Table Of ContentParticle Statistics, Frustration, and Ground-State Energy
Wenxing Nie1, Hosho Katsura2 and Masaki Oshikawa1
1Institute for Solid State Physics, University of Tokyo, Kashiwa 277-8581, Japan
2Department of Physics, Gakushuin University, Tokyo 171-8588, Japan
(Dated: February 5, 2014)
Westudytheconnectionsamongparticlestatistics,frustration,andground-stateenergyinquan-
tum many-particle systems. In the absence of interaction, the influence of particle statistics on
the ground-state energy is trivial: the ground-state energy of noninteracting bosons is lower than
4
that of free fermions because of Bose-Einstein condensation and Pauli exclusion principle. In the
1
0 presence of hard-core or other interaction, however, the comparison is not trivial. Nevertheless,
2 the ground-state energy of hard-core bosons is proved to be lower than that of spinless fermions,
if all the hopping amplitudes are nonnegative. The condition can be understood as the absence of
b
frustration amonghoppings. By mappingthemany-bodyHamiltonian toa tight-bindingmodelon
e
afictitiouslattice, weshowthat theFermistatistics oftheoriginal particles introducesaneffective
F
magnetic flux in the fictitious lattice. The latter can be regarded as a frustration, since it leads to
4 a destructive interference among different paths along which a single particle is propagating. If we
introduce hopping frustration, the hopping frustration is expected to compete with “effective frus-
] tration”, leading to the possibility that the ground-state energy of hard-core bosons can be higher
h
than that of fermions. We present several examples, in which the ground-state energy of hard-core
c
bosonsisprovedtobehigherthanthatoffermionsduetothehoppingfrustration. Thebasicideas
e
m were reported in a recent Letter [W.-X.Nie, H. Katsura, and M. Oshikawa, Phys. Rev. Lett. 111,
100402 (2013)]; moredetails andseveralextensions,includingonetothespinfulcase, arediscussed
-
in the present paper.
t
a
st PACSnumbers: 05.30.Fk,05.30.Jp,71.10.Fd
.
t
a
m I. INTRODUCTION sufficiently high pressure. Rigorously proven examples
of BEC (in the sense of the off-diagonal long-range or-
-
d Inquantummany-bodyproblem,understandingofthe der) in an interacting system are still rather limited2–4.
n groundstate is fundamental. The ground-stateenergy is Even if the occurrence of BEC or the off-diagonal long-
o
aphysicalquantitywhichgovernsthestabilityofthesys- range order is proved in a system of interacting bosons,
c
[ tem, and in principle it is measurable by measuring the it does not necessarily restrict the ground-state energy,
exchangeoftheenergywiththeoutside,duringaprocess becausesingle-particlestateswithhigherenergiescanbe
2 starting from a known initial state. The ground-state partially occupied. In strongly correlated systems, the
v
energy also reflects the statistics of identical particles, influenceofparticlestatisticsontheground-stateenergy
0
9 which pervades all of quantum physics. had not been much explored5,6.
0 In noninteracting systems, the influence of particle In fact, recently we found7 a sufficient condition for
2 statisticsontheground-stateenergyisquitetrivial. The thenaturalinequality(1)tohold,withoutrelyingonthe
. ground-state energy of fermions is simply given by the occurrence of BEC. That is, if all the hopping ampli-
1
0 sum of the lowest single-particle energy eigenvalues, fol- tudes are nonnegative, the ground-state energy of hard-
4 lowing the Aufbau principle. In contrast, in the ground core bosons is still lower than that of the corresponding
1 state of noninteracting bosons, all the bosons condense fermions. Thistheoremisextendedtothespinfulcasein
v: into the lowest single-particle state, as known as Bose- the present paper. Once we relax the condition of non-
i Einstein Condensation (BEC). Therefore, the ground negativehoppingamplitudes, itis possibletoreversethe
X
state energies of bosons and fermions satisfy the “nat- inequality so that the ground-state energy of bosons is
r ural” inequality: higher than that of fermions. We find several concrete
a
models in which such a reversal is realized; and in sev-
EB EF. (1)
0 ≤ 0 eralcasesitisevenprovedrigorously. Moreexamplesand
techniques will be introduced in the present paper, than
Ontheotherhand,thecomparisonoftheground-state
those discussed in Ref. 7. In addition, we also demon-
energiesofbosonsandfermionsis nottrivialinthe pres-
strate the reversedinequality can exist in some interact-
ence of interaction, because the simple argument based
ing systems.
