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Absolute bound of the functional determinant in (QED)2
Volume 91B, number 3,4
PHYSICS LETTERS
21 April 1980
ABSOLUTE BOUND OF THE FUNCTIONAL DETERMINANT IN (QED)2 Keiichi R. ITO 1
Department of Physics, University of BielefeM, 4800 Bielefeld 1, Fed. Rep. Germany Received 4 January 1980
Let S = S(x, y) be the two dimensional euclidean fermion propagator and let K = iXSA, where.4 = ~:~t~A;zand {Au(x)} are external vector fields (real fields). Then the gaussian domination holds for the renormalized Matthews-Salam determinant Idetren(1 + K)I ~
Recently several authors proved diamagnetic inequalities for (euclidean) gauge models [ 1 - 3 ] • This is well stated in the lattice gauge approximation and is very basic to ensure the existence of gauge field theory. In this note, we show that the Matthews-Salam determinant [1,2,4] qY = detren(1 + K) = det(4)(1 + K) e - T
with
.,v(k)-- (8,v k.kq
det(l+p)(1-A) = det[(l-A)
eXP(m~__1 A m / m ) l .
Theorem. Let {Au) be real functions in = L 2 N Lq (q > 2), and let S be the euclidean fermion propagator which satisfies anti-periodic boundary conditions at OA. A = [0,L] ®2. Then [ClYl~< exp[~C(Im X)21[A~I[2] ,
is dominated by a gaussian function o f I[Au[[2 = (fA2(x)d2x) 1/2 in two dimensions for any coupling constant X ~ C. Here K = iXS(x, y)~ (y), S(x, y) = (2S7u au + m ) - l ( x , y ) , (Tu)u=0 1 are euclidean 7 matrices ('~u - 7u' (Tu' ~'u}+ - 28uv),~t - E'YuAu, )t2 f d2k .4a(k)Zv(-k)II~tu(k), r = ~ a(27r) 2
p
(1)
(2)
where C is a constant independent of X and Au.
Remarks 1. (1) The definition of ~ is rather technical. Since X ~ R and {Au} is not a random variable like in refs. [3,5,6], the discussions in refs. [3,5,6] cannot be directly applied. See later discussion. (2) As a special case, one can take A = [ - L / 2 , L/2] ®2 with L -+ ~, namely A = R 2. This corresponds to free boundary conditions. I f L < ~, then the form of lluu is slightly different from that below eq. (1), which is not essential now.
k2 ! X
k 4m 2 tanh- 1 + k2) 1/2 ]_ 1 - k( 4m2 + k2)1/2 (4m2
r
the renormalized second amplitude, and 1 Present address: Res. Inst. Math. Sci., Kyoto Univ., Kyoto 606, Japan. 406
We use lattice theory to prove this. Now let AN = a ( n 0 , n l ) with 0 ~
e0 = a ( 1 , 0 ) ,
e l=a(0,1),
(3)
Volume 91B, number 3,4
PHYSICS LETTERS
y. [Ama(X)a[q a2-q < 0% [aAu,a(X)[ <" al-2/q ea with ea -+ 0 as a ~ 0 for almost all x C A~v (as N -+ oo). Thus although X q~ R,
and for x , y E AN, defineBN(X,y ) and FN(X,y):
BN(X,y ) = (2a - 3 + ma-2)Sx,y - a - 3 7 ( x , y) V(x, y),
(4a)
PN(X,y) = --a -3 [U(x,y) -- 1] V(x,y)T(x,y), where 7 ( x , y ) = ~(1 ~Tu),
y = x +-eu,
= 0,
otherwise,
V(x,y) = exp(+irr/2N),
(4b)
y = x + eu,
(4c)
U(x,y) = exp{+iXaAu, a [(x +- eu/2)]}, y = x + eu , = 0,
(4d)
n), a(no, 0)
(1 +- i)~aAu,a)[ <~C]XaAu,a [2
for almost all x E A ~ as a -+ 0. (This is the typical technique in refs. [3,7] and will be used to prove lemmas 2 and 4.)Au(x ) E L 2 means that T and IlAul[ 2 are well defined andA u E Lq (q > 2) is sufficient to ensure K N -+K in some
Cp--{x;l[xllp=(TrlxlP)l/P
p>2.
