Lyberg, I.

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arXiv:1212.4177v1  [math-ph]  17 Dec 2012A “quantum spherical model” with transversemagnetic field.I. LybergSchool of Theoretical Physics, Dublin Institute for Advanced Studiesilyberg@stp.dias.ieDecember 19, 20121 IntroductionThe Quantum Ising Model with a transverse magnetic field is well known [1][2]. In one dimension it has the HamiltonianHN = −JN∑n=1σxnσxn+1 +BN∑n=1σzn +HN∑n=1σxn, (1)where J > 0 is the coupling constant and σx =(0 11 0), σy =(0 −ii 0)and σz =(1 00 −1)are the Pauli matrices. B and H are transverse andlongitudinal magnetic fields, respectively. The partition function isZN = tr e−βHN (2)where β is the inverse temperature. In the case where H = 0 this model hasbeen exactly solved [1] [3]. The free energy is [2]f(β, J, B) = − limN→∞1βNlogZN= − 12piβ∫ 2pi0log 2 coshβ∆(x) dx (3)where∆(x) =√J2 +B2 − 2BJ cosx. (4)1In particular the ground state energy is given byf∞(J,B) = limβ→∞f(β, J, B) = − 12pi∫ 2pi0∆(x) dx. (5)In this limit there is a critical point in B at B = J . The correlation function〈σxj σxk〉 = limN→∞tr σxj σxke−βHNZN(6)can be written as a Toeplitz determinant of size |j − k| just as the corre-lation function of the two dimensional classical Ising model [4], but only inthe limit β →∞. In fact the correlation function limβ→∞〈σxj σxk〉 is the sameas the diagonal correlation function 〈σjjσkk〉 of the two dimensioanl classi-cal Ising lattice for T < Tc, the ratio B/J in the one dimensional quantummodel corresponding to (sinh 2E1/kBT sinh 2E2/kBT )−1 in the two dimen-sional classical model. (Here E1 and E2 are the coupling constants in thehorizontal and vertical directions, respectively). In particular the limit ofinfinite separation is given by [3]lim|j−k|→∞limβ→∞〈σxj σxk〉 ={{1− (B/J)2}1/4 if B < J,0 if B ≥ J, (7)which is most easily proved using Szego¨’s theorem [5] [6].2 The quantum spherical modelIn analogy with (1) we define a partition function of a (d-dimensional)isotropic quantum spherical model on a lattice Λ as follows:ZN =∫[0,∞)N∫[0,2pi)N∫[0,pi]Ne∑j,k∈Λ : 〈jk〉 βJrj cos θjrk cos θke∑j∈Λ β(Brj sin θj cosϕj+Hrj cos θj)N∏l=1r2l sin θl dNθ dNϕ δ( N∑m=1r2m −N)dNr=∫R3Ne∑〈jk〉 βJzjzk+∑j β(Bxj+Hzj)δ( N∑k=1(x2k + y2k + z2k)−N)d3Nx.(8)Here J > 0, B ≥ 0 and H > 0. δ signifies the Dirac distribution. Thenotation 〈jk〉 means that j and k are nearest neighbors on Λ. Unlike the2Quantum Ising Model with H = 0, in this model the critical point is B = 2Jd(in the limit H → 0). In fact, it will be shown that in this limit the groundstate free energy fH,∞ := − limβ→∞ limN→∞(Nβ)−1 logZN is given byf0,∞ = limH→0fH,∞ = −{Jd+B2/4Jd if B ≤ 2Jd,B if B > 2Jd.(9)We shall now give a proof of (9).2.1 The case B > 2JdWe use the method of steepest descent to prove this result, following thecalculation by Baxter [7]. We let H = 0 in (8). Clearly the integrand in(8) may be multiplied by a factor exp a(∑Nk=1(x2k + y2k + z2k)−N) withoutchanging the partition function ZN . Using the identityδ(x) =12pi∫ ∞−∞eisxds, (10)together with (8) and letting a > 0, we getZN =piN−12∫RN∫ ∞−∞( 1a+ is)NexpN(βB)24(a+ is)exp [∑〈jk〉βJzjzk +∑j(a+ is)(1− z2j )] ds dNz (11)after integrating over x and y. Let V be the symmetric matrix such thatzTVz = (a + is)N∑j=1z2j − βJN∑〈jk〉zjzk. (12)In this way (11) can be written asZN =piN−12∫RN∫ ∞−∞( 1a+ is)NexpN(βB)24(a+ is)exp [−zTVz+N(a + is)] ds dNz. (13)We now choose the constant a so large that all the eigenvalues of V havepositive