on the perfect BEC breaks down. In a system of inter-
acting bosons, it is in fact already a nontrivial question Moreover, our study leads to a novel physical under-
whethertheBECactuallytakesplace. Einstein’soriginal standing of the effects of particle statistics, in terms of
argument depends on the absence of interaction. For in- frustration in quantal phase. This is more general than
teracting bosons, there is no general theorem that BEC the picture based on the perfect BEC, and is indeed ap-
always occurs1. A counterexample is solid 4He phase, plicable to systems with interaction.
where BEC is absent even at zero temperature, under a We can map a quantum many-particle problem to a
2
single-particle problem on a fictitious lattice in higher where j is the label of a site on a finite lattice Λ and
dimensions. When all the hopping amplitudes are non- n c†c is the number of particles on j-th site. Chem-
j ≡ j j
negative and the particles are bosons, the corresponding ical potential µ is the uniform (site independent) part of
single-particle problem also has only nonnegative hop- µ . Forasystemofbosons,weidentifyc withtheboson
j j
ping amplitudes. In such a case, there is no frustration annihilation operator b satisfying the standard commu-
j
in the quantal phase of the wavefunction. On the other tation relations, with the hard-core constraint n = 0,1
j
hand,Fermistatisticsoftheoriginalparticlesgivesanef- at each site. The hard-core constraint may also be im-
fective magnetic flux in the correspondingsingle-particle plemented by introducing the infinite on-site interaction
problem. This implies a frustration in the phase of the U n (n 1), where + . For a system of
wavefunction, induced by the Fermi statistics. When f2erPmijonjs, wje−identify cj wUit→h the∞fermion annihilation
a magnetic flux is introduced in the original quantum operator f satisfying the standard anticommutation re-
j
many-particle problem, it also results in a magnetic flux lations.
in the corresponding single-particle problem, inducing a ThisHamiltonianisverygeneral. Wedonotmakeany
frustration. This hopping-induced frustration and the assumptiononthe dimensionor the geometry ofthe lat-
statisticalfrustrationcansometimespartiallycancelwith tice Λ, or on the range of the hoppings. In addition, the
each other, resulting in the reversed inequality between interaction is also arbitrary, as long as it can be written
the ground-state energies of the hard-core bosons and in terms of V . The interesting aspect of attractive in-
jk
fermions. teraction will be discussed at the end of the Sec. II. We
Thepaperisorganizedasfollows. InSec.II,wepresent note that the Hamiltonian (2) conserves the total parti-
the full proof of the natural inequality for the spinless cle number. Thus the ground state can be defined for
caseandextendthediscussiontothespinfulcase. Based a given number of particles M (canonical ensemble), or
on the proof, in Sec. III, we put forward a unified un- fora givenchemicalpotential µ (grandcanonicalensem-
derstanding of the frustration for bosons and fermions ble). The comparison between bosons and fermions can
in the same manner. As a by-product, a strict version be made in either circumstance.
of the diamagnetic inequality for a generallattice is pre- Now we will present a sufficient condition for the nat-
sented and proved. Several examples, in which the nat- ural inequality (1). Moreover, a sufficient condition for
ural inequality is violated owing to the hopping frustra- the strict inequality EB < EF is provided. The proof is
0 0
tion, are presented in Sec. IV. The examples include a also illuminating for physicalunderstanding of the natu-
simple yet instructive, exactly solvable model of parti- ral inequality in interacting systems.
cles on a one-dimensional ring, two-dimensional systems
of coupled rings, two-dimensional lattice with magnetic Theorem 1. (Natural inequality for spinless case)
flux, and flat band models. Rigorous proof of the re- The inequality (1) holds for any given number of par-
versedinequalityisprovidedformostcases. Conclusions ticles M on a finite lattice Λ with N M sites, if all the
≥
and discussions are presented in Sec. V. hopping amplitudes tjk are real and nonnegative.
