(fundamental) physics of elementary particlesphysics1.howard.edu/~thubsch/fpp1/slides/1107.pdf ·...
TRANSCRIPT
(Fundamental) Physicsof Elementary Particles
QCD: quantum chromodynamics & Feynman rules;Quark-(anti)quark interaction & color
antisymmetrization;
Tristan HübschDepartment of Physics and Astronomy
Howard University, Washington DCPrirodno-Matematički Fakultet
Univerzitet u Novom Sadu
Monday, November 7, 11
Program
Fundamental Physics of Elementary Particles
2
Concrete QCD computationsFeynman rulesGluon loops & interactions
nonlinearitygauge conditions
Quark-quark interactionColor factor computation(qq)3* vs. (qq)6fc(3*|3*), fc(3*′|3*), fc(6|3*), fc(3*|6), fc(6|6), fc(6′|6)
Quark-antiquark interactionColor factor computation(qq*)1 vs. (qq*)8fc(1|1), fc(8|1), fc(8|8), fc(8′|8)
Conclusion: SU(3)c formalism
Monday, November 7, 11
Communication
But, First and ForemostWhen reporting errors in a 344-page document…Help locating the error:
specify page, paragraph & line, table, !gure or equationFor example: p. 123, P. 3, l. 2 (page 123, paragraph 3, line 2) Eq. (4.1), Figure 4.1, Table 4.1, …
State the error/typo:Compare, for example, “Te to the paragraph a"er 4.50” with“Change ‘Since te…’ to ‘Since the…’ four lines a"er Eq. (4.50).”#e la$er version allows for an effective electronic search.
Itemize the suggested corrections.A concatenation of several sentences (even if wri$en clearly)…runs together. (Remember: I need to be able to tell them apart.)
3
Monday, November 7, 11
Feynman rules
Concrete QCD Computations1. Notation:
4-momenta: external = p1, p2, …, internal = q1, q2, …Orientations:
For a spin-½ particle, with 4-momentumFor a spin-½ antiparticle, against 4-momentumGluon lines: external (real) with time, internal (virtual) = arbitrary
Polarizations:
4
) )
Spin-
1
/
2
quark
incoming usf ca s = spin projection = ", #
a = quark color = r, y, bf = quark flavor: u, d, s, . . .
outgoing u f ,s c†
a
Spin-
1
/
2
antiquark
incoming v f ,s c†
a (
⇠= spin-
1
/
2
quark, travels
backwards in time)
outgoing vsf ca
Gluon
incoming eµca eµ pµ = 0 and e0 = 0
outgoing eµ⇤ca⇤
Monday, November 7, 11
Feynman rules
Concrete QCD Computations2. Vertices
Quark-gluon:
3-gluons:
4-gluons:
5
µi, a, f
2
j, b, f1 a
�! �igcggggµ d
f1
f2
( 1
2
la)b
a
�
a, µ
b, n c, r
d, s
�!�ig2
c [ f abe f cde(hµshnr � hµrhns)
+ f ace f dbe(hµshnr � hµnhrs)
+ f ade f bce(hµnhrs � hµrhns)]
a, µ
b, n
r
c�!
�gc f abc[hµn(k1
�k2
)r
+hnr(k2
�k3
)µ
+hrµ(k3
�k1
)n]
Monday, November 7, 11
Feynman Rules
Concrete QCD Computations3. Propagators = internal lines
Quarks:
Gluons:
Recall: internal lines represent virtual particles that are off-shell.4. 4-momentum conservation
Assign to each vertex (2π)4 δ4(Σj kj)5. Integrate over all internal momenta: (2π)–4∫d4qj
6. Read off: –i M (2π)4 δ4(Σj pj)6
quark: qjn, a n0, b
�! idn,n0d
b
a
/qj � mjc= idn,n0
d
b
a
/qj + mjc1
q2
j � m2
j c2
,
h
�
gluon:µ, a n, bqg
�! � ih
µn
q2
gd
ab
Monday, November 7, 11
Feyman rules
Concrete QCD Computations7. Each fermion loop = one (–1) factor8. Amplitudes for partial processes that are related by an exchange of an odd pair of fermions have a relative – sign.As before, we can order the Feynman diagrams by:
counting the orders of gc ,and counting loops.
#ese amplitudes cannot be used as in electromagnetism,… because quarks do not appear as free particles.Nevertheless, they can indicate relative probabilities,…somewhat akin to applying the Wigner-Eckardt theorem.
