component of F2 (x, Q2 )
Overall summary
• Regge theory successfully correlates a large amount of data
• But we do not know how to calculate multiple exchnages