606
On to QCD
Nice image
of what
happens
when two
protons
collide
Let’s go
over this
one by one
http://tex.stackexchange.com/questions/128559/draw-randomized-tree-in-tikz
ATLAS dijet event
607
Another view of the event
608
Some numbers...
Dijet mass = 6.9 TeV
pT of each jet = 3.2 TeV
MET = 46 GeV (impressive job of calorimeter)
609
Trijet event
610
P. Duinker, “Review of e+e- physics at PETRA,” Rev. Mod. Phys. 54 (2), 325-387 (1982), also http://
www.quantumdiaries.org/2011/07/09/in-a-world-without-color-why-do-i-believe-in-gluons/trijet_topology_rhophi_2/
Since quarks/
anti-quarks only
come in pairs,
tri-jet events can
be used as
evidence for
QCD radiation of
gluons
611
QCD production of quarks
e
e
p2
v
q1
p1
e
u
e2
´
rup3qp
2
pp1 ` p2 q
µ
p4
v
p3
u
qvp4qsrvp2q
e
µ up1qs
We evaluated this diagram when discussing
QED. QCD is similar, except that we have to
be careful about the charge of quarks
612
QCD production of quarks
e
e
p2
v
q1
p1
e
u
Qe2
M“´
rup3qp
2
pp1 ` p2 q
p4
v
p3
u
µ
e
qvp4qsrvp2q
µ up1qs
Each vertex gave us a factor of “e” - but u,c
and t quarks have charge Q=2e/3 and d,c,s
quarks have charge Q=-e/3
613
Evaluating matrix element
Qe2
M“´
rup3qp
2
pp1 ` p2 q
M2 “
„
2
1
Qe
4 pp1 ` p2 q2
⇢2
T rr
µ
pp{ 4 ´ mQ q
µ
⌫
qvp4qsrvp2q
pp{ 3 ` mQ qsT rr
µ up1qs
{1
µ pp
` me q
{2
⌫ pp
´ me qs
Where 1/4 comes from averaging over initial
spins and me is mass of electron and mQ is
mass of quarks being collided. That obvious?
614
Let’s expand that out
M2 “
„
2
1
Qe
4 pp1 ` p2 q2
„
⇢2
T rr
⇢2
µ
pp{ 4 ´ mQ q
⌫
pp{ 3 ` mQ qsT rr
{1
µ pp
”
1
Qe
µ
⌫
2 µ ⌫
M2 “
T
r
p
p
´
m
` mQ p
{
{
Q
2
4
3
4 pp1 ` p2 q
”
T r µ p{ 1 ⌫ p{ 2 ´ m2e µ ⌫ ` me p µ p{ 1 ⌫ ´
2
µ
µ
` me q
⌫
´
ı
{2 q
⌫p
p{ 4
{2
⌫ pp
µ
´ me qs
ı
⌫
p{ 3 q ˆ
Product of odd number of gamma matrices
is zero
615
Let’s expand that out
„
2
1
Qe
2
M “
4 pp1 ` p2 q2
„
⇢2
Tr
”
⇢2 ”
1
Qe
M2 “
Tr
2
4 pp1 ` p2 q
2
„
⇢2
2
“
1
Qe
2
M “
p4 p3 T r
4 pp1 ` p2 q2
µ
µ
p{ 4
⌫
p{ 4
⌫
µ
p{ 3 ´ m2Q
p{ 3 ´
⌫
ˆ Tr
ı
”
ˆ Tr
µ ⌫
4m2Q g µ⌫
´
ı
4m2Q g µ⌫
‰
“
”
{ 1 ⌫ p{ 2
µp
{ 1 ⌫ p{ 2
µp
ˆ p1 p2 T r
µ
⌫
µ ⌫
ı
4m2e gµ⌫
ı
´ m2e
´
´
4m2e gµ⌫
‰
616
More plug and chug
„
2
1
Qe
M “
4 pp1 ` p2 q2
2
„
⇢2
“
p4 p3 T r
µ
⌫
´
4m2Q g µ⌫
‰
“
ˆ p1 p2 T r
µ
⌫
´
4m2e gµ⌫
‰
⇢2
“
‰
1
Qe
µ ⌫
µ
⌫
µ⌫
2 µ⌫
M “
p4 p3 4pg g ` g g ´ g g q ´ 4mQ g
ˆ
2
4 pp1 ` p2 q
“
‰
2
p1 p2 4pgµ g⌫ ` gµ g ⌫ ´ gµ⌫ g q ´ 4me gµ⌫
2
2
„
⇢2
“ µ ⌫
‰
Qe
⌫ µ
µ⌫
2 µ⌫
M “4
p 4 p 3 ` p 4 p 3 ´ g p 3 ¨ p 4 ´ mQ g
ˆ
2
pp1 ` p2 q
“
‰
2
p1µ p2⌫ ` p1⌫ p2µ ´ gµ⌫ p1 ¨ p2 ´ me gµ⌫
2
2
More plug and chug
„
617
⇢2
“ µ ⌫
‰
Qe
⌫ µ
µ⌫
2 µ⌫
M “4
p 4 p 3 ` p 4 p 3 ´ g p 3 ¨ p 4 ´ mQ g
