gauge-fixing the su(n) lattice gauge field hamiltonian* abstract

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Rob (uni)bam. - bab6,-2,,. (2.7b). We see .... 6abb,,+; m 9 - Rob (V~)ham. From (3.14) and (3.19), we .... J. Goldstone and R. Jackiw, Phys. Lett. m, 81 (1978). 15.
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GAUGE-FIXING THE SU(N) LATTICE GAUGE FIELD HAMILTONIAN*

BELAL E. BAAQUIE t Stanford Stanford

Linear Accelerator

University,

Stanford,

Center

California,

94305

ABSTRACT We exactly gauge-fix the Hamiltonian

for the SU(N)

lattice gauge field and

eliminate the redundant gauge degrees of freedom. The gauge-fixed lattice Hamiltonian, in particular

for the Coulomb gauge, has many new terms in addition to

the ones obtained in the continuum

Submitted

-

formulation.

to Physical Review D

* Work supported by the Department of Energy, contract DE - AC03 - 76SF00515. t Permanent Address: Department of Physics National University of Singapore, Kent Ridge, Singapore 0511

1. INTRODUCTION The Hamiltonian

for QCD (quantum chromodynamics)

ied using the lattice and continuum

formulations.

has been widely stud-

In a remarkable

paper by

Drell,’ a derivation was given of the running coupling constant of QCD using the continuum

Hamiltonian;

this calculation used weak field perturbation

the Coulomb gauge. The mathematical Hamiltonian

treatment

theory and

of gauge-fixing the Yang-Mills

goes back to Schwinger ;2 the more recent paper by Christ and Lee3

gives a clear and complete treatment

of gauge-fixing the continuum

gauge field

Hamiltonian. The continuum

Hamiltonian

preserves gauge invariance; and Lee,3 a momentum for two-loops

has until now been given no regulation

for the one-loop calculation

invariance and renders the theory non-renormalizable; it is known that dimensional ficient to renormalize dimensional

regularization

the action.

regularization

QCD Hamiltonian

carried out by Drelll

cut-off is sufficient to ensure renormalizability.

and higher it is known that a momentum

which

However,

cut-off violates gauge-

for the action formulation

of the Feynmann

For the Hamiltonian,

diagrams4 is suf-

there is no analog of

and hence it is not clear how to regulate continuum

to all orders.

The lattice Hamiltonian516 is regulated to all orders and could be used for two loops or higher.

involving

Hamiltonian

using weak coupling approximation,

for example the Coulomb gauge. -

If we want to analyze the lattice

calculations

Gauge-fixing

it is necessary to fix a gauge, the action of the lattice gauge

&eory has been solved,7 and in this paper we extend gauge-fixing to the lattice Hamiltonian.

Gauge fixing essentially involves only lattice gauge-field and the

quarks enter only through the quark color charge operator. .

2

So we will essentially

study only the gauge field and introduce the quark fields when necessary. Gauge-fixing

the lattice Hamiltonian

the continuum

Hamiltonian;

formulation.7l8

For the Hamiltonian

is very similar in spirit to gauge-fixing

this similarity

can be clearly seen in the action

we will basically follow the treatment

given

by Christ and Lee.3 There are, however, significant differences between the lattice and continuum

Hamiltonians

both for the kinetic operator and the potential term.

The lattice gauge field is defined using finite group elements of SU(N) fundamental

as the

degrees of freedom whereas the continuum uses only the infinitesimal Th’is d’1ff erence will introduce

elements of SU(N). Given appropriate

generalized interpretation

a lot of extra complications.

of the basic symbols, it will turn out

however that the form of the gauge-fixed continuum

and lattice Hamiltonians

are

very similar. In Sec. 2 we discuss the Hamiltonian electric field operator. we perform freedom.

and give a construction

of the chromo-

We then discuss Gauss’s Law for the system. In Sec. 3

a change of variable and eliminate

the redundant

gauge degrees of

In Sec. 4 we evaluate Gauss’s Law for the new variables and find that

the constrained

variables decouple exactly from the Gauss’s constraint.

3 we evaluate the gauge-fixed lattice Hamiltonian, introduce

the quark charge operator.

In Sec.

discuss operator ordering and

In Sec. 6 we discuss the main feature of

our results.

