Generalization of the Affleck-Kennedy-Lieb-Tasaki Model

for Quantum Ferromagnetism

Isao Maruyama∗
Department of Information and Systems Engineering, Fukuoka Institute of Technology,
3-30-1 Wajiro-higashi, Higashi-ku, Fukuoka 811-0295, Japan

Shin Miyahara
Department of Applied Physics, Fukuoka University,
8-19-1 Nanakuma, Jonan-ku, Fukuoka 814-0180, Japan
(Dated: September 16, 2025)
arXiv:2509.11537v1 [cond-mat.str-el] 15 Sep 2025

We generalize the Affleck-Kennedy-Lieb-Tasaki model to a spin-S ferromagnetic model with

exactly-written ground states, known as the partially-magnetized valence bond solid (VBS) states
with magnetization m = (S − 1)/S. We find that the VBS state and an antiferromagnetic ground
state with magnetization m = 0 are degenerate for S = 3/2 and S = 2 by using the Lanczos method
and the density matrix renormalization group method (DMRG). However, increasing S, the magne-
tization of the ground states is uniquely determined as the fraction m = (S −1)/S. This is not just a
ferromagnet, but a quantum ferromagnet due to quantum entanglement inherent in VBS states. In
the low-energy excitation spectrum, we find the coexistence of the Haldane gap and Goldstone-like
ferromagnetic magnon excitation. This “magnetic chimera” clearly appears under a finite magnetic
field. Finally, we discuss an application to the measurement-based quantum computation and an
extension of the Haldane’s conjecture.

I. INTRODUCTION for example, solvable models13 , spin-liquid states14,15 ,

and resonating-valence-bonds (RVBs)16 , may be embed-
Ferromagnets have been used in many industrial ap- ded in quantum ferromagnets for a large enough spin S.
plications including traditional computers, and are usu- In fact, the ferromagnetic Haldane phase was
ally expressed by fully-polarized Ising states, i.e., clas- discovered3 by applying the theory to the well-known an-
sical states. Meanwhile, antiferromagnets can exhibit tiferromagnetic Haldane phase in the spin-1/2 ladder17 .
quantum entanglement among spins due to quantum spin The numerical calculation on the dynamical structure
fluctuation and have potential for application, for exam- factor S ± (q, ω), one of experimental observables, theo-
ple, in quantum computing. Both properties can coexist retically predicts the co-existence of gapless mode ∆E−
in ferrimagnets with antiferromagnetic coupling; sponta- and gapped mode ∆E+ . As a ferromagnetic property,
neous symmetry breaking in ferrimagnetism due to mixed ∆E− ∝ q 2 is expected in spite of small system size18 .
spins is explained by the Lieb-Mattis theorem1 . Re- As an antiferromagnetic property, ∆E+ is quantitatively
cently proposed “quantum ferromagnet”2,3 can also ex- identical to that of the spin-1 Heisenberg model19 , i.e.,
hibit the coexistence despite single spins without trans- the Haldane gap, ∆E+ > 0.
lational symmetry breaking; spontaneous magnetization The nature of the Haldane gap6 in antiferromag-
satisfies the Oshikawa-Yamanaka-Affleck criterion4 . nets has been revealed by the Affleck-Kennedy-Lieb-
The schematic pictures of the conventional ferromag- Tasaki (AKLT) model20 . Despite nonintegrability of
nets and antiferromagnets in one-dimensional Heisenberg the AKLT model, the ground state is exactly written,
models are illustrated in Fig. 1 (a). In this paper, the and it is called the valence bond solid (VBS) state,
spins in classical ferromagnets are depicted separately, as depicted in the right bottom panel in Fig. 1 (a),
whereas the spins in quantum antiferromagnets are con- and it has recently been referred to as the AKLT state
nected with neighboring spins to illustrate the existence in the context of measurement-based quantum compu-
of quantum entanglement. The quantum ferromagnet tation (MBQC)21,22 . The Haldane phase with entan-
in the spin-S bilinear biquadaratic (BLBQ) model in- gled gapped quantum spin-liquid states23 is a notable
duced by spin-1/2 liquefaction is shown in the top panel example of the symmetry protected topological (SPT)
of Fig. 1 (b), where one can combine a spin-(S − 1/2) phases24,25 . A partially magnetized VBS state, proposed
classical ferromagnet and a spin-1/2 quantum antiferro- by Oshikawa26 , can be a unique ground state under a
magnet by using rigorous “eigensystem embedding”, i.e., magnetic field as a magnetization plateau state4 . How-
the exact eigensystem correspondence between the spin-S ever, because one of the degenerated ground states has
BLBQ model and the spin-1/2 Heisenberg model2 . The a total spin Stot = 0 under a zero magnetic field, their
rigorous correspondence, which might be interesting in model is not appropriate for ferromagnets. Unique total-
the context of quantum many-body scars8–12 , is limited spin Stot of all the ground states is needed to ensure
to the specific point αr in the spin-S BLBQ model but is spontaneous magnetization under a zero magnetic field,
valid on any lattice in any dimension. Then, interesting which is an important aspect of ferromagnets.
research topics for quantum spin-1/2 antiferromagnets, In this paper, to reveal the nature of the ferromag-
 2

FIG. 2. Ferromagnetic AKLT state |Φ⟩ in Eq. (1) writ-

ten with spin-singlets |ϕs ⟩i,i+1 and background ferromagnetic
Ising states |S − 1⟩i,F . Spin-S operator Ŝ i at the i-th site is
decomposed into spin-(S − 1) operator Ŝ i,F and two spin-1/2
operators ŝi,L , ŝi,R . This is identical to Oshikawa’s state26
and is depicted in the lower panel of Fig. 1 (b).

netic AKLT state |Φ⟩ in one dimension under the periodic

boundary condition (PBC) is defined, using symmetriza-
tion mapping operator Ŝ, as
N
Y
|Φ⟩ = Ŝ |ϕs ⟩i,i+1 |S − 1⟩i,F (1)
i=1
FIG. 1. Schematic picture of the ground states (a) in tra-
ditional spin-S chain models and (b) in “quantum” ferro- with the spin-singlet state
magnetism of spin-S chain models. Properties about total
spin Stot of the ground states and low energy excitation ∆E± |ϕs ⟩i,j = |+1/2⟩i,R |−1/2⟩j,L − |−1/2⟩i,R |+1/2⟩j,L (2)
are also summarized. For ∆E+ , the spin-1/2 antiferromag-
net (AF) has ∆E+ ∝ |q|, i.e., des Cloizeaux-Pearson mode5 on the bond i, j.
while the spin-1 AF has the Haldane gap ∆E+ > 06 . On the The state |Φ⟩ is identical to Oshikawa’s ferromagnetic
other hand, ∆E− ∝ q 2 is Goldstone-type gapless one-magnon Ising-VBS state26 , where “ferromagnetic
mode7 . Q Ising” means
a ferromagnetically ordered Ising-state i |S − 1⟩i,F for
the spin-(S −1) sub-system ŝi,F . The term “VBS” comes
netic Haldane phase, we extend the spin- 12 liquefaction2 from an analogy to the valence bond theory for a covalent
to spin-1 liquefaction in the following order. In § II, we bond in chemistry29 ; each bond-singlet state |ϕs ⟩i,i+1 is
define a ferromagnetic AKLT state [the bottom panel of formed by two spin-(1/2)s, which participate from the
Fig. 1 (b)] following Oshikawa26 and we obtain a matrix- i-th and (i + 1)-th spin respectively, i.e., ŝi,R and ŝi+1,L .
product state (MPS) form. We define a ferromagnetic For the S = 1 case,
AKLT Hamiltonian, whose ground state is supposed to
be the ferromagnetic AKLT state in § III, and we give E N
Y
an analytical proof of the ground states in § IV. In § V, Φ(S=1) = Ŝ |ϕs ⟩i,i+1 , (3)
we present the numerical results of the Lanczos method i=1