ˆ
2
pp1 ` p2 q
“
‰
2
p1µ p2⌫ ` p1⌫ p2µ ´ gµ⌫ p1 ¨ p2 ´ me gµ⌫
2
„
Qe2
M2 “ 4
pp1 ` p2 q2
2
⇢2
rpp1 ¨ p4 qpp2 ¨ p3 q ` pp2 ¨ p4 qpp1 ¨ p3 q ´ pp3 ¨ p4 qpp1 ¨ p2 q ´ m2e pp3 ¨ p4 q ` pp1 ¨ p3 qpp2 ¨ p4 q`
pp2 ¨ p3 qpp1 ¨ p4 q ´ pp3 ¨ p4 qpp1 ¨ p2 q ´ m2e pp3 ¨ p4 q ´ pp3 ¨ p4 qpp1 ¨ p2 q ´ pp3 ¨ p4 qpp1 ¨ p2 q`
4pp3 ¨ p4 qpp1 ¨ p2 q ` 4m2e pp3 ¨ p4 q ´ m2Q pp1 ¨ p2 q ´ m2Q pp1 ¨ p2 q ` 4m2Q pp1 ¨ p2 q ` 4m2Q m2e s
„
Qe2
2
M “4
pp1 ` p2 q2
„
⇢2
Qe2
2
M “8
pp1 ` p2 q2
r2pp1 ¨ p4 qpp2 ¨ p3 q ` 2pp2 ¨ p4 qpp1 ¨ p3 q ` 2m2e pp3 ¨ p4 q ` 2m2Q pp1 ¨ p2 q ` 4m2Q m2e s
⇢2
rpp1 ¨ p4 qpp2 ¨ p3 q ` pp2 ¨ p4 qpp1 ¨ p3 q ` m2e pp3 ¨ p4 q ` m2Q pp1 ¨ p2 q ` 2m2Q m2e s
618
Now we pick a frame
„
Qe2
2
M “8
pp1 ` p2 q2
⇢2
rpp1 ¨ p4 qpp2 ¨ p3 q ` pp2 ¨ p4 qpp1 ¨ p3 q ` m2e pp3 ¨ p4 q ` m2Q pp1 ¨ p2 q ` 2m2Q m2e s
|p1|=|p2|=pi
|p3|=|p4|=pf
pi “
a
E 2 ´ m2e
b
pf “ E 2 ´ m2Q
Pick center of
mass frame where
energy of each
object = E
p1
θ
p3
p4
p1 ¨ p2 “ E 2 ` p2i “ 2E 2 ´ m2e
pp1 ` p2 q2 “ p21 ` p22 ` 2p1 ¨ p2 “ 2m2e ` 4E 2 ´ 2m2e “ 4E 2
p3 ¨ p4 “ E 2 ` p2f “ 2E 2 ´ m2Q
p1 ¨ p3 “ E 2 ´ pf pi cos ✓ “ p2 ¨ p4
p1 ¨ p4 “ E 2 ` pf pi cos ✓ “ p2 ¨ p3
p2
619
Plugging it in
„
Qe2
2
M “8
pp1 ` p2 q2
⇢2
rpp1 ¨ p4 qpp2 ¨ p3 q ` pp2 ¨ p4 qpp1 ¨ p3 q ` m2e pp3 ¨ p4 q ` m2Q pp1 ¨ p2 q ` 2m2Q m2e s
p1 ¨ p2 “ E 2 ` p2i “ 2E 2 ´ m2e
pp1 ` p2 q2 “ p21 ` p22 ` 2p1 ¨ p2 “ 2m2e ` 4E 2 ´ 2m2e “ 4E 2
p3 ¨ p4 “ E 2 ` p2f “ 2E 2 ´ m2Q
p1 ¨ p3 “ E 2 ´ pf pi cos ✓ “ p2 ¨ p4
p1 ¨ p4 “ E 2 ` pf pi cos ✓ “ p2 ¨ p3
„
Qe2
2
M “8
4E 2
⇢2
rpE 2 ` pf pi cos ✓q2 ` pE 2 ´ pf pi cos ✓q2 ` p2E 2 ´ m2Q qm2e ` p2E 2 ´ m2e qm2Q ` 2m2Q m2e s
2 4
Q
e
2
4
2 2
2
2
2
2
M “
r2E
`
2p
p
cos
✓
`
2E
pm
`
m
f i
e
Q qs
4
2E
Plugging it in
620
2 4
Q
e
2
4
2 2
2
2
2
2
M “
r2E
`
2p
p
cos
✓
`
2E
pm
`
m
f i
e
Q qs
4
2E
a
pi “ E 2 ´ m2e
b
pf “ E 2 ´ m2Q
Q 2 e4
4
2
2
2
2
2
2
2
2
M “
r2E
`
2pE
´
m
qpE
´
m
q
cos
✓
`
2E
pm
`
m
e
Q
e
Q qs
2E 4
Q 2 e4
2
4
2
4
2 2
2 2
2 2
2 2
2 2
M “
r2E
`
2
cos
✓pE
`
m
m
´
E
m
´
E
m
q
`
2E
m
`
2E
mQ s
e Q
e
Q
e
2E 4
¸
ff
«
˜
2
2
2
2 2
2
mQ
mQ ` me
me mQ
me
2
2 4
2
´ 2 ´ 2 `
M “ Q e 1 ` cos ✓ 1 `
E4
E
E
E2
«
¸
ff
ˆ
˙˜
2
2
2
2
mQ
mQ ` me
me
2
2 4
2
M “ Q e 1 ` cos ✓ 1 ´ 2
1´ 2 `
E
E
E2
2
621
Differential cross section
«
˜
¸
ff
ˆ
˙
2
2
2
2
m
m
`
m
me
e
Q
Q
2
2 4
2
M “ Q e 1 ` cos ✓ 1 ´ 2
1´ 2 `
E
E
E2
Recall that we derived this a long time ago
d
1
|M |2
|pf |
“
d⌦
64⇡ 2 pE1 ` E2 q2 |pi |
Cancel
In region with energy far about masses
1
1
d
2 4
2
“
Q
e
p1
`
cos
✓q
2
2
d⌦
64⇡ 4E
Q 2 e2
d
2
“
p1
`
cos
✓q
2
2
d⌦
256⇡ E
622
How to get at the total cross section?
