Exact solution and interfacial tension of the six-vertex model. M. T. Batchelory, R. J. Baxterzx, M. J. O'Rourkex and C. M. Yungy

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1 Exact solution and interfacial tension of the sixvertex model with antiperiodic boundary conditions PostScript processed by the SLAC/DESY Libraries on 7 Feb 995. M. T. Batchelory, R. J. Baxterzx, M. J. O'Rourkex and C. M. Yungy ydepartment of Mathematics, School of Mathematical Sciences, zschool of Mathematical Sciences, xdepartment of Theoretical Physics, RSPhysSE, Australian National University, Canberra, ACT, Australia January 995 Abstract We consider the sixvertex model with antiperiodic boundary conditions across a nite strip. The rowtorow transfer matrix is diagonalised by the `commuting transfer matrices' method. From the exact solution we obtain an independent derivation of the interfacial tension of the sixvertex model in the antiferroelectric phase. The nature of the corresponding integrable boundary condition on the XXZ spin chain is also discussed. HEPTH954 Short Title: The antiperiodic sixvertex model ANU preprint MRR 95, hepth/954

2 Introduction and main results The sixvertex model and related spin XXZ chain play a central role in the theory of exactly solved lattice models []. Typically the sixvertex model is `solved' by diagonalising the rowtorow transfer matrix with periodic boundary conditions. Several methods have evolved for doing this, including the coordinate Bethe ansatz [, ], the algebraic Bethe ansatz [3, 4], and the analytic ansatz [5]. All of these methods rely heavily on the conservation of arrow ux from row torow of the lattice. In terms of the vertex weights (see gure ) a = sinh ( v); b = sinh ( + v); c = sinh (.) the transfer matrix eigenvalues on a strip of width N are given by [] where (v)(v)=( v)(v+ )+(+v)(v ) (.) = i (.3) (v) = N sinh N ( v) (.4) (v) = ny k= The Bethe ansatz euations follow from (.) as sinh (v v k): (.5) ( v j ) ( + v j ) = (v j ) (v j + ; j=;:::;n: (.) ) The integer n labels the sectors of the transfer matrix. Here we consider the same sixvertex model with antiperiodic boundary conditions. That such boundary conditions should preserve integrability is known through the existence of commuting transfer matrices []. However, the solution itself has not been found previously. In section we solve the antiperiodic sixvertex model by the `commuting transfer matrices' method []. This approach has its origin in the solution of the more general 8vertex model [7], which like the present problem, no longer enjoys arrow conservation. We nd the transfer matrix eigenvalues to be given by (v)(v)=( v)(v+ ) (+v)(v ) (.7) where now (v)= NY k= sinh 4 (v v k): (.8)

3 In this case the Bethe ansatz euations are ( v j ) ( + v j ) = (v j ) (v j + ; j=;:::;n: (.9) ) In contrast with the periodic case the number of roots is xed at N. In section 3 we use this solution to derive the interfacial tension s of the sixvertex model in the antiferroelectric regime. Dening x =e, our nal result is Y +x e s=kbt 4m =x (.) +x 4m m= in agreement with the result obtained from the asymptotic degeneracy of the two largest eigenvalues [, 8]. With antiperiodic boundary conditions on the vertex model, the related XXZ Hamiltonian is H = NX j= j x x + j+ y j y j+ + cosh j z z j+ where x ; y and z are the usual Pauli matrices, with boundary conditions (.) x N+ = x ; y N+ = y ; z N+ = z : (.) This boundary condition has appeared previously and is amongst the class of toroidal boundary conditions for which the operator content of the XXZ chain has been determined by nitesize studies [9]. It is thus an integrable boundary condition, with the eigenvalues of the Hamiltonian following from (.7) in the usual way [], with result E = N cosh NX j= cosh sinh sinh v (.3) j + sinh : We anticipate that the approach adopted here may also be successful in solving other models without arrow conservation. The solution given here can be extended, for example, to the spins generalisation of the sixvertex model/xxz chain []. Exact solution To obtain the result (.7) we adapt where appropriate the derivation of the periodic result (.) (specically, we refer the reader to Ch. 9 of Ref. []). We depict a vertex and its corresponding Boltzmann weight graphically, as w(; j; )= 3