7
sproces 1
sproces 2
=
�
�M1
�
�
2
�
�M2
�
�
2
=
�
�(spin)1
·(isospin)1
·(color)1
·(other)1
�
�
2
�
�(spin)2
·(isospin)2
·(color)2
·(other)2
�
�
2
Monday, November 7, 11
Gluon loops & interactions
Concrete QCD Computations#e gluon Lagrangian involves
which when squared produces:
8
�
�
�
�
�
�
�
�
Fµn := h cigc
[∂µ + igch c Aµ, ∂n +
igch c An] = ∂µAn � ∂nAµ + igc
h c [Aµ, An],
ig= ∂[µAn] +igch c [Aµ, An]
� ig
FµnFµn =�
∂[µAn] +igch c [Aµ, An]
��
∂[µAn] + igch c [A
µ, An]
�
2
� �� �
= ∂[µAn]∂[µAn] + 2igc
h c ∂[µAn][Aµ
, An]� g2
ch2c2
[Aµ, An][Aµ
, An]2
a, µ
b, n
r
c
a
a, µ
b, n c, r
d, s
gluon: µ, a n, bqg
keep in mind, for later…Monday, November 7, 11
Quark-quark interaction
Concrete QCD ComputationsConsider a concrete process, such as p++n0 → p++n0 .Analyze as (uud)+(udd)→(uud)+(udd),
where the strong interaction is dominatingso consider quark-quark interactions(u+u)→(u+u) ≈ (u+d)→(u+d) ≈ (d+d)→(d+d)…up to corrections O(|mu–md|/(mu+md)) ≈ 33%…#en also: (p+p)→(p+p) ≈ (p+n)→(p+n) ≈ (n+n)→(n+n),…as Heisenberg initially observed, introducing isospin.
Consider:
9u-quarkcolor: a
p1
u-quarkcolor: g
p3
q(la)
a
g (lb)b
d
dab
d-quarkcolor: b
p2
d-quarkcolor: d
p4
Monday, November 7, 11
old stuff new stuff
Quark-quark interaction
Concrete QCD Computations#e amplitude computation differs from that in electromagnetism only by color factors:
Re-use the electromagnetism computation,with the ge→gc replacement,Compute the color factor,…for all the different possible cases.Since the EM amplitude would have given…the QCD amplitude will yield
10
Mu+d!u+d = � g2
s2
1
q2
⇥u
3
gg
g
g
µ u1
⇤⇥u
4
gg
g
g
µ
u1
⇤�c
†
3
ll
l
l
ac
1
��c
†
4
ll
l
la c
2
�,
1
4pe0
e2
r=
ae h cr
Vqq(r) = fcas h c
r,
remaining to be determined
fc(3, 4|1, 2) = 1
4
(c†
3
ll
l
l
ac
1
)(c†
4
ll
l
lac
2
) =
Monday, November 7, 11
inout
Quark-quark interaction
Concrete QCD ComputationsSo, consider computing
…for the different possible two-quark in- and out-states.Use the color—tensor—matrix notation translations:
Use also that
11
) = 1
4
c
†
3g
c
†
4d
(la)a
g (la)b
d
c
a
1
c
b
2
.
c
r $ d
a
1
$h
1
0
0
i,
c
y $ d
a
2
$h
0
1
0
i, c
b $ d
a
3
$h
0
0
1
i.
(3⌦3)A = 3⇤ c[a1
cb]2
:= 1p2
(dagd
bd � d
bgda
d ) cg1
cd2
a 6= b, a, b = 1, 2, 3
(3⌦3)S = 6 c(a1
cb)2
:=
⇢
1p2
(dagd
bd + d
bgda
d )
dagd
bd
�
cg1
cd2
n a 6= b,
a = b,
a, b = 1, 2, 3
fc(3, 4|1, 2) = 1
4
(c†
3
ll
l
l
ac
1
)(c†
4
ll
l
lac
2
) =
Monday, November 7, 11
Some SU(3)c representations
Concrete QCD Computations#e fundamental representation
denoted 3, for a complex 3-dimensional vector space,…spanned by (t1, t2, t3): c1t1+ c2t2+ c3t3, i.e., ℂ3={c1, c2, c3}…which are transformed one into another by SU(3)c.
#e antisymmetric product = antisymmetric rank-2 tensormay be identi!ed with 3*:represented by linear combinations of t[12], t[13] and t[23],…which are transformed one into another by SU(3)c.
#e symmetric product = symmetric rank-2 tensormay be identi!ed with 6:represented by linear combinations oft(11), t(22), t(33), t(12), t(13) and t(23),…which are transformed one into another by SU(3)c.