and the density matrix renormalization group method

obtained by omitting |S − 1⟩i,F in Eq. (1), is identical to
(DMRG)27,28 to answer the three questions in the bot-
tom panel in Fig. 1 (b). As a potential application, we the spin-1 VBS ground state20,29 . In other words, |Φ⟩ is
consider the effects of a finite magnetic field to obtain a a natural generalization of the spin-1 case.
unique and gapped ground state in § VI and apply the Because the local spin-singlet |ϕs ⟩i,j can be written
ground state to MBQC in § VII. We present the sum- in a quadratic form with the two-dimensional matrix: !
mary in § VIII.  + 1

0 1

2
|ϕs ⟩i,j = + 12 i,R − 21 i,R j,L
, one
−1 0 − 21 j,L
QN
II. FERROMAGNETIC AKLT STATES can deduce an MPS form of the singlets i=1 |ϕs ⟩i,i+1 =
!
+ 21 i,L   
0 1
QN 
1 1
Tr i=1 + − =
We consider one-dimensional spin-S models. To de- − 12 i,L 2 i,R 2 i,R −1 0
fine a ferromagnetic AKLT state, let us divide spin-S !
QN − + 21 i,L − 12 i,R + 12 i,L + 21 i,R
operator Ŝ i at the i-th site into three kinds of decom- Tr i=1 by using
posed spins, ŝi,L + ŝi,R + ŝi,F : left spin-1/2 operator − − 21 i,L − 12 i,R − 12 i,L + 21 i,R
ŝi,L , right spin-1/2 operator ŝi,R , and front spin-(S − 1) the trace of the two-dimensional MPS, Tr, that comes
operator ŝi,F , as depicted in Fig. 2. The spin-S ferromag- from the PBC. After some calculations on the sym-
 3

metrization mapping operator Ŝ in Eq. (1), one can ob- the positive coefficient 1/6 is artificial in the sense that
tain the MPS form a negative coefficient is natural33 . This artificial Hamil-
tonian is important because both the ground states of
N (S=1) P
Ĥ0 and the spin-1 Heisenberg model i Ŝ i · Ŝ i+1
Y
|Φ⟩ = Tr Ai (4)
i=1
are in the same SPT phase, i.e., the Haldane phase.
The coefficients given as exact fractional values in
with (s)
Eq. (9) come from projection operators, P̂ij , onto the
 |S−1⟩i
 subspace with total spin s on bond i, j. In the projection
− √
2S
|S⟩i operator form, Ĥ0
(S=1)
is written as
Ai =  |S−2⟩i |S−1⟩i . (5)
−√ √
2S
S(2S−1) N
(S=1) (2)
X
Ĥ0 = P̂i,i+1 , (10)
For the S = 1 case,
i=1
|0⟩ !
(S=1) − √2i |1⟩i which can be proved using a general relation34 :
Ai = |0⟩i (6)
−|−1⟩i √
2 2S
(s)
Y Ŝ i · Ŝ j − qn
P̂ij = ,
is equal to that of the previous study30 except for a nor- qs − qn
n=0
malization constant; also for S = 3/2 case31 . n̸=s
In addition, it is easy to obtain the matrix product 2S
(s)
X
operator (MPO)32 form (Ŝ i · Ŝ j )n = qs n P̂ij , (11)
! N s=0
YN Y
|Φ⟩ = Tr Âi |S⟩i (7) with qs = s(s+1) − S(S + 1) for spin-S.
2
i=1 i=1
For one-dimensional models, a higher spin-S general-
with ization has been studied35 :
N
Ŝ −
!
− 2Si 1 (S) (2S)
X
Âi = (8) Ĥ0 = P̂i,i+1 . (12)
(Ŝi− )2 Ŝi−
− 2S(2S−1) 2S
i=1

Even though the ferromagnetic AKLT state |Φ⟩ is a zero-

by using Ŝi− |m⟩i =
p
(S + m)(S − m + 1)|m − 1⟩i . (S)
Q energy ground state of Ĥ0 , other states with a dif-
This MPO, Tr i Âi , is a creation operator of the
ferent total spin Stot are also zero-energy degenerated
state |Φ⟩,
Q where a fully-saturated ferromagnetic state ground states for S ≧ 3/2. For S = 3/2, to stabilize the
|F ⟩ = i |S⟩i plays a role of the vacuum state: i.e.,
 Q † target state Φ(S=3/2) an infinitesimal magnetic field is
N
Tr i=1 Âi |F ⟩ = 0. required4 , where the projection operator P̂i,j in Ĥ0
(3) (S=3/2)

has an additional “bicubic” (BC) term (Ŝ i · Ŝ j )3 :

III. MODEL HAMILTONIANS
(3) 27Ŝ i · Ŝ j 29(Ŝ i · Ŝ j )2 (Ŝ i · Ŝ j )3 11
P̂i,j = + + + . (13)
160 360 90 128
A. AKLT Hamiltonians
Despite the absence of unique total spin in one dimension,
(S=3/2)
Before we define ferromagnetic AKLT Hamiltoni- the spin-3/2 BLBQBC Hamiltonian Ĥ0 on the two-
ans, we review the one-dimensional S = 1 AKLT dimensional honeycomb lattice has a unique ground state
Hamiltonian20 and related models. The one-dimensional with Stot = 020,36 , which has been referred to as the two-
S = 1 AKLT Hamiltonian is defined as dimensional AKLT state in the context of MBQC21,22 .
 2
N 3Ŝ · Ŝ + Ŝ · Ŝ +2
i i+1 i i+1
(S=1)
X
Ĥ0 = , (9) B. Ferromagnetic AKLT Hamiltonians
i=1
6
To realize unique Stot = N (S − 1) ground states in
which corresponds to the specific point α = arctan(1/3)
P one dimension under a zero magnetic field, we define a
of the general BLBQ Hamiltonian i [cos α Ŝ i · Ŝ i+1 + general spin-S “ferromagnetic” AKLT Hamiltonian:
sin α(Ŝ i · Ŝ i+1 )2 ]. The coefficient of the bilinear term
N
" 2S−4
#
3Ŝ i ·Ŝ i+1
is an antiferromagnetic-type coefficient(i.e., X (2S) (2S)
X (s) (s)
6 Ĥ (S) = Ji P̂i,i+1 + Ji P̂i,i+1 , (14)
3/6 = 1/2 > 0). The biquadratic term (Ŝ i · Ŝ i+1 )2 with i=1 s=0
 4

