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THE POKROVSKI-TALAPOV PHASE TRANSITION

arXiv:hep-th/9202082v1 24 Feb 1992CERN-TH.6411/92THE POKROVSKI-TALAPOV PHASE TRANSITIONAND QUANTUM GROUPS ∗Haye HinrichsenUniversit¨at Bonn, Physikalisches InstitutNussallee 12, W-5300 Bonn 1, FRGVladimir RittenbergTheory Division, CERNCH-1211 Geneva 23, SwitzerlandAbstractWe show that the XY quantum chain in a magnetic field is invariant under a two parameterdeformation of the SU(1/1) superalgebra. One is led to an extension of the braid group andthe Hecke algebras which reduce to the known ones when the two parameter coincide.

Thephysical significance of the two parameters is discussed.When both are equal to one, onegets a Pokrovski-Talapov phase transition. We also show that the representation theory of thequantum superalgebras indicates how to take the appropriate thermodynamical limits.CERN-TH.6411/92February 1992∗To appear in the Proceedings of the II International Wigner Symposium, Goslar 1991

1IntroductionThere were several attempts to extend the one-parameter quantum algebras to multiparame-ter ones [1]. As shown however by Reshetikhin [2] the link polynomials depend only on oneparameter.

One can state this result in a different way: if one has a one-dimensional quan-tum chain which is invariant under a multiparameter quantum algebra, one can do a similaritytransformation which eliminates all the parameters but one. As will be shown in this paper,the situation is different in the case of quantum superalgebras.

We will start with a physicalexample. Consider the quantum chainH = ∆qLXi=1σzi + ∆η2L−1Xi=1[(1 + u)σxi σxi+1 + (1 −u)σyi σyi+1] + B + S,(1)where σx, σy and σz are Pauli matrices inserted in the i-th position of the Kronecker productσki = 1 ⊗1 ⊗.

. .

⊗σk|{z}i⊗. .

. ⊗1 ⊗1(i = 1, 2, .

. .

L)(2)[σki , σlj] = 0. (i ̸= j)∆q, ∆η and u are parameters, B and S are boundary and surface terms respectively.

Thischain appears in the domain wall theory of two-dimensional commensurate-incommensuratephase transitions [3, 10] and in Glauber’s kinetic Ising model [4]. In order to make contactwith quantum algebras we will first make an important change of notations, choose B = 0 (noperiodic boundary conditions!) and fix S by∆q = q + q−12,∆η = η + η−12,u = η −η−1η + η−1(3)S= 12 (q−1σz1 + q σzL).With this change of notations we haveH = H(q, η) =L−1Xi=1Hi(q, η)(4)Hi(q, η) = 12 [η σxi σxi+1 + η−1σyi σyi+1 −q σzi −q−1σzi+1].A detailed discussion of the properties of the chain given by eq.

(4) will be given elsewhere [5],here we are going to mention only a few. First, there are the symmetry propertiesH(q, η) .= H(q−1, η) .= H(q, η−1).= H(η, q).

(5)The ”equality” among the Hamiltonians implies that the spectra are identical. The first twoequalities are obvious but not the last one which reminds of duality transformations of quantumchains [6].

In the continuum limit, one has the following phase structure [3, 5]:

∆q ≤1,∆η ≤1:massless-incommensurate∆q ≤1,∆η > 1 or ∆q > 1,∆η ≤1:massive incommensurate∆q > 1,∆η > 1,∆q ̸= ∆η:massive∆q = ∆η,(∆q > 1) :critical Ising type∆q = ∆η = 1 :Pokrovsky-Talapov phase transition [12]We will return to the physical picture later on in the text. It is by now clear that the propertiesof the chain depend on both parameters q and η.We now perform a Jordan-Wigner transformation.

First write σzj=−iσxj σyj and nextdefineτ x,yj= exp(iπ2j−1Xk=1(σzk + 1)) σx,yj(6){τ xi , τ xj } = {τ yi , τ yj } = {τ xi , τ yj } = 0. (i ̸= j)Using Eq.

(4) and (6) we getHj = i2[ ητ yj τ xj+1 −η−1τ xj τ yj+1 + qτ xj τ yj + q−1τ xj+1τ yj+1]. (7)We now observe the following important identity[T X, H(q, η)] = [T Y , H(q, η)] = 0(8)withT X = ∆(τ x) = α−L+12LXj=1αjτ xjT Y = ∆(τ y) = β−L+12LXj=1βjτ yj(9){T X, T X} = 2[L]α,{T Y , T Y } = 2[L]β,{T X, T Y } = 0,(10)where L is the length of the chain andα = qη,β = qη,[L]λ = λL −λ−Lλ −λ−1 .