3

2. DEFINITIONS Consider a d-dimensional Euclidean spatial lattice with spacing a; let Uni, i = 192 , . . . . d, be the SU(N)

link degree of freedom from lattice site

the unit lattice vector in the ith direction) field. The Hamiltonian

for SU(N)

and let &,

n to

n + i (2 is

$,, be the lattice quark

lattice gauge field in the temporal axial gauge

is given by5s6

H =

HYM[U] + HF[&$3u]

- $

c

(2.la)

where H

YM=

v2 (Uni)

n,i

-

$

C

Tr

(U,i

U,+;,juz+;,iU,G)

n,ij

and HF is the quark-gauge field part. operator.

The Hamiltonian

Gauge transformation

acts only on gauge-invariant

Laplace-Beltrami wave-functionals

and the wave-functionals

+

Uni (P)

G PnUni

P;t+;

(2.2)

@ are invariant under (2.2), that is

P-3)

W I = wb)l

- By performing

@.

is given by

Uni

_-.

Note V2 is the SU(N)

an infinitesimal

gauge-transformation

4

(and introducing

the

quark field) we have from (2.3) Gauss’s Law’

{@(Uni)

2



E,L(“n-;S}

-

Pna

[ i

The operators

1

I@)

EF and E,” are first order hermetian

with the commutation

=

(24

O

differential

operators

equation6

(2.5a)

[Ef,

E;]

Et

(U),

E,R(U)=Rab(U)

= -icab,

E,R

Rob (U)=

1

Ef

E,R,

where Rab is the adjoint representation,

Tr(Xa

(2.5b)

UXb U+)

--0

(2.5~)

(2.5d)

Xa the generators and C&c the structure

constants of SU(N). The operator

pna ($3 +, U) is the lattice quark color charge operator6

and

satisfies [ha,

Ptnb]

=

icabc

(2.5e)

Pnc &a

From (2.4) and (2.5~) we have

0 =

C

{ Rab(U,i)Ef’(U,i)

-

Ef(u,-I,i)

Pna

1

i

=-

r cm,i

Dtni

@(urni)

-

5

IQ)

Pna

1IQ>

(2.6a)

(2.6b)

where D~“,i is the lattice covariant backward derivative. of lattice site n and nonabelian

Let In, a) be a ket vector

index a; then, from (2.6) we have the real matrix

Di given by

D&i

=

(n,

aIQlm,

=

Rob (uni)bam

(2.7a)

b)

-

(2.7b)

bab6,-2,,

We see from above that Di performs a finite rotation

Rab on the ket vector and

then displaces it in the backward direction. We write the Hamiltonian

as sum of the kinetic and potential H

= K(U)

+ qJ,

?A q

energy, that is (2.8)

where K = -c

C

(2.9)

V’(Uni)

n,i

and P is the rest of (2.la).

It is known that9 -V2(U)

= c

(2.10)

Ek(U)E,L(U) a

In light of Gauss’s Law and (2.10) we identify tric operator

of the gauge field corresponding

canonical coordinates

E,L(Uni)

as the chromoelec-

to the link variable Uni. Choose

B~i such that Uni =

eXp(iB$ Xa)

-

(2.11)

Then we have, suppressing the lattice and vector indices and summing -

Tepeated nonabelian

on

indices (2.12a)

.

6

(2.12b)

Note e:b(“)

Explicit

(2.13)

= eii(“)

expressions for cab L(R) are given in (3.8).

3. GAUGEFIXING We can see from Gauss’s Law that all the Uni’s are not required to describe the gauge-invariant

wave-functional

a. We gauge-transform

variables Vni which are constrained;

the constrained

Uni to a new set of

variables Vni will decouple

from Gauss’s Law. Consider the change of variables from { Uni} to {pm, Vni}, with {Vni} having one constraint

for each n. That is tin Uni

= (P&9 = Pn

4n Ki

P+

= GaPZ

(3.la)

(3.lb)

n+;

and choosing the Coulomb gauge for the lattice gives xi(Vni)

z Im C

TrXa (Vni - Vn-;,i) = 0

(3.k)

i

In canonical coordinates we have

pn =

Vni = eXp(i&Xa), -

--For small variation

exp(i+iXa)

(3.2)

A” + dAa, we have V(A

+ dA) = V(A)

I+

V+(A)sdA”]

(3.3)

(3=4

[l + iXajz(A)dAb]

= V(A)

where

Define &p)Aa

= j,$@)(A)dAb

V(A

+ dA) = V(A)(l

then + iXa6RA”)

(3.7a)

= (1 + iXaGLA”)V(A)

(3.7b)

It can be shown that L(R) L(R) = (Ijab

eaa

and hence matrix (3.1) from function

Uni

t0

f ab

P-8)

e can be determined

from (3.5). Under the charge of variables

Vni, the potential

energy P in (2.8) can be expressed as a

of only Vnim For the kinetic energy K we need the expression for E:(U).