(s) (2S) (2S−4) (2S−5) (1) (0)

with positive coefficients Ji > 0 for all s and all i. In min(Ji , Ji , Ji , . . . , Ji , Ji ) and with the
(2S−1) (2S−2) P (2S−1)
other words, the absence of terms with Ji , Ji , energy shift i Ji .
(2S−3) For numerical calculations, we limit ourselves to a uni-
and Ji is important for the definition. Due to the
P2S (s) form and simple Hamiltonian. To simplify the general
completeness relation s=0 P̂i,i+1 = 1, we can define a
P P2S (s) (s) Hamiltonian, one can eliminate higher n-th order terms
more general Hamiltonian Ĥ (S) = i s=0 Ji P̂i,i+1 (Ŝ i · Ŝ j )n (n ≥ 5) by properly choosing free parameters
(2S−1) (2S−2) (2S−3) (s)
with the condition Ji = Ji = Ji < Ji > 0. Then, a simple Hamiltonian is given as

N
X
Ĥ (S) (β) = J (Ŝ i · Ŝ i+1 + β)(Ŝ i · Ŝ i+1 − S 2 + 2S)(Ŝ i · Ŝ i+1 − S 2 + 4S − 1)(Ŝ i · Ŝ i+1 − S 2 + 6S − 3), (15)
i=1

which is a bilinear biquadratic bicubic biquartic method. It is enough to program an operator-times-

(BLBQBCBQ) Hamiltonian. Here, J > 0 is an energy vector routine for a general operator αŜ i · Ŝ j + β, where
scale parameter and β is the remaining free parameter the output vector is v out = (αŜ i · Ŝ j + β)v in for an in-
that must satisfy the condition put vector v in . ByQrepeating this routine four times,
4
we obtain v out = k=1 (ak Ŝ i · Ŝ i+1 + bk )v in . After
−S 2 < β < −S 2 + 8S − 6 (16) the summation of the local Hamiltonians, we obtain
(s) v out = Ĥ (S) (β)v in . However, for the DMRG code,
to satisfy the positivity Ji > 0. The lower bound of β
(S) Ĥ (S) (β) is too complex. Thus, we need a simpler Hamil-
> 0, where ϵF (β) = ⟨Ĥ N(β)⟩F is the
(S) (S)
comes from ϵF (β) tonian with a properly chosen free-parameter β.
eigenenergy per site
Q N for the fully-saturated ferromag- A Hamiltonian that is suitable for the DMRG code is
netic state |F ⟩ = i |S⟩i . The explicit form is calculated realized for
as
(S) β = 3(S − 1)(S − 3) =: βS , (20)
ϵF (β) = 6J(S 2 + β)S(4S − 1)(2S − 1) (17)
where the function βS = 3(S − 1)(S − 3) is defined for all
by substituting Ŝ i · Ŝ i+1 with S 2 in Eq. (15). S but the positivity condition Eq. (16) is valid only for
For S = 1 and S = 3/2, the Hamiltonian Ĥ (S) (β) is 2 ≤ S ≤ 4. In the following, we explain simplification by
reduced as follows, βS . For the DMRG code, the Hamiltonian written with
(k)
(S=1) (S=1) SU(k + 1) generators Q̂i is suitable. The Hamiltonian
Ĥ (S=1) (β) = ϵF (β) Ĥ0 , (18)
Ĥ (S) (β) in Eq. (15) for any β is rewritten as
(S=3/2) (S=3/2)
Ĥ (S=3/2) (β) = ϵF (β) Ĥ0 . (19)
4
(S)
X (k) (k)
The coefficients comes from the ferromagnetic energies Ĥ (β) = Ck (β)Q̂i · Q̂j (21)
(S) (S)
⟨Ĥ (S) ⟩F = N ϵF (β) and ⟨Ĥ0 ⟩F = N . Thus, Ĥ (S) (β) k=1
could be considered to be a direct generalization of with coefficient Ck (β) and the operators
(S=1) (S=3/2)
Ĥ0 and Ĥ0 . In addition, for S = 1/2, we have a
†
reduced form Ĥ (S=1/2) (β) = 0 which does not correspond

k Q̂m+;(k−m)z Q̂m+;(k−m)z + h.c.
P (1) (1) (k) (k) X i j
to the general Hamiltonian Ĥ (S=1/2) = i Ji P̂i,i+1 ; Q̂i · Q̂j = ,
however, this is because there is no β satisfying −1/4 < m=0
2
β < −9/4, which is Eq. (16). In general, for S ≥ 2, the
h i
n c + m z n z n + m
l;m,n (Ŝi ) (Ŝi ) + (Ŝi ) (Ŝi )
biquartic term (Ŝ i · Ŝ i+1 )4 cannot be reduced in Eq. (14).
X
Q̂m+;nz
i = .
This is equivalent to the fact that the biquadratic term 2
l=0
(Ŝ i · Ŝ i+1 )2 in Eq. (10) cannot be reduced for S ≥ 1.
(1)
In this sense, the BLBQBCBQ Hamiltonian Ĥ (S) (β) is Note that Q̂i = Ŝ i for k = 1. One important point is
essential for S ≥ 2. that the Hamiltonian Ĥ (S) (βS ) has no third-order term
(3) (3)
Q̂i · Q̂j , i.e., C3 (βS ) = 0. As a result, the DMRG
C. Simplification for DMRG code for Ĥ (S) (βS ) requires only 10=2+3+5 operators:
two SU(2)-operators Ŝiz = Q̂0+;1zi and Ŝi+ = Q̂1+;0zi ,
P Q4 0+;2z 1+;1z 2+;0z
The Hamiltonian Ĥ (S) (β) written in i k=1 (ak Ŝ i · three SU(3)-operators Q̂i , Q̂i , and Q̂i , and
Ŝ i+1 + bk ) is complex but suitable for the Lanczos five SU(5)-operators Q̂0+;4z
i , Q̂ 1+;3z
i , Q̂2+;2z
i , Q̂3+;1z
i , and
 5