(11)The equalities (8) come from the existence of a fermionic zero mode for any q and η. Theequations (10) together with the coproduct (9) give a representation of a Hopf algebra. Beforewe proove this statement let us consider the case α = β = q.

2The η = 1 case. Mathematics.We first notice that in this case Sz = 12PLi=1 σzi also commutes with H(q, η).

We now remindthe reader the Uα[SU(1/1)] algebra [7]. With A± = 12(T X ± iT Y ) we have{A±, A±} = 0,{A+, A−} = [E]α,[Sz, A±] = ±A±(12)[E, Sz] = [E, A±] = 0with the coproduct∆(α, A±) = αE/2 ⊗A± + A± ⊗α−E/2∆(α, Sz)= Sz ⊗1 + 1 ⊗Sz(13)∆(α, E)= E ⊗1 + 1 ⊗E.The fermionic representations correspond to take E = 1, Sz = 12σz, A± = a± and {a+, a−} = 1in Eq (13).

In this representation E in Eq. (12) is equal to L (the number of sites).

Comparingnow (12), (13) with Eqs. (9,10) we observe [8] that the quantum chain (4) with η = 1 isinvariant under Uα[SU(1/1)] transformations.

It was also shown by Saleur [8] that the quantitiesUj = ∆q −Hj(q, 1) are the generators of the Hecke algebraU2j = 2 ∆q UjUjUj±1Uj −Uj = Uj±1UjUj±1 −Uj±1(14)UiUi±j = Ui±jUi. (j ̸= 1)Actually they correspond to a quotient of this algebra since the generators satisfy also therelations [9]UjUj+2Uj+1(2∆q −Uj)(2∆q −Uj+2) = 0.

(15)The generators ˇRj = q−q−12+ Hj(q, 1) satisfy the braiding relationsˇRj ˇRj±1 ˇRj = ˇRj±1 ˇRj ˇRj±1(16)withˇR2j = (q −q−1) ˇRj + 1. (17)Considering the matrices Rj = P ˇRj (P is the graded permutation operator) we have(see Eq.

(13))R∆(α)R−1 = ∆(α−1). (18)

3The η = 1 case. Physics.Before persuing our mathematical developements let us pause and discuss some physical impli-cations.

First we notice the very unusual role of the operator E for the quantum chain. It doesnot behave like an usual symmetry operator in quantum mechanics (like the angular momen-tum) which commutes with the Hamiltonian and helps in its diagonalisation.

Since E simplycounts the number of sites it plays a different role that we clarify now. From Eq.

(12) we seethat for α generic (α ̸= eiπ rs), Uα[SU(1/1)] has two-dimensional irreducible representations andone one-dimensional irreducible representation where A± = E = Sz = 0. If α is not generic(α = eiπ rs), notice that{A+, A−} = sin( πrLs )sin( πrs )(19)and that for L = ns one has only one-dimensional irreducible representations.

This implies thatfor a given value of q, changing L one can reach pathological situations. As shown in Ref.

[5],if L = ns one has not only one zero mode but two which makes the degeneracies larger andnot smaller as one would expect from the fact that we have only one-dimensional irreduciblerepresentations. In order to avoid this type of problems and to keep the normalisations of thezero-mode operator (i.e.

A and A+), if one wants to take the thermodynamical limit, one hasto take sequences likeL = ns + t(t = 0, 1, . .

. , s −1; n ∈Z+)(20)and the results will depend on the sequence.

The necessity of taking sequences for the quantumchain (1) with periodic boundary conditions is already known [11] but now we understand itsorigin. The same observation applies when we have two parameters (see Eq.

(9)) and one orboth of them are not generic [5].A more detailed discussion of the physical meaning of the parameter q as well as the con-nection of the model with the experimental data [13] can be found in Ref. [5].4The η ̸= 1 case.As suggested by Eqs.

(9,10) we define the two parameter deformation of the SU(1/1) algebraas follows:{T X, T X} = 2 [E]α,{T Y , T Y } = 2 [E]β(21){T X, T Y } = 0[E, T X] = [E, T Y ] = 0with the coproduct∆(α, β; T X) = αE/2 ⊗T X + T X ⊗α−E/2∆(α, β; T Y ) = βE/2 ⊗T Y + T Y ⊗β−E/2(22)∆(α, β; E) = E ⊗1 + 1 ⊗E.

Notice that Sz does not appear in the algebra anymore. We denote this quantum algebra byUα,β[SU(1/1)].

It is a Hopf algebra for the same reasons as the Uα[SU(1/1)]. If we take thefermionic representations E = 1,τ x = (a+ + a−),τ y = −i(a+ −a−), from Eq.

(22) wederive Eqs. (9,10).