Note, using the chain rule and formula

(3.8)

w-4 =C

f,R,(Ani)z mi

n,i

Therefore, -

-

(3.10)

efp(Ani)$-+... ni

from (2.12) and (3.10)

Ef(umj)=’

(3.11)

iSL B~j

We now evaluate the coefficient functions .

a

of above equation.

The constraint

Eq.

(3.1~)

is valid under variations

of A~i to Aa,i + Dali,

i.e. (3.12)

0 = x;(A) = x&i

(3.13)

+ dA)

Hence, from (3.12) and (3.13) C

IT&i(A)

6RAki

(3.14)

= 0

m,i

where, for constraint

(3.lb) we have I?ab nma. =

(n, alI’ilm, b)

(3.15a) (3.15b)

= ~X~I~LA~i =

W$6,m

-

WlL;

ibn-;

9

,

(3.15c)

m

where from (3.1~) W$

The constraint

=

Tr (xavnixb

(3.16)

+ xbv$xa)

(3.14) on A~i determines 6p/6B.

Consider from (3.lb),

the

following variation

Jk(A

+ dA) = CP:(~ + d4)Uni(B

+ dB)pn+;(4

+ d4) .

(3.17)

which yields from (3.7a) bRA:i

--

= 6R4:+2

-

=- c D

&i

Rab(VL)JR$f:

bR6fn

+ Rab

+ Rab(‘Pi+;)6RB%i

(3.18)

(3.19)

m

From (3.18) and (3.19), we have the lattice covariant forward derivative operator .

9

Di given

by ab Dnmr . ='(n,alDilm, b) = 6abb,,+; 9m -

(3.20) (3.21)

Rob (V~)ham

From (3.14) and (3.19), we have

c

(%alriDilm,b)6R&,,

-I- c

m,i,b

(n, alriRT[m,

b)6RBki

(3.22)

= 0

m,i,b

where T stands for transpose and

bvlRilm,b)

(3.23)

= barn Rab (r~,+;)

Hence, from (3.22) we have (3.24) where (I’ - D)- ’ is the inverse of operator

Ci

I'iDi. We

also have from (3.19) and

(3.24) ‘RAii

_

- -

‘LBRj

1

n,a (

II

PiI--D

I’jRT

(3.25)

- RT&j

Hence, from (3.11), (3.24) and (3.25)

(3.26)

-

-

Equation moelectric .

(3.26) provides the solution for expressing the unconstrained

operator

6/6~B

in terms of the new constrained 10

operator

6/6~A

chroand

S/6,4.

the gauge transformation

i

In essence, this solves the problem of guage-

fixing the lattice Hamiltonian. Note that from (3.25) we have the identity

C(L,cll?iln,a) $$$

=0

(3.27)

mi

n,i

as expected. We have from (3.14)

C(n,alWd$$-- = 0

(3.28)

mi

m,i

Hence, from (2.12) and (3.28) 6 SL Aa,i ’

Ainj

] =

(6mrnbijJac

-

(%

a ( r?

&

rj(

m, C))

efb (Amj)

(3.29)

4. GAUSS’S LAW _

We check that constrained variables

Vni

decouple from Gauss’s Law. Recall

from (2.7) and (3.26), we have

-

-

-

c( I

1 r&-D?

I) ec

6

3 3 3 ’ s,4;

n,am

n,ij

(4-l)

From the definitions .

of

Di

and

Pi given

in (2.7) and (3.21) respectively, we have 11

the crucial operator identity

(4.2) where h

alRIm,

b) = Jnm

(4.3)

Rab(%)

Hence, from (4.2) we see that the first term in (4.1) is zero and we have

(44 -=

We see that

Vni

(4.5)

6R4;

has decoupled from Gauss’s constraint,

and we have from

(2.6) and (4.5) 6 + Pna ibR4;

IQ> = O

(4.6)

1

Solving (4.6)) we have from (13.1)~ qih

$9 U) =

&

w-$4%)

e-E,pn.4=

(4.7)

cp(