12
ta = #abg t[bg]
Monday, November 7, 11
Quark-quark interaction
Concrete QCD ComputationsCases of fc(3,4|1,2) to be examinedwhere (1,2) and (3,4) range over:
two copies of the same element of 3*: fc(3*|3*), e.g., [13]|[13];two different elements of 3*: fc(3*′|3*), e.g., [12]|[13];one element of 3* & one of 6: fc(6|3*), e.g., (11)|[12], (33)|[12], (13)|[13] & (12)|[13];two copies of the same element of 6: fc(6|6), e.g., (11)|(11);two different elements of 6: fv(6′|6). e.g., (11)|(33).
#ere are plenty of other choices, but they may all be transformed into one of the eight above, by SU(3)c.It then suffices to work with the above eight representatives.
13
Monday, November 7, 11
Quark-quark interaction
Concrete QCD ComputationsConsider a representative of f(3*|3*):
14
n
1
4
(c†
3g c†
4d)3 (la)a
g (la)bd (ca
1
cb2
)3⇤o
� �
o
� 1
4
1p2
�
d1
gd3
d � d1
d d3
g
�
(la)ag (la)b
d1p2
�
da1
db3
� db1
da3
�
,
1
⇥ a 1 3 a 1 3 a 3 1 a 3 1
⇤
�2
�
��
2
�
�
= 1
8
⇥
la1
1 la3
3 � la3
1 la1
3 � la1
3 la3
1 + la3
3 la1
1
⇤
1
⇥ a 1 3 a 1 3
⇤
8
⇥
� �= 1
4
⇥
la1
1 la3
3 � la3
1 la1
3
⇤
.
1 3
llll1
=h
0 1 0
1 0 0
0 0 0
i
, llll2
=h
0 �i 0
i 0 0
0 0 0
i
, llll3
=h
1 0 0
0 �1 0
0 0 0
i
, llll4
=h
0 0 1
0 0 0
1 0 0
i
,
llll5
=h
0 0 �i0 0 0
i 0 0
i
, llll6
=h
0 0 0
0 0 1
0 1 0
i
, llll7
=h
0 0 0
0 0 �i0 i 0
i
, llll8
= 1p3
h
1 0 0
0 1 0
0 0 �2
i
.
13
= ∂[µAn]∂[µAn] + 2igc
h c ∂[µAn][Aµ
, An]� g2
ch2c2
[Aµ, An][Aµ
, An] (1.115)
quark:
qjn, a n0, b �! idn,n0
dba
/
qj � mjc= idn,n0
dba
/
qj + mjc1
q
2
j � m2
j c2
, (1.116)
gluon: µ, a n, bqg
�! � ihµn
q
2
gdab
(1.117)
1
4pe0
e2
r=
ae h cr
(3⌦3)A = 3⇤ c[a1
cb]2
:= 1p2
(dagd
bd � d
bgda
d ) cg1
cd2
a 6= b, a, b = 1, 2, 3
(3⌦3)S = 6 c(a1
cb)2
:=
⇢
1p2
(dagd
bd + d
bgda
d )
dagd
bd
�
cg1
cd2
n a 6= b,
a = b,
a, b = 1, 2, 3
ta = #abg t[bg]
f (3⇤A|3⇤A) e.g.
1
4
1p2
(d1
gd3
d � d3
gd1
d) (la)a
g(la)bd 1p
2
(da1
db3
� db3
da1
)
f (3⇤ 0A|3⇤A) e.g.
1
4
1p2
(d1
gd2
d � d2
gd1
d) (la)a
g(la)bd 1p
2
(da1
db3
� db3
da1
)
n
1
4
(c†
3g c†
4d)3 (la)a
g (la)bd (ca
1
cb2
)3⇤o
� 1
4
1p2
�
d1
gd3
d � d1
d d3
g
�
(la)ag (la)b
d1p2
�
da1
db3
� db1
da3
�
,
= 1
8
⇥
la1
1 la3
3 � la3
1 la1
3 � la1
3 la3
1 + la3
3 la1
1
⇤
= 1
4
⇥
la1
1 la3
3 � la3
1 la1
3
⇤
.
la3
1 6= 0 6= la1
3
: a = 4, 8
= 1
4
⇥
l8
1
1 l83
3 � l4
3
1 l41
3 � l5
3
1 l5
1
3
⇤
= 1
4
⇥
1p3
·�2p3
� 1·1 � i·(�i)⇤
= � 2
3
.