«
˜
¸
ff
ˆ
˙
2
2
2
2
m
m
`
m
me
e
Q
Q
2
2 4
2
M “ Q e 1 ` cos ✓ 1 ´ 2
1´ 2 `
E
E
E2
Start again with
d
1
|M |2
|pf |
“
d⌦
64⇡ 2 pE1 ` E2 q2 |pi |
“
“
ª 2⇡
0
d
ª⇡
0
1
sin ✓
64⇡ 2
ª
d
sin ✓d✓d
d⌦
”
Q e 1 ` cos2 ✓p1 ´
2 4
m2e
E 2 qp1
4E 2
´
m2Q
E2 q
`
m2Q `m2e
E2
ıd
E 2 ´ m2Q
E 2 ´ m2e
d✓
623
Total cross section calculation
ª 2⇡
ª⇡
”
Q e 1 ` cos2 ✓p1 ´
m2e
E 2 qp1
´
1
“
d
sin ✓
64⇡ 2
4E 2
0
0
”
m2Q
m2e
2 4
2
ª⇡
1 Q e 1 ` cos ✓p1 ´ E 2 qp1 ´ E 2 q `
“
sin ✓
32⇡
4E 2
0
ª⇡
0
2 4
sin ✓d✓ “
⇡
r´ cos ✓s0
ª⇡
m2Q
E2 q
`
m2Q `m2e
E2
m2Q `m2e
E2
ıd
ıd
E 2 ´ m2Q
E 2 ´ m2e
E 2 ´ m2Q
E2
´
m2e
d✓
d✓
“ p´p´1q ` 1q “ 2
cos2 sin ✓d✓
0
u “ ´ cos ✓, u2 “ cos2 ✓, du “ sin ✓d✓
ª⇡
ª
1
1
2
2
2
3
⇡
cos sin ✓d✓ “ u du “ r´ cos ✓s0 “ p1 ` 1q “
3
3
3
0
624
Total cross section calculation
”
m2e
2 4
2
ª⇡
1 Q e 1 ` cos ✓p1 ´ E 2 qp1 ´
“
sin ✓
32⇡
4E 2
0
d
E 2 ´ m2Q
ª⇡
m2Q
E2 q
`
m2Q `m2e
E2
ıd
E 2 ´ m2Q
E 2 ´ m2e
d✓
m2Q ` m2e
m2Q
m2e
2
sin ✓p1 `
q ` sin ✓ cos ✓p1 ´ 2 qp1 ´ 2 qd✓
2
2
2
E ´ me 0
E
E
E
d
˜
¸
2
2
2
2
2
E ´ mQ
mQ ` me
mQ
Q 2 e4
2
m2e
2`2
“
` p1 ´ 2 qp1 ´ 2 q
128E 2 ⇡ E 2 ´ m2e
E2
3
E
E
d
˜
¸
2
2
2
2
2
2
2 4
E ´ m Q 8 4 mQ ` me
Q e
2 me mQ
“
`
`
128E 2 ⇡ E 2 ´ m2e 3 3
E2
3 E4
d
˜
¸ˆ
˙
2
2
2
2 4
2
E
´
m
m
Q e
me
Q
Q
1
`
“
1
`
48E 2 ⇡ E 2 ´ m2e
2E 2
2E 2
Q 2 e4
“
128E 2 ⇡
Total cross section calculation
d
˜
¸ˆ
˙
2
2
2
2
2 4
E ´ mQ
mQ
me
Q e
1`
1`
“
2
2
2
2
48E ⇡ E ´ me
2E
2E 2
625
Clear that energy can’t be less than quark
mass or electron mass or calculation makes no
sense (good!) If energy is large enough, this is
approximated by
Q 2 e4
“
48E 2 ⇡
Total cross section calculation
626
Q 2 e4
“
48E 2 ⇡
As we crank up the energy, we expect the
ee→qq cross section to be flat until we reach
another kinematic regime where a new quark is
allowed to be produced. Have to be careful about
two things:
1) Q charges not all the same! (Evidence for
charges of quarks!)
2) If we compute R, the rate relative to muonantimuon production, in region where mass
effects are unimportant, we need a factor of 3x
for color. Evidence for quark color!
