Table Of ContentCOUNTING THE LOCAL FIELDS IN SG THEORY.
Feodor A. Smirnov
Laboratoire de Physique Theorique et Hautes Energies,
4 place Jussieu, Paris 75005, FRANCE
and
5
Steklov Mathematical Institute,
9
9 Fontanka 27, St. Petersburg 191011, RUSSIA
1
n Abstract. In terms of the form factor bootstrap we describe all the local fields
a in SG theory and check the agreement with the free fermion case. We discuss the
J interestingstructureresponsibleforcounting thelocalfields.
8
1
2 1. Introduction.
v
9 In the present paper we solve the longstanding problem of the description of all
5 the local fields in SG theory in terms of the form factor bootstrap approach. The
0
results of this paper add an important piece of information to the general picture,
1
so,this papercanbe consideredasanadditionalchaptertothe book[13]. Itmight
0
5 besurprisingthatthesolutionoftheproblemtooksuchalongtime,butitrequired
9 deeperunderstandingofthestructureofSGformfactorsthantheauthorhadwhen
/ the book [13] was written. The most important contributions to the field during
h
t the last several years were added in the papers [5,14,6,15]where the algebraic and
-
p theanaliticpropertiesofthe formfactorswereoutlined. Itshouldbealsosaidthat
e the necessity of a more general construction for the form factors became clear to
h the author while thinking of the form factor bootstrap for massless S-matrices [18]
:
v formulated in [4].
i Let us formulate the problem. It is known that the form factors of any local
X
operator in the integrable field theory are subject to the following axioms:
r
a 1. Riemann-Hilbert problem.
f(β ,··· ,β ,β ,··· ,β )S(β −β )=f(β ,··· ,β ,β ,··· ,β ),
1 i i+1 n i i+1 1 i+1 i n
f(β ,··· ,β ,β +2πi)=f(β ,β ,··· ,β ), (1)
1 n−1 n n 1 n−1
2. Residue condition
2πi res f(β ,··· ,β ,β ,β )= (2)
βn=βn−1+πi 1 n−2 n−1 n
=f(β ,··· ,β )⊗s I −S(β −β )···S(β −β )
1 n−2 n−1,n n−1 1 n−1 n−2
(cid:0) (cid:1)
whereusualconventionsaredone[13],S(β)isatwoparticleS-matrix. Forthesake
of simplicity we do not consider the problem of bound states.
TypesetbyAMS-TEX
1
Theclassificationofallthelocaloperatorsinthetheoryisequivalenttocounting
allthesolutionstothisinfinitesystemofequations. Letusaskourselveswhatisthe
besttheorytoattacktheproblem. Thefirstpossibleideaisthatonehastoconsider
the theory with the simplest kind of S-matrix, for example, we can take the scalar
S-matrix (the theory in which particles do not have internal degrees of freedom).
For such a model the equation (1) is satisfied trivially, so the only problem left is
to satisfy (2) which is reduced to a certain recurrence relation for polynomials in
e±βj. The problem was solved for several scalar S-matrices in [3,8,9]. Notice that
the S-matrices considered in [8,9] are nontrivial ones.
Inthepresentpaperwetaketheoppositedirection. Weconsiderthetheorywith
a complicated S-matrix, which is the soliton S-matrix for SG model:
S = (∂ φ)2+m2cos(βφ) d2x
µ
Z
(cid:0) (cid:1)
In what follows we shall use only the renormalized coupling constant:
8π
ξ =
8π−β2
In that case the equations (1) are very restrictive. Actually, they can be viewed as
a deformation of level 0 Knizhnik-Zamolodchikov equations with trigonometric r-
matrix. The complete solutionto this equationis available,moreoverthe solutions
are parametrized by a simple and understandable structure (deformed cycles on
hyperelliptic surface). So, to solve the equation (2) one has to combine these
solutions. Weshallseethatthisproblemcanbesolvedcompletely,andthusallthe
local operators in SG theory will be described. The space of operators obtained
in this way is complete because it coincides with the space of local operators at
the free fermion point. Moreover, we shall see that the structure responsible for
counting the fields does not depend on the coupling constant. We shall comment
on this important point later.