4 where the bond `spins' ; ; ; are each + if the corresponding arrow points up or to the right and if the arrow points down or to the left. Thus in terms of the parametrisation (.) the nonzero vertex weights are w(+; +j+; +) = w( ; j; ; )=a w(+; j ; +) = w( ; +j; +; )=b (.) w(+; j+; )=w( ;+j; ;+) = c The rowtorow transfer matrix T has elements X X T = ::: N N+ N N N (.) where = f ;:::; N g, =f ;:::; N g, and the antiperiodic boundary condition is such that N+ =.Nowconsider an eigenvector y of the form y = g g g N (.3) where g i ( i ) are twodimensional vectors. From (.) the product Ty can be written as (Ty) =Tr[G ( )G ( )G N ( N )S] (.4) where G i () are matrices with elements X G i () = g i (): (.5) The appearance of the spin reversal operator S A (.) in (.4) is the key dierence with the periodic case. However, it does not eect much of the working. Using (.) and (.5) we have G i (+) ag i(+) cg i ( ) bg i (+) A G i ( )=@ bg i( ) cg i (+) A : (.7) ag i ( ) In particular, there still exist the same matrices P ;:::;P N such that G i () =P i H i ()P i+ (.8) 4

5 where P i and H i are of the form P i p i(+) p i ( ) A H i () g i() gi () gi () A : (.9) As for the periodic case, (.8) follows from the local `pairpropagation through a vertex' property, i.e. the existence of g i ();g i();p i ();p i+ () such that X ; w(; j; )g i ()p i+ ( )=g i()p i () (.) for ; =. The available parameters are [] g i (+) = ; g i ( )=r i e (+v)i= g i(+) = a; g i( )= ar i e (3+v)i= (.) p i (+) = ; p i ( )=r i where i = and r i =( ) i re (+:::+i ) : (.) However, p N+ needs to be dierent from the periodic case (where p N+ = p ). The antiperiodicity suggests that we p N+(+) p N+ ( ) A = p ( ) p (+) A (.3) where h is some scalar. Since we already reuire p i (+) = and p i ( )=r i,wemust have r = h = r and r N+ = h = r : (.4) In addition, r =( ) N e (+:::+N ) : (.5) To proceed further, we write P and P N+ in full, P p (+) (+) p ( ) ( ) A P N+ hp ( ) N+ (+) hp (+) N+ ( ) A : (.) Then P N+ SP = N+( ) N+ (+) detp N+ hp (+) hp ( =h h detp detp N+ 5 A A : (.7)

6 Putting the pieces together we then have (Ty) = Tr P H ( )H N ( N )P N+ S However, as for the periodic case, we have which follows from (.7) to (.9). Thus = h g ( ) g N( N ) h detp detp N+ g ( ) g N( N ): (.8) g i ( i )=ab g i ( i )detp i+ g i( i )detp i (.9) (Ty) = rg ( ) g N( N )+ r (ab)n g ( ) g N( N ) g ( )g N( N ) : (.) At this point it is more convenient to write y(v) =h (v)h (v)h N (v) (.) where we have dened h i (v) r i e (+v)i A : (.) The result (.) can then be more conveniently written as T (v)y(v)= ra N y(v + )+ r bn y(v ): (.3) Also let y (v) = exp( with r = exp( P N i= i). Then from (.3) we have v 4 NX i= i ) y() (.4) T(v)y (v)=( v)y (v+ )(+v)y (v ): (.5) To proceed further, let Q R(v) be a matrix whose columns are a linear combination of y with dierent choices of ( N altogether). It follows from (.5) that T (v)q R(v)=( v)q R(v+ )(+v)q R(v ): (.) One can show that the transpose of the transfer matrix has the property T ( v) = t T(v). With Q L(v) = t Q R( v) it follows from (.) that Q L(v)T (v)=( v)q L(v+ )(+v)q L(v ): (.7)

