LPENSL-TH-02/02 Emptiness formation probability of the XXZ spin-1 2 2 0 Heisenberg chain at ∆ = 1 0 2 2 n a J 7 1 N. Kitanine1, J. M. Maillet2, N. A. Slavnov3, V. Terras4 1 v 4 3 1 1 0 Abstract 2 0 / h Using a multiple integral representationfor the correlationfunctions, we com- t - pute theemptiness formationprobabilityoftheXXZ spin-1 Heisenbergchain p 2 1 e at anisotropy ∆ = . We prove it is expressed in term of the number of 2 h alternating sign matrices. : v i X r a The Hamiltonian of the XXZ spin-1 Heisenberg chain is given by 2 M H = σxσx +σy σy +∆(σz σz 1) . (1) m m+1 m m+1 m m+1− m=1 X (cid:0) (cid:1) 1Graduate School of Mathematical Sciences, University of Tokyo, Japan, [email protected] On leave of absence from Steklov Instituteat St. Petersburg, Russia 2 Laboratoire dePhysique,UMR 5672 du CNRS,ENS Lyon,France, [email protected] 3 Steklov Mathematical Institute,Moscow, Russia, [email protected] 4Department of Physics and Astronomy, RutgersUniversity,USA,[email protected] On leave of absence from LPMT, UMR 5825 du CNRS,Montpellier, France 1 Here ∆ is the anisotropy parameter, σx,y,z denote the usual Pauli matrices acting on the quan- m tum space at site m of the chain. The emptiness formation probability τ(m) (the probability to find in the ground state a ferromagnetic string of length m) is defined as the following expectation value m 1 σz τ(m) = ψ − k ψ , (2) g g h | 2 | i k=1 Y where ψ denotes the normalized ground state. In the thermodynamic limit (M ), this g | i → ∞ quantity can be expressed as a multiple integral with m integrations [1, 2, 3, 4]. Recently, in the article [5], a new multiple integral representation for τ(m) was obtained; for ∆ = cosζ, 0 < ζ < π, one has τ(m) = lim τ(m, ξ ), (3) j ξ1,...ξm→−i2ζ { } where ∞ 1 Z ( λ , ξ ) i τ(m, ξ ) = m { } { } det dmλ, (4) { j} m!−Z∞ m sinh(ξa ξb) m 2ζsinhπζ(λj −ξk)! − a<b Q with m m det −isinζ Z ( λ , ξ ) = sinh(λa−ξb)sinh(λa −ξb−iζ) m sinh(λj−ξk)sinh(λj−ξk−iζ) . (5) m { } { } sinh(λ λ iζ) · (cid:16) m (cid:17) aY=1bY=1 a− b− sinh(ξa−ξb) a>b Q In this letter, we consider the particular case ∆ = 1 (ζ = π/3). Recently several interesting 2 conjectureswereobtainedforthegroundstateofthemodelatthisspecialvalueoftheanisotropy parameter ∆ [6, 7, 8, 9]. Note that the unitary transformation UH∆U−1 = H−∆, U = − M 2 σz relates our Hamiltonian (1) for ∆ = 1 to the case ∆ = 1 in [7]. In particular, it was j=1 2j 2 −2 conjectured in [7] that, in this case, the emptiness formation probability is equal to Q √3 3m2 m Γ(k 1)Γ(k+ 1) τ(m) = − 3 3 . (6) 2 ! Γ(k 1)Γ(k+ 1) kY=1 − 2 2 The aim of this letter is to give the proof of this conjecture using the representations (3)–(5). Observe first that for ζ = π/3, m2−m m ( 1) 2 sinh3(ξb ξa) Z ( λ , ξ ) = − − m { } { } 2m2+m sinh(ξ ξ )sinh(ξ ξ ) b a a b aY>b − − 2 det 1 det 1 m(cid:18)sinh(λj−ξk+i3π)(cid:19). (7) m × sinh(λj −ξk)sinh(λj −ξk −iζ)! detm sinh3(1λj−ξk) (cid:16) (cid:17) Here we have used the identities, n sinh(x x )sinh(y y ) j k k j 1 − − a>b det = , (8) nsinh(x y ) Q n j − k