llll1
=h
0 1 0
1 0 0
0 0 0
i
, llll2
=h
0 �i 0
i 0 0
0 0 0
i
, llll3
=h
1 0 0
0 �1 0
0 0 0
i
, llll4
=h
0 0 1
0 0 0
1 0 0
i
, (1.118)
llll5
=h
0 0 �i0 0 0
i 0 0
i
, llll6
=h
0 0 0
0 0 1
0 1 0
i
, llll7
=h
0 0 0
0 0 �i0 i 0
i
, llll8
= 1p3
h
1 0 0
0 1 0
0 0 �2
i
. (1.119)
13
= ∂[µAn]∂[µAn] + 2igc
h c ∂[µAn][Aµ
, An]� g2
ch2c2
[Aµ, An][Aµ
, An] (1.115)
quark:
qjn, a n0, b �! idn,n0
dba
/
qj � mjc= idn,n0
dba
/
qj + mjc1
q
2
j � m2
j c2
, (1.116)
gluon: µ, a n, bqg
�! � ihµn
q
2
gdab
(1.117)
1
4pe0
e2
r=
ae h cr
(3⌦3)A = 3⇤ c[a1
cb]2
:= 1p2
(dagd
bd � d
bgda
d ) cg1
cd2
a 6= b, a, b = 1, 2, 3
(3⌦3)S = 6 c(a1
cb)2
:=
⇢
1p2
(dagd
bd + d
bgda
d )
dagd
bd
�
cg1
cd2
n a 6= b,
a = b,
a, b = 1, 2, 3
ta = #abg t[bg]
f (3⇤A|3⇤A) e.g.
1
4
1p2
(d1
gd3
d � d3
gd1
d) (la)a
g(la)bd 1p
2
(da1
db3
� db3
da1
)
f (3⇤ 0A|3⇤A) e.g.
1
4
1p2
(d1
gd2
d � d2
gd1
d) (la)a
g(la)bd 1p
2
(da1
db3
� db3
da1
)
n
1
4
(c†
3g c†
4d)3 (la)a
g (la)bd (ca
1
cb2
)3⇤o
� 1
4
1p2
�
d1
gd3
d � d1
d d3
g
�
(la)ag (la)b
d1p2
�
da1
db3
� db1
da3
�
,
= 1
8
⇥
la1
1 la3
3 � la3
1 la1
3 � la1
3 la3
1 + la3
3 la1
1
⇤
= 1
4
⇥
la1
1 la3
3 � la3
1 la1
3
⇤
.
la3
1 6= 0 6= la1
3
: a = 4, 8
= 1
4
⇥
l8
1
1 l83
3 � l4
3
1 l41
3 � l5
3
1 l5
1
3
⇤
= 1
4
⇥
1p3
·�2p3
� 1·1 � i·(�i)⇤
= � 2
3
.
llll1
=h
0 1 0
1 0 0
0 0 0
i
, llll2
=h
0 �i 0
i 0 0
0 0 0
i
, llll3
=h
1 0 0
0 �1 0
0 0 0
i
, llll4
=h
0 0 1
0 0 0
1 0 0
i
, (1.118)
llll5
=h
0 0 �i0 0 0
i 0 0
i
, llll6
=h
0 0 0
0 0 1
0 1 0
i
, llll7
=h
0 0 0
0 0 �i0 i 0
i
, llll8
= 1p3
h
1 0 0
0 1 0
0 0 �2
i
. (1.119)
Monday, November 7, 11
Quark-quark interaction
Concrete QCD ComputationsSo: fc(3*|3*), represented by fc([13]|[13]), = –⅔ (a$ractive!)
Similarly, fc(3*′|3*) represented by fc([12]|[13]),
fc(6|3*), represented by fc((11)|[12]), = 0fc(6|3*), represented by fc((33)|[12]), = 0fc(6|3*), represented by fc((13)|[13]), = 0fc(6|3*), represented by fc((13)|[13]), = 0fc(6|6), represented by fc((11)|(11)), = +⅓ (repulsive!)fc(6′|6), represented by fc((11)|(33)), = 0
15
� 1
4
1p2
�d
1
g
d
2
d
� d
1
d
d
2
g
�(la)
a
g (la)b
d
1p2
�d
a
1
d
b
3
� d
b
1
d
a
3
�,
1 a 2 1
⇤1
⇥ a 1 2 a 1 2
⇤ ��
�
= 1
8
⇥l
a1
1
la3
2 � l
a3
1
la1
2 � l
a1
2
la3
1 + l
a3
2
la1
1
⇤�
� ��⇤
= 1
4
⇥l
a1
1
la3
2 � l
a3
1
la1
2
⇤= 0,
(4.60c)
}— not happening!
— not happening!
— not happening!
Monday, November 7, 11
Quark-quark interaction
Concrete QCD ComputationsTo summarize:#e quark-quark 1-gluon-exchange interaction is
a!ractive when the quarks’ colors are antisymmetrized— and stay in the same particular state,repulsive when the quarks’ colors are symmetrized— and stay in the same particular state,forbidden (verboten) in all other cases.
More-gluons’ exchange interaction does follow this pa$ern.In a baryon, there are three quarks.