It has to be said that the problem of description of all the local operators in a
situation similar to the SG one can be considered in the framework of [6] (see also
[11]). In the terminology accepted in [6] the solution of the problem is ”a local
operator is everything commuting with the vertex operator of the second kind”.
We think that the alternative answer given in the present paper is more direct. It
isimportanttounderstandtherelationbetweenthesetwopossibilitiesofanswering
the same question.
Let us describe the plan of the paper. The second and the third sections are
of introductory character: they describe the necessary solutions to KZ equations
before and after the deformation. The forth section is the central one where the
general construction of the local fields is given. In the fifth section we explain the
agreementwith the free fermioncaseand discuss the possibility of takinginto con-
siderationthedisordertypefieldsandthefieldswithgeneralizedstatistics. Finally,
the last section contains the discussion of further problems related to the subject
of the paper.
2. KZ equations for the trigonometric r-matrix on level 0.
Consider the equations
d
b f(b ,··· ,b )=f(b ,··· ,b ) r(b ,b ) (3)
i 1 2n 1 2n i j
db
i (cid:0)Xj6=i (cid:1)
2
where f(b ,··· ,b ) ∈ ((C2)⊗2n)∗, the trigonometric r-matrix r(b ,b ) acts in the
1 2n i j
tensor product of i-th and j-th spaces as follows:
r(b ,b )= 1bi+bjσ3σ3+ bibj (σ1σ2+σ2σ1)
i j 2b −b i j bp−b i j i j
i j i j
We consider only the uncharged sector:
2n
f(b ,··· ,b )Σ3 =0, Σ3 = σ3
1 2n i
Xi=1
Moreover, we are interested in those solutions annihilated by either of the two
operators:
2n
Σ+ = (σ3)j b±41σj3 σ+
± j j j
Xj=1
Let us denote the set of integers {1,··· ,2n} by U. Different components of the
vectors f are counted by the partitions U =T ∪T¯, where to be precise we require
1∈T. The solutions to these equations can be described as follows.
Consider the function:
G (a ,··· ,a )=
T 1 n−1
1 1
= det(cid:12) (a −b )(cid:12)
(b −b ) (cid:12)a −b k j (cid:12)
i j (cid:12) k i j∈YT,j6=i (cid:12)
16=i<Qj,i,j∈T (cid:12) (cid:12)k=1,···,n−1; i∈T,i6=1
(cid:12) (cid:12)
(cid:12) (cid:12)
Then the solutions are given by
f±{γ1,···,γn−1}(b1,··· ,β2n)T = (bi−bj)41 (4)
Yi<j
±1 ∓1 da1 da1
× b 4 b 4 ··· G (a ,··· ,a )
i i Z P(a ) Z P(a ) T 1 n−1
iY∈T iY∈T¯ γ1 p 1 γn−1 p n−1
where {γ ,··· ,γ } are arbitrarycycles on the hyperelliptic surfacef2 = P(a)
1 n−1
for P(a)= (a−b ). p
j
LetuscoQuntthenumberofdifferentsolutions. Atthefirstglancewehave2Cn−1
2n−2
of them: 2 is for ± and Cn−1 is the number of choices of cycles. However this is
2n−2
greater than the dimension of the space where f live: 2(Cn −Cn−1). How to
± 2n 2n
resolve this contradiction? There must be a linear dependence between different
solutions. Let us explain where it comes from.
Thedifferentialswhichwehavein(4)areobviouslylinearcombinationsofthose
of the first and of the second kind, i.e. they do not have residues on the surface.
For the periods of such differentials we have Riemann bilinear identity:
n−1
ω ω − ω ω = res(Ω ω )≡ω ◦ω
1 2 1 2 1 2 1 2
(cid:18)Z Z Z Z (cid:19)
Xi=1 αi βi αi βi pXoles
3
whereΩ isaprimitivefunctionforω ;α ,β isacanonicbasiswiththeintersection
1 1 i i
numbers:
α ◦α =β ◦β =0, α ◦β =δ
i j i j i j ij
It can be chosen as α = δ , β = i δ for δ drawn around the branching
i 2i i k=1 2k−1 i
points bi and bi+1. P
Consider the differentials from (4):
1 da
ζ = (a−b ) , i∈T\1,
T,i j
a−bi j∈YT,j6=i P(a)
p
for them we have
ζ ◦ζ =0
T,i T,j
because their poles and zeros cancel. For that reason one has the following linear
dependence between different solutions
n−1
f{αi,βiγ3,···,γn−1}(b ,··· ,β )=0
± 1 2n
Xi=1
for arbitrary {γ ,··· ,γ }. So, the real number of solutions is
3 n−1
2(Cn−1 −Cn−3 )=2(Cn −Cn−1)
2n−2 2n−2 2n 2n
which resolves the contradiction.
