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Submitted on 1 Jan 1988
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THE S = 1 QUANTUM SPIN CHAIN WITH PURE
BIQUADRATIC EXCHANGE
J. Parkinson
To cite this version:
JOURNAL DE PHYSIQUE
Colloque C8, Suppl6ment au no 12, Tome 49, decembre 1988
THE
S=l
QUANTUM SPIN CHAIN WITH PURE BIQUADRATIC EXCHANGE J. B. ParkinsonMathematics Department, UMIST, P O Boz 88, Saclcville St, Manchester M60 1 QD, G.B.
Abstract. - The S = 1 quantum spin chain with pure biquadratic exchange has 2-strings that behave in the same way as 1-strings (magnons) in other chains. An exact partial mapping exists from this system to an integrable S = 112 XXZ system. The ground state is not included in the mapping for finite N, but appears t o be the same in the thermodynamic limit. The elementary excitations may also be deduced.
There has been great interest recently in quantum spin chains with S = 1 and various terms in the Hamil- tonian, such as isotropic Heisenberg and biquadratic exchange, anisotropic exchange, single-ion anisotropy and magnetic field. Much of this work was stimulated by Haldane's [l] discussion of the isotropic Heisenberg model for which he predicted the existence of a dis- ordered ground state (i.e. one in which the correla- tions decay exponentially) and a gap to the first ex- cited state a t k = n.
The class containing just isotropic Heisenberg and biquadratic exchange is particularly interesting be- cause as well as the model studied by Haldane, it con- tains two integrable models: the Tabhtajan-Babujian [2, 31 model with opposite signs but equal magni- tude for the bilinear and biquadratic terms and the Schrodinger permutation model studied by Lai [4] and Sutherland [5] in which the two terms have the same sign. One other member of the class in which the bi- quadratic term has magnitude 1/3 of the bilinear term has recently been shown [6] to have a valence bond ground state whose energy and correlations can be cal- culated exactly.
In this paper a further member of the same class is studied which has zero bilinear. term, described by the Hamiltonian
with periodic boundary conditions S ~ + N Si. This pure biquadratic model is ferromagnetic for J 2
>
0and antiferromagnetic for J 2
<
0. We consider the lat- ter case explicitly. The fully aligned state (ferromag- netic ground state) has energy Eo = NJz and all other energies will be given relative to this. The "spin-wave" states with one deviation are of the non-propagating kind and are degenerate with the aligned state.States with two deviations can also be calculated. If the two deviations are on the same or adjacent sites the eigenstates are in the form of propagating 2-strings with spectrum
E2 = 5 2 (3
+
2 cos k),
(2)while if they are on sites further apart they are non- propagating and again degenerate with the aligned state. The form (2) is notable both for its simplicity and also for its resemblance to the spectrum of nor- mal spin-waves in a variety of quantum spin chains. In fact an identical spectrum is obtained for the fol- lowing S = 1/2 XXZ model
provided J = -352, and p = -2/3.
In normal Hamiltonians it is possible to obtain the eigenstates with two deviations in a Bethe anzatz form even when the Hamiltonian is not of the integrable type [7]. The states are either interacting spin-waves or bound states (2-strings). Because the 2-string eigen- states of (1) have a similar spectrum to single devia- tion states of (3), it seemed possible that four devia- tion states of (1) could be found of the Bethe anzatz form corresponding to interacting 2-strings and bound states of 2-strings.
The details of the calculation are rather lengthy
[a],
but the result has the expected form. Four deviation states of (1) exist with energywith Nkl = 2x11
+
4,
Nkz = 2x12-
4, (5) and -3 sin (q/2) cot (q5/2) = 2 cos (k/2)+
3 cos (q/2) ' (6) Here, 11, 12 are integers, k = kl+
k2 and q = kl-
k2.Now it may be observed that equations (4), (5) and (6) are precisely the Bethe anzatz equations of the spin-l/2 Hamiltonian (3), which is known to in- tegrable, i.e. states with an arbitrary number of de- viations can be expressed in the Bethe enzatz form [9]. This suggests that states of (1) with an arbitrary number of 2 strings may also have this form. Ana- lytic proof that this is so is difficult to obtain but a
C8
-
1414 JOURNAL DE PHYSIQUEnumerical comparison of the energies of states with r deviations of (3) with the energies of states with 2r deviations of (1) shows that these are identical, al- though with additional degeneracy in the latter. There is, however, one exception: when T = N/2 the energies
do not match. These results have been verified numer- ically for all even N
5
14. The reason for this failure of the mapping for the T = N/2 case is not known atpresent.
Since r = N/2 includes the ground state this would be rather unfortunate. However, if the energies of the two ground states are plotted as a function of 1/N, there is strong evidence that they do indeed become identical in the limit N -+ m. The exact ground state energy of (3) is given by Orbach [9] and Walker [lo] and is NJze, where e = 1.796864. Furthermore, if the ground states of the two Hamiltonians are the same in the thermodynamic limit then one may conjecture that the elementary excitations with T = N/2 are also
the same. (The excitations with T
<
N/2 are the samefor all N of course.) A detailed description of the ex- citations of (3), (the so-called antiferromagnetic spin waves) has been given by des Cloiseaux and Gaudin [ll]. They found an interesting structure in which the r = N/2 excitations form a gapless spectrum, some- what similar in shape t o the sine curve of the S = 1/2 isotropic Heisenberg model (but with a different ana- lytical form). The r = N/2
-
l states have an identical shape but are separated by a gap which is independent of k. The value of the gap can be written [12] as EG=2&1~2l dn (T)
.
The other main advantage of the integrability of a
Hamiltonian is that the thermodynimics can be stud- ied, as was done by Gaudin [12] for (.3). However, these results cannot be used directly for (:I), even if one as- sumes that the mapping becomes exact in the thermo- dynamic limit. This is because the degeneracies of the states in the two models are different. An additional contribution t o the degeneracies connes from the states with r odd which have not been discussed above but in fact give no new eigenvalues. The thermodynamics of the pure biquadratic model is an interesting prob- lem, which in the light of the above results should be tractable.
[I] Haldane, F. D. M., Phys. Lett. A 93 (1983) 464; Phys. Rev. Lett. 50 (1983) 1153.
[2] Takhtajan, L. A., Phys. Lett.
A
87 (1982) 479. [3] Babujian, H. M., Phys. Lett. A 90 (1982) 479;Nucl. Phys. B 215 (1983) 317.
[4] Lai, C. K., J. Math. Phys. 15 (1974) 1675. [5] Sutherland, B., Phys. Rev. B 12 (1975) 3795. [6] Affleck, I., Kennedy, T., Lieb, E. H., Tasaki, H.,
Phys. Rev. Lett. 59 (1987) 793.
[7] Hodgson, R. P., Parkinson, J. B., J. Phys. C 18 (1985) 6385.
[8] Parkinson, J. B., J. Phys. C 21 (1988) 3793. [9] Orbach, R. L., Phys. Rev. 11:: (1958) 309. [lo] Walker, L. R., Phys. Rev. 116; (1959) 1089. [ l l ] Des Cloiseaux, J., Gaudin, M., J. Math. Phys. 7
(1966) 1384.