7 Now let Q R (v) =Q + R(v) and Q L (v) =Q + L(v). Then we can show that the \commutation relations" Q L (u)q R (v)=q L (v)q R (u) (.8) hold for arbitrary u and v. This result follows if we can prove that F = t y ( u)y + (v) is a symmetric function of (u; v) for all choices of ;. Using (.4), (.) and (.) this function reads F = exp u 4 8 < NX i= i v 4 NX i= i! NY h P N ( ) i= (i+ i ) j= exp : ( u) j + ( + v) j 4 N X i=j ( i + i) j X i= ( i + i) 393 = 5 ; 5 : (.9) Now suppose that in ; there are p pairs ( ik ;i k ) where ik +i k = with k =;:::;p. The terms in F which involve these ik (in the prefactor and in the j = i k terms) are manifestly symmetric in (u; v). The remaining terms are exactly of the form (.9) with N! N p after relabelling of sites. We can thus restrict ourselves to the case where i = i ;i =;:::;N, for all N.To prove this case we proceed inductively. From (.8) we have F = NY j= h e 4 (u v)j ( ) N e 4 (v u)j e (N +:::+j+) e (j +:::+) i : (.3) Let us now denote F = F N ( ;:::; N ). By inspection, F ( ) and F ( ; ) are symmetric in (u; v). Suppose F N ( ;:::; N ) is symmetric in (u; v) and furthermore that k + k+ = for some k. Then from (.3) wehavef N ( ;:::; k ; k ;:::; N )=F N ( ;:::;^ k ;^ k+ ;:::; N ) times a symmetric function of (u; v), which is therefore symmetric in (u; v). This is true for all k N. The only case left to consider is therefore = = :::= N. But from (.3) we havef N ( ; ;:::; N ; )=F N ( ;:::; N ) times a symmetric function of (u; v), which is again symmetric. Thus by induction on N, the assertion (.8) follows. dene As in the periodic case, we assume that Q R (v) isinvertible at some point v = v Then from (.7) and (.8) we obtain and Q(v) =Q R (v)q R (v )=Q L (v )Q L (v): (.3) T (v)q(v)=q(v)t(v)=( v)q(v+ ) (+v)q(v ) (.3) and Q(u)Q(v) =Q(v)Q(u). This allows T (v);q(v) and Q(v ) to be simultaneously diagonalised, yielding the relation (.7) for their eigenvalues. The precise functional form of Euivalent results are obtained using the other choice of sign. 7

8 the eigenvalue (v) ofq(v), given in (.8), follows from (.3) by noting that T (v +i) = T (v) and considering the limits v!. 3 Interfacial tension In this section we derive the interfacial tension by solving the functional relation (.7) and integrating over the band of largest eigenvalues of the transfer matrix []. We consider the case where N, the number of columns in the lattice, is even. The partition function of the model is expressed in terms of the eigenvalues (v) of the rowtorow transfer matrix T (v) as X Z= [(v)] M (3.) where the sum is over all N eigenvalues. The interfacial tension is dened as follows. Consider a single row of the lattice. For a system with periodic boundary conditions, in the!limit we see from (.) that the vertex weight c is much greater than the weights a and b, so in this limit, the row can be in one of two possible antiferroelectrically ordered ground states. These are made up entirely of spins with Boltzmann weight c, and are related to one another by arrowreversal. When we impose antiperiodic boundary conditions, this groundstate conguration is no longer consistent with N even. To ensure the antiperiodic boundary condition, vertices with Boltzmann weight c must occur an odd number of times in each row. Thus the lowestenergy conguration for the row in the!limit will consist of N vertices with weight c, and one vertex of either types a or b. This dierent vertex can occur anywhere in the row. As we add rows to form the lattice, the a or b vertex in each row forms a \seam" running approximately vertically down the lattice; it can jump from left to right but the mean direction is downwards. y Atypical lowestenergy conguration is shown in gure. The extra free energy due to this seam is called the interfacial tension. This will grow with the height M of the lattice, so we expect that for large N and M the partition function of the lattice will be of the form Z exp [( NMf Ms)=k B T ] (3.) where f is the normal bulk free energy, and s is the interfacial tension. We introduce the variables x = e ; z = e v= : (3.3) y This is the analogue of the antiferromagnetic seam in the Ising model []. 8

9 Expressing the Boltzmann weights in terms of z and x, from (.) the model is physical when z and x are real, and z lies in the interval x = z x = : (3.4) We consider in order that the Boltzmann weights are nonnegative, so we must have x. Let ~Q(z) = where z j = e vj =, j =;:::;N, and NY j= (z z j ) (3.5) V (z) =(v)(z ) N ( ) N= : (3.) In terms of these variables, the functional relation (.7) becomes ~Q(z)V (z) =( z x ) N Q( ~ zx) ( z x) N Q( ~ zx ): (3.7) Both terms on the right hand side of (3.7) are polynomials in z of degree 3N, but the coecients of and z 3N vanish, so z V (z) is a polynomial in z of degree N. We know how to solve euations of this form for both V (z) and Q(z) ~ using WienerHopf factorisations (see references [7, 8] and [3]). We shall need some idea where the zeros of the polynomials Q(z) ~ and V (z) lie in order to construct the WienerHopf factorisations. From the antiperiodicity oft(v)we see that V (z) is an odd function of z, V ( z) = V(z) (3.8) so its zeros and poles must occur in plus{minus pairs. To locate the zeros in the zplane, we consider z to be a free variable, and vary the parameter x, in particular looking at the limit x!. We nd the following; in the x! limit, N of the N zeros of Q(z) ~ lie on the unit circle, the other two lying at distances proportional to x = and x =.For V (z), there is the simple zero at the origin, and two zeros on the unit circle. The remaining N 4 zeros of V (z) are divided into two sets, with N of them that approach the origin and N that approach as x!. The N zeros of the two polynomials that lie on the unit circle are spaced evenly around the circle. As x is increased, the zeros of Q(z) ~ and z V (z) will all shift. We assume that the distribution of the zeros mentioned above does not change signicantly as x increases. Thus the zeros that lie at the origin in the x! limit move out from the origin as x increases, 9

