The phase-diagram of the Blume-Capel-Haldane-Ising spin chain

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by  Christian Boudreault

UdeM-GPP-TH-15-246
The phase-diagram of the Blume-Capel-Haldane-Ising spin chain
Christian Boudreaulta,c, S. A. Owerrea,b, and M. B. Paranjapea,b
∗ † ‡
aGroupe de physique des particules, Département de physique, Université de Montréal,
C.P. 6128, succursale centre-ville, Montréal, Québec, Canada, H3C 3J7
bPerimeter Institute for Theoretical Physics, 31 Caroline Street North Waterloo, Ontario, Canada N2L 2Y5 and
6
cCollège militaire royal de Saint-Jean, Case Postale 100,
1
0
2 succursale Bureau-chef, Richelain, Québec, Canada, J0J 1R0

n
a
ABSTRACT
J

2
2
Weconsidertheone-dimensionalspinchain forarbitraryspin sonaperiodicchainwith N sites,

]
l thegeneralization of thechain that was studied by Blume and Capel [1]:
e
-
r N
t.st H =Xi=1(cid:0)a(Siz)2+bSizSiz+1(cid:1).
a
m TheHamiltonian onlyinvolvesthez component ofthespinthusitisessentially anIsing[2]model.
-
d TheHamiltonianalsofiguresexactlyastheanisotropicterminthefamousmodelstudiedbyHaldane
n
o [3]ofthelargespinHeisenbergspinchain[4]. ThereforewecallthemodeltheBlume-Capel-Haldane-
c
[ Isingmodel. AlthoughtheHamiltonianistriviallydiagonal, itisactuallynotalwaysobviouswhich