627
R
down
strange
«up
ˆ ˙2 ˆ
˙2 ˆ
˙2 ff
2
´1
´1
R“3
“2
`
`
3
3
3
up
down strange
charm
«ˆ ˙
˙2 ˆ
˙2 ˆ ˙2 ff
ˆ
2
2
´1
´1
2
R“3
“ 3.3
`
`
`
3
3
3
3
up down
charm
«ˆ ˙
˙2 ˆ
˙2 ˆ ˙2 ˆ
˙2 ff
ˆ
2
2
´1
´1
2
´1
R“3
“ 3.7
`
`
`
`
3
3
3
3
3
strange
bottom
below
charm
mass
when we
reach
charm
threshold
when we
reach
bottom
threshold
628
Total cross section ratio from PDG
10
10
10
10
10
10
10
-2
-3
-4
-5
-6
-7
-8
10
3
10
2
1
10
10
1
10
10
2
10
1
10
Can see falling
(1/E2) cross
section, and also
evidence for
charm quark and
then bottom
quark!
-1
2
http://pdg.lbl.gov/2007/hadronic-xsections/
hadronicrpp_page6.pdf
629
On R
Pretty nice agreement
between prediction and
observation, though this is a
simplified, leading order
calculation. There are loop/
higher-order effects, and the
model cannot account for
bound states/resonances, for
taus, and especially not for
Drell-Yan/Z boson
production
Form factors
Interesting discussion of
form factors in Griffiths, but
we’ll skip it - hopefully it
makes for fun reading :) If
we have time we can return
to it
630
631
QCD
We haven’t really used it, but
as an alternative to electric
charge e (or ge as Griffiths
uses), we can define a
coupling constant for QCD:
?
e “ 4⇡e
?
gs “ 4⇡↵s
↵e „ 1{137
↵s „ 1
“Strong” force!
632
Color for quarks
psq
psq
ppq Ñ u ppqc
¨ ˛
1
cpredq “ ˝ 0 ‚
0
¨ ˛
0
cpgreenq “ ˝ 1 ‚
0
¨ ˛
0
cpblueq “ ˝ 0 ‚
1
u
Spinors now get an
associated color vector!
Of course, remember that
“red”, “green” and “blue”
are just convenient names
and nothing more than that
Color for gluons
633
Gluons are spin-1
bosons and carry
two color quantities
- one unit of color
and one unit of
anti-color. Here, a
blue quark emits a
blue/anti-red
quark, and
becomes a red
quark (color is then
conserved)
634
Gluons
Naively would
predict nine gluons
- a color octet and
a color singlet. But
the singlet has not
been observed
(only 8 gluons).
Difference
between SU(3)
and U(3) color
symmetry
Gluons carry two
color quantities one unit of color
and one unit of
anti-color.
635
Gluons
Color octet
?
?
|1 °“ prb ` brq{ 2 |5 °“ ´iprg ´ grq{ 2
?
?
|2 °“ ´iprb ´ brq{ 2 |6 °“ pbg ` gbq{ 2
?
?
|3 °“ prr ´ bbq{ 2 |7 °“ ´ipbg ´ gbq{ 2
?
?
|4 °“ prg ` grq{ 2 |8 °“ prr ` bb ´ 2ggq{ 6
Color singlet (not observed!)
?
|9 °“ prr ` bb ` ggq{ 3
Range of strong force
Observed states (proton and
neutron, for example) are
color singlets. If they could
exchange color singlet gluons
then QCD would be a longrange force!
Color singlet (not observed!)
?
|9 °“ prr ` bb ` ggq{ 3
636
Gauge conditions
Think back to your E&M (we
skipped this) - have some
gauge freedom. Given gluon
4-vector p and polarization ε
Common to choose:
Lorentz condition
✏0 “ 0 Ñ ✏ ¨ p “ 0 Coulomb gauge
✏ µ pµ “ 0
637
638
Color state of gluon
¨
˚
˚
˚
˚
˚
|1 °“ ˚
˚
˚
˚
˚
˝
¨
˚
˚
˚
˚
˚
|2 °“ ˚
˚
˚
˚
˚
˝
¨
˚
˚
˚
˚
˚
|5 °“ ˚
˚
˚
˚
˚
˝
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
˛
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‚
˛
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‚
˛
‹
‹
‹
‹
‹
‹
‹
‹
‹
‹
‚
Column vector a represents
the color state of the gluon
(one of the 8 possible states)
Reminder that gluons selfcouple! These are both
valid diagrams/vertices
639
Gell-Mann Lambda matrices
¨
˛
¨
˛
¨
˛
1 0 0
0 1 0
0 ´i 0
1
“ ˝ 1 0 0 ‚ 2 “ ˝ i 0 0 ‚ 3 “ ˝ 0 ´1 0 ‚
0 0 0
0 0 0
0 0 0
˛
˛
¨
¨
˛
¨
0 0 0
0 0 ´i
0 0 1
4
“˝ 0 0 0 ‚ 5“˝ 0 0 0 ‚ 6“˝ 0 0 1 ‚
0 1 0
1 0 0
i 0 0
¨
˛
¨
˛
0 0 0
1 0 0
1
7
“ ˝ 0 0 ´i ‚ 8 “ ? ˝ 0 1 0 ‚
3
0 i 0
0 0 ´2
Compare with...
?
?
|1 °“ prb ` brq{ 2 |5 °“ ´iprg ´ grq{ 2
?
?
|2 °“ ´iprb ´ brq{ 2 |6 °“ pbg ` gbq{ 2
?