Dealing with the differentials of the first and of the second kind on the hyperel-
liptic surface it is convenient to move the singularities to the infinity. The basis of
the differentials with singularities at the infinity can be taken as
ap−1
η = da, (5)
p
P(a)
p 2n−p
1
ζ = (−1)k+p(k−p)ak−1σ (b ,··· ,b )da
p 2n−p−k 1 2n
P(a)
kX=p
p
wherep=1,··· ,n−1(recallthatthe genusofthe surfaceisn−1),σ (b ,··· ,b )
i 1 2n
are the elementary symmetric polynomials. This basis is canonic:
η ◦η =ζ ◦ζ =0, η ◦ζ =δ
i j i j i j ij
The differentials from (4) can be rewritten in terms of (5). Up to total derivatives
one has
n−1
1
G (a ,··· ,a )∼
T 1 n−1
P(a )da
iY=1 i i
p 1
∼ det|ζ (a )|
(b −b ) T,i k k=1,···,n−1; i=1,···,n−1
i j
i∈TQ,j∈T¯
4
where
A (a)da
T,i
ζ (a)=
T,i
P(a)
p
with the following polynomials:
(a−bi) (a−bi)
AT,i(a)= (a−bi)iQ∈T¯an−i + (a−bi)iQ∈Tan−i
iY∈T + iY∈T¯ +
where [] means that only the polynomial part of the expression in brackets is
+
taken. Considering ζ (a) as components of a vector one has
T,i
ζ (a)=ζ(a)+Cη(a)
T
where C is certain(n−1)×(n−1)matrix. It is easy to construct the differentials
η (a)dualtoζ (a)andthustoobtainasymplecticmatrixconnectingη (a), ζ (a)
T T T T
with η(a), ζ(a) from which I and C are two blocks.
Letussummarizethediscussionofthissection. Weseethatthemostimportant
ingredient of the solutions to (3) is the symplectic matrix composed of all the
periods of all the differentials of the first and of the second kind
P P
1 2
(cid:18)P3 P4(cid:19)
where P are (n−1)×(n−1) matrices:
i
(P ) = η , (P ) = ζ ,
1 i,j j 2 i,j j
Z Z
αi αi
(P ) = η , (P ) = ζ
3 i,j j 4 i,j j
Z Z
βi βi
The solutionsthemselves canbe consideredasmaps fromdifferent polarizations
ofdifferentials(ζ (a)foranyT givessuchapolarization)intodifferentpolarizations
T
of cycles (different half-bases of homologies). Let us also mention that the above
considerationsarenot verydifferent fromthose presentedin [15]for the rationalr-
matrix,forthereasonthatinourparticularcase(level0andspecialsymmetry)the
solutionfortrigonometricandrationalr-matricesareinone-to-onecorrespondence.
3. Solutions to the deformed equations.
Consider the equations
f(β ,··· ,β ,β ,··· ,β )S(β −β )=f(β ,··· ,β ,β ,··· ,β ),
1 i i+1 2n i i+1 1 i+1 i 2n
f(β ,··· ,β ,β +2πi)=f(β ,β ,··· ,β ), (6)
1 n−1 2n 2n 1 2n−1
5
which can be regarded as a deformation of (3), the SG S-matrix is [17]
∞
sin(kβ)sh(π−ξk)
S (β)=S (β)S (β), S (β)=−exp −i 2 ,
ξ ξ,0 ξ ξ,0 Z ksh(πk)ch(ξk)
(cid:0) 2 2 (cid:1)
b 0
shπ(β−πi) 0 0 0
ξ
1 0 shπβ −shπ2i 0
S (β)= ξ ξ ,
bξ shπξ(β−πi) 00 −sh0πξ2i sh0πξβ shπ(β0−πi)
ξ
Similarly to the classical case we consider only those solutions of (6) which are
invariantwith respect to one of the two quantum groups existing in the SG theory
2n
[12]whichmeansthat they areannihilatedbyΣ3 = σ3 andby eitherofthe two
i
iP=1
operators:
2n
Σ+ = (σ3)j e±2πξβjσj3 σ+
± j j
Xj=1
The variables exactly similar to b from the previous section are the following
i
bi =e2ξπβi
The solutions to the equations (6) constitute a linear space over the field of quasi-
constants: symmetric 2πi-periodic functions, i.e. those depending on
B =eβi
i
In what follows we shall use all this notations (β , b , B ).