1
eigenstate is the ground state. In this paper we establish which state is the ground state for all
v
5
9 regionsoftheparameterspaceandthusdeterminethephasediagramofthemodel. Weobservethe
1
6 existence of solitons-like excitations and we show that the size of the solitons depends only on the
0
. ratioa/bandnotonthenumberofsitesN. Thereforethesizeofthesolitonisanintrinsicproperty
1
0
of thesoliton not determined byboundary conditions.
6
1
:
v
i
X
r I. INTRODUCTION
a
Blume and Capel [1] studied a model that corresponded to an Ising spin chain with a nearest neighbour exchange
interactionandaneasy-plane/easy-axisinteraction. Isingspinchainmeansthatonlythez componentofthespinwas
∗Electronicaddress: [email protected]
†Electronicaddress: [email protected]
‡Electronicaddress: [email protected]
2
dynamical, and they only considered the model for spin 1. The corresponding Hamiltonian was ab initio diagonal.
We will study the model for arbitrary large spin s. The Hamiltonian is still diagonal. The exchange interaction
is minimized if neighbouring spins are maximal, S = s, and anti-aligned for positive coupling, but aligned for
z
±
negative coupling. The easy-plane/easy axis term, on the other hand, is minimized if the spins are maximal, for
negative coupling but minimal (zero if possible) for positive coupling. Thus there can be a competition between the
twotermsandavarietyofphaseswhenonetermorthe otherisdominant. We aresomewhatsurprisedtofindsecond
order phase transitions between the various phases described by the Hamiltonian. Indeed, the two competing terms
commute, and thus, in principle we expect only first order phase transitions. However, here we find clearly that the
criticalpointscorrespondtothe appearanceofnegativeeigenvaluesofthe energyfunctional,whichisjustaquadratic
form, and the crossing of the ground state energy eigenvalue from positive to negative signals a second order phase
transition.
For antiferromagnetic coupling, with odd N, there must be a defect in the Néel state on a periodic chain. The
defect is localized between two adjacent spins for easy-axis coupling, a<0. However, for easy-plane coupling, a>0,
we find that the defect spreads out to maximal size as the anti-ferromagnetic coupling is weakened. The defect is a
soliton in the Néel state, and corresponds to a finite energy excitation. We calculate the size of the soliton and find
that it is independent of the number of sites N but depends on the ratio a/b. Hence the size is a characteristic that
is not an artefact of the odd number of sites, the same solitons could be excited in the periodic chain with an even
number of spins or the finite or infinite, open chain.
Haldane [3] considered the Heisenberg model with an anisotropy which corresponds exactly to the Blume-Capel
model, with Hamiltonian
N
H = J S~ S~ +λSzSz +µ(Sz)2 (1)
| | i· i+1 i i+1 i
Xi=1
(withperiodicboundaryconditions,S~ =S~ ),forlargespin S~ =s 1butforsmallanisotropy0<(λ µ)1/2 1,
N+1 1
| | ≫ − ≪
λ>µ. Howeverthe complete phase diagramof the model, for all values of the couplings is still of much interest. We
define a = J µ and b = J λ, and consider the model for J = 0 but a and b finite. This is in the extreme opposite
| | | | | |
limit to Haldane’s considerations, however, it is still of import to Haldane’s considerations. The anisotropy, however
smallinHaldane’swork,picksthe anti-ferromagneticNéelorderedgroundstatethatisalignedinthez direction. For
different parts of the parameter space in the anisotropy,it is possible and indeed true that a different groundstate is
indicated.
3
Thus we consider the Hamiltonian
N
H(Sz,...,Sz )= (a(Sz)2+bSzSz ). (2)
1 N i i i+1
Xi=1
This Hamiltonian is trivially diagonal, the eigenstates can be written as s , ,s , where s
1 N i
| ··· i ∈
[ s, s+1, ,s 1,s] is the z component of the ith spin, written in spin eigenstates that diagonalize the to-
− − ··· −
tal spin and the z component of the spin, for each individual spin. The corresponding energy eigenvalue is
E(s , ,s )= N (a(s )2+bs s ). Which eigenstate has the minimum energyi.e. which is the groundstate,is
1 ··· N i=1 i i i+1
P
not always clearly evident.
II. BLUME-CAPEL-HALDANE-ISING MODEL WITH ARBITRARY SPIN s
We consider a one-dimensionalperiodic “lattice”, a spin chain with N sites N 2 sites. The Hamiltonian, (2) can be
≥
written as
1
H = STAS (3)
2
where ST =(Sz,Sz, ,Sz ).
1 2 ··· N
A. The ferromagnetic case (b<0) for all N.
The ferromagnetic case is easily dealt with. We write the state s ,...,s as rα ,...,α where r = s2
| 1 Ni | 1 Ni k k
pP
and α2 =1. Then the energy is given as
k k
P
