Post on 03-Apr-2018
(Fundamental) Physicsof Elementary Particles
Non-abelian gauge symmetry; QCD Lagrangian, color-conservation law and equations of motion
Tristan HübschDepartment of Physics and Astronomy
Howard University, Washington DCPrirodno-Matematički Fakultet
Univerzitet u Novom Sadu
Tuesday, November 1, 11
Program
Fundamental Physics of Elementary Particles
2
A gauge (local) symmetry principle recap!e partial derivative vs. the gauge-covariant derivativeGeneral transformation “rules”
!e SU(3)c transformationsColor as a 3-dimensional chargeMatrix-valued phases and local symmetryMatrix representations of SU(3)
!e SU(3)c-invariant Lagrangian!e curvature tensor and the Bianchi identityEquations of motionColor conservation and equation of continuity
Tuesday, November 1, 11
external
internal
Partial vs. gauge-covariant derivatives
Gauge (Local) Symmetry Principle
First and foremost:!e mathematical object, Ψ(r, t), used to represent a particledepends on (is a function of)
the position in space and time,the phase (as a complex function),additional degrees of freedom
spinisospincolor…
Gauge (local symmetry) principle:internal coordinates are free to depend on external ones.
3
…all of which are “coordinates,” ina broad sense ofthe word.}
base
!ber
!ber bundles/sheaves/…
Tuesday, November 1, 11
connexion 1-form
Gauge (Local) Symmetry Principle
Derivatives compute the rate of changeSo, if Ψ(r, t) also depends on a (r, t)-dependent phase,then the rate of change stems from
varying Ψ(r, t) explicitly, andvarying Ψ(r, t) implicitly, via varying its phase.
If “gauge” refers to “#xing” that phase,…then a “gauge-covariant” derivative…must contain two terms: Dμ = ∂μ + Aμ(r, t) ,where Aμ(r, t) is the gauge potential, a.k.a. connexion.Mathematicians: dxμDμ = dxμ∂μ + dxμAμ(r, t)
is now the (external) coordinate-independent de!nition.
4
Partial vs. gauge-covariant derivatives
Tuesday, November 1, 11
General transformation “rules”
Gauge (Local) Symmetry Principle
!e unitary gauge (local symmetry) transformation is of the form Uφ = exp{i φ·Q}, (Uφ† = Uφ–1)
where φ is the (array of) gauge parameter(s),where Q is the (array of) gauge transformation generator(s).
!en,Ψ(r, t) → Uφ Ψ(r, t);Ψ†(r, t) → Ψ†(r, t) Uφ–1;O(r, t) → Uφ O(r, t) Uφ–1;…and therefore also: Dμ → Uφ Dμ Uφ–1.
us, ∂μ + Aμ(r, t) → Uφ (∂μ + Aμ(r, t)) Uφ–1 implies that Aμ(r, t) → Uφ (–i (∂μφ) + Aμ(r, t)) Uφ–1 .
5
Tuesday, November 1, 11
Color as a 3-dimensional charge
The SU(3)c TransformationsRecall:
Δ++ = (uuu),Δ– = (ddd),Ω– = (sss).
It follows that:either quarks are not fermions (O.W. Greenberg, 1964),
…or…6
} Spin-³/₂ baryonsS-states; no orbital angular momentumSpatially symmetric wave-functions
⇥bi , b †
j⇤= dij,
⇥bi , b j
⇤= 0 = [b †
i , b †
j ], bosons,�†
� † †
�˜fi,a ,
˜f †
j,a = dij,
�˜fi,a ,
˜f j,a = 0 =
�˜f †
i,a ,
˜f †
j,a
,⇥˜fi,a ,
˜f †
j,b⇤= dij,
⇥˜fi,a ,
˜f j,b⇤= 0 =
⇥˜f †
i,a ,
˜f †
j,a⇤, a 6= b,
)para-fermions,
⇥ ⇤ ⇥ ⇤�fi , f †
j = dij,
�fi , f j
= 0 =
�f †
i , f †
j
, fermions,
Tuesday, November 1, 11
The SU(3)c TransformationsQuarks are fermions,…but have an additional degree of freedom.