j j j
Exactly as in the classical case, the most nontrivial part of the solutions is
described by the deformed matrix of the periods of hyperelliptic differentials. The
definition is as follows.
For two given polynomials Q(a) and L(A) (which can also depend on b and on
j
B respectively as on parameters) we consider the paring hQ(a),L(A)i defined by
j
the integral:
∞
2n
hQ(a),L(A)i≡ ϕ˜(α,βj) Q(e2ξπα)L(eα) e2ξπα dα (7)
Z
−∞ jY=1
where
1 π
ϕ˜(α,β)=ϕ(α−β) exp(− ( +1)(α+β))
2 ξ
the function ϕ is given by
∞
sin2kα shπ+ξk
ϕ(α)=C exp−2 2 2 dk
Z k shξk shπk
2
0
6
where C is certain constant [13]. The integral (7) is defined for 1 ≤ deg(L(A)) ≤
2n−2, and with a proper regularization [13] can be defined for an arbitrary poly-
nomial Q(a). We must understand the polynomial Q(a) as the one defining a
deformeddifferentialwhile the polynomialL(A)defines a deformedcycle. There is
a canonic basisof deformeddifferentials whichis definedby the deformationofthe
polynomials involved into (5):
R (a)=ap−1,
p
2n−p
1
S (a)= (−1)k+p(τk−p−τp−k)ak−1σ (b ,··· ,b )
p τ −τ−1 2n−p−k 1 2n
kX=p
where
2π2i
τ =e ξ
The most important property of the deformed periods is that they can be com-
bined into a symplectic matrix. Namely, consider the matrix
P P
P = 1 2
(cid:18)P3 P4(cid:19)
where P are the following (n−1)×(n−1) blocks:
i
(P ) =hR (a),A2ii, (P ) =hS (a),A2ii,
1 i,j j 2 i,j j
(P ) =hR (a),A2i−1i, (P ) =hS (a),A2i−1i
3 i,j j 4 i,j j
It is shown in [16] that
P P −Pt Pt X 0
1 2 4 2 =
(cid:18)P3 P4(cid:19)(cid:18) P3t −P1t(cid:19) (cid:18) 0 X(cid:19)
where X is the symmetric matrix composed of quasiconstants:
∞
A2(i+j)
X = dA
i,j Z (A2+B2)
j
−∞
Q
We could redefine the basis of the deformed cycles combining them with quasicon-
stans in order to eliminate X:
α =A2i, β =(X−1) A2j−1
i i ij
We hope that α and β here (deformed cycles) will not be confused with the
i i
rapidities and the integration variables.
The deformed period matrix allows a great simplification at the free fermion
point ξ = π. In fact, the calculation of the period matrix at this point is not
absolutely trivial, one should be careful with the regularization of the integral (7)
explained in [13]. It can be shown that in the limit ξ =π+ǫ, ǫ→0 the matrix P
has to be rescaled as follows:
1 0
P˜ = P
(cid:18)0 ǫ(cid:19)
7
The matrix P˜ is finite in the limit:
0 I
P˜| = (8)
ξ=π (cid:18)X Y (cid:19)
where
2n 2n−j
1
Y = (k−j)B2i+2k−1σ (B2,··· ,B2 )
ij ′(B2−B2) l 2n−j−k 1 2n
Xl=1 k l k Xk=j
Q
By a simple symplectic transformation of the deformed cycles this matrix can be
transformed into the unit matrix. This is an important fact the meaning of which
for the SG model will be explained later.