E(s ,...,s )= a+b α α r2 =C(αˆ)r2 (4)
1 N k k k+1
(cid:0) P (cid:1)
where expicitly,
C(αˆ)= a+b α α . (5)
k k k+1
(cid:0) P (cid:1)
As (α α )2 0, we have
k k± k+1 ≥
P
1 α α 1. (6)
k k+1
− ≤ ≤
Xk
Thus for b <a, using Eqn.(6) in Eqn.(5) we have C(αˆ)>0 andhence the minimum energy configurationis realized
| |
exactly for r =0 corresponding to the state, 0,...,0 with corresponding energy E =0. For half odd integer spin,
0
| i
the state 0,...,0 is not permitted. Then in this case, one of the states closest to the origin, 1/2,..., 1/2
| i |± ± i
4
(with uncorrelated signs)willbe the minimalenergyconfiguration. Since the firstterminthe energydoesnotcare
±
whether the spin is 1/2 and since b<0, the energy is minimized at the “little” ferromagnetic states
±
1/2,...,1/2 or 1/2,..., 1/2 (7)
| i |− − i
with energy E =(1/4)N2(a b).
0
−| |
For all other cases, b > a, (including a negative), the factor C(αˆ) becomes negative for certain directions, and
| |
in particular for αˆT = 1 (1,...,1) the upper bound Eqn.(6), for the sum α α is saturated. These are the
±√N k k k+1
P
only two states for which the bound is saturated, and here C(αˆ)=a+b=a b. C(αˆ) is negative and minimal for
−| |
this direction. Theextremecorners (s,...,s)ofthehypercube [ s,s]N areattainedalongthisdirection. Hencer is
± −
maximal, and correspondingly, the energy is minimal at the two corners. Thus the two corresponding ferromagnetic
states, which we will write as ,..., and ,..., , are the ground states in the regime b > a (and we will use
| ↑ ↑i | ↓ ↓i | |
the notation and when the corresponding spin is maximally up, s, or maximally down, s, respectively).
↑ ↓ −
Fora= b thestates m,...,m aredegenerateforanym [ s,s]thusthegroundstateis2s+1folddegenerate
| | ±| i ∈ −
with E =0, and the system passes through a highly degenerate critical point. In the large s limit this is veritably a
0
massless continuum.
B. The antiferromagnetic case (b>0) for N even
The case of N even and N odd are substantially different, thus we will treatthem separately. For N even the Néel
state closes periodically as the lattice is bi-partite, but for N odd, the Néel state does not close without frustration,
one has a defect in the ground state. For N odd we will see that the ground state is highly degenerate.
For 0<b<a continuing with the notationofEqn. (4), we find C(αˆ)>0 for alldirections αˆ. Hence the minimum
of the energy occurs at r = 0, i.e. for the state 0,...,0 . Therefore, for integer spin, the ground state is non-
| i
degenerate and given by 0,...,0 with ground state energy E =0. Now for half odd integer spins, this state is not
0
| i
permitted, andoneofthe closestnearbystates,attheverticesofthe hypercubeofcentredontheoriginofsidelength
1, 1/2,..., 1/2 (the signs are not correlated) becomes the groundstate. For the case under consideration, N
|± ± i ±
even, it is easy to verify the minimum energy states correspond to the two “little” Néel states
1/2, 1/2,1/2, 1/2,...,1/2, 1/2 and 1/2,1/2, 1/2,..., 1/2,1/2 . (8)
| − − − i |− − − i
Noticeforthis state,onthe periodiclattice sinceN is even,the Néelpatterncloseswithoutfrustration. Thusforhalf
odd integer spin, the ground state is doubly degenerate and given by the two “little” Néel states, with ground state
5
energy E =(1/4)N2(a b).
0
−
Forallothervalues,a<bthe factorC(αˆ)is negativeinsomedirections,andinparticularit ismaximallynegative
for the states αˆ = 1 (1, 1,...,1, 1). For these states the inequality Eqn.(6), which continues to hold for the
±√N − −
anti-ferromagnetic case, is saturated with α α = 1. Then the energy is equal to E =r2(a b) which is the
k k k+1 − −
P
groundstateenergy. ThusthegroundstateofthesystemisdoublydegenerateandcorrespondstothetwoNéelstates
, , , ..., , and , , , ,..., , .
|↑ ↓ ↑ ↓ ↑ ↓i |↓ ↑ ↓ ↑ ↓ ↑i
1. Phase diagram for N even
Fig. (1) givesthe phase diagramfor the case N evenfor alla and b. For half oddinteger spin the regionsa> b is
| |
doubly degeneratewith the groundstates givenby Eqn.(8) or Eqn.(7) depending on whether b is positive ornegative
but non-degenerate with ground state 0,0,...,0 for integer spin.
| i
We furthermore note that defining the staggered spin operators S¯ ( 1)jS we find a mapping between the
j j
≡ −
ferromagnetic and the anti-ferromagnetic cases
Hb>0(S¯ ,...,S¯ )=Hb<0(S ,...,S ) (9)
1 N 1 N
and vice versa. There is an obvious symmetry in the phase diagram between b b. The ground states in the two
→−
cases are related by the transformation S S¯ , which can be seen from the Fig. (1).
j j
←→
FIG. 1: Phase diagram for N even and arbitrary spin s
b
a=b=|b|
|↑,↓,...,↑,↓i
|0,...,0i integerspin