Q̂4+;0z
i . The Hamiltonian’s coefficient Ck (β) and coeffi- IV. ANALYTICAL PROOF
cients cl;m,n of Q̂m+;nz
i will be detailed in elsewhere.
Our first task in this section is to prove that the ferro-
magnetic AKLT state |Φ⟩ defined in § II is a zero-energy
D. Summary of Models ground state for the general Hamiltonian Ĥ (S) of Eq. (14)
in § III:
Before we move on to next section, we summarize the
models defined in §III. All the Hamiltonians defined in Ĥ (S) |Φ⟩ = 0. (23)
this section are SU(2) symmetric. Then, the total spin
z
Stot and its z-component Stot are good quantum numbers Because Ĥ (S) is composed of positive semidefinite oper-
(s)
to label eigenstates. Here, the total spin operator Ŝ tot is ators P̂i,i+1 , the lowest energy is non-negative. Then,
defined as Eq. (23) means that |Φ⟩ is not only a zero energy eigen-
state but also a ground state. The following are two
N
X proofs for Eq. (23), but both proofs are simple: just a
Ŝ tot = Ŝ i . (22)
two-site problem. That is, Ĥ (S) is frustration free. We
i=1
are considering PBC but it is easy to consider an open
Because a ferromagnetic state with the maximum to- boundary.
tal spin Stot = N S is an excited state due to P̂i,i+1 ,
(2S) The proof based on Eq. (1) is natural. For two neighbor
Stot < N S is expected for the ground states. In addi- sites i and i + 1, because there exist spin 0 for |ϕs ⟩i,i+1 ,
(S)
tion, unlike the AKLT Hamiltonian Ĥ0 in Eq. (12), the spin 2S − 2 for |S − 1⟩i |S − 1⟩i+1 , and a pair of free spin-
(0) (1) (2S−4) 1/2s for ŝi,L and ŝi+1,R , the spin composition of de-
additional terms P̂i,i+1 , P̂i,i+1 , . . . P̂i,i+1 in the general
composed spins becomes 0 ⊗ (2S − 2) ⊗ 1/2 ⊗ 1/2 =
Hamiltonian Ĥ (S) may lift up the small Stot states among (2S − 1) ⊕ (2S − 2) ⊕ (2S − 2) ⊕ (2S − 3). Then, the
(S)
degenerated ground states in Ĥ0 . Then, the realiza- state |Φ⟩ becomes a zero-energy eigenstate of the projec-
tion of ground states having a unique Stot = N (S − 1) is (2S−1) (2S−2) (2S−3)
tion Hamiltonian without P̂i,i+1 , P̂i,i+1 , or P̂i,i+1 .
naively expected. This is a natural proof for Eq. (23).
The BLBQBCBQ Hamiltonian Ĥ (S) (βS ) at β = βs The other proof based on the MPS form in Eq. (5)
(S=1)
for 2 ≤ S ≤ 4 are summarized in Table I with Ĥ0 is straight-forward. The proof is composed of two
(S=3/2) (s)
and Ĥ0 . Here, the coefficients of bilinear terms are parts, P̂i,i+1 Ai Ai+1 = 0 (0 ≤ s ≤ 2S − 4) and
negative (ferromagnetic) for S ≥ 2 while those for S < 2 (2S)
P̂i,i+1 Ai Ai+1 = 0, where matrix product Ai Aj becomes
are positive (antiferromagnetic). Due to the ferromag-  − − − − −

(aŜi −Ŝj )Ŝj −Ŝi +Ŝj
netic bilinear terms for S ≥ 2, ferromagnetic states are 2S(2S−1) 2S
Ai Aj =   |S⟩ |S⟩ with
favorable but due to the positive biquartic term (ĥ4i ) the Ŝi− (Ŝi− −Ŝj− )Ŝj− Ŝi− (−Ŝi− +aŜj− ) i j
maximum total-spin states are not favorable. This is a 4S 2 (2S−1) 2S(2S−1)
naive understanding of fractional magnetization under a a = 2S−12S . The four states in the MPS Ai Aj have
z
zero magnetic field. Si,j = Siz + Sjz ≥ 2S − 3, which gives the lower bound
of total spin as Si,j ≥ 2S − 3 for the four states; thus,
(s)
we obtain P̂i,j Ai Aj = 0 (0 ≤ s ≤ 2S − 4). The re-
(S=1) (S=3/2)
TABLE I. Hamiltonian Ĥ0 , Ĥ0 , and Ĥ (S≥2) (βS ) (2S)
maining task is to prove P̂i,j Ai Aj = 0. First, let
in Eq. (14) and Eq. (20) normalized by the coefficient of the
us classify the four states in Ai Aj with parity for the
bilinear term, ĥi = Ŝ i · Ŝ i+1 .
swap of indices i, j; three elements, (Ai Aj )1,2 , (Ai Aj )2,1 ,
Hamiltonian normalized Hamiltonian with ĥi = Ŝ i · Ŝ i+1 and (Ai Aj )1,1 − (Ai Aj )2,2 , have odd parity, whereas the
remaining state (Ai Aj )1,1 + (Ai Aj )2,2 has even parity.
ĥ2
 
(S=1) P i 2
Ĥ0 i ĥi + 3 + 3 The former three states with odd parity cannot have
(S=3/2) P

116ĥ2 i 16ĥ3i 55
 total spin 2S, because the highest total spin 2S state
Ĥ0 i ĥi + 243 + 243 + 108 must have even parity. For the latter state with even
P  ĥ2 ĥ3 ĥ4

Ĥ (S=2) (β2 ) −ĥ i − i
+ i
+ i parity, we must√calculate that (Ai Aj )1,1 + (Ai Aj )2,2 ∝
i 5 9 45

(S=5/2) P  178ĥ2 80ĥ3 16ĥ4 11385

 |S − 2⟩i |S⟩j − 2a|S − 1⟩i |S − 1⟩j + |S⟩i |S − 2⟩j is or-
i −ĥi − 503 + 503 + 503 + 8048
i i i
Ĥ (β5/2 ) thogonal to the total spin Si,j = 2S state with the
ĥ2 5ĥ3 ĥ4
= 2S − 2 written as (Ŝi− + Ŝj− )2 |S⟩i |S⟩j ∝
P  
z
Ĥ (S=3) (β3 ) i −ĥi − 3 + 36 + 36
i i i same Si,j
p
P  134ĥ2 80ĥ3 16ĥ4

|S − 2⟩i |S⟩j + 2/a|S − 1⟩i |S − 1⟩j + |S⟩i |S − 2⟩j .
Ĥ (S=7/2) (β7/2 ) i −ĥi − 717 + 2151 + 2151 − 3824
i i i 2415
− s
P  73ĥ2 5ĥ3 ĥ4
 The other ground states are written as (Ŝtot ) |Φ⟩ with
Ĥ (S=4) (β4 ) i −ĥi − 293 + 293 + 293 + 293
i i i 360
− P − −
Ŝtot= Ŝi . The eigenequation Ĥ (Ŝtot )s |Φ⟩ = 0
(S)
− s − − s−1
is derived as Ĥ (S) (Ŝtot ) |Φ⟩ = Ŝtot Ĥ (S) (Ŝtot ) |Φ⟩ =
 6

− s (S)
· · · = (Ŝtot ) Ĥ |Φ⟩ = 0 by using which becomes a non-integer if states having a different
z
±
Stot are degenerated during the Stot = 0 sector calcu-
[Ĥ (S) , Ŝtot ] = 0. (24) lations. In other words, the non-integer Stot provides
numerical evidence of the degeneracy. Since the ground
Here, Ĥ (S) |Φ⟩ = 0 is Eq. (23). Using the property37 of
state |Φ⟩ has Stot = N (S − 1) as shown in § IV, we also
the symmetrization mapping operator Ŝ define the shift in the total spin as
 
