We discuss the Lagrange-multiplier method in a many-body system, and how to apply the Lagrange multiplier correctly to realize quantum constraint in a Hamiltonian for a quantum system. It is pointed out that the discussion in Yanagisawa’s paper [Phys. Rev. B 57, 6208 (1998)] lacks a basic knowledge of mathematics, and that his conclusion is generally wrong.published_or_final_versio

Shen, SQ

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Title Comment on 'Lagrange-multiplier method in correlated-electronsystems: Exact diagonalization study'Author(s) Shen, SQCitation Physical Review B (Condensed Matter and Materials Physics),1999, v. 59 n. 19, p. 12689-12691Issued Date 1999URL http://hdl.handle.net/10722/43263Rights Creative Commons: Attribution 3.0 Hong Kong LicensePHYSICAL REVIEW B 15 MAY 1999-IVOLUME 59, NUMBER 19COMMENTSComments are short papers which criticize or correct papers of other authors previously published in Physical Review B. EachComment should state clearly to which paper it refers and must be accompanied by a brief abstract. The same publication schedule asfor regular articles is followed, and page proofs are sent to authors.Comment on ‘‘Lagrange-multiplier method in correlated-electron systems:Exact diagonalization study’’Shun-Qing ShenDepartment of Physics, The University of Hong Kong, Pokfulam Road, Hong Kong, China~Received 30 March 1998!We discuss the Lagrange-multiplier method in a many-body system, and how to apply the Lagrange mul-tiplier correctly to realize quantum constraint in a Hamiltonian for a quantum system. It is pointed out that thediscussion in Yanagisawa’s paper @Phys. Rev. B 57, 6208 ~1998!# lacks a basic knowledge of mathematics, andthat his conclusion is generally wrong. @S0163-1829~99!02720-4#The Lagrange-multiplier method is often used to mini-mize a function with a condition. In quantum many-bodyphysics it is also applied to realize some quantum constraintwhen we treat the Schro¨dinger equation or the Hamiltonianof a quantum system. A famous example is the use of thechemical potential in a Hamiltonian of a many-body quan-tum system to adjust the number of particles in grand canoni-cal ensembles. Although the method is well established,there still exists some misunderstanding of its application. Inhis recent paper, Yanagisawa investigated the Lagrange-multiplier method in quantum systems by an exact diagonal-ization study1 ~referred to as Ref. 1 in this Comment!. Hisdiscussion is based on a misunderstanding of the method,and lacks a rigorous mathematics basis. His work also con-tains some responses to my recent comment on his previouswork.2 In this Comment, I clarify several problems and makea response to his comment on how to introduce Lagrangemultipliers into a Hamiltonian correctly.To compare with Yanagisawa’s work, I use the same no-tations in this Comment as in Ref. 1. A Schro¨dinger equationwith quantum constraint~s! is written asHuC&5EuC&, ~1!QiuC&50, ~ i51, . . . ,M !, ~2!where H is the Hamiltonian, E is the eigenvalue, uC& is thewave function, and Qi is the operator for the constraints. Inorder to solve Eqs. ~1! and ~2!, we can introduce theLagrange multipliers to form a generalized Hamiltonian torealize the quantum constraints in Eq. ~2!.~1! The Lagrange multiplier l iQi : One way to introducethe Lagrange multiplier is to generalize the Hamiltonian inEq. ~1! toHL15H1(i51Ml iQi ~3!PRB 590163-1829/99/59~19!/12689~3!/$15.00to replace the M constraints @Eq. ~2!#. It is expected that theconstraint could be realized by minimizing the expectationenergy with respect to the Lagrange multipliers l i :ddl i^CuHL1uC&50. ~4!The resulting equations areS H1(il iQiD uC&5EuC&, ~5!^CuQiuC&50, ~ j51, . . . ,M !. ~6!From the notations, we should note the differences betweenEqs. ~2! and ~6!: the averages of Qi in the state uC& replacethe original equations. Generally speaking, Eqs. ~6! areweaker than Eqs. ~2!. They are not equivalent except forsome special cases. For example, if Qi is semipositive defi-nite, the Lagrange multipliers can realize the constraints be-cause, in this case, Eqs. ~6! are equivalent to Eqs. ~2!. An-other available case is that if we consider only the ground-state properties and know that the ground state isnondegenerate, Equations. ~2! and ~6! are also equivalent.However, we cannot use this if we want to prove the nonde-generacy of the ground state of a system.~2! The Lagrange multiplier l iQi†Qi : If Qi are not semi-positive definite, or we have to consider the complete set ofsolutions, the introduction of the Lagrange multipliers in Eq.