|↓,↑,...,↓,↑i
|−1,1,...,−1,1i
2 2 2 2  half-oddspin
|1,−1,...,1,−1i
2 2 2 2 
a
|1,...,1i 
2 2  half-oddspin
|↑,...,↑i |−21,...,−21i
|↓,...,↓i |0,...,0i integerspin
a=−b=|b|
6
C. The antiferromagnetic case (b>0) for N odd
For the case of N odd, the most interesting phenomena takes place for a < b, (note b > 0 here). For N even this
case had the groundstates correspondingto the Néel states. But now, the closest states to Néel states correspondto
configurationswithadjacentspins maximallyanti-aligned,(s, s,s, s, ,s)or( s,s, s,s, , s)exceptforthe
− − ··· − − ··· −
lastspinwiththefirstspinsinceN isodd. Forsuchastate, α α = 1+ 2 > 1. Thesearethegroundstates
k k k+1 − N −
P
for a < 0. However, we will find that these states are not always the ground states. We will show that for a > 0, as
theratioa/bchanges,thedirectionwhichminimizesC(αˆ)doesnotcorrespondtooneofthecornersofthehypercube
and the minimum of the energy is attained for a value of r that is not maximal. As a is increased from a = 0, the
groundstate rαˆ changesdiscretelytor′αˆ′ withr′ <r. The finaltransitionoccursfora/bwhentheminimumvalue
of C(αˆ) becomes positive and the ground state becomes 0,...,0 for integer spins and the corresponding little Néel
| i
states 1/2, 1/2,...,1/2, 1/2,1/2 and 1/2,1/2,..., 1/2,1/2, 1/2 ,for half odd integer spin.
| − − i |− − − i
The energy of a general state s ,...,s is of course given by
1 N
| i
N
1
E(s ,...,s )= (as2+bs s )= sTAs. (10)
1 N i i i+1 2
Xi=1
The critical points are obtained by solving the linear system of equations
N
∂E(s ,...,s )
1 N
=(2as +b(s +s )= A s =0, (11)
i i+1 i 1 ij j
∂si − Xj=1
whichhastheevidentsolutions =0forallgenericpointswhereAisinvertible. Thenthecorrespondingstate 0,...,0
i
| i
will be the ground state if the energy is at a minimum at this point. The matrix A, which is the Hessian matrix of
second derivatives of the energy with respect to s and s , is a circulant matrix, [19], with first row (2a,b,0,...,0,b).
i j
Subsequent rows correspond to the previous row permuted circularly by one, so for example, the second row is
(b,2a,b,0,...,0), and so on. The eigenvalues and eigenvectors of circulant matrices are well known, the eigenvalues
areλk =2(a+bcos2Nπk),k =0,1,...,N−1withcorrespondingnormalizedeigenvectorssTk = √1N(1,ωk,ωk2,...,ωkN−1)
i2πk
where ωk is the kth, Nth rootof unity, ωk =e N . The eigenvalues are doubly degenerate except for λ0. For N odd,
the Hessianisa positivedefinite matrixforthe entireregiona>bcos π, whichcorrespondstothe pointatwhichthe
N
smallest eigenvalue at k = [N/2] or k = [N/2]+1 vanishes and then becomes negative for smaller a. (For N even,
k = N/2 is allowed and the Hessian is positive definite only for a > b.) Thus for a > bcos π, the minimum of the
N
energyis triviallyobtainedto be at 0,...,0 ,whichis then, ofcourse,the groundstate forintegerspin. Forhalfodd
| i
integer spin, the ground state must be chosen from the corners of the (smallest) hypercube of side length 1 centered
on the origin, with vertices ( 1/2, 1/2, 1/2), with uncorrelated signs. It is easy to see that then the little
± ± ···± ±
7
Néel states, with one defect, chosen for example between the first and last sites, 1/2, 1/2,...,1/2, 1/2,1/2 and
| − − i
1/2,1/2,..., 1/2,1/2, 1/2 ,correspondto the ground states. Each one is then in fact N fold degenerate,since
|− − − i
the position of the defect can be placed at any of the N different sites.
For the region 0 < a < cos π, the Hessian is indefinite, and the critical point at s = 0 is a saddle point. Thus
b N i
the minimum energy is not attained at the critical point. Then Fermat’s theorem for the extrema of a differentiable
functiondefinedonacompactsetimpliesthat,sincetheminimumdoesnotoccurandaninternalpoint,itmustoccur
on the boundary of the hypercube and we must look for the minimum on its surface. Without loss of generality, we
gotothe boundarybytakings =s. Takings = sis clearlyalsopossible,butcorrespondssimplyto the solution
N N
−
that we will find if we flip all the spins, s s . The energy, taking the face s =s, is then
i i N
→−
N 1 N 2
− −
E =a s2+as2+b s s +bs(s +s ). (12)
i i i+1 N 1 1
Xi=1 Xi=1 −
Varying with respect to s to find the critical points gives the set of equations
i
2as +b(s +s )=0 i=2, ,N 2,
i i+1 i 1
− ··· −
2as +b(s +s)=0, and 2as +b(s +s)=0. (13)
1 2 N 1 N 2
− −
Thesecanbeeasilysolved,howeveritisnotparticularlyuseful,sincethecorrespondingHessianisnotpositivedefinite.
The Hessian is an N 1 N 1 matrix, which we will call B , which is a tridiagonal Toeplitz matrix with the
N 1
− × − −
three non-zero diagonals given by
2a b
 