Ŝi− Ŝ = Ŝ ŝ−
i,L + ŝ −
i,R + ŝ −
i,F (25) ∆S = Stot − N (S − 1). (28)
for Eq. (1), one can show The magnetization m of a state is given by the expec-
N N
tation value of Stot as
Y Y
− s
(Ŝtot ) |Φ⟩ = Ŝ |ϕs ⟩i,i+1 (ŝ−
tot,F )
s
|S − 1⟩i,F (26) Stot
i=1 i=1 m = . (29)
NS
because (ŝ− −
i,R + ŝi+1,L )|ϕs ⟩i,i+1 = 0. Here, ŝ− tot,F = For example, the fully saturated value, m = 1, is obtained
P −
ŝi,F . It should be noted that (ŝN,R + ŝ−
− QN
1,L )|ϕ ⟩
s N,1 = 0 by the fully-polarized ferromagnetic Ising state i=1 |S⟩i .
due to PBC. Since the ferromagnetic background state
QN In the spin-1/2 BLBQ chain2,3 , fractional magnetization
(ŝ−
tot,F )
s
i=1 |S − 1⟩i,F has a total spin stot,F = N (S − m = S−1/2 = 2S−1
S 2S under a zero magnetic field has been
1) and z-component sztot,F = N (S − 1) − s, the ground numerically observed: m = S−1/2 corresponds to Stot =
− s S
state (Ŝtot ) |Φ⟩ has the same total spin Stot = N (S − 1) N (S − 1/2) due to spin-1/2 liquefaction.
z
and the same z-component Stot = N (S −1)−s. Here, s is In addition, with using DMRG, to obtain a specific
in the range [0, 2N (S −1)]; i.e., the number of degeneracy Stot = n states for a given n, we consider the specific
z
is 2N (S − 1) + 1. Stot = n sector of a Hamiltonian Ĥα = Ĥ (S) (βS )+αŜ tot ·
For S = 1, the number of degeneracy is 1 and Stot = 0: Ŝ tot for α > 0 which can lift large Stot states. Due
i.e., the unique ground state Φ(S=1) in Eq. (3). In to the SU(2) symmetry of the Hamiltonian, a desired
other words, the above discussion is a simple generaliza- Stot = Stotz
= n state becomes a ground state of Ĥα
tion of previous studies for S = 1. Then, as a future for a large enough α in an Stot z
= n subspace. Then,
research topic, one could consider anomalous features we calculate the energy of the original Hamiltonian for a
of the S = 1 system even in the ferromagnetic AKLT given n by using the ground state of Ĥα as
model, for example, correlation functions including the
string order parameters26,38 , hidden Z2 × Z2 symmetry EStot =n = ⟨ Ĥ (S) (βS ) ⟩α . (30)
revealed by the Kennedy-Tasaki transformation39 , high
dimensions20 , large spin VBS states34,40 , recent topolog- We perform the DMRG finite-size method under the
ical indices41–43 , SPT for larger S 31 . periodic boundary condition using a ladder configuration
z
Despite the rigorous proof of the ground states, there with and without the conserved value Stot . The max-
− s imum number of finite-size sweeps is 20. The number
is a possibility that another ground state than (Ŝtot ) |Φ⟩
exists. Then, we need numerical results to show that the of remaining basis in the block is χ ≤ 800, where the
ferromagnetic AKLT state is a unique ground state in maximum memory is approximately 100GB.
the next section.
A. Unique Total Spin of Ground States
V. NUMERICAL RESULTS
In this subsection, we discuss whether the ferromag-
In this section, we present the numerical results of us- netic AKLT states with Stot = N (S − 1) are unique
ing the Lanczos method and DMRG to answer the three ground states or not for Ĥ (S) (βS ), as defined in § III.
questions in the lower panel of Fig. 1 (b). Because the As a numerical result, the ferromagnetic AKLT states
simplified Hamiltonian Ĥ (S) (βS ) still has both transla- with Stot = N (S − 1) are unique under the finite-size
tional symmetry and SU(2) symmetry, each state is la- gap for 5/2 ≤ S ≤ 4 but are not unique for S = 2.
beled by quantized numbers: wave number q = 2πn/N , In the latter case, the expectation value of Stot becomes
z a non-integer value due to the degeneracy, for both the
total spin Stot , and its z component Stot . Here, n is an
z z
integer and N is the system size. The Stot and Stot can ED and the DMRG results in the Stot = 0 sector. To
be calculated both in the Lanczos method and DMRG, increase S is inevitable. Although the upper limit, S ≤
whereas the wave number q can be calculated only in the 4, is introduced in our calculation through the specially
Lanczos method. With both methods, the total spin is chosen βS in Eq. (20), it is naively expected that the
calculated as uniqueness of Stot will hold for S > 4 with using β in
q Eq. (16).
4⟨Ŝ tot · Ŝ tot ⟩ + 1 − 1 In addition, we conclude that another ground state
Stot = , (27) for S = 2 has Stot = 0. To obtain direct evidence of
2
 7

100 states have Stot = N (S − 1) uniquely. ItQis expected that

ES
tot=0
Stot a triplet excitation of the spin-singlets i |ϕs ⟩i,i+1 in the
10-1
ground state, Eq. (1), can define the Haldane gap. More
10-2 precisely, the Haldane gap is defined as the lowest energy
z
E+ in the sector Stot = N (S −1)+1. Because the ground
10-3
E/E+,N=4, Stot

states have zero energy, the lowest excitation energy ∆E+

10-4 from the ground state energy becomes

10-5 ∆E+ = E+ − 0 = E+ , (32)

10-6
which corresponds to the ∆E+ depicted in Fig. 1.
(S=1)
10
-7 For Ĥ0 , by using the single mode approximation
5
(SMA)34 , a triplet dispersion ESM A (q) = 27 (5 + 3 cos q)
10-8
0 50 100 150 200 has been obtained; the wave number q = π gives the low-
# loop (finite size-method) est energy ESM A (π) = 10
27 > 0, which is identified as the
Haldane gap ∆E+ = E+,q=π .
As a result for 5/2 ≤ S ≤ 4, the lowest-energy state in
z
FIG. 3. EStot =0 and Stot as a function of the number of loop the sector Stot = N (S − 1) + 1 has a gapped excitation
in the finite size method for S = 2, and N = 24 obtained by energy ∆E+ > 0 and a total spin Stot = N (S − 1) + 1,
DMRG with χ = 100. All data points are positive: EStot =0 > (∆S = +1). Based on the Lanczos method, the lowest-
0 and Stot > 0. The unit of energy E+,N =4 will be given by energy state has wave number q = π with quadratic dis-
Eq. (33) in the next section § V B. persion around q = π as in the case of S = 1. In addition,
the number of degeneracy for the states with the lowest
excitation energy ∆E+ is 2Stot + 1 = N (S − 1) + 3. For
the zero energy ground state with Stot = 0 by DMRG, S = 1, the number of degeneracy becomes 2Stot + 1 = 3,
we calculate the energy EStot =0 by using the projection which are the spin-triplet excitations.
method in Eq. (30) by sweeping α. The typical results As an analytical result for N = 4, obtained using
for the system size N = 24 are shown in Fig. 3. The pos- Mathematica, we found the explicit formula
itivity of energy EStot =0 > 0 comes from the variational
principle holding at every step of the finite-size DMRG (S)
ϵF (βS )
method, which means that we can choose the lowest en- E+,N =4 = , (33)
ergy in the finite-size-method loop. Then, Fig. 3 shows (4 − 1/S)
the numerical evidence of the zero energy ground state (S)
with Stot = 0 for the system size N = 24 with a very where ϵF is ferromagnetic energy per site in Eq. (17).
small error < 10−6 . We use E+,N =4 as a unit of energy in this paper.
We briefly discuss the S = 1 and S = 3/2 cases. For Figure 4 shows the Haldane gap ∆E+ = E+ obtained
the S = 1 case, the ground state Φ(S=1) in Eq. (3) using DMRG for a large system-size N . In Fig. 4, one can
with Stot = 0 is unique, whereas for the S = 3/2 case, find the system-size independence of each S for N > 10
E+
an infinitesimal magnetic field is required to stabilize the and the small spin-S dependence of E+,N =4
including
z 44
ground state with Stot = Stot = N/24 . We also calcu- S = 1. For S = 1, the previous study reported that
late the S = 3/2 and N = 12 case and discover another the Haldane gap was ∆E+ ≃ 0.350, which corresponds
ground state with Stot = 0. to E+ /E+,N =4 = 1.05 in Fig. 4 because the S = 1 AKLT
To conclude this subsection, the finite-size numerical Hamiltonian Eq. (10) has the energy unit E+,N =4 = 1/3.
calculation has shown that the ground states have Another previous study45 reported 2∆E+ ≃ 0.71, i.e.,
E+ /E+,N =4 = 1.065, which seems to be an over estima-
Stot = N (S − 1), S ̸= 3/2, 2 (31) tion relative to Fig. 4.
Based on the above results, we conclude the existence
uniquely, and the number of degenerated ground states of the Haldane gap at q = π,
is 2Stot +1 = 2N (S −1)+1. This conclusion is an answer
to the first question in Fig. 1 (b). In the remaining part ∆E+ = ∆E+,q=π > 0, (34)
of this paper, we focus on 5/2 ≤ S ≤ 4 and we ignore
S = 3/2, 2 due to the lack of unique Stot of the ground as an answer to the third question in Fig. 1 (b).
states.