Lemma 1. [5,7]. I f A u @ c~, then limN~oodet(1 + K N ) = q.~.
otherwise,
[a(0, n) and a(2N, identified], and
]e+-iXaAu, a --
This is implicit in ref. [5] and one can prove this by slight modifications (see also refs. [6,7]).
otherwise,
= 0,
21 April 1980
and
a(no, 2 N )
The key role of V(x, y) is to introduce anti-periodic boundary conditions at OAN [3,8], which is inevitable for considering the transfer matrix. Note that
are
a/2 a/2 Au'a(X) = a-2 =af/2 Zaf/2 Au(x + r/)d2r/'
det(t +KN) = det[RN(Au) ]/det[RN(O)] ,
(5) x 6 A~V.
These are operators on
(8)
where R N = ,O-I(BN + P N ) ' P = m + 2/a. Following refs. [3,8] we introduce the transfer matrix. First consider the g = 0 direction. Then
det[RN(Au) ] = Tr ToUoT 1 U 1 ... T2N_ 1U2N_I. (9) ~ N = {{f(x); x E AN}; Ilfl[ 2 = 0 2
If(x)12},
Here (TI, Ul}2N-1 are operators on the 24N-dimen sional Hilbert space ~ constructed by operating fermion creation operators {c+(l), d+(/)} 2N-1 on a cyclic vacuum ~ E 9 , and {TI, Ul) are explicitly represented in terms of the fermion operators (see refs. [3, 8] for the construction):
xEA N
and one formally finds:
a2
~
BN(X,y)f(y)-+(Tu8 u +m)f(x),
Y~A N
a2
~
2N- 1
P N ( x , y ) f ( y ) ~ i L ~ ( x ) f ( x ),
Ut=ex p lax ~
YEA N
Ao,a[a(l+½),an ]
n=0
asa ~ 0 (or a s N ~ oo). Let
(10)
× [c+(n)c(n) - d+(n)d(n)]l
SN = BN 1 = p2 UN ,
(6)
p
I
On the other hand, T 1 depends only on {A 1,a [al, a (n 5YJJn =0 and is positive when X E R, and analytic for X in a neighborhood co of X = 0 (co -+ C as a -+ 0) [3]. Then Tl(X)* = TI(X),
+ 1 ~]12N-1
where PN > 0, U~ = UN 1, and define
KN=--PNUNFNPN,
K:iXPU~P.
(7)
Now (Au(x)} are usual functions and unfortunately we do not assume X E R. An easy technical assumption is to set °M = L z ( d Z x ) N Lq(dZx), q > 2. Since 407
Volume 91B, number 3,4
PHYSICS LETTERS
2N-1 leq.(9)l ~< l-I [Tr(TIUIU[T[)N] 1/2N /=0 2N-1 = FI {Tr[Tl(X)Ul(2i Im X)TI(X)IN} 1/2N, l=O
Lemma 2. LetA u E c~. Then _ er 2~4. 2 K(2),r N t22 ~< const, u L S "~u
(the constant is independent of (m, n)) as a + 0, where e > 0 and
by H61der's inequality. Repeat the same discussion for /a = 1 for each Tr[TIUIU~T~]N: Idet(l+KN)l<~ l-Ildet[l+KN(m,n)][
21 April 1980
K(N2),r = PN UN['(N2),rPN . Proof.
1/4N2,
(11)
mn
where K N (m, n) = PN UN FN (m, n)PN and PAr(m, n) is the F N function given by the followingAu(x), x @ A~v (with coupling constant Im X):
K(2) r 2 ~< iiF(2),rll~lgmll4 N ~2 (H61der), where IIPNII4 ~< coast. L 2 In a - 1 as a + 0 (see the proof of lemma 3), and IIF(N2)'rll2 ~< coast. (aau)2~4Z 2 .
A o ( x )=
2iA0,
x 0 = a ( e + ~),
x 1 =a(e),
= -2iA0,
x 0 = a ( e + {),
x 1 = a(o),
= 0,
otherwise
Al(X)=
2iA 1,
Q.E.D.
Lemma 3. Let K(N1) = PNUNF(N1)PN. Then
x0=a(e),
xl=a(e+{),
=-2iAl,
x 0 = a(o),
x 1 = a(o +1),
=
otherwise.