10 but not so far out as the unit circle, and similarly for the zeros that lie at. Also, the zeros that lie on the unit circle are assumed to stay in some neighbourhood of the unit circle as x increases (we will show that these zeros remain exactly on the unit circle as x increases, which is what happens in the periodic boundary condition case). Bearing in mind the above comments, we write ~Q(z) = Q ~ (z)(z )(z ) (3.9) where ~ Q (z) is a polynomial of degree N whose zeros are O() as x!, and ; = O(x = ), so lies inside the unit circle, outside. Also, let V (z) =z(z t )(z t )A(z)B(z), where A(z) and B(z) are both polynomials of degree N, the zeros of A(z) being all the zeros of V (z) that lie inside the unit circle, B(z) containing all those that lie outside, and t and t are the zeros that lie on the unit circle. Since V (z) is an odd function, both A(z) and B(z) must be even functions of z, and we must have t = t, so letting t = t = t, we write V (z) =z(z t )A(z)B(z): (3.) Draw the contours C + and C in the complex zplane, both oriented in the positive direction, with C outside the unit circle, C + outside C, and such that there are no zeros of ~Q(z) orv(z) on the boundary of or inside the annulus between C and C +. Then and all the zeros of B(z) lie outside C + (see gure 3). Dene r(z) as the uotient of the two terms in the RHS of the functional relation (3.7); r(z) = ~ Q( zx )( z x) N ~Q( zx)( z x ) N (3.) (r(z) has no zeros or poles on or between the curves C + and C ). Then in the x! limit, we see that jr(z)j=z N, so when jzj >, jr(z)j <. Thus ln[ + r(z)] can be chosen to be singlevalued and analytic when z lies in the annulus between C and C +.We can therefore make a WienerHopf factorisation of + r(w) by dening the functions P + (z) and P (z) as ln P (z) = i I C ln [ + r(z )] dz z z (3.) Then P + (z) is an analytic and nonzero (ANZ) function of z for z inside C +, and P (z) isan ANZ function of z for z outside C.Asjzj!,we note that P (z)!. When z is inside the annulus between C and C +,wehave, by Cauchy's integral formula +r(z)=p + (z)p (z)= V(z) ~ Q(z) ~Q( zx)( z x ) N : (3.3)

11 We then dene the functions V (z); V + (z) = P + (z) ~ Q( zx)=(z ) (3.4) V (z) = P (z)( z x ) N.h ~ Q (z)(z ) i (3.5) where V + (z) is an ANZ function of z for z inside C +, V (z) an ANZ function of z for z outside C.Wehave split V (z) into two factors, V + (z) and V (z), with V (z) =V + (z)v (z) when z is between C + and C. Euating (3.) with the expression V (z) =V + (z)v (z)wehave V + (z) B(z) = A(z) V (z) z(z t ): (3.) The LHS (RHS) is an ANZ function of z inside C + (outside C ), which is bounded as jzj! and so the function must be a constant, c say. Thus V + (z) = c B(z) (3.7) V (z) = c z(z t )A(z): (3.8) When jzj <, we proceed the same way. Draw the curves C and + C, C + inside the unit circle, C inside C +, and with and all the zeros of A(z) inside C. In the limit x!, j=r(z)j z N,soj=r(z)j <. Thus ln[ + =r(z)] can be chosen to be singlevalued and analytic between and on C + and C.We can then WienerHopf factorise +=r(z) by dening the functions P +(z) and P (z) as ln P (z) = i I C ln + dz r(z ) z z ; (3.9) where P + (z) is ANZ inside C +, P (z) is ANZ for z outside C. As jzj!,p (z)!. When z is in the annulus between C + and C, Cauchy's integral formula now implies + r(z) =P +(z)p (z)= V(z) ~ Q(z) ~Q( zx )( z x) N : (3.) Dene V + (z) and V (z) as follows: V + (z) = P + (z)( z x) N.h ~ Q (z)(z ) i (3.) V (z) = P (z) Q( ~ zx )=(z ): (3.) We havenow factorised V (z) into two factors, V+(z) which is is ANZ for z inside C+, and V (z) which is ANZ for z outside C. When z is in the annulus between C + and C,wehave the euality V (z) =V+(z)V (z).