?
|3 °“ prr ´ bbq{ 2 |7 °“ ´ipbg ´ gbq{ 2
?
?
|4 °“ prg ` grq{ 2 |8 °“ prr ` bb ´ 2ggq{ 6
Why are we introducing the lambda matrices?
¨
˛
¨
˛
¨
640
˛
1 0 0
0 1 0
0 ´i 0
1
“ ˝ 1 0 0 ‚ 2 “ ˝ i 0 0 ‚ 3 “ ˝ 0 ´1 0 ‚
0 0 0
0 0 0
0 0 0
˛
˛
¨
¨
˛
¨
0 0 0
0 0 ´i
0 0 1
4
“˝ 0 0 0 ‚ 5“˝ 0 0 0 ‚ 6“˝ 0 0 1 ‚
0 1 0
1 0 0
i 0 0
¨
˛
¨
˛
0 0 0
1 0 0
1
7
“ ˝ 0 0 ´i ‚ 8 “ ? ˝ 0 1 0 ‚
3
0 i 0
0 0 ´2
r
↵
,
s “ 2if ↵
f↵
“ ´f
↵
“ ´f ↵
f 123 “ 1, f 147 “ f 246 “ f 257 “ f 345 “ f 516 “ f 637 “ 1{2
?
458
678
f
“f
“ 3{2
Plus commutations (and rest are zero)
641
Example QCD diagram
q
q
q
q
Here we have one of
the diagrams
contributing to
qqbar→qqbar
scattering. Quark and
anti-quark in initial
state must be the
same flavor. Same
for final state.
Obvious why? Let’s
move on to the matrix
element calculation
Feynman rules for QCD (1)
q
p1
q
p3
q
p2
q
p4
q
642
Label all incoming
and outgoing lines
with p1, p2, ... pn
Internal lines q can
go either way
Use arrows to keep
track of what is going
in and out (of course
we have arrows on
the anti-particles, but
that is labeling
something different).
643
Feynman rules for QCD (2)
q p1
vc†
q1
p2
q
p3
uc
p4
q
vc
uc†
q
Incoming
(outgoing) quarks
get a factor of
uc(uc†),
outgoing
(incoming) antiquarks get a factor
of vc(vc†). Spin
implicit here and
not labeled
Color factors to be grouped together!
Feynman rules for QCD (3)
q p1
vc†
p3
q1
p2
q
p4
quc
εµ(p)aα
q
vc
†
uc
q
q
644
Incoming
(outgoing) external
gluons with color
label a get a factor
of εµ(p)aα
(ε*µ(p)aα*)
645
Feynman rules for QCD (4)
Add factors of
q p1
vc†
´igs
2
↵ µ
p2
q
uc
p3
q1
q
vc
´igs
2
p4
´igs
2
uc†
q
⌫
↵ µ
at each quark-gluon
vertex. Lambda
matrices define the
gluon that is
exchanged (can be
any of the 8),
though only some
will contribute
646
Feynman rules for QCD (5)
k3,𝜆ᵞ
k3,𝜆ᵞ
k2,βν
k2,βν
k1,αµ
k1,αµ
k4,δρ
3 gluon
vertex
4 gluon
vertex
´igs2
“
f
↵ ⌘
f
´gs f ↵
⌘
rgµ⌫ pk1 ´ k2 q ` g⌫ pk2 ´ k3 qµ ` g
pgµ g⌫⇢ ´ gµ⇢ g⌫ q ` f
↵ ⌘
f
⌘
pgµ⌫ g
⇢
´ gµ g⌫⇢ q ` f
µ pk3
↵ ⌘
f
´ k1 q⌫ `s
⌘
pgµ⇢ g⌫
Can see that QCD can easily get
tricky, and that’s without loops!
‰
´ gµ⌫ g ⇢ q
647
Feynman rules for QCD (6)
q p1
p3
vc†
´igs
2
↵ µ
pq2
uc
`
q
´igµ⌫
q2
˘
i q{ ` m
q 2 ´ m2
q
vc
´igs
2
↵
p4
uc†
q
⌫
For each internal
gluon line add a
factor for the
propagator (delta
function ensures
conservation of
color!)
