QFT 3 : Problem Set 2
1.) Peskin & Schroeder 16.1 Arnowitt-Fickler gauge
In this problem we are supposed to look at the Faddeev-Popov (FP) quantization of Yang-Mills (YM)
theory in the Arnowitt-Fickler gauge:
LYM
A
3a
1 a 2
= − (Fµν
) + ψ(i6D − m)ψ
4
= 0
(1)
(2)
Let us consider a 4-vector ηµ so as to write our Gauge condition in a covariant way:
ηµ
≡ (0, 0, 0, 1)
(3)
µa
= 0
(4)
ηµ A
In order to work out the FP quantization of the YM theory, we consider the Lagrangian for the gauge
field and a generalized gauge condition:
1 a 2
L = − (Fµν
)
4
µa
G(A) = ηµ A (x) − ω(x)
(5)
(6)
Following the FP quantization process then we obtain the following Lagrangian:
LF P
1
1 µ a 2
= LYM − c ( η µ Dµ ) c −
(η Aµ )
g
2ζ
(7)
The equation-of-motion for the ghost field c is:
η µ Dµ c = 0
⇒ η µ ∂µ c = 0
(8)
(9)
⇒ ∂3 c = 0
(10)
where we have used the gauge condition A3a = 0 and that η µ = (0, 0, 0, 1). Evidently the ghost field
doesn’t have a propagating solution. The gauge condition A3a = 0 restricts the polarization vectors to
be perpendicular to the z-axis. Using the residual gauge degree of freedom of Aµa (a gauge parameter
independent of z, but dependent on the other variables) one can remove one more polarization. This
reduces the gauge field to two positive metric degrees of freedom.
Feynman Rules: Notice that the ghost field is completely decoupled as we dont have any propagating
ghost solutions. The Feynman Rules are same as normal for Yang-Mills theory in any other covariant gauge except for the gauge boson propagator which gets contributions from the extra term in the
Lagrangian. The gauge boson propagator takes the form:
kµ ην + kν ηµ
iδ ab
kµ kν
−
− 2 g µν −
(11)
k
(η · k)2
η·k
2.) Peskin & Schroeder 16.2 Scalar field with non-Abelian charge.
(a.)
The Lagrangian for the non-Abelian gauge theory coupled to a scalar field may be written as:
1
1 a 2
) − (∂ µ Aaµ )2 − ca ∂ µ Dµac cc + Dµ φ∗i Dµ φi − m2 φ∗i φi
L = − (Fµν
4
2ζ
1
(12)
One can read off the Feynman Rules for the scalar field from the above Lagrangian. The scalar field has
the usual propagator. Also there are couplings of 2 scalars (momenta p and p 0 ) + a gauge boson, and a
4-point coupling (2 scalars + 2 gauge bosons):
P ropagator
:
3 − P oint V ertex :
4 − P oint V ertex :
iδij
p2 − m2 + i
ig(p + p0 )µ taij
ig 2 g µν ta , tb jk
(13)
(14)
(15)
Evidently there are minor modifications over the Abelian results displayed in P &S 9.1.
(b.)
The standard way to obtain the (1-loop) β-function is to calculate the counter-terms at one-loop level.
The corrections to the gauge-boson propagator come from 5-diagrams (scalar, ghost and gauge loops;
scalar and gauge bubbles.) Since we are interested in UV-divergences we may take the limit in which
the scalar mass goes to zero. One obtains the following counter-term upon evaluating the one loop level
correction to the gauge boson propagator:
δA = g 2
5 C2 (G) − C(r)
24π 2 (16)
Next we need to find corrections to the scalar field propagator. Only 2 diagrams (gauge rainbow and a
gauge bubble) contribute. We introduce a mass term for the gauge field and take limit of vanishing gauge
boson mass. We obtain the counter-term for the scalar propagator:
δφ =
g 2 C2 (r)
4π 2 (17)
Finally we need the corrections to the 3-point vertex (2 scalars and a gauge boson) which is needed for
one loop corrections to the coupling constant. Once again there are 5 diagrams (2 triangular diagrams
with 2 gauge bosons and a scalar or 2 scalars and a gauge boson forming the 3 sides of the triangle; 2
diagrams with a gauge rainbow on a scalar external leg with the rainbow ending at the vertex; a gauge
boson tadpole with 2 external scalar legs on the gauge boson loop.) Evaluating these diagrams one finds
the counter-term for the 3-point vertex:
δg = −
g 3 µ/2
(C2 (G) − 2 C2 (r))
8π 2 (18)
Finally we use logg0 = (1 + δA )−1/2 (1 + δφ)−1 (gµ/2 + δg ) where g0 being the bare coupling is a constant.
This gives us the β-function:
β(g)
∂g
∂(logµ)
g3
1
11
C2 (G) − C(r)
= −
(4π)2 3
3
=
(19)
(20)
Notice that the only contribution from the scalar field to the β-function comes from the correction to the
gauge boson propagator. The correction (due to the scalar field) to the scalar propagator and the vertex
function cancel each other and hence don’t show up in the final β-function.