January 1965: Boris V. Struminsky, Dubna (Moscow, Russia)…then with N. Bogolyubov + Albert TavchelidzeMay 1965, A. Tavchelidze: ICTP, Trieste (Italy)December 1965, Moo-Young Han + Yoichiro Nambu
integrally charged, colored quarks + 8 (color-anticolor) gluonsFinal version (w/fractionally charged quarks):1974, William Bardeen, Harald Fritzsch & Murray Gell-MannQuark: Ψn
αA(r, t), where:n = u, d, s, c, … indicates the $avorα = red, blue, yellow indicates the “color”A = 1, 2, 3, 4 indicates the component of the Dirac spinor
P.S.: Greenberg subsequently proved equivalence…7
Color as a 3-dimensional charge
meson baryon
Tuesday, November 1, 11
Matrix-valued phases and local symmetry
The SU(3)c TransformationsWithout spelling out the Dirac components,
…where n = u, d, s, c, b, t indicates the “&avor.”Arranging the colors in a matrix format,
the quark wave-function phase becomes 3×3 matrix-valued,as does the unitary phase-transformation operator Uφ.
where Qa are 3×3 matricesHermitian, so Uφ would be unitary,traceless, so Uφ would be unimodular.
8
Yn(x) ! eigcjj
j
j
(x)/h̄ Yn(x), jj
j
j
(x) := j
a(x)Qa,
Diagonal phase-transformationpertains to electromagnetism…
Yn(x) = ˆe
a
Ya
n(x) = ˆerYrn(x) + ˆeyYy
n(x) + ˆebYbn(x) =x) =
"Yr
n(x)Yy
n(x)Yb
n(x)
#,
Tuesday, November 1, 11
The SU(3)c TransformationsGauge (local symmetry) transformations
Notice the multi-component-ness:
…which is usually suppressed in notation.In general:
where however the form of Qa depends on what it acts upon.9
Matrix-valued phases and local symmetry
[ih̄ /D � mc]Yn(x) = 0 ! [ih̄ /D0 � mc]Y0n(x) = 0,
0 �1
! �D
µ
! D0µ
:= Ujj
j
j
Dµ
U�1
jj
j
j
,
ig /h̄U
jj
j
j
:= eigcjj
j
j
/h̄,
/DYn ⌘ gg
g
g
µDµ
Yn, ( /DYn)a ⌘ gg
g
g
µDµ
a
b
Yb
n, ( /DYn)aA ⌘ (gµ)A
BDµ
a
b
YbBn ,
Dµ
:= 1 ∂
µ
+ igch̄ c Aa
µ
Qa,
a a 1 h̄ c
Tuesday, November 1, 11
a,b,c = 1,…,8
Matrix representations of SU(3)
The SU(3)c TransformationsAs obtained in the “general” formalism:
…where
…and where
are the 3×3 matrices that act upon the (quark) color 3-vector.But, what about the Qa’s acting on the 8 Aμ
a’s or 8 φa’s?
10
h̄ c
A0aµ
Qa = Aaµ
Ujj
j
j
QaU�1
jj
j
j
+ h̄ cigc
Ujj
j
j
(∂µ
U�1
jj
j
j
) = Aaµ
Ujj
j
j
QaU�1
jj
j
j
� c(∂µ
j
a)Qa,
i.e. 0 �1 : a
A0µ
= Ujj
j
j
Aµ
U�1
jj
j
j
� c(∂µ
jj
j
j
), Aµ
:= Aaµ
Qa,
[Qa, Qb] = i fabc Qc.
d Aaµ
= �(Dµ
jj
j
j
)a := �c(∂µ
j
a) + igch̄ c Ab
µ
( eQb)ca
j
c =a a ac = �(∂
µ
j
a)� gch̄ c Ab
µ
fbca
j
c,
.
µ
�( eQb)c
a = i fbca,
Tuesday, November 1, 11
nonlinear
The curvature tensor and the Bianchi identity
The SU(3)c-invariant Lagrangian
Notice the differences: D′μ = Uφ Dμ Uφ–1 implies
Also,
But, for the non-abelian (matrix-valued) case:
Recall however that
11
for electromagnetism,
Fµn
(A0) = Fµn
(A),
�∂
µ
(A0)an
� ∂
n
(A0)aµ
�6= (∂
µ
Aan
� ∂
n
Aaµ
),
� ��∂
µ
(A0)an
� ∂
n
(A0)aµ
�6= U
jj
j
j
(∂µ
Aan
� ∂
n
Aaµ
)U�1
jj
j
j
.