0,
Here A 0 = A O,a [a(m + ~ ), an], A 1 = A 1,a [am, a(n + ~)], and "e" and " o " mean even and odd, respectively. For simplicity let KN(m, n) = K N, PN(m, n) = PN = PN,0 + PAr, 1 and ~"= Im X, where PN, u ( x , y ) = 0 unless y = x + e u and
IIK(N1)II2 ~< const. ~'2L2A /z" 2
Proof. Let F(N1)v,1 be the term in F (N1)uwhich contains "),u. Then
=(a2) 2
-
PN, u(x, x + eu) = a-3sinh ~aAux
~
eikx-ik'YF(Nl?v,l(x,y)
x,y~AN 1 2 . - i k ' e +in/2N
~TuAtaiL (~
ts
+
eikev-iTr/2N)
X exp(~aA u + irr/2N)(-7 u + 1),
(+eu) (e, e),
× [*kv,G+~ + (~--' -~)]
X exp(-~aA u + izr/2N)(Tu - 1),
(+eu) (e, o),
x [5k.,i,; + 8k.,k;+~+ (~ + --01,
X exp(~aA u - izr/2N)(-7 u - 1),
(--e~)
(o, e),
X e x p ( - ~ a A u - izr/2N)(7 u + 1),
(-e.)
(o, o),
otherwise.
XO,
where k, k E AN and ~ = 7r/a = 2Nrr/L (--)-0% as N -+,,~). The other term in P~)u takes a similar form after Fourier transformation except for ~/u and signs. Thus it suffices to prove
(12) Here (e, o), for example, means the evenness and oddness of a - l x ~ and a - l x v , respectively. Let F(N1) (respectively F(N1)u) be the term obtained by replacing sinh(a~Au) exp(+a~Au)/a by ~A u, and let "lVP(2)'r(re"
(47r 2/L 2) k ~ x" f f ~ ( k ) ' f f 2 ( k ~ + ~) <~ const., (independent of N), where ~ = (~, 0) and "fiN(k) = a 4
X Y,,x,yEA N e ik(x -Y)PN (x, y)/L2:
spectively [,(2),r~ be the remainder term: F N = 1-(1) ~N N,~ j p(2), r
+ -N
, etc.
\
+ (1/a 2)
~sinea(k
u - rr/L)
) 408
~ 1~4
Volume 91B, number 3,4
PHYSICS LETTERS
21 April 1980
It is easy to see [5] that, whenever k ~ ~W
[[KN(m, n) + KN(m, n)[I 1 ~< C ( I m ~)2L2A 2(m, n),
PN(k) <~e[m2 + k 2 ] - 1 / 4 ,
where C is independent of X, L, (m, n) and N (or a).
PN(k + ~) <~e [m 2 + ([k0[ - ~)2 + k2]-1/4, with some constant c (independent of N). Then we prove that
Thus lemmas (2), (3) and (4) mean (a 2 = L2/4N 2) [det(1 + K N ) det(1 + K N ) ] ~
(4zr2/L 2
~
(m 2 + k2)-1/2
k~.,Ikl<~ X [m 2 + (Ik0l - ~)2 + k 2] - 1 / 2 is bounded by a constant which is independent of = n/a, where A = (2n/L)Z 2. Rewrite this as flkl<~ d2k [ ] + error, and show that the integral part is finite for all ~ ~< oo (then the error term -+ 0 as a -+ 0). Let k = r(cos 0, sin 0), and use r(r 2 + m 2 ) - l / 2 ~< 1 and [(r - ~lcos 01) 2 + ~2 sin20 + m 2] -1/2 ~< 31/2(1r- ~lcos 01[ + ~[sin 0l + m) - 1 .