Itis possibleto introduce the deformationofthe integralofak-differentialform
over a k-chain:
∞ ∞
k 2n
hQ(a ,··· ,a ),L(A ,··· ,A )i= dα ··· dα ϕ˜(α ,β )
1 k 1 k 1 k i j
Z Z
−∞ −∞ iY=1jY=1
×Q(e2ξπα1,··· ,e2ξπαk)L(eα1,··· ,eαk)e2ξπΣαi
The polynomials Q and L are supposed to be antisymmetric; however, later we
shall take only Q as antisymmetric then L will automatically antisymmetrize itself
under the integral.
We wouldlike to emphasize that contraryto the usualhomologieswhichconsti-
tuteanexterioralgebraovertheringofintegersthedeformedhomologiesconstitute
an exterior algebra over the ring of quasiconstants.
Letusreturntotheequations(6). Theirsolutionscanbeconstructedasfollows:
1
fk1,···,kn−1(β ,··· ,β ) = ζ(β −β )
± 1 2n T i j shπ(β −β )
Yi<j T′∈UX,#T=n ξ i j
i∈TQ,j∈/T
×en2Σβj f˜±k1,···,kn−1(β1,··· ,β2n)T′ w(β1,··· ,β2n)TT′
where ζ(β) is certain special function [13], w(β ,··· ,β )T′ is a special basis in
1 2n T
((C2)⊗2n)∗ [13]. The mostinterestingpartofthe answeristhe functionf˜k1,···,kn−1
±
which is given by
f˜k1,···,kn−1(β ,··· ,β ) =hF (a ,···a ),Ak1···Akn−1)i (9)
± 1 2n T T 1 n−1 1 n−1
where
F (a ,···a )=det|A (a )|
T 1 n−1 T,i j (n−1)×(n−1)
the polynomials A (a) are given by
T,i
i−1
AT,i(a)= (a−τbq) (1−τ2i−2k)ai−k−1(−τ)kσk(bT′)
qY∈T Xk=0
i−1
+τ2i (a−τ−1b ) (1−τ2i−2k)ai−k−1(−τ)kσ (b )
q k T
qY∈T′ Xk=0
8
where b denotes the subset {b } with j ∈ T. There is another useful formula for
T j
f˜:
f˜k1,···,kn−1(β ,··· ,β ) =det hA (a),Akji
± 1 2n T T,i i,j=1,···,n−1
It is known that the functions f satisfy (6) [13]. The arguments of the previous
section concerning the counting of solutions apply perfectly well to the deformed
case. Indeed, it can be shown that
A =S+CR
T
where I and C are blocks of a symplectic matrix. This fact together with the
deformedRiemannbilinearidentityimpliesthefollowinglineardependencebetween
the solutions:
n−1
(X−1) f˜2i,2j−1,k3,···,kn−1(β ,··· ,β )=0 (10)
ij ± 1 2n
Xi=1
forarbitraryk ,··· ,k . RecallthatthematrixX iscomposedofquasiconstants.
3 n−1
It can be shown that the solutions f as functions of β do not have other
2n
singularities for 0 ≤ Imβ ≤ 2π but the simple poles at β = β +πi. If we
2n 2n j
impose this requirement generally and also require that the solutions do not grow
faster than exp(x|β |) for some x as β →±∞ then the totality of solutions to (6)
j j
is generated by
∞
exp t ( elβj) fk1,···,kn−1(β ,··· ,β ) (11)
l ± 1 2n
(cid:0)l=X−∞ X (cid:1)
modulo the relation (10).
4. Local operators in SG theory.
In order to construct a local operator one has to find a solution to the system
composed of (6) and
2πi res f(β ,··· ,β ,β ,β )= (12)
β2n=β2n−1+πi 1 2n−2 2n−1 2n
=f(β ,··· ,β )⊗s I −S(β −β )···S(β −β )
1 2n−2 2n−1,2n 2n−1 1 2n−1 2n−2
(cid:0) (cid:1)
In the previous section we constructed all the solutions to (6), so, our goal is to
learnhow to combine them inorderto satisfy (12). Let us explainthe generalidea
abouttheorganizationofthespaceoflocalfieldsintermsofformfactorbootstrap.