C. Goldstone-type Gapless Excitation

B. Haldane Gap
The remaining question in Fig. 1 (b) is about ∆E− .
In this subsection, we generalize the concept of the In this subsection, we clarify the nature of a one-magnon
Haldane gap to the ferromagnetic case, where the ground branch ∆E−,q . Before we discuss the one-magnon branch
 8

1.07
S=1 0.45 S=5/2 N=12,14
S=5/2 S=3 N=8,10,12
1.06 S=3 S=7/2 N=8,10
0.4
S=7/2 S=4 N=8,10
1.05 S=4
0.35

1.04 0.3

E-,q /E+,N=4
E+/E+,N=4

0.25
1.03
0.2
1.02
0.15
1.01
0.1
1
0.05

0.99 0
0 0.05 0.1 0.15 0.2 0.25 0.00π 0.25π 0.50π 0.75π 1.00π
1/N q

FIG. 4. System-size N dependence of the Haldane gap

∆E+ = E+ of Ĥ (S) (βS ) defined in Eq. (32) with the unit FIG. 5. Low energy excitation ∆E−,q = E−,q as a function
(S=1)
of energy E+,N =4 in Eq. (33). Using Ĥ (S=1) (β1 ) ∝ Ĥ0 , q with ∆S = −1 (Stot = N (S − 1) − 1), obtained using the
the S = 1 data (solid line) was obtained from a table in a Lanczos method. For each S, larger point indicates a larger
previous study44 and the S = 1 data points without a solid system size N . The solid lines are fitted curves using the
line were calculated by DMRG additionally. dispersion function in Eq. (35). E+,N =4 is the unit of energy
in Eq. (33).

in the ferromagnetic AKLT model, we summarize the 0.16

S=5/2
nature of the magnon in conventional ferromagnets. In S=3
0.14
general, for fully-saturated ferromagnetic ground states S=7/2
S=4
with the total spin Stot = N S, (2N S +1)-fold degeneracy 0.12 fit(Lanczos S=5/2)
fit(Lanczos S=3)
reflects the spatial rotational symmetry O(3) on sponta- fit(Lanczos S=7/2)
neous magnetization; one can spatially rotate magnetic 0.1 fit(Lanczos S=4)
E- /E+,N=4

moments with infinitesimally small energy for infinite 0.08

system size. Then, there is gapless excitation with a small
wave number q, which corresponds to a spin twist with 0.06
long distance, i.e., the Goldstone mode. In fact, for fer-
0.04
romagnetic Heisenberg chains, the gapless one-magnon
dispersion ∆E−,q ∝ 1 − cos q is obtained exactly7 . For 0.02
small q, the low energy excitation is approximated as
∆E−,q ∝ q 2 , as depicted in Fig. 1 (a). Except for 0
0 0.02 0.04 0.06 0.08 0.1 0.12
q = 0, the low energy excitation ∆E−,q has a total spin 1/N
Stot = N S − 1, which is decreased by one spin from a
fully-saturated total spin Stot = N S. FIG. 6. System size N dependence of the projected energy
In the ferromagnetic AKLT model, the lowest energy EStot =N (S−1)−1 in Eq. (30), obtained using DMRG. Solid lines
z
excited state in the sector Stot = N (S − 1) − 1 has a total are E−,q=2π/N of the fitted dispersion curves that were deter-
spin Stot = N (S − 1) − 1 (∆S = −1), except for very mined in Fig. 5 without using the DMRG results. E+,N =4 is
small system-sizes. Low energy excitation ∆E−,q = E−,q the unit of energy in Eq. (33).
as a function of q with ∆S = −1 (Stot = N (S − 1) − 1)
are obtained by the Lanczos method as shown in Fig. 5.
The dispersion E−,q is approximated very well by fourth- fitted dispersion curves that were determined in Fig. 5
order cosine bands with the fitting parameters vn defined show good agreement with the DMRG data in Fig. 6.
as Then, it can be expected that the lowest energy in the
sector Stot = N (S − 1) − 1 has q = 2π/N , even for a large
4
X system size N .
∆E−,q = E−,q = vn (1 − cos nq), (35) Based on the small q expansion of Eq. (35), we con-
n=1 clude that the answer to the second question in Fig. 1 (b)
where the system-size independence for a small system is
size N ≤ 14 is shown in Fig. 5. Through DMRG ∆E− ∝ q 2 . (36)
with Eq. (30), the lowest energy in the sector Stot =
N (S − 1) − 1 is obtained for a larger system size N . Al- It should be emphasized here that the gapless excitation
though wave number q is not obtained in DMRG, the ∆E− has ∆S = −1 and does not exist for S = 1 because
 9