For the color of each pair to be antisymmetrized,…the triple color factor has to be fully antisymmetrized.(3⊗3⊗3)A = 1, i.e., (tα tβ tγ)A ∝ εαβγ , which is an SU(3)-invariant.Ψbaryon = [Ψ(space)·χ(spin)·χ("avor)]S · χA(color)
16
Monday, November 7, 11
Quark-antiquark interaction
Concrete QCD ComputationsA 1-gluon exchange:
produces the amplitude
with the color factor
17
u-quarkcolor: a
p1
u-quarkcolor: g
p3
q(la)a
g (lb)db
dab
d antiquarkcolor: anti-b
p2
d antiquarkcolor: anti-d
p4
Mu+d!u+d = � g2
c4q2
[u3
ggggµu1
][v2
ggggµv4
]�
c†
3
lllla c1
��
c†
2
lllla c4
�
,
p is the 4-momentum exchange, and the result differs from the electrody-gcthe color factor, fc(3, 4|1, 2) = 1
4
(c†
3
llllac1
)(c†
2
llllac4
), is inserted.
Monday, November 7, 11
Quark-antiquark interaction
Concrete QCD Computations#e incoming and outgoing quarks may now have the colors in the
color-singlet (SU(3)c-invariant) stateor the (traceless hermitian matrix) color-octet state:
Symbolically:3⊗3* = 1 ⊕ 8tα⊗sβ = [⅓ δαβ (tγsγ)] + [tαsβ – ⅓ δαβ (tγsγ)]
18
, where (c1
c†
2
)ab = da
b ˚cccc is a multiple of the unit matrix,Normalization is again quantum-mechanical, so
246 Chapter 4. Non-Abelian Gauge Symmetries and Interactions
The amplitude of a single-gluon exchange has a contribution from only one Feynmandiagram:
u-quarkcolor: a
p1
u-quarkcolor: g
p3
q(la)a
g (lb)db
dab
d antiquarkcolor: anti-b
p2
d antiquarkcolor: anti-d
p4
(4.65)
Following procedure 4.1, p. 238 and analogously to the result for the first part of (3.146b),we have
Mu+d!u+d = � g2
c4q2
[u3
ggggµu1
][v2
ggggµv4
]�
c†
3
lllla c1
��
c†
2
lllla c4
�
, (4.66)
where q = (p1
�p3
) is the 4-momentum exchange, and the result differs from the electrody-namics one only in that
1. ge is replaced by gc,2. the color factor, fc(3, 4|1, 2) = 1
4
(c†
3
llllac1
)(c†
2
llllac4
), is inserted.
The color factor for the incoming quark-antiquark pair again may belong to one of the twovector spaces:
1. 8, Hermitian octet of states, i.e., the 8-dimensional vector space spanned by the colorfactors
�
c12
ab =
q
1+ 1
2
dab(c
a1
c†
2b � 1p3
dab ˚cccc), a, b = red, yellow, blue = 1, 2, 3
, (4.67)
=n
p
3
2
(da1
d1
b� ˚cccc),p
3
2
(da2
d2
b� ˚cccc),p
3
2
(da3
d3
b� ˚cccc),
(da1
d2
b), (da1
d3
b), (da2
d1
b), (da2
d3
b), (da3
d1
b), (da3
d2
b)o
,
which form a traceless Hermitian matrix, where ˚cccc := 1p3
Tr(c1
c†
2
) = 1p3
(ce1
c†
2
e).2. 1, where (c
1
c†
2
)ab = da
b ˚cccc is a multiple of the unit matrix, i.e., the SU(3)c-invariant.Normalization is again quantum-mechanical, so kc
12
abk2 = 1 for every choice a, b.
Similarly to (4.59), for u + d ! u + d we have
fc(3, 4|1, 2) = fc(8|8), fc(80|8), fc(8|1), fc(1|8), fc(1|1). (4.68)
Also, just as in electrodynamics, the gluon exchange gives rise to a potential of the form
Vqq(r) = � fcac h c
r, (4.69)
where the sign is now negative, since the color charges of a quark and an antiquark are“opposite”: one is the (chromodynamics) “color” the other the “anti-color”14.14 In electrodynamics, of course, there is only one kind of charge—electric—and the opposite charge is simply
the negative charge. For chromodynamics colors, “anti-color” is not simply negative “color”, but the opposite“color”—i.e., the color which with the original one together produces a colorless, i.e., an SU(3)c-invariantwhole. This we may write, e.g., (ca (red))
† = (c†)a (green). We will not use this notational possibility, as itadditionally complicates the tensor algebra rules and necessitates printing in color; with the current convention,computations may be followed even in monochromatic printout.