for internal quarks/anti-quarks
648
Feynman rules for QCD (7)
p2⇡q4 4 pp1 ` p2 ´ q1 q
q p1
vc†
´igs
2
↵ µ
pq2
uc
p3
q
´igµ⌫
q2
q
vc
´igs
2
↵
p4
†
uc
q
p2⇡q4 4 pq1 ´ p3 ´ p4 q
⌫
Impose
conservation of
energy and
momentum at
each vertex with
4d Dirac Delta
function (with
appropriate 2pi
normalization)
649
Feynman rules for QCD (8)
p2⇡q4 4 pp1 ` p2 ´ q1 q
q p1
vc†
´igs
2
↵ µ
pq2
uc
Integrate over
4-momentum
q
1
p3
4
d
vc p2⇡q4 q of internal
q
lines with
´ig
appropriate
2
´igµ⌫ ↵
q2
†
2pi
uc
p4 q
normalization
factor
p2⇡q4 4 pq1 ´ p3 ´ p4 q
s
⌫
650
Feynman rules for QCD (9)
p2⇡q4 4 pp1 ` p2 ´ q1 q
i
Cancel
remaining
q p1
q
1
p
4
3
†
d
vc
vc p2⇡q4 q delta function
q
and add a
´ig
´ig
2
factor of i, and
2
´igµ⌫ ↵
q2
†
you have the
uc
pq2
p4 q
matrix
uc
element
4 4
p2⇡q pq1 ´ p3 ´ p4 q
(phew)
s
↵ µ
s
⌫
651
Feynman rules for QCD (10)
p2⇡q4 4 pp1 ` p2 ´ q1 q
Add minus sign
between diagrams
q p1
q
differing only in
1
p3
4
†
d
vc
vc p2⇡q4 q exchange of two
q
incoming or two
´igs ↵ µ
´igs
⌫
outgoing
2
2
´igµ⌫ ↵
fermions, or
q2
†
uc
pq2
incoming fermion
p4 q
uc
and outgoing antifermion (or vice
p2⇡q4 4 pq1 ´ p3 ´ p4 q
versa)
i
652
Feynman rules for QCD
p2⇡q4 4 pp1 ` p2 ´ q1 q
q p1
vc†
´igs
2
↵ µ
pq2
uc
p3
q
´igµ⌫
q2
q
1
4
d
vc p2⇡q4 q
´igs
2
↵
p4
i
uc†
q
p2⇡q4 4 pq1 ´ p3 ´ p4 q
⌫
PHEW
Example calculation (qqbar→qqbar)
q
q
653
Can’t do a full calculation,
q since we don’t have here
the ability to calculate the
hadronization steps and
formation of bound states
(not to mention higher
order effects). But there
q
are still interesting things
to see...
Example calculation (qqbar→qqbar)
p2⇡q4 4 pp2 ` q ´ p4 q
q
p2
vc†
´igs
2
q
p1
q uc
´igs
2
⌫
´igµ⌫
q2
↵
i
p4 q
1
4
d
vc p2⇡q4 q
p3
↵ µ
p2⇡q4 4 pp1 ´ q ´ p3 q
uc†
q
654
655
Example calculation (qqbar→qqbar)
p2⇡q4 4 pp2 ` q ´ p4 q
q
p2
vc†
´igs
2
q
p1
q uc
´igs
2
i
⌫
´igµ⌫
q2
↵ µ
p4 q
1
4
d
vc p2⇡q4 q
↵
As always, follow
fermion lines
p3
backwards to get
uc†
q
grouping right!
p2⇡q4 4 pp1 ´ q ´ p3 q
M“
ª
gs
:
irup3qc3 sr´i
2
↵
´igµ⌫
µ
srup1qc1 s
q2
↵
gs
:
rvp2qc2 sr´i
2
4
d
q
4 4
4 4
p2⇡q pp2 ` q ´ p4 qp2⇡q pp1 ´ q ´ p3 q
p2⇡q4
⌫
srvp4qc4 s
656
Example calculation (qqbar→qqbar)
M“
ª
gs
:
irup3qc3 sr´i
M“
↵
2
´igµ⌫
µ
srup1qc1 s
q2
↵
gs
:
rvp2qc2 sr´i
2
4
d
q
4 4
4 4
p2⇡q pp2 ` q ´ p4 qp2⇡q pp1 ´ q ´ p3 q
p2⇡q4
ª
´gs2
rup3qc:3 sr
4
↵
:
rvp2qc
2 sr
2
pp1 ´ p3 q
d4 q
4 4
p2⇡q pp2 ` q ´ p4 q
p2⇡q4
↵ µ
srup1qc1 s
⌫
srvp4qc4 s
µ srvp4qc4 s
So matrix element is:
´gs2
rup3qc:3 sr
M“
4
↵ µ
srup1qc1 s
Or simplifying last delta:
´gs2
:
M“
rup3qc
3 sr
2
4pp1 ´ p3 q
↵
:
rvp2qc
2 sr
2
pp1 ´ p3 q
↵ µ
srup1qc1 srvp2qc:2 sr
µ srvp4qc4 s
↵
µ srvp4qc4 s
657
Example calculation (qqbar→qqbar)
Matrix element is
´gs2
:
M“
rup3qc
3 sr
2
4pp1 ´ p3 q
↵ µ
srup1qc1 srvp2qc:2 sr
↵
µ srvp4qc4 s
Compare with e+ e- scattering:
e2
ps3q
M“´
pp3q
ru
2
pp1 ´ p3 q
µ ps1q
u
pp1qsrv ps2q pp2q
µb
ps4q
pp4qs
Similar matrix elements except for (ignoring gs vs
e) a color factor:
1 :
f “ pc3
4
↵
c1 qpc:2
↵
c4 q
658
What does this tell us?
quark-antiquark scattering is the same form as
electron-antielectron scattering, so the potential
between the quarks must be of the same form,
too! But only true up to some additional factor
1 :
f “ pc3
4
↵
c1 qpc:2
↵
c4 q
Let’s look at color octet case first. Let’s pick the
incoming quark to be red and the incoming antiquark to be anti-blue, just as an example. Then
outgoing quark must also be red and outgoing
anti-quark must be anti-blue (since there is no
other source of QCD color)
659
What does this tell us?