~3! usually enlarges the Hilbert space. HL1 are usually nolonger equivalent to the original problem. Some unexpectedsolutions may exist in Eqs. ~5! and ~6!. Hence the Lagrangemultipliers cannot realize the constraints as expected. To thisend one of the correct ways to introduce the Lagrange mul-tiplier to realize the quantum constraint is to use l iQi†Qiinstead of l iQi in Eq. ~3!:12 689 ©1999 The American Physical Society12 690 PRB 59COMMENTSHL25H1(il iQi†Qi, ~7!which was, to my best knowledge, first proposed in mypaper.2 ~Reference 1 did not mention any reference, whilethe method was commented upon.! The resulting conditionsare^CuQi†QiuC&50 ~8!instead of Eqs. ~6!. Due to the semipositive definiteness ofQi†Qi , the conditions are reduced to Eqs. ~2!. This provesthat the problem of HL2 is equivalent to the original problem.Mathematically, it also proves the validity of the method inEq. ~7!. Generally speaking, any mathematical transforma-tion should guarantee that the transformed problem isequivalent to the original problem. However, it is anticipatedthat the transformed problem could be treated in a simplerway. If the transformed problem is not equivalent to theoriginal one, it is hard to justify the transformation.Reference 1 proposed that square terms cannot be used ina real analysis. This is not true. Consider the same problemin Ref. 1; i.e., minimizing x21y2 under the condition x1y51, the function with a Lagrange multiplier isf ~x ,y ,l!5x21y22l~x1y21 !2. ~9!Based on the variational principle we have x5y5 12 and l51` with l(x1y21)5 12 . It is apparent that Ref. 1 ne-glected this set of solutions. The same minimum of x21y2 isobtained as in the usual way. l is determined by the Eulerequations. If we assume that l is purely imaginary here, wecannot find a solution no matter which way we introduce theLagrange multiplier.3 It is worth emphasizing that whether l iare complex or imaginary is a pseudoscientific problem.From the principle of mathematics, l i must be determinedby solving the resulting equations simultaneously. In thequantum system we discussed, the expectation value of H isenergy, which should be real. The author of Ref. 1 performedan exact diagonalization for a finite system by setting l iimaginary. It should be noted how to compare a real eigen-value with a complex one. We cannot say simply that thestate with the lowest real part of the eigenvalue is the groundstate. Thus the data listed in Tables I and II in Ref. 1 do notmake sense.~3! A two-site example: To compare the results of the twoways to introduce the Lagrange multiplier, let us re-examinethe two-site S5 12 Heisenberg model,H15S1S2 . ~10!There are four eigenstates of H1: one is the spin singlet statewith E52 34 , and others are the spin triplet with E5 14 . Inorder to justify the different Lagrange-multiplier methods,we solve the problem in the fermion representation, wherethe single occupancy of a fermion on a site is expected to berealized by the Lagrange multiplier. The spin operator can beexpressed in terms of the fermion operator c,Si15ci ,"† ci ,# , ~11!Si25ci ,#† ci ," , ~12!Siz5 12 ~ni ,"2ni ,#!, ~13!with the constraints of single occupancy,~ni ,"1ni ,#21 !uC&50. ~14!ni ,s5ci ,s† ci ,s (s5" ,#). Usually the quantum constraint isvery difficult to realize. If we introduce the Lagrange multi-plier as in Ref. 1,H2512 ~c1,"† c1,#c2,#† c2,"1c1,#† c1,"c2,"† c2,#!114 ~n1,"2n1,#!3~n2,"2n2,#!1 (i51,2l i~ni ,"1ni ,#21 !, ~15!the physical space has been enlarged so that the condition ofsingle occupancy is replaced by^Cu~ni ,"1ni ,#21 !uC&50. ~16!In the case the number of particle per site is no longer alwaysequal to 1, and will be determined by the Schro¨dinger equa-tion of H2 and the condition of Eq. ~16!. The conditioncannot guarantee single occupancy. For example, a localstateuf i&51A2~12ci ,"† ci ,#† !u0&,consisting of empty and double occupancies, satisfies thecondition^f iu~ni ,"1ni ,#!uf i&51,where u0& is the vacuum state.If we focus on a fixed number of particle, say 2, and the zcomponent of total spin 0, the wave function isC5~ac1,"† c2,#† 1cc2,"† c1,#† 1bc1,"† c1,#† 1dc2,"† c2,#† !u0& .~17!From Eq. ~16!, we have ubu5udu. There are two sets of so-lutions: ~1! E52 34 with a5cÞ0 and b5d50, and E5 14with a52cÞ0 and b5d50. The two l i’s can be any finitevalues. These two solutions correspond to the two states withtotal spin 0 and 1. ~2! E50 with a5c50, b56d , and l15l2. Solutions ~2! satisfy the condition @Eq. ~16!