DR
AFT
—co
ntac
tdir
ectly
Tris
tan
Hub
sch,
thub
sch@
how
ard.
edu,
with
any
com
men
ts/
sugg
estio
ns/
corr
ectio
ns;t
hank
you!
—D
RA
FT
Monday, November 7, 11
Quark-antiquark interaction
Concrete QCD ComputationsSince the color charge of an antiquark is oppositeof the color of the corresponding quark,…the 1-gluon exchange gives rise to the potential
Need to compute fc(3,4|1,2) for:fc(8|8), represented by fc(13|13),fc(8′|8), represented by fc(31|13),fc(8|1), represented by fc(13|1),fc(1|1), represented by fc(1|1).
Proceed as before:
19
Vqq(r) = � fcac h c
r,
––
7!
� 1
4
(d1
gdd3
) (la)ag (la)d
b (da1
d3
b),
�
= 1
4
la1
1 la3
3 = 1
4
l8
1
1 l83
3 = 1
4
1p3
�2p3
= � 1
6
,
Monday, November 7, 11
a$ractive!
Quark-antiquark interaction
Concrete QCD ComputationsObtain:
fc(8|8), represented by fc(13|13), = –⅙fc(8′|8), represented by fc(31|13), = 0fc(8|1), represented by fc(13|1), = 0fc(1|1), represented by fc(1|1):
#e quark-antiquark 1-gluon exchange potential is:a!ractive for in- and out-state color-singlets,repulsive for in- and out-state (same!) color octets,forbidden (verboten) otherwise.
20
1
4
(c†
3g cd4
)1 (la)a
g (la)db (ca
1
c†
2b)11 1
1 2 3 a= 1
4
1p3
(d1
gdd1
+ d2
gdd2
+ d3
gdd3
) (la)ag (la)d
b1p3
(da1
d1
b + da2
d2
b + da3
d3
b),
1
a1
a b1
ab1 4
3
= 1
12
laa
g lagaa = 1
12
dab Tr(lllla llllb) =
where we used the relation (A.73). This coefficient,
4.2. Concrete Calculations 247
Example 4.3: To compute the functions fc(8|8), fc(80|8) and fc(1|1), we pick the simplest partic-ular cases for each; the diligent Student will convince themselves by direct computation that allcases give the same quantitative results.
For fc(8|8), the incoming and the outgoing quark-antiquark pair have the same combina-tion of color-anticolor; take, e.g., red-antiblue (8 7! (d1
gdd3
)) combination:�
1
4
(c†
3g cd4
)8 (la)a
g (la)db (ca
1
c†
2b)8
� 1
4
(d1
gdd3
) (la)ag (la)d
b (da1
d3
b),
= 1
4
la1
1 la3
3 = 1
4
l8
1
1 l83
3 = 1
4
1p3
�2p3
= � 1
6
, (4.70)
since only the eighth Gell-Mann matrix has (la1
1 6= 0 6= la3
3). For fc(80|8) we take, e.g.,8 7! (d1
gdd3
) and 80 7! (d3
gdd1
):�
1
4
(c†
3g cd4
)80 (la)a
g (la)db (ca
1
c†
2b)8
� 1
4
(d3
gdd1
) (la)ag (la)d
b (da1
d3
b),
= 1
4
la1
3 la1
3 = 1
4
(l4
1
3 l41
3 + l5
1
3 l5
1
3) = 1
4
(1·1 + (�i)·(�i)) = 0, (4.71)
Since the representation 1 has only one dimension, for fc(1|1) there is a single contribution:
1
4
(c†
3g cd4
)1 (la)a
g (la)db (ca
1
c†
2b)1
= 1
4
1p3
(d1
gdd1
+ d2
gdd2
+ d3
gdd3
) (la)ag (la)d
b1p3
(da1
d1
b + da2
d2
b + da3
d3
b),
= 1
12
laa
g laga = 1
12
dab Tr(lllla llllb) = 1
12
dab 2dab = 1
6
8 = 4
3
, (4.72)
where we used the relation (A.73). This coefficient, fc(1|1) = 4
/
3
, has shown up in the rela-tion (2.102).
Direct computation shows also that fc(8|1), fc(1|8) = 0, and we have that:
Conclusion 4.6 These results show the single-gluon exchange15 between a quark andan antiquark preserves the color state: incoming and outgoing quark-antiquark pairshave the same color combination. Besides, the chromodynamics force (4.69) betweena quark and an antiquark is:
1. attractive when both the incoming and the outgoing pair is in the SU(3)c-invariant state, and
2. repulsive otherwise.