We expect that there are infinitely many local operators which are counted by
the following data:
1. We fix the number of particles (2m) in the minimal form factor. If we fixed 2m
it means that for the given operator the form factors with 2n particles vanish
for n<m.
2. The minimal 2m-particle form factor can be taken in infinitely many possible
ways due to a possible multiplication of the solution to (6) by quasiconstant.
Keeping in mind the general formula (11) we write the following generating
function for the minimal 2m-particle form factor:
∞
(eβi +eβj) exp t ( elβj) fk1,···,kn−1(β ,··· ,β ) (13)
l ± 1 2n T
Yi<j (cid:0)l=X−∞ X (cid:1)
9
modulorelation(10). Thisformulaneedssomecomments. Theproduct (eβi+
i<j
eβj) is put in order to cancel the poles of f±k1,···,kn−1(β1,··· ,β2n) whichQis needed
for the form factor to be the minimal one: the minimal form factor should not
be related to those with lower number of particles by (12). Different minimal
form factors are obtained from (13) applying any differential polynomial in t and
l
putting t = 0 afterwards. The times t play a very different role for l odd and
l l
even. The times t can be identified with those related to the local integrals of
2p+1
motion, in particular t = z, t = z¯ (z and z¯ are usual euclidian coordinates) .
1 −1
∞
The multiplication of the form factors by exp( t2p+1( e(2p+1)βj)) is always
p=P−∞ P
possible because it does not spoil the residue condition. The times t are far
2p
more nontrivial and, hence, more interesting: they do not correspond to any local
symmetry of the theory, but still they must be considered. To summarize, in the
formula (13) we have three structures counting the minimal form factors: two
bosonic (related to t and t respectively) and one fermionic (related to the
2p+1 2p
counting by k ,··· ,k ). We shallshow thatto every minimal formfactor (13)a
1 n−1
local operator can be related.
Let us consider the problem of calculation of the residue (12). Generally the
formulaeforresidue offk1,···,kn−1(β ,··· ,β ) cannotbe written interms ofthe
± 1 2n T
functions of the same kind. There exists, however, one nice possibility which we
are going to describe. It is useful to generalize the previous notations considering
the functions
fLn(β ,··· ,β )
± 1 2n T
where L (A ,··· ,A ) is an arbitrary polynomial whose degree in any vari-
n 1 n−1
able is between 1 and 2n − 2. The generalization corresponds to the insertion
of L (A ,··· ,A ) instead of Ak1···Akn−1 in the formula (9). The polynomial
n 1 n−1 1 n−1
L (A ,··· ,A ) can be taken as an antisymmetric one, but we prefer to have it
n 1 n−1
partly antisymmetric, namely, antisymmetric in the variables A ,··· ,A for some
1 k
givenk≤n−1. The polynomialL (A ,··· ,A )canalsodependsymmetrically
n 1 n−1
on B as on parameters, we shall explicitly write this arguments when needed.
j
The only situation where a good formula for the residue at β = β +πi
2n 2n−1
exists is the following one
L (A ,··· ,A ,A ··· ,A |B ,··· ,B ,B ,B )| =
n 1 k k+1 n−1 1 2n−2 2n−1 2n B2n−1=−B2n≡B
k
= (−1)i (A2+B2)L{i}(A ,··· ,A ,A ··· ,A |B ,··· ,B |B)
l n 1 k k+1 n−1 1 2n−2
Xi=1 Yl6=i (14)
for some polynomials L{i}. Provided the equation (14) holds we have
n
2πi res fLn(β ,··· ,β ,β ,β )=
β2n=β2n−1+πi ± 1 2n−2 2n−1 2n
=fL˜+n−1(β ,··· ,β )⊗s +
± 1 2n−2 2n−1,2n
+fL˜−n−1(β ,··· ,β )⊗s S(β −β )···S(β −β )
± 1 2n−2 2n−1,2n 2n−1 1 2n−1 2n−2
where
L˜± (A ,··· ,A ,A ··· ,A |B ,··· ,B |B)=
n−1 1 k−1 k n−2 1 2n−2
=L{k}(A ,··· ,A ,±iB,A ··· ,A |B ,··· ,B |B) (15)
n 1 k−1 k n−2 1 2n−2
10