f~odr [ ] is easily done and yields a logarithmic function of cos 0 and sin 0 which is integrable for all ~ ~<~. Q.E.D. In the continuum limit, detrem(1 + K) ~ detren(1 - K) (Furry's theorem [9]). L e t K N and PAr (respectively KN(m , n) and IN(m , n)) be the operators obtained by replacing X by - X in K N and PN (respectively KN(m , n) and IN(m , n)), respectively. Since lim det(1 + KN) = lim det(1 + KN) = detren(1 + K), we consider Idet{1 + [KN(m, n) + [£N(m, n)]
+K N (m, n)KN(m, n)}l ~< exp [llgN(m, n) + [£N(m, n)lll
+ IIgN(m, n)ll211KN(m, n)ll2] • (See ref. [10] .) See eq. (12). Thus IIFN + I~NII~ ~< const, a ( I m X)2A2(m, n) for almost all (m, n) a s N -+ 0% where Au(m, n) =A. a [a(m, n) + {eu]. In addition IIPNII2 = 2Y.k~XN"ff~N(k) <~const. ZZ(1/a). Thus we have
Lemma 4. Let A u E ~ and let N be large enough. Then
as a --* 0, which, together with lemma 1, completes the proof.
Remarks 2. (1) Introducing space and m o m e n t u m cutoffs into Au, one can regard Au(x ) as gaussian random variables. Assume that the vector field has a large mass/~ > 0. Then Z = fdetren(1 + K) d/l is analytic in { h E C: 1~[
References [1] D. Brydge, J. Fr6hlich and E. Seller, Construction of gauge fields I, II, IHES preprints (1978, 1979). [2] R. Seiler and R. Schrader, Commun. Math. Phys. 61
(1978) 169. [3] D.H. Weingarten, Continuum limit of (QED) 2 on a lattice II, Indiana preprint (1979). 409
Volume 91B, number 3,4
PHYSICS LETTERS
[4] G. Parisi, Phys. Lett. 66B (1977) 387. [5] D.H. Weingarten and J.L. ChaUifour, Continuum limit of (QED) 2 on a lattice I, to be published in Ann. Phys. [6] K.R. Ito, Estimation of functional determinant in QFT, RIMS preprint RIMS-284 ( 1979).
410
21 April 1980
[7] K.R. Ito, Construction of (QED) 2 via lattice gauge theory I, paper in preparation. [8] M. Ltischer, Commun. Math. Phys. 54 (1977) 284. [9] E. Seiler, Commun. Math. Phys. 42 (1975) 163. [10] B. Simon, Adv. Math. 24 (1977) 244. [11] F.J. Dyson, Phys. Rev. 85 (1952) 631.
PHYSICS LETTERS
21 April 1980
ABSOLUTE BOUND OF THE FUNCTIONAL DETERMINANT IN (QED)2 Keiichi R. ITO 1
Department of Physics, University of BielefeM, 4800 Bielefeld 1, Fed. Rep. Germany Received 4 January 1980
Let S = S(x, y) be the two dimensional euclidean fermion propagator and let K = iXSA, where.4 = ~:~t~A;zand {Au(x)} are external vector fields (real fields). Then the gaussian domination holds for the renormalized Matthews-Salam determinant Idetren(1 + K)I ~
Recently several authors proved diamagnetic inequalities for (euclidean) gauge models [ 1 - 3 ] • This is well stated in the lattice gauge approximation and is very basic to ensure the existence of gauge field theory. In this note, we show that the Matthews-Salam determinant [1,2,4] qY = detren(1 + K) = det(4)(1 + K) e - T
with
.,v(k)-- (8,v k.kq
det(l+p)(1-A) = det[(l-A)
eXP(m~__1 A m / m ) l .
Theorem. Let {Au) be real functions in = L 2 N Lq (q > 2), and let S be the euclidean fermion propagator which satisfies anti-periodic boundary conditions at OA. A = [0,L] ®2. Then [ClYl~< exp[~C(Im X)21[A~I[2] ,
is dominated by a gaussian function o f I[Au[[2 = (fA2(x)d2x) 1/2 in two dimensions for any coupling constant X ~ C. Here K = iXS(x, y)~ (y), S(x, y) = (2S7u au + m ) - l ( x , y ) , (Tu)u=0 1 are euclidean 7 matrices ('~u - 7u' (Tu' ~'u}+ - 28uv),~t - E'YuAu, )t2 f d2k .4a(k)Zv(-k)II~tu(k), r = ~ a(27r) 2
p
(1)
(2)
where C is a constant independent of X and Au.