Note, however, that both in electrodynamics and in chromodynamics the derivatives: D0
µ
= Ujj
j
j
Dµ
U�1
jj
j
j
,
) A0µ
= Aµ
� (∂µ
j)
) (A0)aµ
= Aaµ
� (Dµ
j
a) = Aaµ
� (∂µ
j
a) + gch̄ c Ab
µ
fbca
j
c.
Tuesday, November 1, 11
The curvature tensor and the Bianchi identity
The SU(3)c-invariant Lagrangian
In electrodynamics:
It must be a commutator, so that the result would not be a differential operator, but an “ordinary” function.A commutator also computes the mismatch in… well,…commuting.In general: [D, D] = (torsion)·D + (curvature).So:
12
⇥D
µ
, Dn
⇤=
⇥∂
µ
+ iqh̄ c A
µ
, ∂
n
+ iqh̄ c A
n
⇤= + iq
h̄ c (∂µ
An
� ∂
n
Aµ
) = iqh̄ c F
µn
.
Fµn
:= h̄ cigc
⇥D
µ
, Dn
⇤= h̄ c
igc
⇥∂
µ
+ igch̄ c Ab
µ
Qb , ∂
n
+ igch̄ c Ac
n
Qc⇤,
ig
⇥ ⇤ ⇥ ⇤= (∂
µ
Aan
� ∂
n
Aaµ
)Qa + h̄ cigc
( igch̄ c )
2 Abµ
Acn
[Qb, Qc] = Faµn
Qa,
a a g a b c
�
Faµn
:= (∂µ
Aan
� ∂
n
Aaµ
)� gch̄ c f a
bc Abµ
Acn
,
Tuesday, November 1, 11
Curvature and torsion
DigressionPicture the result of computing [D, D′] = DD′–D′D…in the total space of a #ber bundle:
13
#ber
base
base
D′
D
D D′
D′base
Dbase
Dbase
D′base
torsion
curvature
torsion
Generally, the rate-of-change operators (D) will fail to commute both in the#ber-space direction(= curvature: at the same “place” but of a different “value”) and in thebase-space direction(= torsion: not even at the same “place”).
Tuesday, November 1, 11
The curvature tensor and the Bianchi identity
The SU(3)c-invariant Lagrangian
Of course, we have that
Independently,
And, for all SU(n):
14
Fµn
! F0µn
:= ih̄ cgc
⇥D0
µ
, D0n
⇤= ih̄ c
gc
⇥U
jj
j
j
Dµ
U�1
jj
j
j
, Ujj
j
j
Dn
U�1
jj
j
j
⇤1 ⇤
= ih̄ cgc
Ujj
j
j
⇥D
µ
, Dn
⇤U�1
jj
j
j
,
c
⇥= U
jj
j
j
Fµn
U�1
jj
j
j
.
Dµ
(Fnr
) = [Dµ
, Fµn
] = h̄ cigc
⇥D
µ
, [Dn
, Dr
]⇤.⇥
A , [ B , C ]⇤+
⇥B , [C , A ]
⇤+
⇥C , [ A , B ]
⇤⌘ 0,
#
µnrsDµ
(Fnr
) = h̄ cigc
#
µnrs
⇥D
µ
, [Dn
, Dr
]⇤= 0,
Tr[Fµn
] = Fkµn
Tr[Qk] = 0.
Tuesday, November 1, 11
Equations of motion
The SU(3)c-invariant Lagrangian
Since the matrix-valued (su(3) algebra-valued) curvaturetransforms by similarity transformation,…with respect to which the trace function is invariant,
…so one chooses:
15
Tr[Fµn
Fµn] ! Tr[F0µn
F0 µn] = Tr[Ujj
j
j
Fµn
U�1
jj
j
j
Ujj
j
j
FµnU�1
jj
j
j
] =
Tr
µn] = Tr[Fµn
FµnU�1
jj
j
j
Ujj
j
j
],
(4.22)= Tr[Fµn
Fµn]
LQCD = Ân
Tr
⇥
Yn(x) [ih̄ c/
D�mnc2]Yn(x)⇤
� 1
4
Tr
⇥
FµnFµn⇤,
h
� �
i
⇥ ⇤ ⇥ ⇤
= Ân
Ya n(x)h
iggggµ�h̄ cdab∂µ+igc Aa
µ( 1
2
la)a
b�
�mnc2dab
i
Ybn(x)
� 1
4
FaµnFµn
a .