Furthermore, if the lattice Λ is connected, and has a
site directly connected to three or more sites, and if the
number of particles satisfies 2 M N 2, the strict
II. NATURAL INEQUALITY inequality EB <EF holds. ≤ ≤ −
0 0
The natural inequality (1) holds trivially for nonin- Proof. To write the matrix elements of the Hamilto-
teracting bosons and fermions with the same form of nian (2), we choose the occupation number basis φa
tshtaeteHathmaitltothneianE.q.N(o1w) whoeldpsreesevnenttfhorreehathrde-ocroermesb,owshoincsh, i|n{ngaj}ij,nwajhe=reMM. Tishtehme taottraixl neluemmbenertsooffptahretincluems sb|aetrisiofpy≡--
provided that all the hopping amplitudes are nonnega- eratPor nj are the same for hard-core bosons and spinless
tive. A brief overview appeared in Ref. 7, but here we fermionsinthisbasis. Webeginbydefiningtheoperator
give a more detaileddiscussion, and also anextension to
B,F B,F+C1. (3)
the spinful case. K ≡−H
For convenience,we added an identity matrix with large
enough diagonal elements C such that all the eigenval-
A. Natural inequality for spinless case uesκB,F ofmatrix B,F andthusallthediagonalmatrix
elements B,F areKpositive. The relation of the matrix
Kaa
First we consider the comparisonof spinless hard-core elementsforbosonicandfermionicoperatorscanbesum-
bosons with spinless fermions. We assume the system of marized as
bosonsorfermionsisdescribedbythesameHamiltonian,
F (a=b)
B = |Kab| 6 = F . (4)
Kab (cid:26) F (a=b) |Kab|
= t c†c +H.c. µ n + V n n , Kaa
H −Xj6=k(cid:16) jk j k (cid:17)−Xj j j Xj,k jk j k Thedifferencebetweenbosonsandfermionsisthat,given
(2) nonnegativehoppingamplitudest ,thematrixelements
jk
3
ofthebosonicoperator B isnonnegative,whilethoseof (0) (1) (2)
the fermionic operator KF can be negative. This differ-
K
ence in signs generically leads to different ground-state
energies between bosons and fermions. (3) (4) (5) (6)
ThegroundstateoftheHamiltonian B,Fcorresponds
H
to the eigenvector belonging to the largest eigenvalue
κB,F of B,F. Let Ψ = ψ φa be the normal-
max K | 0iF a a| iF
ized ground state for fermioPns. The trial state for the FIG.1. Aschematicexampleoftwo-particleexchangeprocess
bosons can be assumed as Ψ = ψ φa , where in six steps.
| 0iB a| a|| iB
φa BisthebasisstateforbosonscorPrespondingto φa F.
| i | i
Then, by a variational argument,
expansion of Eq. (6), at least from the invariant process
κBmax ≥BhΨ0|KB|Ψ0iB = |ψa||ψb|KaBb cj = a in which no particle moves in n steps. Thus, the
Xab strict inequality B > F holds in this case.
Laa |Laa|
≥Xab ψa∗ψbKaFb =κFmax (5) canWbheenretahcehleadttbicyecΛonisseccountniveceteadp,palincyatbioanssisosftatthee|φhaoipB-
ping term in B, and thus the matrix B satisfies the
thhouldssp,rimovpeldy.inAgsEa0Bsi≤mEpl0Fe.cTorhoellafirrys,ttphaertgorofuTnhde-osrteamte1enis- c(aonndnetchtuivsitayl.soKTogBe)thiesrawPitehrrtohne-FprroobpeenritKuysabKmaBabtr≥ix80., AKpaBb-
ergiesforagivenchemicalpotentialµalsosatisfyEq.(1). plying a corollaLryabof the Perron-Frobenius theorem9 we
In order to prove the strict version of the natural in- find κB > κF and hence the latter part of the theo-
equality, let us consider S S n, where S = B,F, max max
rem follows.
L ≡ K
for a positive integern. Inthe o(cid:0)ccup(cid:1)ationnumber basis,
the matrix element of is expanded as
We note in passing that, a consequence of the Perron-
L
Frobeniustheoremisthatthegroundstateofbosonshas
LSab = KaSc1KcS1c2KcS2c3...KcSn−1b, (6) anonvanishingamplitudeB φa Ψ0 B withadefinite(say,
c1,.X..,cn−1 positive) sign for every bashis s|tatei φa B. This may be
| i
understood as a lattice version of the “no-node” theo-
in which each term in the sum represent a particle hop-
rem10,11.
ping process among the connected sites.
Fromthedefinitionof S andtherelationbetween B
and F denoted by Eq.L(4), we have the inequalityKfor
matrKix elements of B,F: B. Natural inequality for spinful case
L
LBab = KaBc1KcB1c2KcB2c3...KcBn−1b (7) Let us now discuss the spinful case. Here we compare
c1,.X..,cn−1 spinful hard-core bosons and spinful fermions on a finite
= F F F ... F lattice,withspin-1/2. Thisispseudospin-1/2forbosons.
c1,.X..,cn−1|Kac1Kc1c2Kc2c3 Kcn−1b| Namely, we consider the Hamiltonian
≥|c1,.X..,cn−1KaFc1KcF1c2KcF2c3...KcFn−1b|=|LFab|.(8) H=−Xj6=kXσ (cid:16)tjkc†jσckσ +H.c.(cid:17)−Xjσ µjnjσ
Thisapplies,inparticular,tothe diagonalelementswith
+ Vjknjσnkσ′ + Ujnj↑nj↓, (9)
b=a.
