Renormalization of Hamiltonian QCD
Renormalization of Hamiltonian QCD
A. Andraši∗
’Rudjer Bošković’ Institute, Zagreb, Croatia
John C. Taylor†
Department of Applied Mathematics and Theoretical Physics,
University of Cambridge, UK
April, 10 2007
Abstract
We study to one-loop order the renormalization of QCD in the Coulomb gauge
using the Hamitonian formalism. Divergences occur which might require counter-
terms outside the Hamiltonian formalism, but they can be cancelled by a redef-
inition of the Yang-Mills electric field.
PACS: 11.15.Bt; 11.10.Gh
Keywords: Coulomb gauge, Hamiltonian, renormalization
∗e-mail:aandrasi@rudjer.irb.hr
†e-mail:J.C.Taylor@damtp.cam.ac.uk
http://arxiv.org/abs/0704.1420v1
1 Introduction
We study the renormalization of QCD in the Coulomb gauge Hamiltonian for-
malism. By Hamiltonian form, we mean that the Lagrangian contains only
first order terms in time derivatives, and depends upon the conjugate momen-
tum field Eai as well as the (transverse) gluon field A
i (here a is the colour
index and i = 1, 2, 3 is a 3-vector index). This form has a number of attractive
features:
(i) As a Hamiltonian exists, the theory is explicitly unitary, without the
necessity to cancel unphysical degrees of freedom with ghosts.
(ii) The Lagrangian form of the Coulomb gauge has “energy divergences” in
some of its Feynman integrals, that is integrals of the form (we use K for the
spatial part of the 4-vector k)
d3Kdk0f(K, k0) (1)
where f does not decrease as k0 → ∞ (for fixed K). These divergences cancel
between different Feynman graphs [1], but this cancellation has to be organized
“by hand”. In the Hamiltonian form, each individual Feynman graph is free of
such divergence. Formally ’energy divergent’ integrals such as
(2π)3
p20 − P 2 + iη
(P −K)2
are assigned the value zero.
(iii) It has been argued [2] that the Coulomb gauge throws light on con-
finement. Certainly it is known [3] that, in the Coulomb gauge, the source of
asymptotic freedom lies in the Coulomb potential.
In spite of (i) above, to 2-loop order, mild energy-divergences remain [4], [5],
[6] which result in ambiguities which have to be resolved by a prescription. This
is connected with questions of operator ordering [7].
For other applications of the Coulomb gauge, for example to lattice QCD,
see [8], [9].
The question addressed here is the following. Ultra-violet divergences exist
which seem to require the existence of counter-terms containing second order
terms in time derivatives, (∂Aai /∂t)
2. Do these take us out of the Hamiltonian
form? We argue that this does not happen because the divergences concerned
can be cancelled by a redefinition of the Eam field.
We do not use quite the strict Hamiltonian formalism. We retain the auxil-
liary field Aa0 , which contains no time derivatives and should be integrated out
to give a nonlocal Coulomb potential term in the real Hamiltonian. It seems
to be convenient, for the purposes of renormalization, to retain Aa0 in the La-
grangian. Because of this, there is a ghost field, but it has an instantaneous
propagator, and so is not relevant to unitarity. Its purpose is only to cancel out
closed loops in the Aa0 field.
2 The Feynman rules
The Lagrangian for the Coulomb gauge is
L′ = L− 1
2 (3)
(where α will eventually tend to zero to go to the Coulomb gauge),
L = −1
Fij · Fij −
2 +Ei ·F0i
∗∂ic+ g∂ic
∗ · (Ai ∧ c)
+ui · [∂ic+ g(Ai ∧ c)]
+u0 · [∂0c+ g(A0 ∧ c)]
gK · (c ∧ c) + gvi · (Ei ∧ c) (4)
where we use a colour vector notation, and
F aij = ∂iA
j − ∂jAai + gfabcAbiAcj
(Ai ∧ c)a = fabdAbicd (5)
Here c, c∗ are the ghost fields, and the sources ui,vn and K are inserted for
future use in formulating the BRST identities. The conjugate momentum (elec-
tric) field Em could be integrated out to obtain the ordinary Lagrangian for-
malism, but for the Hamiltonian formalism it must be retained.