4.2.3 Quark-Antiquark Annihilation
The single-gluon exchange amplitude now has two contributions, corresponding to the twoFeynman diagrams:
Mu+u!u+u =
1
3
2
4
�
1
3
2
4
(4.73)
15 See footnote 13, p. 244.
DR
AFT
—co
ntac
tdir
ectly
Tris
tan
Hub
sch,
thub
sch@
how
ard.
edu,
with
any
com
men
ts/
sugg
estio
ns/
corr
ectio
ns;t
hank
you!
—D
RA
FT
4.2. Concrete Calculations 247
Example 4.3: To compute the functions fc(8|8), fc(80|8) and fc(1|1), we pick the simplest partic-ular cases for each; the diligent Student will convince themselves by direct computation that allcases give the same quantitative results.
For fc(8|8), the incoming and the outgoing quark-antiquark pair have the same combina-tion of color-anticolor; take, e.g., red-antiblue (8 7! (d1
gdd3
)) combination:�
1
4
(c†
3g cd4
)8 (la)a
g (la)db (ca
1
c†
2b)8
� 1
4
(d1
gdd3
) (la)ag (la)d
b (da1
d3
b),
= 1
4
la1
1 la3
3 = 1
4
l8
1
1 l83
3 = 1
4
1p3
�2p3
= � 1
6
, (4.70)
since only the eighth Gell-Mann matrix has (la1
1 6= 0 6= la3
3). For fc(80|8) we take, e.g.,8 7! (d1
gdd3
) and 80 7! (d3
gdd1
):�
1
4
(c†
3g cd4
)80 (la)a
g (la)db (ca
1
c†
2b)8
� 1
4
(d3
gdd1
) (la)ag (la)d
b (da1
d3
b),
= 1
4
la1
3 la1
3 = 1
4
(l4
1
3 l41
3 + l5
1
3 l5
1
3) = 1
4
(1·1 + (�i)·(�i)) = 0, (4.71)
Since the representation 1 has only one dimension, for fc(1|1) there is a single contribution:
1
4
(c†
3g cd4
)1 (la)a
g (la)db (ca
1
c†
2b)1
= 1
4
1p3
(d1
gdd1
+ d2
gdd2
+ d3
gdd3
) (la)ag (la)d
b1p3
(da1
d1
b + da2
d2
b + da3
d3
b),
= 1
12
laa
g laga = 1
12
dab Tr(lllla llllb) = 1
12
dab 2dab = 1
6
8 = 4
3
, (4.72)
where we used the relation (A.73). This coefficient, fc(1|1) = 4
/
3
, has shown up in the rela-tion (2.102).
Direct computation shows also that fc(8|1), fc(1|8) = 0, and we have that:
Conclusion 4.6 These results show the single-gluon exchange15 between a quark andan antiquark preserves the color state: incoming and outgoing quark-antiquark pairshave the same color combination. Besides, the chromodynamics force (4.69) betweena quark and an antiquark is:
1. attractive when both the incoming and the outgoing pair is in the SU(3)c-invariant state, and
2. repulsive otherwise.
4.2.3 Quark-Antiquark Annihilation
The single-gluon exchange amplitude now has two contributions, corresponding to the twoFeynman diagrams:
Mu+u!u+u =
1
3
2
4
�
1
3
2
4
(4.73)
15 See footnote 13, p. 244.
DR
AFT
—co
ntac
tdir
ectly
Tris
tan
Hub
sch,
thub
sch@
how
ard.
edu,
with
any
com
men
ts/
sugg
estio
ns/
corr
ectio
ns;t
hank
you!
—D
RA
FT
Vqq(r) = � fcac h c
r,
Mesons must be SU(3)c-invariant.
— repulsive!
Monday, November 7, 11
Concrete QCD ComputationsHow about the possible (virtual) annihilation + re-creation?