sinh(xj yk) − a,b=1 Q and sinh(3x) = 4sinh(x)sinh(x+iπ/3)sinh(x iπ/3). Substituting (7) into (4), we obtain − τ(m, ξ ) = 3i m (−1)m22−m m sinh3(ξb−ξa) m sinh−1(ξ ξ ) { j} 4π 2m2m! sinh(ξ ξ ) a− b (cid:18) (cid:19) aY>b b− a aYa,b6==b1 ∞ 1 1 dmλ det det . (9) ×−Z∞ m sinh(λj −ξk + i3π)! m sinh(λj −ξk)sinh(λj −ξk − i3π)! Due to the symmetry properties of the integrand, we can replace the first determinant with the product of its diagonal elements multiplied by m!. Then, we insert each of these diagonal elements into thecorrespondinglineof theseconddeterminant. Bythis procedure,theintegrals over the variables λ are decoupled and we can integrate each line of the determinant separately. Let us set ξ = ε iπ/6. We obtain k k − m m τ(m, εj ) = ( 1)m22−m3m2−m2 sinh3(εb−εa) 1 { } − sinh(ε ε ) sinh(ε ε ) aY>b b− a aaY,b6==b1 a− b ∞ dλ det . (10) × m 4πcosh(λ ε )sinh(λ ε iπ)sinh(λ ε + iπ) −Z∞ − j − k − 6 − k 6 The computation of the integral over λ in (10) leads to τ(m, ε ) = (−1)m22−m m sinh3(εb−εa) m 1 det 3sinh εj−2εk . (11) { j} 2m2 aY>b sinh(εb−εa) aaY,b6==b1 sinh(εa−εb) · msinh3(εj2−εk) 3 To obtain the emptiness formation probability (2), one has to take the homogeneous limit ε 0. Using the fact that j → m−1 det f(x y ) lim m j − k = (n!)−2det f(j+k−2)(z) , z = x y, (12) xj→x m m − yk→y (xa−xb)(yb−ya) nY=0 h i a>b Q we finally obtain τ(m) = (−1)m22−m3m22+m2−m2m−1(n!)−2detm"∂∂xjj++kk−−22ssiinnhh3x2x# . (13) n=0 2 x=0 Y The determinant in (13) can be computed using the following identity [10], m m 1 sinhα(j +k 1) sinh(α+β(j k)) det − = 2m2−m − , (14) m sinh2β(j −k) · msinhβ(j +k−1) jY=1kY=1 sinhβ(j +k−1) j>k Q (see [10] for the proof). The determinant (13) is a particular case of (14). Indeed, we can consider the case β = 3α, α 0, and apply (12) for x = αj, y = α(1 k). Then, j k → − m−1 ∂j+k−2 sinhx m m j k+ 1 (n!)−2det 2 = 3m2−m − 3 m"∂xj+k−2sinh3x# j +k 1 nY=0 2 x=0 jY=1kY=1 − m−1 m2−m −m+m2 (3k+1)! = ( 1) 2 3 2 . (15) − (m+k)! k=0 Y Substituting these expressions into (13), we finally obtain 1 m2m−1 (3k+1)! τ(m) = . (16) 2 (m+k)! (cid:18) (cid:19) k=0 Y Observe that the quantity A = m−1(3k + 1)!/(m + k)! is the number of alternating sign m k=0 matrices of size m [11]. Using Q 1 √π Γ(3z) = 33z−1/2Γ(z)Γ(z+1/3)Γ(z +2/3), Γ(k+1/2) = (2k 1)!!, (17) 2π 2k − one can easily check the equivalency of (16) and (6). Thus (6) is proved. 4 The asymptotic behavior of τ(m) for m can be evaluated using the Stirling formula → ∞ [7]: 3m2 √3 − 5 τ(m) c m 36, m , (18) → 2 ! → ∞ with ∞ 5e−t sinh 5t sinh t dt c = exp 12 12 . (19) "Z0 36 − sinh2 2t ! t # Acknowledgments N. K. would like to thank the University of York, the SPhT in Saclay, JSPS and the Tokyo University for financial support. N. S.is supportedby thegrants INTAS-99-1782, RFBR-99-01- 00151, Leading Scientific Schools 00-15-96046, the Program Nonlinear Dynamics and Solitons and by CNRS. J.M. M. is supported by CNRS. V. T is supported by DOE grant DE-FG02- 96ER40959 and by CNRS. N. K, N. S. and V. 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