#, but donot satisfy the condition of the single occupancy. These twostates do not exist in the original Heisenberg model. Thisillustrates that the Lagrange multipliers in Eq. ~15! do notrealize the single occupancy, as expected. It enlarges thephysical space, and some unphysical states are included.If we introduce the Lagrange multipliers l i(ni ,"1ni ,#21)2 in Eq. ~15! instead of l i(ni ,"1ni ,#21),H3512 ~c1,"† c1,#c2,#† c2,"1c1,#† c1,"c2,"† c2,#!114 ~n1,"2n1,#!3~n2,"2n2,#!1 (i51,2l i~ni ,"1ni ,#21 !2, ~18!we have the solutions E52 34 with a5cÞ0 and b5d50,and E5 14 with a52cÞ0 and b5d50. The two l i’s can bePRB 59 12 691COMMENTSany finite values. These are the two physical states. The othertwo unphysical states are excluded spontaneously.Comparison of the results of this simple example clearlyillustrates that the two ways to introduce the Lagrange mul-tipliers will lead to different results. H2 in Eq. ~15! containstwo unphysical states that H1 in Eq. ~10! does not. Hence wecannot say that the Lagrange multiplier in Eq. ~15! is correct.Although some other conditions can be used to exclude theunphysical states for a few-body system, those conditions areneither attached to the method itself nor have a solid math-ematical foundation. They have gone beyond the method it-self. It is common sense that a conclusion drawn from aconcrete example cannot be naively extended to a generalcase without a solid mathematical foundation, especiallywhen we discuss a general method.~4! Compatible H and Qi: When Qi are not semipositivedefinite, but all Qi and H are compatible, i.e., @H ,Qi#50 and@Qi ,Q j#50, are Eqs. ~1! and ~2! equivalent to Eqs. ~5! and~6!? The two-site problem we just discussed has given anegative answer. It is well known that two compatible opera-tors can be diagonalized simultaneously, but this does notmean that eigenstates of H are always the eigenstates of Qi .Assume uc1& and uc2&, two eigenstates of H with energyeigenvalues E1 and E2. From the commutators of H and Qi ,we obtain^c1u@H ,Qi#uc2&5~E12E2!^c1uQiuc2&50. ~19!If E1ÞE2 , ^c1uQiuc2& must be zero. But if E15E2 ,^c1uQiuc2& can be nonzero. Therefore, if all eigenstates of Hare nondegenerate, the eigenstates of H are automatically theeigenstates of Qi . In this case the Lagrange multipliers inEq. ~3! can realize the constraints in Eq. ~2!. However, if partof the eigenstates of H are degenerate, the eigenstates of Hare no longer always the eigenstates of Qi . Thus Eqs. ~1!and ~2! are not equivalent to Eqs. ~4! and ~5!. The latter cancontain more solutions that do not appear in Eqs. ~1! and ~2!.In the example of a two-site problem, the Hamiltonian in Eq.~7! has two unphysical states which are degenerate. That iswhy we have two unphysical solutions in the example.Therefore, the compatible properties of H and Qi do notguarantee that the Lagrange multipliers in Eq. ~3! can realizethe constraints.Although we can justify very easily whether the Qi op-erators are semipositive definite, it is extremely difficult todetermine whether the eigenstates of H for a many-body sys-tem are nondegenerate or not. Especially when we want toprove the nondegeneracy of the ground state of a Hamil-tonian ~see references in Ref. 2!, l iQi is not a correct way torealize the quantum constraints. In many-body physics, thereare many examples to introduce the Lagrange multipliers.One of the most famous examples in a many-body systems isthe chemical potential to adjust the number of particles. In agrand canonical ensemble the dispersion of the particle num-bers^~DN !2&5^N2&2~^N&!2}^N&.This is acceptable, or reasonable, since the relative disper-sion A^(DN)2&/^N& approaches zero in the thermodynamiclimit. In a quantum many-body system, for example, thespin-12 Heisenberg model in the fermion representation, it isquite difficult to realize the condition of single occupancy ofa fermion at each lattice site. On the Lagrange multipliersl iQi†Qi we have taken into account the strong correlationsbetween particles. How to make approximation to treat themultipliers goes beyond the scope of this Comment. Whenwe introduce one type of Lagrange multiplier, at least weshould guarantee that the method is correct for a solvablecase before we make other approximations such as a mean-field approximation.ACKNOWLEDGMENTThis work was supported by a CRCG research grant ofthe University of Hong Kong.1 T. Yanagisawa, Phys. Rev. B 57, 6208 ~1998!.2 S. Q. Shen, Phys. Rev. Lett. 79, 1781 ~1997!.3 Since the condition x1y2150 is absolutely equal to the condi-tion (x1y21)250, it is not strange that we obtain the sameresult in two different ways.

Comment on 'Lagrange-multiplier method in correlated-electron systems: Exact diagonalization study'

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