Remarks 1. (1) The definition of ~ is rather technical. Since X ~ R and {Au} is not a random variable like in refs. [3,5,6], the discussions in refs. [3,5,6] cannot be directly applied. See later discussion. (2) As a special case, one can take A = [ - L / 2 , L/2] ®2 with L -+ ~, namely A = R 2. This corresponds to free boundary conditions. I f L < ~, then the form of lluu is slightly different from that below eq. (1), which is not essential now.
k2 ! X
k 4m 2 tanh- 1 + k2) 1/2 ]_ 1 - k( 4m2 + k2)1/2 (4m2
r
the renormalized second amplitude, and 1 Present address: Res. Inst. Math. Sci., Kyoto Univ., Kyoto 606, Japan. 406
We use lattice theory to prove this. Now let AN = a ( n 0 , n l ) with 0 ~
e0 = a ( 1 , 0 ) ,
e l=a(0,1),
(3)
Volume 91B, number 3,4
PHYSICS LETTERS
y. [Ama(X)a[q a2-q < 0% [aAu,a(X)[ <" al-2/q ea with ea -+ 0 as a ~ 0 for almost all x C A~v (as N -+ oo). Thus although X q~ R,
and for x , y E AN, defineBN(X,y ) and FN(X,y):
BN(X,y ) = (2a - 3 + ma-2)Sx,y - a - 3 7 ( x , y) V(x, y),
(4a)
PN(X,y) = --a -3 [U(x,y) -- 1] V(x,y)T(x,y), where 7 ( x , y ) = ~(1 ~Tu),
y = x +-eu,
= 0,
otherwise,
V(x,y) = exp(+irr/2N),
(4b)
y = x + eu,
(4c)
U(x,y) = exp{+iXaAu, a [(x +- eu/2)]}, y = x + eu , = 0,
(4d)
n), a(no, 0)
(1 +- i)~aAu,a)[ <~C]XaAu,a [2
for almost all x E A ~ as a -+ 0. (This is the typical technique in refs. [3,7] and will be used to prove lemmas 2 and 4.)Au(x ) E L 2 means that T and IlAul[ 2 are well defined andA u E Lq (q > 2) is sufficient to ensure K N -+K in some
Cp--{x;l[xllp=(TrlxlP)l/P
p>2.
Lemma 1. [5,7]. I f A u @ c~, then limN~oodet(1 + K N ) = q.~.
otherwise,
[a(0, n) and a(2N, identified], and
]e+-iXaAu, a --
This is implicit in ref. [5] and one can prove this by slight modifications (see also refs. [6,7]).
otherwise,
= 0,
21 April 1980
and
a(no, 2 N )
The key role of V(x, y) is to introduce anti-periodic boundary conditions at OAN [3,8], which is inevitable for considering the transfer matrix. Note that
are
a/2 a/2 Au'a(X) = a-2 =af/2 Zaf/2 Au(x + r/)d2r/'
det(t +KN) = det[RN(Au) ]/det[RN(O)] ,
(5) x 6 A~V.
These are operators on
(8)
where R N = ,O-I(BN + P N ) ' P = m + 2/a. Following refs. [3,8] we introduce the transfer matrix. First consider the g = 0 direction. Then
det[RN(Au) ] = Tr ToUoT 1 U 1 ... T2N_ 1U2N_I. (9) ~ N = {{f(x); x E AN}; Ilfl[ 2 = 0 2
If(x)12},
Here (TI, Ul}2N-1 are operators on the 24N-dimen sional Hilbert space ~ constructed by operating fermion creation operators {c+(l), d+(/)} 2N-1 on a cyclic vacuum ~ E 9 , and {TI, Ul) are explicitly represented in terms of the fermion operators (see refs. [3, 8] for the construction):
xEA N
and one formally finds:
a2
~
BN(X,y)f(y)-+(Tu8 u +m)f(x),
Y~A N
a2
~
2N- 1
P N ( x , y ) f ( y ) ~ i L ~ ( x ) f ( x ),
Ut=ex p lax ~
YEA N
Ao,a[a(l+½),an ]
n=0
asa ~ 0 (or a s N ~ oo). Let
(10)
× [c+(n)c(n) - d+(n)d(n)]l
SN = BN 1 = p2 UN ,
(6)
p
I
On the other hand, T 1 depends only on {A 1,a [al, a (n 5YJJn =0 and is positive when X E R, and analytic for X in a neighborhood co of X = 0 (co -+ C as a -+ 0) [3]. Then Tl(X)* = TI(X),
+ 1 ~]12N-1
where PN > 0, U~ = UN 1, and define
KN=--PNUNFNPN,
K:iXPU~P.