Tuesday, November 1, 11
The SU(3)c-invariant Lagrangian
Variation by Aaμ yields:
But, while
implies
the same is not true of Dμ Fa μν.Instead:
16
Equations of motion
(DµFµn = ∂µFµn) = geYA(gn)A
BYA,
∂
n
jn
e =4pe
0
c4p
∂
n
∂
µ
Fµn ⌘ 0, since Fµn
= �Fnµ
Dn ja n(q) = Dn Dµ Fa µn = � 1
2
[Dµ, Dn]Fa µn = � 1
2
f abcFb
µnFc µn = 0,
Dµ
Fa µn = gc Ân
YnaA(gn)A
B( 1
2
l
a)a
b
YbBn ,
ja µ
(q) := gc Ân
YnaA(gµ)A
B( 1
2
l
a)a
b
YbBn .
However, the left-hand side of Eq. (4.25) contains terms non-linear in
Tuesday, November 1, 11
Color conservation and equation of continuity
The SU(3)c-invariant Lagrangian
!is does not lead to a conserved color:
…as the additional right-hand side term doesn’t vanish.However, use that
…so both quarks and gluons contribute to the color charge:
17
0 = Dµ ja µ(q) = ∂µ ja µ
(q) �gch̄ c f a
bc Abµ jc µ
(q) ,
) d
dt
⇣
Z
Vd
3~r ja 0
(q)
⌘
= �I
∂Vd
2~r ·~ja(q) +
gch̄ c f a
bc
⇣
Z
Vd
3~r Abµ jc µ
(q)
⌘
,
Dµ
Fa µn = ∂
µ
Fa µn � gch̄ c fbc
a Abµ
Fc µn
,
Dµ
Fa µn = ja n
(q) ) ∂
µ
Fa µn = Ja n
(c) , ) ∂
n
Ja n
(c) = 0,
Ja n
(c) := ja µ
(q) +gch̄ c fbc
a Abµ
Fc µn
,Z ZQa
(c) :=Z
d
3~r Ja 0
(c) = gc
Zd
3~r�Ân[Yn gg
g
g
µ
1
2
ll
l
l
a Yn] + 1
h̄ c fbca Ab
µ
Fc µn
�,
Tuesday, November 1, 11
Color conservation and equation of continuity
The SU(3)c-invariant Lagrangian
!is changes the analogues of Gauss-Ampère laws.Consider the ν = 0 case of the equation Dμ Fa μν = j(q)
aν:
…and de#ne:
!en,
andit is impossible to write analogues of Maxwell’s equationswith no reference to the gauge potentialsboth quarks and gluons serve as “sources” for the color force-#eldthe equations are nonlinear.
18
∂
µ
Fa µ0 � gch̄ c f a
bc Abµ
Fc µ0 = ja 0
(q) ,
~Ea := ˆeiFa i0, r
a(q) := ja 0
(q) ,
~Aa := �ˆe
i Aai ,
~r·~Ea = r
a(q) �
gch̄ c f a
bc ~Ab·~Ec,
Tuesday, November 1, 11
Color conservation and equation of continuity
The SU(3)c-invariant Lagrangian
To sum up:
are the matrix-valued analogues ofthe Maxwell’s equationsthe conserved color-current & color-charge.
19
Dµ
Fµn = Jn
(q) and #
µnrsDµ
(Fnr
) = 0,
∂
µ
Fµn = Jn
(c),
RJn
(c) := Jn
(q) +igch̄ c [Aµ
, Fµn],
∂
n
Jn
(c) = 0,
d
dt
ZV
d
3~r J0
(c) = �I
∂Vd
2~s ·~J(c).
It should be clear that the application to an abelian (commutative) group where
Jn
(q) := gc
⇣Ân
YnaA(gµ)A
B( 1
2
l
a)a
b
YaAn
⌘Qa
Tuesday, November 1, 11
Thanks!
Tristan HubschDepartment of Physics and Astronomy
Howard University, Washington DCPrirodno-Matematički Fakultet
Univerzitet u Novom Sadu
http://homepage.mac.com/thubsch/
Tuesday, November 1, 11