real part. This allows us to change the order of integration, and wemay now write (13) asZN =pi3N/2−12∫ ∞−∞( 1a+ is)N(detV)−1/2exp[N(βB)24(a+ is)+N(a + is)]ds. (14)3We need to calculate the eigenvalues of V. Since V is cyclic, this is easilydone. We let the lattice be d-dimensional hypercubic, so thatN = Ld (15)for some positive integer L. It now follows from (12) that the eigenvalues areλ(ω1, ..., ωd) = a + is− βJd∑j=1cosωj (16)where each ωj takes the values {2pik/L}L−1k=0 , and a > βJd. The determinantof V is the product of its eigenvalues, solog detV =∑ωj : 1≤j≤dlog λ(ω1, ..., ωd). (17)ClearlyZN =βJ2pii(piβJ)3N/2 ∫ c+i∞c−i∞eNφ(w)dw, (18)whereφ(w) = βJw − 12g(w) + (βB)2/4βJw, (19)c = (a− βJd)/βJ andg(z) = 2 logw +1N∑ωjlog (w −∑jcosωj). (20)Since φ approaches +∞ as w approaches 0 or +∞ along the real line, φ hasa minumum at some w0, 0 < w0 <∞. Thus ℜφ has a maximum at w0 alongthe line (w0− i∞, w0+ i∞). Since B > 2Jd, we may choose c = w0. We nowuse the method of steepest descent (see for instance Murray [8]), by lettingN approach infinity. In this way, the free energy isf = −β−1 limN→∞N−1 logZN= − 32βln (pi/βJ)− β−1φ(w0). (21)Nowlimβ→∞w0 = B/2J, (22)and thus the ground state energy islimβ→∞f = − limβ→∞β−1φ(w0)= −B. (23)42.2 The case B ≤ 2JdIn this case we let H > 0, so instead of (13) we haveZN =piN/2−12∫RN∫ ∞−∞( 1a+ is)NexpN(βB)24(a+ is)exp [−zTVz+ hTz+N(a + is)] ds dNz, (24)where h = βH(1, ..., 1). We change variables to t = z− 12V−1h, and rotatethe axes in (t1, ..., tN ) to make V diagonal. Thus we getZN =piN/2−12∫ ∞−∞( 1a + is)N(detV)−1/2 expN(βB)24(a+ is)exp [hTV−1h/4 +N(a + is)] ds. (25)ThusZN =piN/22pii∫ c+i∞c−i∞eNφ(w) dw, (26)where a + is− βJd = βJw andφ(w) = βJ(w + d) +(βH)24βJw+(βB)24βJ(w + d)− log βJ(w + d)− 12∑ωjlog (βJ(w + d)− βJ∑jcosωj). (27)We proceed in the same way as before, taking the limit N → ∞ and thenβ →∞. In this case w0 → 0 as H → 0. The free energy is thusf = −Jd− B24Jd. (28)This ends the proof.3 DiscussionComparison of (5) and (9) shows that the susceptibilites of the two mod-els at B = 0 are equal when d = 1; that is −∂2f∞/∂B2|B=0 = 1/2J and−∂2f0,∞/∂B2|B=0 = 1/2Jd. While the Quantum Ising Model has only beenexactly solved in the one dimensional case, the quantum spherical modelcanbe solved in any finite dimension.Acknowledgements The authour would like to thank Prof. T. Dorlas formany discussions.5References[1] E. Lieb, T. Schultz and D. Mattis. Two Soluble Models of an Antiferro-magnetic Chain. Ann. Phys., 16, 407 (1961).[2] D. C. Mattis. The theory of magnetism made simple. World Scientific,New Jersey (2006).[3] B. M. McCoy. Spin Correlation Functions of the X-Y Model. Phys. Rev.,173, 531 (1968).[4] B. Kaufman and L. Onsager. Crystal statistics. III. Short-range orderin a binary Ising model. Phys. Rev. 76, 1244 (1949).[5] E. W. Montroll, R. B. Potts and J. C. Ward. Correlations and sponta-neous magnetization of the two dimensional Ising model. J. Math. Phys.,4, 308 (1963).[6] U. Grenander and G. Szego¨. Toeplitz Forms and Their Applications.University of California Press. Berkeley (1958).[7] R. J. Baxter. Exactly solved models in statistical mechanics. AcademicPress, London (1982).[8] J. D. Murray. Asymptotic Analysis. Springer-Verlag, New York (1984).6

A “quantum spherical model” with transverse magnetic field

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A “quantum spherical model” with transverse magnetic field

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