(7)
Now (Au(x)} are usual functions and unfortunately we do not assume X E R. An easy technical assumption is to set °M = L z ( d Z x ) N Lq(dZx), q > 2. Since 407
Volume 91B, number 3,4
PHYSICS LETTERS
2N-1 leq.(9)l ~< l-I [Tr(TIUIU[T[)N] 1/2N /=0 2N-1 = FI {Tr[Tl(X)Ul(2i Im X)TI(X)IN} 1/2N, l=O
Lemma 2. LetA u E c~. Then _ er 2~4. 2 K(2),r N t22 ~< const, u L S "~u
(the constant is independent of (m, n)) as a + 0, where e > 0 and
by H61der's inequality. Repeat the same discussion for /a = 1 for each Tr[TIUIU~T~]N: Idet(l+KN)l<~ l-Ildet[l+KN(m,n)][
21 April 1980
K(N2),r = PN UN['(N2),rPN . Proof.
1/4N2,
(11)
mn
where K N (m, n) = PN UN FN (m, n)PN and PAr(m, n) is the F N function given by the followingAu(x), x @ A~v (with coupling constant Im X):
K(2) r 2 ~< iiF(2),rll~lgmll4 N ~2 (H61der), where IIPNII4 ~< coast. L 2 In a - 1 as a + 0 (see the proof of lemma 3), and IIF(N2)'rll2 ~< coast. (aau)2~4Z 2 .
A o ( x )=
2iA0,
x 0 = a ( e + ~),
x 1 =a(e),
= -2iA0,
x 0 = a ( e + {),
x 1 = a(o),
= 0,
otherwise
Al(X)=
2iA 1,
Q.E.D.
Lemma 3. Let K(N1) = PNUNF(N1)PN. Then
x0=a(e),
xl=a(e+{),
=-2iAl,
x 0 = a(o),
x 1 = a(o +1),
=
otherwise.
0,
Here A 0 = A O,a [a(m + ~ ), an], A 1 = A 1,a [am, a(n + ~)], and "e" and " o " mean even and odd, respectively. For simplicity let KN(m, n) = K N, PN(m, n) = PN = PN,0 + PAr, 1 and ~"= Im X, where PN, u ( x , y ) = 0 unless y = x + e u and
IIK(N1)II2 ~< const. ~'2L2A /z" 2
Proof. Let F(N1)v,1 be the term in F (N1)uwhich contains "),u. Then
=(a2) 2
-
PN, u(x, x + eu) = a-3sinh ~aAux
~
eikx-ik'YF(Nl?v,l(x,y)
x,y~AN 1 2 . - i k ' e +in/2N
~TuAtaiL (~
ts
+
eikev-iTr/2N)
X exp(~aA u + irr/2N)(-7 u + 1),
(+eu) (e, e),
× [*kv,G+~ + (~--' -~)]
X exp(-~aA u + izr/2N)(Tu - 1),
(+eu) (e, o),
x [5k.,i,; + 8k.,k;+~+ (~ + --01,
X exp(~aA u - izr/2N)(-7 u - 1),
(--e~)
(o, e),
X e x p ( - ~ a A u - izr/2N)(7 u + 1),
(-e.)
(o, o),
otherwise.
XO,
where k, k E AN and ~ = 7r/a = 2Nrr/L (--)-0% as N -+,,~). The other term in P~)u takes a similar form after Fourier transformation except for ~/u and signs. Thus it suffices to prove
(12) Here (e, o), for example, means the evenness and oddness of a - l x ~ and a - l x v , respectively. Let F(N1) (respectively F(N1)u) be the term obtained by replacing sinh(a~Au) exp(+a~Au)/a by ~A u, and let "lVP(2)'r(re"
(47r 2/L 2) k ~ x" f f ~ ( k ) ' f f 2 ( k ~ + ~) <~ const., (independent of N), where ~ = (~, 0) and "fiN(k) = a 4
X Y,,x,yEA N e ik(x -Y)PN (x, y)/L2:
spectively [,(2),r~ be the remainder term: F N = 1-(1) ~N N,~ j p(2), r
+ -N
, etc.