4 4
1.6 1.6

1.4 2 1.4 2

1.2 Lanczos 1.2

DMRG 0 0
1 E+,q 1

∆E/E+,N=4
E/E+,N=4

E-,q
E-2,q -2 -2
0.8 E-3,q 0.8

0.6 0.6
-4 Lanczos -4
DMRG
0.4 0.4 E+,q
E-,q
-6 E-2,q -6
0.2 0.2 E-3,q

0 -8 0 -8
0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3
q q

FIG. 8. Effect of Zeeman splitting on Fig. 7 when the mag-

FIG. 7. Eigenenergy E as a function of wave number q for netic field h is equal to half of the Haldane gap E+,π , where
spin S = 3, depicted with total-spin shift ∆S from -8 to 4 the ground state is |Φ⟩ in Eq. (II) and the ground state energy
in the color scale. Larger point of the Lanczos data indicates is E0 = −hN (S − 1).
larger system size N in N = 8, 10, and 12. E+,q is the Hal-
dane gapped dispersion and E−m,q is the Goldstone-type m-
magnon dispersion. E−,q = E−1,q is the same data in Fig. 5 is not captured by the Lanczos method within N ≤ 12.
and Fig. 6. In the light-blue shaded region, there should be
Note that, for a fixed system size N , since the smallest
many excitations with negative ∆S, which are not captured
numerically. E+,N =4 is the unit of energy in Eq. (33).
wave number of the m-magnon mode is q = 2πm/N ,
there exists the relationship E−,q=2π/N < E−2,q=4π/N <
E−3,q=6π/N ; thus, the first excitation energy becomes
the S = 1 AKLT “antiferromagnetic” state has a total E−,q=2π/N , except for a very small system size N .
spin Stot = 0 and there is no possibility for ∆S = −1. In The existence of E−m,q for a large m means there
other words, the existence of ∆E− is a specific feature of are more low energy excitations for a small q. In fact,
ferromagnets—not only for quantum ferromagnets but the Goldstone-type m-magnon gapless dispersion shows
also for classical ferromagnets—as shown in Fig. 1. In E−,q > E−2,q > E−3,q at a fixed q > 0. Then, in
this sense, we call the dispersion of ∆E− a Goldstone- Fig. 7, we shade the low energy region with negative ∆S
type gapless mode. in light blue. Similarly, in the light-blue shaded region
for 0.6 ≲ E/E+,N =4 ≲ 1.4, there should be many exci-
tation data points with a small total spin, and there are
D. Summary of Calculations not captured by the Lanczos method. In other words,
for Stot ≥ N (S − 1) + 1 (∆S ≥ 1), there is no uncap-
tured state below the Haldane gap E/E+,N =4 ≃ 1.053.
As a short summary of this section, we have answered
This means that there is a gapped structure for Stot ≥
all three questions in Fig. 1 (b): 1) unique total spin
N (S − 1) + 1 in the light-blue shaded region, which plays
Stot = N (S − 1) of the ground states under the finite size
an important role under a magnetic field, as discussed in
gap, 2) the first excitation energy with q = 2π/N comes
the next section.
from the Goldstone-type gapless mode ∆E− ∝ q 2 , and
3) a generalized Haldane gap ∆E+ > 0. In general, how-
ever, many low energy states exist. For example, there is
the multi-magnon dispersion of the Goldstone-type gap- VI. ENERGY GAP CONTROLLED BY
less mode, which is considered to be a generalization of MAGNETIC FIELD
m-magnon modes ∆E−m ∝ (1 − cos q)/m for spin-1/2
ferromagnetic Heisenberg chains7 . In this section, we introduce a magnetic field term into
Figure 7 shows all the low energy states obtained us- the Hamiltonian as
ing the Lanczos method and DMRG for S = 3 . The
fitted lines of the Haldane gapped dispersion E+,q and Ĥ (S) (βS ) − hŜtot
z
. (37)
the Goldstone-type m-magnon gapless dispersion E−m,q
with ∆S = −m are also depicted in Fig. 7, where these The finite magnetic field h splits the (2Stot + 1)-fold de-
were obtained by fitting Eq. (35) with the lowest energies generated eigenenergy by the addition of Zeeman energy
z
at a fixed q in the fixed sector Stot = N (S − 1) − m, as −hStot , −hStot + h, −hStot + 2h, . . . , +hStot ; this is the
calculated using the Lanczos method for m ≤ 3. The dis- Zeeman splitting. If a full energy diagram is classified
persion E−m,q for m ≥ 4 must exist but its q-dependence by total spin Stot under a zero magnetic field h = 0, one
 10

can depict the energy diagram at finite h without any where we have the gapped and unique ground state |Φ⟩.
calculation.
Figure 8 shows the energy diagram for spin S = 3
when the magnetic field h is equal to one half of the Hal- VII. APPLICATION TO MBQC
dane gap E+,π . Compared with Fig. 7 covered with the
light-blue region, the structure of low energy excitation As the gapped and unique ground state |Φ⟩ was estab-
is clarified with a finite gap. In the following, we discuss lished in § VI, a generalization of MBQC for the spin-1
this clarification induced by the magnetic field. AKLT model47 is straightforward. Four-fold degenerated
For Goldstone-type m-magnon gapless excitations ground states under the open boundary condition (OBC),
E−m,q which were connected to E = 0 at q = 0, as i.e., Ĥ (S) |J (s) =0 in Eq. (14), are written in the MPS as
shown in Fig. 7, the lowest excitation energy with Zee- N

man energy becomes ∆E−m,q = E−m,q − h[N (S − 1) − N

!
m] + hN (S − 1) = E−m,q + mh. This result indicates the t
Y R0
|L, R⟩[1:N ] = (L0 , L1 ) Ai . (39)
emergence of the gap mh for the gapless mode E−m,q . In i=1
R1
z
fact, the lowest E−,q dispersion in Fig. 8 has ∆Stot = −1
E+,π The left and right edge states can be determined as two-
and is connected to ∆E = h = 2 at q = 0. In ad- !
dition, both the second lowest E−,q dispersion and the L0
dimensional complex vectors L = and R =
z L1
lowest E−2,q dispersion have ∆Stot = −2 and are con-
nected to ∆E = 2h = E+,π at q = 0. In this way, a
!
R0
state with a small Stot < N (S − 1) gains a large gap due . One key idea in MBQC is that the unitary
to the Zeeman splitting. Using the Lanczos method, we R1
calculated a small enough Stot z
sector to depict Fig. 8. transformation of a two-dimensional complex “q-bit” vec-
For the Haldane-gap branch with E+,q and ∆S = tor can be realized by projection measurement onto the
1, the lowest excitation energy with Zeeman energy is corresponding edge state. In fact, using an appropriate
∆E+,q = E+,q − h[N (S − 1) + 1] + hN (S − 1) = E+,q − h, basis |pm ⟩, (m = 1, . . . 2S + 1), the projection measure-
which indicates the shrinking of the Haldane gap E+ = ment of the N -th site on the right-edge can generate a
E+,q=π . Then, it is expected that a transition from the new state |L, Um R⟩[1:N −1] |pm ⟩N with a determined state
original ground state |Φ⟩ with ∆S = 0 and q = 0 to |pm ⟩N at the N -th site, where Um is a two-dimensional
the Haldane-gap state with ∆S = 1 and q = π occurs matrix and R is the q-bit vector.
under the finite magnetic field h ≥ E+,π . To confirm As a generalization of S = 1 MBQC, we define the
this transition, the ground state energy under the mag- (unnormalized) orthogonal basis
netic field h is calculated by DMRG without considering p √
z
the Stot preservation. As a result, the transition point |p1 ⟩ = |S⟩ − S(2S − 1)|S − 2⟩ + C S − 1|S − 3⟩,
p
is hc = E+,π for a system size up to N = 24, as ex- |p2 ⟩ = −i|S⟩ − i S(2S − 1)|S − 2⟩
pected. In other words, there is a stable magnetic plateau √
of Stot = N (S − 1) for |h| < hc . The magnetic tran- (2S + 1) S − 1
−i |S − 3⟩,
sition at hc = E+,π from ∆S = 0 to ∆S = 1 means √ C
that a gapped structure also exists for high energy sec- |p3 ⟩ = − 2S|S − 1⟩, (40)
tors with ∆S > 1. If a state with ∆S has energy E at
h = 0, its excitation energy at h = hc = E+,π is ∆E = with a free parameter C, which is required for⟨p1 |p2 ⟩ =
E −hc [N (S −1)+∆S]−hc N (S −1) = E −E+,π ∆S. Then, 0. Here, for S = 1, |S − 3⟩ becomes non-physical
√ but
the above magnetic transition requires E > E+,π ∆S for does not appear due to the zero coefficient, S − 1 = 0.
∆S > 1, which is the gapped structure that is required For m = 1, 2, and 3, the corresponding unitary matrices
to stabilize the ground state of ∆S = 0 in its mag- U1 , U2 , U3 become the Pauli matrices as
netic plateau. The situation is the same for the S = 1 !
case, which has been established both theoretically and 0 1
U1 = ,
experimentally46 . The major difference from the S = 1 1 0
case is the controllability of lowest-excitation type; the !
lowest excitation comes from the one-magnon branch, 0 −i
U2 = ,
E−,q + h, for |h| < E+,π /2 and the Haldane-gap branch, i 0
E+,q − h, for |h| > E+,π /2. Such a controllable coexis- !
tence is common with that of the magnetization plateau 1 0
U3 = , (41)
state in spin systems. 0 −1
To summarize this section, the low energy spectrum
revealed by the magnetic field indicates the stable mag- whose proof is simply a calculation on
netic plateau  
− N⟨pm√|S−1⟩
2S
N
N⟨pm |S⟩N
Stot S−1 Um = N⟨pm |AN =  N⟨pm |S−2⟩ N⟨pm |S−1⟩N
.
m= = , |h| < hc = E+,π , (38) − √ N
√
2S
S(2S−1)
NS S
 11