The quantum chain H(q, η) is thus invariant under the quantum algebraUα,β[SU(1/1)]. We would like to see what replaces the relations (14-18) when we have twoparameters.

We first notice a remarkable identity satisfied by the Hj(q, η)[HjHj±1Hj −Hj±1HjHj±1 + (ν −1)(Hj −Hj±1)] (Hj −Hj±1) = µ(23)H2j = ν,whereν = α + α−12! β + β−12!= q + q−12!2+ η + η−12!2−1(24)µ = α + α−12−β + β−12!2= 4 q −q−12!2 η −η−12!2.We can now define a generalised Hecke algebra takingUi = √ν −Hi(p, q)(UiUi±1Ui −Ui±1UiUi±1 −Ui + Ui±1) (Ui −Ui±1) = µ(25)U2i= 2 √ν Ui.Notice that when η = 1,µ = 0 and we have representations which coincide with those ofthe original Hecke algebra (For a detailed discussion of the representation theory of (25) seeRef.

[16]). We did not have the patience to find the equivalent of Eq.

(15) which gives thequotient of the generalised Hecke algebra (25) corresponding to the chain given by Eq. (4).Another quotient is however suggested by the structure of Eq.

(25):(UiUi±1Ui −Ui) (Ui −Ui±1) = µ2 . (26)For µ = 0 one gets in this case representations of the Temperley-Lieb algebra UiU±iUi = Ui.

Wenow turn our attention to the generalised braid group algebra. We take ˇRi = Hi(q, η) + √ν −1and get( ˇRi ˇRi±1 ˇRi −ˇRi±1 ˇRi ˇRi±1) ( ˇRi −ˇRi±1) = µ(27)withˇR2i = 1 +√ν −1 ˇRi .

(28)

In the basis where the σzi are diagonal (see Eq. (4)) we haveˇRi =√ν −1 + q+q−1200η−η−120√ν −1 −q−q−12η+η−1200η+η−12√ν −1 + q+q−120η−η−1200√ν −1 −q+q−12.

(29)We take the graded permutation matrix PP =10110−1(30)and define the matrix Ri = P ˇRi. We now write the coproduct (22) in the original language ofPauli matrices∆(α, β; T X) = α−1/2(σy ⊗1) −α1/2(σz ⊗σy)∆(α, β; T Y ) = β−1/2(σx ⊗1) −β1/2(σz ⊗σx)(31)∆(α, β; E) = 1 ⊗1 + 1 ⊗1.It is trivial to check that similar to Eq.

(18) we getR ∆(α, β) R−1 = ∆(α−1, β−1). (32)5Are more-parameter deformations possible?In this section we would like to show that for Uα,β[SU(1/1)] one can introduce more than twoparameters (as in the Lie algebra case when we had more than one).

The most general chainwhich has a zero mode for all its values of the parameters is [5]Hi=12 {Θ + Θ−12(ηζσxi σxi+1 + η−1ζ−1σyi σyi+1)+Θ −Θ−12(ηζ−1σxi σyi+1 + η−1ζσyi σxi+1)(33)+q σzi + q−1σzi+1) }.Hi depends on four parameters. One can check however that Eq.

(23) holds withν=q2 + q−24+ (η2 −η−2)(ζ2 −ζ−2)8+ (η2 + η−2)(ζ2 + ζ−2)(Θ2 + Θ−2)16,

µ= (η2 −η−2)(ζ2 −ζ−2)4+ (η2 + η−2)(ζ2 + ζ−2)(Θ2 + Θ−2)8−1! (34)×(q −q−1)22+(Θ −Θ−1)2( ηζ −ζη)28which means that we are back to two parameters.

This means that there is a similarity trans-formation which connects the Hamiltonian with four parameters and the one with two (seeEq. (4)).

In order to illustrate this point we consider the ”two-parameter deformation” of Ref.[14]. It corresponds to the choiceζ = eiπ/4,η = e−iπ/4,q =qQP,θ =sQP(35)in Eq.

(33) where Q and P are the two parameters given in [14]. From Eq.

(34) we get µ = 0which implies that we are back to the Uα[SU(1/1)] case. From Eq.

(33) we deriveˇRi=12(qQP −1√QP ) +sQP σ+i σ−i+1 +sPQ σ−i σ+i+1 +(36)12qQP σzi + 121√QP σzi+1 .We now do the similarity transformation [15]σ+i →(qQP)i−1 σ+i ,σ−i →(√PQ)i−1 σ−i ,σzi →σzi(37)and recover Eq. (4) with η = 1 and q = √QP, which means that the two-parameter deformationis a one-parameter deformation.References[1] E. E. Demidov, Yu.

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