12 When z is inside this annulus, we euate (3.) with V (z) =V +(z)v (z) to get V +(z) B(z)(z t ) = za(z) V (z) (3.3) where now the LHS (RHS) is an ANZ function of z for z inside C + (outside C ). Thus both sides of the euation are constant, c say, and we have V +(z) = c (z t )B(z) (3.4) V (z) = c za(z): (3.5) From euations (3.7), (3.4) and (3.8), (3.5), we have the following V +(z) = (c =c )V + (z)(z t ) (3.) V (z) = (c =c )V (z)(z t ): (3.7) To evaluate the constant c =c, consider (3.7) in the limit z!;we noted earlier that P (z), P (z)! asz!, so from (3.5), (3.5) and (3.) we deduce that c =c =: (3.8) We may use euations (3.) and (3.7) to derive recurrence relations satised by ~ Q(z), which we can solve explicitly in the N!limit. From euations (3.4), (3.) and (3.), we deduce the recurrence relation ~Q(z) Q( ~ zx) =( z x) N(z )(z ) P+(z) (z t ) P + (z) (3.9) valid when z is inside C+. In the limit N!, the P + and P + functions!, so we nd that Q(z) ~ isgiven by Y ~Q(z) = Q() ~ z x 4m 3N ( z t x 4m ) ( z x m ) ( zx m ) z x 4m ( z t x 4m 4 ) ( + z x m ) ( + zx m ) : (3.3) m= This still contains the parameters t, and. From (3.9) in the N!limit, setting z = we note that h ~Q() i = t : (3.3) From euations (3.5), (3.) and (3.7), we get the recurrence relation ~Q(z) Q( ~ zx )=( z x ) N(z )(z ) P (z) (z t ) P (z) (3.3)

13 which isvalid for z outside C. Taking the limit N!once more, so that the functions P (z) and P (z)!, we get Y z ~Q(z) =z N x 4m 3 N ( z t x 4m ) ( z x m ) ( z x m ) z x 4m ( z t x 4m 4 ) ( + z x m ) ( + z x m ) : m= (3.33) To derive an expression for V (z) valid between C + and C, using euation (3.7), we have V(z) = V + (z)v (z)(z t ) = ~ Q( zx) ~ Q( zx )(z t ) (z )(z ) : (3.34) We use (3.3) for ~ Q( zx) and (3.33) for ~ Q( zx ), and substitute into euation (3.34). This produces a lengthy expression for V (z) involving the parameters ; and t, which simplies when one considers the oddness of V (z). The poles of V (z) must occur in pairs, and this is only possible if and are related by = x: (3.35) Substituting this in, the innite products involving and cancel, and we get, from (3.) and (3.34) where (v) =G(z=t)(=x) N Y m= G(z) =ix = (z z ) Y m= z x 4m z x 4m+ z x 4m z x 4m+ N (3.3) z x 4m z x 4m z x 4m z x 4m : (3.37) This expression for the eigenvalue is still dependent on the parameter t, dierent values of t corresponding to dierent eigenvalues of the transfer matrix. All we know about t so far is that it is bounded as x!, and that it lies on the unit circle in the x! limit. We shall now show that it in fact remains exactly on the unit circle as x increases. We substitute into the functional relation (3.7), using euations (3.3) and (3.33) to get an expression for the product ~ Q(z)V (z) which isvalid when z is in the annulus between C + and C. Substituting into (3.7), the function on the right hand side is eual to zero when z is one of the N zeros of ~ Q (z), or when z = t. For the latter case, substituting z = t and t, and dividing the resulting euations, we arrive at the following relation between ; x; and t = t x (3.38) which means that t must satisfy [(t)] N = (3.39) 3