We will use indices m,n, ... = 1, 2, 3 to denote the (spatial) components of
E, so the seven fields are (Aai , A
0 , E
n). We will use indices I, J, .. to denote the
seven indices (i, 0, n). The bilinear part of the Lagrangian in momentum space
is a 7× 7 matrix
SIJδab =
−K2(Tij + Lij/α) 0 −ik0δin
0 0 iKn
ik0δmj −iKm −δmn
where
Tij ≡ δij − Lij , Lij ≡ KiKj/K2,
k2 = k20 −K2. (6)
For the propagators, we need the inverse
S−1IJ δab =
Tij/k
2 − αLij/K2 αk0Ki/(K2)2 −ik0Tin/k2
αk0Kj/(K
2)2 1/K2 + αk20/(K
2)2 iKn/K
ik0Tmj/k
2 −iKm/K2 TmnK2/k2
 . (7)
We can now let α → 0, to obtain the Coulomb gauge. From this, and the
interaction terms in the Lagrangian (4), we can read off the Feynman rules. We
represent the Ai field by dashed lines, the En field by continuous lines, and the
A0 field by dotted lines. With this notation, we now list the rules (a factor
(2π)4i
is to be included for each propagator, and a factor of (2π)4i for each
vertex). If we choose the propagators in fig.1 to be the negative of the matrix
(7), the extra factors of 1
(2π)4i
for the propagator and (2π)4i for the vertices
cancel.
3 The ultra-violet divergences
The divergent graphs with 2 and with 3 external lines are shown in Figures 4
till 31. Examples of the method of evaluation of divergent parts are given in
Appendices A and B.
The ultra-violet divergent parts of these graphs are, in terms of the divergent
constant (using dimensional regularization in 4− ǫ dimensions)
CGΓ(ǫ/2), (8)
(where the superfix (4), (5) etc. refers to the corresponding figure and Πij ,
Π0i...Πmn denote self-energies, Vijk, V0in...V0in vertices and Λ stands for dia-
grams with external ghost lines), are:
(4)ab
ij = ic[
k20δij +K
2δij −KiKj]δab (9)
(5)ab
i0 = −
ick0Kiδab (10)
(6)ab
icK2δab (11)
(7)ab
mi = 0 (12)
(8)ab
m0 = −
ic[iKiδab] (13)
Π(9)abmn = −
icδmnδab (14)
(10)abc
ijk (p, q, r) = −
cgfabc[(Q− P )kδij + (R −Q)iδjk + (P −R)jδik] (15)
(11)abc
ijk (p, q, r) = −
cgfabc[(Q− P )kδij + (R −Q)iδjk + (P −R)jδik] (16)
(12)abc
ijk (p, q, r) = −
cgfabc[(Q− P )kδij + (R −Q)iδjk + (P −R)jδik] (17)
(13)abc
ijk (p, q, r) =
cgfabc[(Q− P )kδij + (R−Q)iδjk + (P −R)jδik] (18)
(14)abc
i00 (p, q, r) =
cgfabc(R−Q)i (19)
(15)abc
i00 (p, q, r) = −
cgfabc(R −Q)i (20)
(16)abc
i00 (p, q, r) =
cgfabc(R−Q)i (21)
(17)abc
i00 (p, q, r) =
cgfabc(R−Q)i (22)
(18)abc
0jl (p, q, r) = 0 (23)
(19)abc
0jl (p, q, r) =
cgfabc(R −Q)0 (24)
(20)abc
0jl (p, q, r) = −
cgfabc(R −Q)0 (25)
(21)abc
0jl (p, q, r) = 0. (26)
Graphs involving external Em line are
(29)abc
im0 (p, q, r) =
icgfabcδim (27)
(30)abc
im0 (p, q, r) = −
icgfabcδim (28)
(31)abc
im0 (p, q, r) = 0. (29)
All other graphs involving external Em -lines are convergent. The divergent
parts of graphs with open ghost line are
Λ(22)ab(q) = −4
icQ2δab (30)
(23)ab
i (q) = −
cQiδab (31)
Λ(24)abc(p, q) = 0 (32)
(25)abc
k (p, q, r) = 0 (33)
(26)abc
0 (p, q) = 0 (34)
(27)abc
i (p, q) = 0 (35)
Λ(28)abcn (p, q) = 0. (36)
4 Counter-terms
d4xL(x) (37)
be the original action, Γ be the complete effective action, and let Γ1 be the
effective action to one-loop order. The complete BRST identities are
Γ ∗ Γ ≡ ∂Γ
= 0. (38)
So to one-loop order
Γ1 ∗ Γ0 + Γ0 ∗ Γ1 ≡ ∆Γ1 = 0 (39)
where
∆2 = 0. (41)
One class of solutions to this equation is of the form
1 = ∆G, (42)
where the allowed form of G is, in terms of constants a5, ...a11,
G = a5Ai · (ui + ∂ic∗) + a6A0 · u0 + a7c ·K+ a8Ei · vi
+a9vi · ∂iA0 + a10vi · ∂0Ai + a11vi · (A0 ∧Ai). (43)
Other solutions of equation (39) are the explicitly gauge-invariant terms
1 = a1(Fij)