21
Quark-antiquark interaction
Mu+u!u+u =
1
3
2
4
�
1
3
2
4
Mu+u!u+u = � g2
c4(p
1
� p3
)2
[u3
ggggµu1
][v2
ggggµv4
](c†
3
llllac1
)(c†
2
llllac4
)
+g2
c4(p
1
+ p2
)2
[v2
ggggµu1
][u3
ggggµv4
](c†
2
llllac1
)(c†
3
llllac4
),
Monday, November 7, 11
Concrete QCD ComputationsHow about the possible (virtual) annihilation + re-creation?#e color factors are now:
fc(8|8):
fc(8′|8):
fc(1|1):
22
Quark-antiquark interaction
e
�
1
4
(c†
3g cd4
)8 (la)a
b (la)dg (ca
1
c†
2b)8
� 1
4
(d1
gdd3
) (la)ab (la)d
g (da1
d3
b),
= 1
4
la1
3 la3
1 = 1
4
(l4
1
3 l4
3
1 + l5
1
3 l53
1) = 1
4
(1·1 + (�i)·(i)) = 1
2
,
e
�
1
4
(c†
3g cd4
)80 (la)a
b (la)dg (ca
1
c†
2b)8
� 1
4
(d3
gdd1
) (la)ab (la)d
g (da1
d3
b),
= 1
4
la1
3 la1
3 = 1
4
(l4
1
3 l41
3 + l5
1
3 l5
1
3) = 1
4
(1·1 + (�i)·(�i)) = 0,
e
1
4
(c†
3g cd4
)1 (la)a
b (la)dg (ca
1
c†
2b)1
= 1
4
1p3
(d1
gdd1
+ d2
gdd2
+ d3
gdd3
) (la)ab (la)d
g1p3
(da1
d1
b + da2
d2
b + da3
d3
b),
= 1
12
laa
a lagg = 1
12
Tr(lllla)Tr(lllla) = 0,
Monday, November 7, 11
Concrete QCD ComputationsHow about the possible (virtual) annihilation + re-creation?#e algebraic sum (actually difference) of the two amplitudes is
An SU(3)c-invariant quark-antiquark pair cannot decay into a single gluon—even virtually—by color-conservation.Similarly, (SU(3)c-invariant) hadrons can neither emit nor absorb a single gluon—by color-conservation.All hadron-hadron interaction must be mediated by SU(3)c-invariant objects: (n≥2)-gluons and/or quark-antiquark pairs.
23
Quark-antiquark interaction
Mu+u!u+u = � g2
c(p
1
� p
3
)2
⇢
� 1
6
+ 4
3
�
[u3
ggggµu1
][v2
ggggµv4
]
+g2
c(p
1
+ p
2
)2
n
1
2
0
o
[v2
ggggµu1
][u3
ggggµv4
], if
⇢
cccc12
⇢ 8,
cccc12
= 1.
Monday, November 7, 11
Concrete QCD ComputationsSo, in a n0 + π–→ n0 + π– sca$ering, 1-gluon exchange could happen as follows:
…except that two SU(3)c-invariant hadrons cannot exchange an SU(3)c-variant gluon and stay SU(3)c-invariant.So, the processes depicted in (a) must additionally involve an exchange of at least one more gluon, or a d-quark…
24
Quark-antiquark interaction
(a) (b)
(d,d,u)
(d,d,u)
ud
ud
3⇥ 2 = 6
combinationsd
d u
dd u
ud
ud
Monday, November 7, 11
Concrete QCD ComputationsSo, in a n0 + π–→ n0 + π– sca$ering, 1-gluon exchange will include
…which is still O(gc2), but is signi!cantly complicated by the d-quark exchange. #e mediating particle effectively becomes another hadron (π0, or its P-wave excitation, ρ0, or…).
25
Quark-antiquark interaction
(d,d,u)
(d,d,u)
ud
ud
2⇥ 3⇥ 2 = 12
combinations
Monday, November 7, 11
Concrete QCD ComputationsGenerally speaking,the QCD interactions must proceed so as to…not change the color-invariance of the hadrons involved…nor any other (real) intermediate state.
26
QCD QCD QCD
Conclusions
Monday, November 7, 11
Conclusions
Concrete QCD ComputationsQCD interactions favor color-antisymmetrization:In a baryon, the three quarks a$ract each other by way of QCD precisely if they form an SU(3)c-invariant state.
#at is the color factor must be totally antisymmetric.In a meson, the quark-antiquark pair a$ract each other by way of QCD precisely if they form an SU(3)c-invariant state.Two SU(3)c-invariant hadrons cannot exchange an SU(3)c-variant gluon and stay SU(3)c-invariant.#us, two hadrons can interact only by exchanging
SU(3)c-invariant objects, consisting of 2 or more of…gluons and/or quark-antiquark pairs.#e hadron-hadron force is thus a “remnant” (à la van der Waals).
27
Monday, November 7, 11
Conclusions
Concrete QCD Computations#e 1-gluon exchange produces a reasonable qualitative statement (antisymmetrization ⇔ a$raction).But, it is indicative of:
neither large-distance (≥10–15m) con"nementnor short-distance (≪ 10–15m) asymptotic #eedom (next time)
Con!nement is a large-distance featureakin to the Coulomb (static) !eld in EM…formed as a condensate of inde!nitely many quanta…essentially a non-perturbative phenomenon
Asymptotic freedom is a perturbative result1973, David Gross & Frank Wiczek, & David Politzer…a year before the “November (1974) revolution.”
28
Monday, November 7, 11
Thanks!
Tristan HubschDepartment of Physics and AstronomyHoward University, Washington DC
Prirodno-Matematički FakultetUniverzitet u Novom Sadu
http://homepage.mac.com/thubsch/
Monday, November 7, 11