\
+ (1/a 2)
~sinea(k
u - rr/L)
) 408
~ 1~4
Volume 91B, number 3,4
PHYSICS LETTERS
21 April 1980
It is easy to see [5] that, whenever k ~ ~W
[[KN(m, n) + KN(m, n)[I 1 ~< C ( I m ~)2L2A 2(m, n),
PN(k) <~e[m2 + k 2 ] - 1 / 4 ,
where C is independent of X, L, (m, n) and N (or a).
PN(k + ~) <~e [m 2 + ([k0[ - ~)2 + k2]-1/4, with some constant c (independent of N). Then we prove that
Thus lemmas (2), (3) and (4) mean (a 2 = L2/4N 2) [det(1 + K N ) det(1 + K N ) ] ~
(4zr2/L 2
~
(m 2 + k2)-1/2
k~.,Ikl<~ X [m 2 + (Ik0l - ~)2 + k 2] - 1 / 2 is bounded by a constant which is independent of = n/a, where A = (2n/L)Z 2. Rewrite this as flkl<~ d2k [ ] + error, and show that the integral part is finite for all ~ ~< oo (then the error term -+ 0 as a -+ 0). Let k = r(cos 0, sin 0), and use r(r 2 + m 2 ) - l / 2 ~< 1 and [(r - ~lcos 01) 2 + ~2 sin20 + m 2] -1/2 ~< 31/2(1r- ~lcos 01[ + ~[sin 0l + m) - 1 .
f~odr [ ] is easily done and yields a logarithmic function of cos 0 and sin 0 which is integrable for all ~ ~<~. Q.E.D. In the continuum limit, detrem(1 + K) ~ detren(1 - K) (Furry's theorem [9]). L e t K N and PAr (respectively KN(m , n) and IN(m , n)) be the operators obtained by replacing X by - X in K N and PN (respectively KN(m , n) and IN(m , n)), respectively. Since lim det(1 + KN) = lim det(1 + KN) = detren(1 + K), we consider Idet{1 + [KN(m, n) + [£N(m, n)]
+K N (m, n)KN(m, n)}l ~< exp [llgN(m, n) + [£N(m, n)lll
+ IIgN(m, n)ll211KN(m, n)ll2] • (See ref. [10] .) See eq. (12). Thus IIFN + I~NII~ ~< const, a ( I m X)2A2(m, n) for almost all (m, n) a s N -+ 0% where Au(m, n) =A. a [a(m, n) + {eu]. In addition IIPNII2 = 2Y.k~XN"ff~N(k) <~const. ZZ(1/a). Thus we have
Lemma 4. Let A u E ~ and let N be large enough. Then
as a --* 0, which, together with lemma 1, completes the proof.
Remarks 2. (1) Introducing space and m o m e n t u m cutoffs into Au, one can regard Au(x ) as gaussian random variables. Assume that the vector field has a large mass/~ > 0. Then Z = fdetren(1 + K) d/l is analytic in { h E C: 1~[
References [1] D. Brydge, J. Fr6hlich and E. Seller, Construction of gauge fields I, II, IHES preprints (1978, 1979). [2] R. Seiler and R. Schrader, Commun. Math. Phys. 61
(1978) 169. [3] D.H. Weingarten, Continuum limit of (QED) 2 on a lattice II, Indiana preprint (1979). 409
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PHYSICS LETTERS
[4] G. Parisi, Phys. Lett. 66B (1977) 387. [5] D.H. Weingarten and J.L. ChaUifour, Continuum limit of (QED) 2 on a lattice I, to be published in Ann. Phys. [6] K.R. Ito, Estimation of functional determinant in QFT, RIMS preprint RIMS-284 ( 1979).
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[7] K.R. Ito, Construction of (QED) 2 via lattice gauge theory I, paper in preparation. [8] M. Ltischer, Commun. Math. Phys. 54 (1977) 284. [9] E. Seiler, Commun. Math. Phys. 42 (1975) 163. [10] B. Simon, Adv. Math. 24 (1977) 244. [11] F.J. Dyson, Phys. Rev. 85 (1952) 631.