14 where (t) is given by (t) =t Y m= t x 4m t x 4m 3 t x 4m 3 t x 4m : (3.4) This implies that t lies on the unit circle for all x, there being N possible choices for t. The partition function depends on t only via t, so there are only N distinct eigenvalues. The right hand side of (3.7) also vanishes when z is a zero of ~ Q (z) so in the same way we show that the zeros of ~ Q (z) lie exactly on the unit circle for all x. As the zeros lie exactly on the unit circle, we may shift the curves C and C + so that they just surround the unit circle. Hence our expressions for ~ Q(z) are valid all the way up to the unit circle; (3.3) is valid for jzj <, and (3.33) is valid for jzj >. We now evaluate the partition function, as dened in (3.), in the largelattice limit. When v is real, the eigenvalues (3.3) are complex, so as N!, the partition function, a sum over the N eigenvalues dened by (3.39), becomes an integral over all the allowed values of t, Z = I (t)[(v)] M dt (3.4) where the integral is taken around the unit circle, and (t) is some distribution function, independent of N and M. Substituting (3.34) into (3.4) then gives an expression for Z. (The number of rows M is even to ensure periodic boundary conditions vertically, and so the sign in (3.37) is irrelevant.) The eigenvalue (3.3) contains two distinct types of factors; those that are powers of N, and those that are not. The terms that increase exponentially with N contribute to the bulk part of the partition function, the free energy per site in the thermodynamic limit. factor is also independent of t, and can be taken out of the integral (3.4). The integral is then independent of N, so we have, from (3.) e f=kbt =(=x) Y m= z x 4m z x 4m+ z x 4m z x 4m+ This (3.4) for the free energy per site in the thermodynamic limit. This result agrees with the result for periodic boundary conditions (euations (8.9.9) and (8.9.) of Ref. []). From euation (3.), the other factors in (3.34) make up the interfacial tension, given by e Ms=kBT = I (t)[g(z=t)] M dt: (3.43) For M suciently large, we may evaluate this integral using saddlepoint integration. The integral is given by the value of the integrand at its saddle point, together with some multi 4

15 plicative factors that we can disregard as M!. The function G satises the relation G(z) =G( =z) (3.44) which implies that the function has a saddle point when z = i. Hence the integrand in (3.43) is maximised when t = t saddle = iz: (3.45) As z is arbitrary, this point may lie o the unit circle; it will however lie inside the annulus between C + and C because of the restriction (3.4), and so we will be able to deform the contour to pass through this saddle point. Hence we arrive at the nal result Y +x e s=kbt 4m =x = +x 4m : (3.4) Acknowledgements MTB and CMY thank the Australian Research Council for nancial support. m= 5

16 References [] Baxter R J 98 Exactly Solved Models in Statistical Mechanics (London: Academic) [] Lieb E H and Wu F Y 97 in Phase Transitions and Critical Phenomena vol, ed C Domb and M S Green (London: Academic) p 3 [3] Sklyanin E K, L Takhtajan and L Faddeev 979 Theor. Maths. Phys [4] Kulish P P and Sklyanin E K 98 in Lecture Notes in Physics vol 5, ed J Hietarinta and C Montonen (Berlin: SpringerVerlag) p [5] Reshetikhin N Yu 983 Sov. Phys. JETP 57 9 [] de Vega H J 984 Nucl. Phys. B [7] Baxter R J 97 Ann. Phys. (N.Y.) 7 93 [8] Baxter R J 973 J. Stat. Phys. 8 5 [9] Alcaraz F C, Baake M, Grimm U and Rittenberg V 988 J. Phys. A L7 [] Yung C M and Batchelor M T 995 Exact solution for the spins XXZ uantum chain with nondiagonal twists, ANU preprint [] Johnson J D, Krinsky S and McCoy B M 973 Phys. Rev. A 8 5 [] Onsager L 944 Phys. Rev. 5 7 [3] Noble B 958 Methods Based on the WienerHopf Techniue (London: Pergamon)

17 a a b b c c Figure : Standard vertex congurations and corresponding weights. Figure : A typical lowestenergy state of the system with N even and antiperiodic boundary conditions. The dotted line indicates the interface dividing the lattice into two domains, each of which is an antiferroelectrically ordered ground state. 7

18 + t + + C C α + + t + + β Figure 3: The complex zplane; the curves C + and C are indicated, with the unit circle lying inside C. The zeros of Q ~ are indicated by and the zeros of z V (z) by +. There are no zeros of either function in the annulus between the contours C + and C. 8