2 + a2Ei · F0i + a3(F0i)2 + a4(Ei)2. (44)
Finally, by differentiating (38) with respect to the coupling constant g and
specialising to one-loop order, we see that
(iii)
1 = 0 (45)
where (a0 being another divergent constant)
(iii)
i = a0g
. (46)
Combining these three contributions, we obtain
Γ1 = Γ
1 + Γ
1 + Γ
(iii)
d4xL(x) (47)
where
L1 = a1(Fij)2 + (a2 + a8 + a9)Ei ·F0i
+(a3 − a9)(F0i)2 + (a4 − a8)(Ei)2
+a5Fij · ∂jAi − (a5 +
a0)gFij · (Ai ∧Aj)
−(a0 + a5 + a6)gEi · (Ai ∧A0) +Ei · (a5∂0Ai − a6∂iA0)
−a5(ui + ∂ic∗) · ∂ic+ a0g∂ic∗ · (Ai ∧ c)
−a6u0 · ∂0c+ a0gu0 · (A0 ∧ c)
−a7(ui + ∂ic∗) · {∂ic+ g(Ai ∧ c)}
+a0gui · (Ai ∧ c)− a7u0 · {∂0c+ g(A0 ∧ c)}
g(a7 − a0)K · (c ∧ c) + (a0 − a7)gvi · (Ei ∧ c). (48)
The conditions coming from the vanishing ghost graphs Figs. 24, 25, 26, 27
and 28 are particularly simple. They fix
a9 = −a10
a11 = −ga9
a0 = a7 = −a6. (49)
In order for the counter-terms to cancel the divergences in the other graphs, we
require the conditions
4a1 − 2a5 = −c
4a1 − 3a5 − a0 =
a3 − a9 = −
a6 − a5 =
a5 + a7 = −
a4 − a8 =
a2 + a5 + a8 + a9 = 0. (50)
These equations do not fix the constants uniquely. We are free to make
some choices. The term (F0i)
2 in Γ
1 eq.(44) is not present in the original
Hamiltonian form of the Lagrangian (4), so we choose
a3 = 0. (51)
We can also arrange for the combination
2 +Ei ·F0i (52)
to appear in L(ii)1 as it does in L0. This requires (from (50))
a1 = −
a2 = c− 2a5
a4 = −
c+ a5
c+ a5
a7 = −
c− a5
a8 = −
c+ a5
a0 = −
c− a5 (53)
and so
L(ii)1 = −4a1[−
(Fij)
2 − 1
2 +Ei ·F0i] (54)
proportional to the non-ghost part of the original Lagrangian (3).
Equation (54) does not come from the BRST identities, it just emerges from
the numerical values of the divergent integrals. It may be a consequence of some
hidden Lorentz invariance.
The constants a0, a1, ... are still not uniquely fixed. There are two particu-
larly simple choices.
(i) Choose a0 = 0 with a5 = − 43c. Then we find
a1 = −
a4 = −
a6 = a7 = 0
a8 = −
c. (55)
(ii) The second choice is a1 = 0 with a5 =
c. Then
a0 = −
a2 = 0
a4 = 0
a7 = −
a8 = −
c. (56)
Note that a0 has the expected value for coupling constant renormalization.
The counter-terms in either case are
L1 = −
c(Fij)
2 − 4
cFij · ∂jAi +
cgFij · (Ai ∧Aj)
c(F0i)
c(Ei)
cEi · F0i
cgEi · (Ai ∧A0)−
cEi · ∂0Ai +
c(ui + ∂ic
∗) · ∂ic. (57)
The counter-terms in a5, a6, a7, a8 and a9 are involved in a rescaling of the
fields. Defining
i = (1 + a5)Ai
0 = (1 + a6)A0
m = (1 + a8)Em − a9F0m
i = (1− a5)ui
0 = (1− a6)u0
′ = (1− a7)c
′ = (1 + a7)K
g′ = (1 + a0)g
′∗ = (1 − a5)c∗
′ = (1− a8)v, (58)
we have from (48) that
L0 + L1 = (1− 4a1)L0(g′,A′i,A′0,E′, c′, c′∗,u′i,u′0,K′). (59)
Note that a6 which determines the renormalization of the Coulomb field A
has the same numerical value as a0.
We have not calculated the divergences in graphs with four external lines.
We assume they will be cancelled by the same counter-terms.
5 Comments
We conclude that there is no difficulty to one-loop order in renormalizing the
Hamiltonian form of the Coulomb gauge. We guess that the renormalization
would formally go through to higher orders, but then there is the problem men-
tioned in [4], [5], [6] of combining the renormalization of ultra-violet divergences
with the resolution of energy-divergence ambiguities.
It is not quite obvious how the renormalization would be formulated if the
Aa0 field had been eliminated to give the non-local colour Coulomb potential
(note the non-zero value of the Aa0 field renormalization constant a6 in (56)).