1 : ↵
Let’s pick the incoming
f “ pc3 c1 qpc:2 ↵ c4 q
4
quark to be red and the
¨ ˛
incoming anti-quark to be
1
cpredq “ ˝ 0 ‚
anti-blue, and the outgoing
0
quark red and outgoing anti¨ ˛
quark anti-blue
0
¨
˛
0
c2 “ c4 “ ˝ 0 ‚
1
¨ ˛
1
c1 “ c3 “ ˝ 0 ‚
0
cpgreenq “ ˝ 1 ‚
0
¨ ˛
0
cpblueq “ ˝ 0 ‚
1
660
Calculating color factor
1 :
f “ pc3
4
↵
c1 qpc:2
¨
˛
0
c2 “ c4 “ ˝ 0 ‚
1
¨ ˛
1
c1 “ c3 “ ˝ 0 ‚
0
»
1–
p1 0 0q
f“
4
¨
↵
c4 q
λα is describing the
possible types of
exchanged gluons (for any
of the 8 values of α)
˛fi »
1
↵˝
0 ‚fl –p0 0 1qs
0
¨
˛fi
0
1
↵˝
‚
fl
0
“
4
1
↵ ↵
11 33
661
Calculating color factor
¨
˛
¨
˛
˛
¨
1 0 0
0 1 0
0 ´i 0
1
“ ˝ 1 0 0 ‚ 2 “ ˝ i 0 0 ‚ 3 “ ˝ 0 ´1 0 ‚
0 0 0
0 0 0
0 0 0
˛
˛
¨
¨
˛
¨
0 0 0
0 0 ´i
0 0 1
4
“˝ 0 0 0 ‚ 5“˝ 0 0 0 ‚ 6“˝ 0 0 1 ‚
0 1 0
1 0 0
i 0 0
¨
˛
¨
˛
0 0 0
1 0 0
1
7
“ ˝ 0 0 ´i ‚ 8 “ ? ˝ 0 1 0 ‚
3
0 i 0
0 0 ´2
»
¨
˛fi »
1
1–
p1 0 0q ↵ ˝ 0 ‚fl –p0 0 1q
f“
4
0
¨
˛fi
0
1
↵˝
‚
fl
0
“
4
1
↵ ↵
11 33
𝛌8 is only matrix with entries in 11 and 33
662
Calculating color factor
»
¨
˛fi »
1
1–
p1 0 0q ↵ ˝ 0 ‚fl –p0 0 1q
f“
4
0
1
f“
4
8 8
11 33
¨
˛fi
0
1
↵˝
‚
fl
0
“
4
1
↵ ↵
11 33
1
1 1 ´2
“ ? ? “´
4 3 3
6
Compared to e+e- potential, which is
attractive, here we have an extra minus sign,
indicating that color octet is repulsive! So no
binding occurs (pions do not have any color)
663
Color factor for singlet
1 :
f “ pc3
4
↵
: ↵
c1 qpc2 c4 q
Let’s switch to the color singlet case:
?
|9 °“ prr ` bb ` ggq{ 3
So out-going q/qbar are in a singlet state, and
in-coming quarks are also in a singlet state
Start with incoming ones (c1, c2)
»
1 1
f “ ? –c:3
4 3
¨
˛fi
»
¨
˛fi
1
0
1
1
:
↵˝
0 ‚fl rp1 0 0q ↵ c4 s ` ? –c3 ↵ ˝ 1 ‚fl rp0 1 0q
4 3
0
0
»
¨ ˛fi
0
1 1 – : ↵ ˝ ‚fl
?
0
rp0 0 1q ↵ c4 s `
c3
4 3
1
↵
c4 s `
664
Color factor for singlet
Out-going q-qbar (c3,c4) must also be in a
singlet state
˛fi »
˛fi
¨
»
1
1
1 1 1
1 1 1
f “ ? ? –p1 0 0q ↵ ˝ 0 ‚fl –p1 0 0q ↵ ˝ 0 ‚fl ` ? ? –p0 1 0q
4 3 3
4 3 3
0
0
˛fi »
˛fi
»
¨
¨
»
¨
1
0
1 1 1 –
1 1 1
? ?
p0 0 1q ↵ ˝ 0 ‚fl –p1 0 0q ↵ ˝ 0 ‚fl ` ? ? –p1 0 0q ↵ ˝
4 3 3
4 3 3
0
1
»
¨
»
¨
˛fi »
¨
˛fi
0
0
1 1 1 –
1 1 1
? ?
p0 1 0q ↵ ˝ 1 ‚fl –p0 1 0q ↵ ˝ 1 ‚fl ` ? ? –p0 0 1q ↵ ˝
4 3 3
4 3 3
0
0
»
¨
»
¨
˛fi »
¨
˛fi
0
1
1 1 1 –
1 1 1
? ?
p1 0 0q ↵ ˝ 0 ‚fl –p0 0 1q ↵ ˝ 0 ‚fl ` ? ? –p0 1 0q ↵ ˝
4 3 3
4 3 3
1
0
»
¨
˛fi »
¨
˛fi
0
0
1 1 1 –
? ?