b 2a b
 
 b 2a b 
 
 
BN−1 = · · ·  (14)
 
 · · · 
 
 · · · 
 b 2a b
 
 
 b 2a
 (N 1) (N 1)
− × −
Again it is well understood how to find the eigenvalues and eigenvectors of a Toeplitz matrix, [20]. The eigenvalues
are given by λ = 2 a+bcoskπ , k = 1,2,...,N 1. Thus, this matrix is again indefinite in the entire region
k N −
(cid:0) (cid:1)
0< a <cos π andwe must go to the boundaryof the face, s =s, ofthe hypercube by setting one ofthe remaining
b N N
s ’s equal to s. This s can be chosen adjacent to s or separated from it. We consider first the case s = s,
i i N N
s = s and return to the separated case later.
N 1
− ±
8
Taking s =s, s = s we get the energy
N N 1
− −
N 2 N 3
− −
E =a s2+2as2+b s s +bs( s+s ) bss . (15)
Xi=1 i Xi=1 i i+1 − 1 − N−2
with the corresponding critical point given by the, now inhomogeneous equations, analogous to Eqn. (13),
B s= bt (16)
N 2
− −
where B , which is as defined in Eqn.(14), is the Toeplitz matrix of dimension N 2 N 2, as usual s =
N 2
− − × −
(s ,s ,...,s )T and t = (s,0,...,0, s)T. The equation for the critical point can be generally solved however,
1 2 N 2
− −
this is againnotnecessaryfor the full region0< a <cos π, since the Hessianmatrix is B whose eigenvaluesand
b N N−2
eigenvectorsare easily found. The Hessian is B with eigenvalues λ =2 a+bcos kπ , k =1,2,...,N 2. Now
N−2 k (cid:0) N−1(cid:1) −
the Hessian is positive definite for cos π < a < cos π, therefore in this range the solution for the critical point,
N 1 b N
−
Eqn.(16) is the minimum energy configuration. The solution is obtained by inverting the Toeplitz matrix B ,
N 2
−
s= b(B ) 1t, explicitly from [20]
N 2 −
− −
sin(N 1 k)θ sinkθ
s =( 1)ks − − − , k =1,...,N 2, (17)
k
− (cid:18) sin(N 1)θ (cid:19) −
−
where cosθ =a/b. For π <θ < π
N N 1
−
sin(N 1 k)θ sinkθ 2sin(N−1−2k)θ cos(N−1)θ
− − − = | 2 2 |
(cid:12)(cid:12) sin(N −1)θ (cid:12)(cid:12) 2sin(N−21)θ cos(N−21)θ
(cid:12) (cid:12)
(cid:12) (cid:12)= |sin(N−12−2k)θ| = |sin N2−1 −k θ|
sin(N−21)θ si(cid:0)n N2−1 θ(cid:1)
(cid:0) (cid:1)
sin N 1 1 π sin π π
< | s(cid:0)in2−N−1 (cid:1)πN−1| = | sin(cid:16)2π− Nπ−1(cid:17)| (18)
2− N 2 − 2N
(cid:0) (cid:1) (cid:0) (cid:1)
cos π
= | (cid:16)N−1(cid:17)|
cos π
2N
(cid:0) (cid:1)
<1,
where in the first equality we have used a standard trigonometric identity, then we have used the fact that the
expression is symmetric for k N 1 k, hence we can restrict k 1,2,..., N 1 on which domain the sine
→ − − ∈ 2−
(cid:2) (cid:3)
function is monotone increasing. Thus at the first inequality we replace the argument in the numerator with the
largest possible value and the argument in the denominator with its smallest value and the second inequality is
obvious. Therefore we have s <s.
k
| |
For the case s =s=s (or with s = s for N even), a similar calculation gives
N N 1 N 1
− − −
sin(N 1 k)θ+sinkθ
s =( 1)ks − − , k =1,...,N 2. (19)
k
− (cid:18) sin(N 1)θ (cid:19) −
−
9
Now the sum in the numerator, instead of the difference that we had before, forces s >s, which is not permitted,
k
| |
thereforethis solutionisunacceptable. Howeveritis easyto determine the lowestenergyconfigurationsubjectto the
boundaryconditions =s=s . Thisboundaryconditionmakesthechaincompletelyequivalenttoanopenchain
N N 1
−
obtainedbycuttingthe periodicchainbetweentheNthandN 1thsites,andimposingtheboundaryconditionthat
−
the spins at the ends are equal to s. The contribution to the energy of these end spins is just a constant hence the
minimum energy configuration of this chain will not be affected. We can easily prove using mathematical induction
that, for anopen chain oflength N, even or odd, the Néel states are the states of minimum energyfor the parameter
range that we are in, π <θ < π , when no boundary condition is imposed. Thus for an odd number of sites, there
N N 1
−
is a Néel state that will satisfy the boundary condition s =s=s . This chain has an odd total number of spins
N N 1
−
andsubjectto these boundaryconditionsthe Néelconfigurationfits exactlywithoutdefectandthe state ...
|↓↑ ↓↑↑i