Acknowledgements
A.A. wishes to thank the Royal Society for a grant and DAMTP for hospi-
tality. We are grateful to Dr. G. Duplančić for drawing the figures. The work
was supported by the Ministry of Science and Technology of the Republic of
Croatia under contract No. 098-0000000-2865.
Appendix A
Here we give as an example the evaluation of the ultra-violet divergent part
of the graph in Fig. 20.
(20)abc
0jk (q,−q, 0) = ig
(p2 + iη)2
(q + p)2 + iη
×Trz(P )Tzv(P )Tru(Q+P )[(−2Q−P )vδuj+(Q−P )uδjv+(2P+Q)jδvu]. (A1)
Applying the integral
(p2 + iη)2
(q + p)2 + iη
dyy(1− y){(P + yQ)2 + y(1− y)(−q2 − iη)}− 52 (A2)
and power counting to (A1)
(20)abc
0jk (q,−q, 0) = −4g
πq0Γ(
dyy(1− y)
d3−ǫPPjPk{(P + yQ)2 + y(1− y)(−q2 − iη)}−
2 , (A3)
leading to
(20)abc
0jk (q,−q, 0) = −
cgfabcq0δjk. (A4)
Appendix B
Example of self-energy evaluation Π
(6)ab
00 in eq.(11). Let p, q be internal and
k external momentum, p− q = k. The sum of two graphs is
(2π)−4
Tij(P )Tji(Q)
(P 2 +Q2)− (ip0)(iq0)]δab (B1)
where we have symmetrized the first term in P,Q. the minus sign in the second
term comes from the opposite order of the fabc factors at the two vertices. Doing
the p0 integration by Cauchy, we get
(2π)−4(2πi)
TijTji
(P +Q)2 − k20
(P +Q)[P 2 +Q2 − 2PQ]δab. (B2)
The last factor (P − Q)2 is approximately (P · K)2/P 2. With this factor, the
integral is only logarithmically divergent, and to get the divergent part we can
put Q = P everywhere. We use Tij(P )Tji(P ) = 2. Then we get
(2π)−4
d3−ǫP
(P 2 +m2)5/2
. (B3)
So the divergent part is1
icK2δab. (B4)
References
[1] R. N. Mohapatra, Phys. Rev. D4, 22, 378, 1007 (1971)
[2] D. Zwanziger, Nucl. Phys. B 485, 185 (1997)
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1Note that there was an error of sign in Eur. Phys. J. C37, 307-313(2004) which however
did not influence the final result.
http://arxiv.org/abs/hep-lat/0612004
Figure 1: Feynman rules for the propagators in the Coulomb gauge.
q; b; j
r; 
; k
p; a; i
a; i b; j

; k a;m
(Q� P )
(R�Q)
(P �R)
Figure 2: Feynman rules for the vertices in the Coulomb gauge. The arrows
denote the directions of the momenta.
p; j; b
Figure 3: Feynman rules for ghosts and sources in the Coulomb gauge. Doubled
lines denote ghosts. The black arrows distinguish between ghosts and antighosts.
Momenta flow into the vertex.
Figure 4: The transverse gluon self-energy graphs.
Figure 5: The AiA0 two-point function.
Figure 6: The time-time component of the gluon self-energy.
Figure 7: The transition between the transverse gluon field and its conjugate
field Ei.
Figure 8: The transition between the Coulomb field A0 and the conjugate field
Figure 9: The conjugate field self-energy.
Figure 10: Graph contributing to the three-gluon vertex function.
Figure 11: There are three graphs in this class with permutations of the vertices.
Figure 12: Graph representing a class of 6 diagrams.
Figure 13: There are 3 graphs in this class of diagrams.
Figure 14: Graph with two external Coulomb lines (there are 3 diagrams in
this class).
Figure 15: There are two graphs in this class.
Figure 16: There are two graphs in this class.
Figure 17: The graph with two external Coulomb lines and one three-gluon
vertex.
Figure 18: Graphs contributing to the(AiAjA0) three-point function.
Figure 19: Graph contributing to the (AiAjA0) three-point function which
contains a three-gluon vertex.
Figure 20: The (AiAjA0) graph with a three-gluon vertex.
Figure 21: The (AiAjA0) graph with a four-gluon vertex.
Figure 22: The ghost self-energy.
Figure 23: Ghost and the ui source graph.
Figure 24: The ghost vertex graph with a K source.
Figure 25: Graph with external Ai, ghost and anti-ghost lines.
Figure 26: Graph with u0 source, Ei and c lines.
Figure 27: Graph with ui source, Ai and c lines.
Figure 28: Diagram with vn source, A0 and c lines.
Figure 29: Graph contributing to the (AiEjA0) vertex function.
Figure 30: Graph with external gluon, Coulomb and E-field.
Figure 31: Graph in the (AiEjA0) vertex function.
	Introduction
	The Feynman rules
	The ultra-violet divergences
	Counter-terms
	Comments