If we work in the background field gauge (as discussed in class), our work is reduced to computing the
1
corrections to δA (from which we get δg by Zg ZA2 = 1: Zg = 1 + δg defines the running of g). [Background
field method beyond 1-loop reference: L.F.Abbott, Nucl. Phys. B185, 189(1981)]
2
Alternatively, one could also evaluated functional determinants directly to read off β(g) directly, as
in the text (the background field gauge is used here) :
1
nf
3
β(g) = g
cG,1 − cG,0 −
cr,1/2
(21)
2
2
with
cr,0
=
cr,1/2
=
cr,1
=
1
C(r)
(+ )
2
(4π)
3
C(r)
8
(− )
2
(4π)
3
C(r)
20
(− )
2
(4π)
3
(22)
(23)
(24)
(25)
3.) QCD β-function from supersymmetric gauge theory
a)
We note the following results :
σ2 ~σ σ2
∗
iγ 2 ψ1,2
=
=
ta = − (ta )∗
−~σ ∗ = −~σ T
ψ1,2
(26)
= − (ta )T
; real representation
The Dirac fermion kinetic term may be written as
† µ
† µ
ψi6Dψ = iψR
σ Dµ ψR + iψL
σ Dµ ψ L
(27)
The two Majorana fermion terms may be written as
ψ 1 i6Dψ1
ψ 2 i6Dψ2
=
† µ
T
∗
iψL
σ Dµ ψL + iψL
σ2 σ µ Dµ σ2 ψ L
(28)
=
† µ
iψR
σ Dµ ψ R
(29)
+
T
∗
+iψR
σ2 σ µ Dµ σ2 ψ R
The second terms in the above expressions may be further simplified using Eq.(27) and the anti-commutation
properties of the fermion fields (the simplification is perhaps done most conveniently using gauge and
lorentz indices). It then turns out that they are identical to the first terms in the above expression and
we get the desired result. For example,
T
∗
iψL
σ2 σ µ Dµ σ2 ψ L
∗
µ
a
∗ β
= i(ψL )a (σ2 σ µ σ2 )ab ∂µ (ψL
)b + gAaµ (ψL )α
a (σ2 σ σ2 )ab (t )αβ (ψL )b
=
=
i(−1)(−1)(ψL )∗b (σ2 σ µ σ2 )Tba ∂µ (ψL )a
† µ
iψL
σ Dµ ψ L
+
(30)
µ
T
a T
α
(−1)gAaµ (ψL )∗β
b (σ2 σ σ2 )ba (t )βα (ψL )a
where in the first term of the second step the two negative signs are from the anticommutation of the
fermion fields and integration by parts, and the negative sign in the second term is because of the
anticommutation of the fermion fields. In the last step we have used Eq.(27).
b)
We are given with the fact that the β-function for the full N = 4 supersymmetric theory SU (N c )
gauge theory vanishes. The supermultiplet consists of one massless vector field, four massless Majorana
fermions, and six real (three complex) scalar fields all of them belonging to the adjoint representation
3
of SU (Nc ). We may thus write the total β-function as a sum of contributions from the individual
components:
βg = β1 + 4 β1/2 + 6 β0 = 0
(31)
where β1/2 and β0 are the respective contributions due to one Majorana fermion and one real scalar field,
and β1 is the contribution due to the vector field. In Problem 2 we saw that a complex scalar field in the
representation r contributes to the β-function:
β0c (r) =
g 3 C(r)
(4π)2 3
(32)
Hence a real scalar field in the adjoint rep of SU (Nc ) should give:
β0
=
=
=
1
β0c (G)
2
g 3 C(G)
(4π)2 6
g 3 Nc
6(4π)2
(33)
(34)
(35)
Furthermore we know that the contribution to the β-function from a Dirac fermion field in the adjoint
rep of SU (Nc )is given as (Refer to P &S16.85) :
β1/2D
g 3 4 C(G)
(4π)2
3
= 2 β1/2 = 8 β0
=
(36)
(37)
Using Equations 21, 25 and 27 we find:
β1
= − 22 β0
g 3 11
Nc
= −
(4π)2 3
(38)
(39)
This result may also be reasoned from the general expression for β(g) obtained using the background
field method (spin-1 adjoint comes with the coefficient 11/3)and the fact that for an SU (N ) adjoint
C(G) = C2 (G) = N
(40)
[Background field method beyond 1-loop reference: L.F.Abbott, Nucl. Phys. B185, 189(1981)]
In fact using the background field method one can work out the β-function in a supersymmetric theory.
For a non-supersymmetric theory we have


X
2
1 X
g 3  11
− C2 (G) +
(41)
C(r) +
C(r)
β(g) = +
(4π)2
3
3
3
i f ermion
j scalar
where the first term is from the gauge bosons in the adjoint representation (with Casimirs C 2 (G)), the
second term is from the fermions in reperesetation ‘r 0 and the third term from scalars in representation
‘r0 (C(r) are the represetation constants). Note that that the coefficient for fermions is taken as 2/3
instead of 4/3 since we count the Weyl fermions separately.
In a supersymmetric theory every gauge boson has a fermionic superpartner, the gaugino, in the adjoint
representation and every fermion has a scalar superpartner, the sfermion, in the same representation.
Hence using the fact that for an adjoint C2 (G) = C(G) we get
"
#
X
g3
SU SY
β(g)
=
−3C2 (G) +
C(r)
(42)
(4π)2
i
4
Thus in a supersymmetric SU (Nc ) theory with Nf fermions the above gives
β(g)SU SY = −
g3
[3Nc − Nf ]
(4π)2
which is a well know result.
5
(43)