p0 0 1q ↵ ˝ 0 ‚fl –p0 0 1q ↵ ˝ 0 ‚fl
4 3 3
1
1
»
¨
˛fi »
1
↵˝
0 ‚fl –p1 0 0q
0
˛fi »
¨
0
1 ‚fl –p0 1 0q ↵ ˝
0
˛fi »
¨
0
1 ‚fl –p0 1 0q ↵ ˝
0
˛fi »
¨
0
0 ‚fl –p0 0 1q ↵ ˝
1
¨
˛fi
0
↵˝
1 ‚fl `
0
˛fi
1
0 ‚fl `
0
˛fi
0
0 ‚fl `
1
˛fi
0
1 ‚fl `
0
¨
Each of 9 terms is a multiplication of λij and λji,
which simplifies this
665
Color factor for singlet
˛fi »
˛fi
¨
»
1
1
1 1 1
1 1 1
f “ ? ? –p1 0 0q ↵ ˝ 0 ‚fl –p1 0 0q ↵ ˝ 0 ‚fl ` ? ? –p0 1 0q
4 3 3
4 3 3
0
0
˛fi »
˛fi
»
¨
¨
»
¨
1
0
1 1 1 –
1 1 1
? ?
p0 0 1q ↵ ˝ 0 ‚fl –p1 0 0q ↵ ˝ 0 ‚fl ` ? ? –p1 0 0q ↵ ˝
4 3 3
4 3 3
0
1
»
¨
»
¨
˛fi »
¨
˛fi
0
0
1 1 1 –
1 1 1
? ?
p0 1 0q ↵ ˝ 1 ‚fl –p0 1 0q ↵ ˝ 1 ‚fl ` ? ? –p0 0 1q ↵ ˝
4 3 3
4 3 3
0
0
»
¨
»
¨
˛fi »
¨
˛fi
0
1
1 1 1 –
1 1 1
? ?
p1 0 0q ↵ ˝ 0 ‚fl –p0 0 1q ↵ ˝ 0 ‚fl ` ? ? –p0 1 0q ↵ ˝
4 3 3
4 3 3
1
0
»
¨
˛fi »
¨
˛fi
0
0
1 1 1 –
? ?
p0 0 1q ↵ ˝ 0 ‚fl –p0 0 1q ↵ ˝ 0 ‚fl
4 3 3
1
1
»
¨
˛fi »
1
↵˝
0 ‚fl –p1 0 0q
0
˛fi »
¨
0
1 ‚fl –p0 1 0q ↵ ˝
0
˛fi »
¨
0
1 ‚fl –p0 1 0q ↵ ˝
0
˛fi »
¨
0
0 ‚fl –p0 0 1q ↵ ˝
1
¨
˛fi
0
↵˝
1 ‚fl `
0
˛fi
1
0 ‚fl `
0
˛fi
0
0 ‚fl `
1
˛fi
0
1 ‚fl `
0
¨
Each of 9 terms is a multiplication of λij and λji,
which simplifies this (go ahead and write it out
yourself if you want to check)
1 ÿ
f“
T rp
12 ↵
↵ ↵
q
666
Color factor for singlet
˛
0 1 0
1
“˝ 1 0 0 ‚
0 0 0
¨
˛
0 0 1
4
“˝ 0 0 0 ‚
1 0 0
¨
0 0
7
˝
0 0
“
0 i
¨
˛
˛
¨
1 0 0
0 ´i 0
2
“ ˝ i 0 0 ‚ 3 “ ˝ 0 ´1 0 ‚
0 0 0
0 0 0
˛
˛
¨
¨
0 0 0
0 0 ´i
5
“˝ 0 0 0 ‚ 6“˝ 0 0 1 ‚
0 1 0
i 0 0
˛
¨
˛
0
1 0 0
1 ˝
8
‚
´i
0 1 0 ‚
“?
3
0
0 0 ´2
¨
˛
¨
0
1
2
2
0 ‚ p q “˝ 0
0
0
˛
¨
1
0 0
5 2
‚
˝
0
0 0
p q “
0 1
0
¨
˛
0 0 0
7 2
˝
0 1 0 ‚ p 8 q2
p q “
0 0 1
¨
1
1 2
p q “˝ 0
0
¨
1
4 2
˝
0
p q “
0
0
1
0
1
f“
p2 ¨ 8q “ 4{3
12
1 ÿ
f“
T rp
12 ↵
˛
¨
1
0
3
2
0 ‚ p q “˝ 0
0
0
˛
¨
0
0 0
6 2
‚
˝
0
0 0
p q “
0
0 1
¨
˛
1{3 0
0
˝
0 1{3 0 ‚
“
0
0 4{3
0
1
0
0
1
0
0
1
0
↵ ↵
˛
0
0 ‚
0
˛
0
0 ‚
1
Color singlet is
attractive!
q
We now skip a bit
Interesting
discussion in Griffiths
on pair annihilation in
QCD. Please read
it :)
667
Asymptotic Freedom
QCD screening effects:
QCD coupling varies as a
function of the momentum
transfer (ie how close you
probe the quarks), just like
in QED
668
Asymptotic Freedom and gluons
But now we have to
account for virtual gluon
loops too, since gluons selfcouple! These anti-screen
the coupling and compete
with quark loops
669
Asymptotic Freedom and gluons
↵s pq 2 q “
q`
↵s pµ2 q
↵s pµ2 q
12⇡ p11n
´ 2f q ln p|q 2 |{µ2 q
n=3 is number of colors, f=6
is number of quarks, so
anti-screening dominates.
Can’t use µ=0 as reference,
so need it as a parameter
that defines the baseline for
renormalization!
670