This is a direct generalization of spin-1 MBQC47 . For tional magnetization is called as the Haldane plateau by
S > 1, other generalizations with using |m⟩ (m < S − 3) Sakai and Okamoto57 .
can be possible and might be suitable for measurement
Moreover, combined with the previous study on unique
along rotated spin axis.
magnetization m = (S − 1/2)/S in spin-S BLBQ
Compared with the spin-1 MBQC, a finite magnetic
models2,3 , the spin parity effect on the existence of the
field, which is not required for the spin-1 MBQC, is re-
Haldane gap can be generalized to ferromagnets. This
quired to realize the gapped and unique ground state.
spin parity effect for ferromagnets does not depend on
The magnetic field also affects edge states, which can be
spin S but depends on the macroscopic shrinking of Stot
a drawback. However, there can be new properties which
as shown Fig. 1 (b). After we quantify the shrinking of
do not exist in the spin-1 antiferromagnetic MBQC,
Stot as a variable s = S − Stot /N in
such as the magnetic field control of spontaneously-
magnetized ground states, inter-edge interaction through
the Goldstone-type one magnon modes, edge states at
Stot = N (S − s), (42)
domain-wall boundaries.

the spin-s liquefaction of ferromagnetic moment due to

VIII. SUMMARY quantum spin fluctuation gives rise to des Cloizeaux-
Pearson mode (∆E− ∝ |q|) for s = 1/22 and the Haldane
In summary, we have generalized the spin-1 AKLT gap (∆E− > 0) for s = 1, as shown in Fig. 1. This mech-
model to spin-S ferromagnetic AKLT models, and we anism for fractional magnetization m = (S − s)/S in uni-
have presented an application to MBQC. Except for form spin-S systems differ from the Lieb-Mattis theorem1
S = 3/2 and S = 2, the ground states, which are identical for ferrimagnetism in antiferromagnetically-coupled al-
to Oshikawa’s states26 , have the unique and macroscopi- ternating spins. In addition, the generalized spin-parity
cally large total-spin Stot = N (S − 1), i.e., the fractional effect that depends on the amount of spin liquefaction s
magnetization m = (S − 1)/S, even under a zero mag- is completely different from the spin-S dependent spin-
netic field. Then, the spontaneous symmetry breaking parity effect on ferromagnets58 .
can happen in ferromagnetic AKLT models, unlike an-
From the view point of the spin-s liquefaction in spin-S
tiferromagnets with magnetization plateau induced by a
ferromagnets, the traditional spin-1/2 quantum antifer-
finite magnetic field.
romagnetic chains having des Cloizeaux-Pearson mode
The low energy excitation is composed of both gapless
correspond to the full spin-(s = 1/2) liquefaction of
∆E− ∝ q 2 and gapped ∆E+ > 0 together; the former is a
the spin-(S = 1/2) ferromagnet; thus, the ground state
one-magnon mode of the Goldstone-type multi-magnon
has Stot = N (1/2 − 1/2) = 0. In addition, the spin-
gapless modes that are characteristic to ferromagnets,
1 quantum antiferromagnetic chains exhibiting the Hal-
whereas the latter is a generalization of the Haldane
dane gap correspond to the full spin-(s = 1) liquefaction
gap, which is an antiferromagnetic character of integer
of the spin-(S = 1) ferromagnet; the ground state has
spin chains. In short, this quantum ferromagnet is a
Stot = N (1−1) = 0. In other words, the spin parity effect
“magnetic chimera” of a classical ferromagnet and quan-
for quantum antiferromagnets, i.e., the Haldane’s conjec-
tum antiferromagnet, as shown in Fig. 1. The magnetic
ture, is considered as the S = s cases of the generalized
chimera having the ability to break the spin symmetry
spin-parity effect. Because spin-(s ≥ 3/2) liquefaction
spontaneously, hidden behind continuous magnetic ex-
has not been established yet, the generalized spin-parity
citations under a zero magnetic field (Fig. 7), appears
effect is a conjecture at this stage. On the other hand, the
clearly in the energy gap when a magnetic field is ap-
two cases of s = 1/2 and s = 1 are based on two differ-
plied (Fig. 8). In addition, one can control its excitation
ent rigorous theories: 1) “eigensystem embedding”2 for
by tuning the magnetic field h through the opposing Zee-
s = 1/2 and 2) exactly written ground states of the ferro-
man shift, i.e., ∆E− +h for the ferromagnetic branch and
magnetic AKLT models, which has been studied above,
∆E+ − h for the antiferromagnetic branch.
for s = 1. This conjecture for SU(2) symmetric ferromag-
Just as the original AKLT model has been a rigorous
nets is completely different from the SU(n) generalization
starting-point to explore a wide class of SPT phases the-
of the Haldane’s conjecture59 .
oretically and Haldane materials experimentally, the fer-
romagnetic AKLT models will play the same role in the- The ground state of quantum ferromagnets after the
oretical studies and experimental observations. A future spontaneous symmetry breaking has quantum entangle-
work is topological classification between the ferromag- ment coming from the quantum entanglement in spin-
netic AKLT models and known materials, for example, s antiferromagnets because the classical ferromagnetic
ferrimagnetism in the verdazyl-based salts48,49 and m = background cannot contribute to quantum entanglement.
1/3 plateau31,50–52 in antiferromagnet Na2 Cu3 Ge4 O12 53 . The above theoretical results not only abolish the prej-
Especially, ferrimagnetism exhibits similar combination udice that ferromagnetism is classical but also will open
of gapless ferromagnetic and gapped antiferromagnetic another frontier of “quantum ferromagnetism” in this
branches studied by Yamamoto, et al.54–56 , and its frac- new era of quantum computer science.
 12

ACKNOWLEDGMENTS

This work was supported by Japan Society for the

Promotion of Science (JSPS) KAKENHI Grants No.
JP22H01171. The computation was partly carried out
using the computer resources offered by the Research In-
stitute for Information Technology, Kyushu University.

∗
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