corresponds to the configuration of minimum energy subject to the boundary condition. We note of course that this
stateisnotthegroundstateoftheperiodicchainwiththesamenumberofspins,asonecaneasilycheck,forexample,
... 0 has lower energy (as a > 0). The actual ground state corresponds to the boundary conditions s = s,
N
| ↓↑ ↓↑ i
s = s as analyzed first, with the soliton of Eqn.(17) interpolating through the spins 1 to N 2, and of course
N 1
− − −
the states obtained by translation of the soliton.
Itwillbe usefulinoursubsequentanalysistorecordherewhathappens forthe caseofevennumber ofspins. Here,
withboundaryconditions =sets = s,itfollowsthattheminimumenergyconfigurationisthecorresponding
N N 1
− −
Néel state, it fits without frustration. Howeverwith the boundary condition s =s=s for N even, it is an easy
N N 1
−
repetition of the previous analysis to see that we get the same solution s as in Eqn. (17), but now since N is even,
k
the first and last spins both point in the same direction. We summarize our finding in Table (I).
TABLE I: Ground states for the open chain with boundary conditions.
|↑,s2,...,sN−1,↓i |↑,s2,...,sN−1,↑i
N odd soliton Eqn.(17) Néel
N even Néel soliton Eqn.(17)
Thus the case where the second spin is adjacent to the first one is completely understood. If the two spins are up,
for a chain with an odd number of sites, the rest of the chain is in the appropriate Néel state, while if the chain has
an even number of sites, the rest of the chain is in the soliton configuration given by of Eqn.(17). If the two spins
10
are opposite then for an odd number of sites we find the soliton state of Eqn.(17) and for an even number of sites we
have the appropriate Néel state.
Next we will consider,andeliminate, the case where the secondspinis notadjacentto the first fixedspin. We first
found that we must go to the boundary, the face corresponding to taking s = s, but then we found we must go to
N
the boundary of this face. We will now do so by taking s = s with specifically j = N 1 or 1, so that the second
j
6 −
fixed spin is not adjacent to the first. Thus we fix j to be an integer in the interval [2,N 2] and put s = s and
′N
−
s = s (the case s = +s will follow). This will give rise to a (N 2) (N 2) system of equations, the
′N j ′N j
− − − − × −
solutionofwhichwillbe the appropriatecriticalpointifitcorrespondsto aminimum. Thuswe willfirstconsiderthe
Hessian at this point, which is a matrix that is block diagonal
B 0
N j 1
C = − − =B B , (20)
N 2 N j 1 j 1
− − − ⊕ −
 0 Bj 1
 − 
theeigenvaluesaregivenbyλ =2 a+bcos kπ ,k =1,2,...,N j 1andλ =2 a+bcosk′π ,k =1,2,...,j 1.
k N j − − ′k j ′ −
(cid:0) − (cid:1) (cid:0) (cid:1)
It is a straightforwardexercise to verify that these are positive definite as we are in the region cos π < a <cos π.
N 1 b N
−
Thus the critical point is the minimum and designates the putative ground state, which we identify through the
following analysis. The spin s = s divides the sequence s ,...,s , s,s ,...,s into two parts,
′N j ′1 ′N j 1 ′N j+1
− − | − − − − i
one of which must be even. Without loss of generality, we take s,s ,...,s sequence to be even. Then.
′N j+1
|− − i
consulting Table (I), the energy is minimized on this segment by the Néel state, s,+s,..., s,+s , and the
| − − i
minimum energy sequence on the remaining odd length segment is given by the solution s = (s ,...,s )T of
′ ′1 ′N j 1
− −
the equation
BN j 1s′ = bt, (21)
− − −
where t=(s,0,...,0, s)T with solution as before given by
−
sin(N j k)θ sinkθ
s′k =( 1)ks − − − , k =1,...,N j 1. (22)
− (cid:18) sin(N j)θ (cid:19) − −
−
But this state, s ,...,s , s,+s,..., s,+s cannot be the lowest energy state. Since s = s, and we
| ′1 ′N−j−1 − − i ′N−1 −
have already found the state that minimizes the energy when s =s= s , the true minimum energy state s is
N N 1
− − | i
given by Eqn. (17). Thus the case where the second fixed spin s = s, j [2,N 2] is not adjacent to s = s
N j N
− − ∈ −
gives a higher, minimum energy state than that obtained when the second spin is adjacent, j = 1. Therefore the
second spin must be adjacent to the first.