Xj6=kXσ,σ′ Xj
From Eq. (4), the matrix elements of F and thus the
K
amplitudes of the process in Eq. (6) can be negative for whichisageneralizationofEq.(2)withtheintroduction
fermions, while they are nonnegative for bosons. The of the spin degrees of freedom σ = , .
↑ ↓
differencebetweenbosonsandfermionsshowsupexactly Letus firstdiscussthe caseinwhichallU ’sarefinite.
j
when two particles are exchanged. To make two-particle Then the following simple generalization of Theorem 1
exchangeprocesspossible,letusconsideralatticewitha holds:
“branching”sitedirectlyconnectedtothreeormoresites.
Ifthenumberofparticlefallsintherange2 M N 2, Theorem 2. (Natural inequality for spinful case with
≤ ≤ −
two particles canbe exchangedfroman initial state φa finite U ’s)
j
| i
and back to the same state in 6 hoppings. An example For any set of finite U ’s, if all the hopping amplitudes
j
of particle exchange on a lattice with a branching site is t are real and nonnegative, the inequality (1) holds for
jk
demonstrated schematically in Fig. 1. The contribution any given number of particles M 2N on a finite lattice
tothe diagonalelementsofbosons B isalwayspositive ΛwithN sites. Furthermore,ifth≤elatticeΛisconnected,
at n=6, while the contribution toLaFa is negative when and has a site directly connected to three or more site,
Laa
two particles are exchanged. On the other hand, there and if the number of particles satisfies 3 M 2N 3,
is always a positive contribution to B and F in the the strict inequality holds. ≤ ≤ −
Laa Laa
4
The proof, which is a generalization of Theorem 1, is for spinless case, the strict inequality EB < EF follows
0 0
given as follows. from the Perron-Frobenius theorem.
Proof. Since the total number operator M = jσnjσ Now let us discuss the case Uj = +∞. The first
and total magnetization Sz = 1/2 j(nj↑ −njP↓) com- ihtayl,froefmTahinesoruenmaff2e,cttheednboyn-tsatkriicntgvUersi=on+of t.heHionweqeuvaerl-,
mute with the Hamiltonian (9), onePcan diagonalize the j ∞
the latter half of Theorem 2, the strict inequality, is af-
Hamiltonian in each sub-Hilbert space with fixed val-
fected. The proof of the strict inequality is based on
ues of M and S . Each sub-Hilbert space has defi-
z
the Perron-Frobenius theorem, which requires the irre-
nite numbers of up-spin and down-spin particles. Let
φµ nµ (µ = 1,2, ,u) be the occupation ducibilityofthematrix. Forspinlessparticlesandspinful
| i↑ ≡ |{ j↑}i ··· particles with finite U ’s, when the lattice is connected,
number basis for up-spin particles, and ψν nν j
| i↓ ≡ |{ j↓}i anypairofoccupationnumberbasisstates Φa and Φb
(ν = 1,2, ,v) be the occupation number basis for | i | i
··· of the many-particle problem are connected by consec-
down-spin particles. Then, we can take the direct prod-
utive application of particle hoppings. This implies the
uct Φa = ψν φµ , where a = 1,2, ,uv, as the
| i | i↓ ⊗| i↑ ··· irreducibility of the matrix representing the many-body
basis of the sub-Hilbert space mentioned above.
Hamiltonian.
The Hamiltonian can be rewritten as:
However,in the case of spinful systemwith U =+ ,
j
∞
connectivity of the lattice does not guarantee the irre-
= + , (10)
t int
H H H ducibility of the many-body Hamiltonian matrix. An il-
Ht =1↓⊗Ht↑+Ht↓⊗1↑, (11) lustrative example is the Hubbardmodel with Uj =+∞
at half filling. Each site is occupied by a particle with
where σ = (t c† c +H.c.). The matrix ele-
Ht − j6=k jk jσ kσ either spin up or spin down; there are many basis states
ments of the nuPmber operator njσ are the same in this corresponding different spin configurations. However,
basis, for hard-core bosons and fermions. We introduce
since there is no empty site and double occupancy with
the operator B,F B,F +C1 with a constant C.
spin up and down particles is forbidden, each basis state
K ≡ −H
Choosing C large enough, we make all the eigenvalues
is not connected by hopping to any other basis state.
and all the diagonal matrix elements of B,F positive.
Therefore, in order to prove the strict inequality, we
K
The matrix elements of bosonic and fermionic Hamilto-
need some additional condition which guarantees the ir-
nians obey the relation:
reducibility of the Hamiltonian matrix. In fact, the irre-
ducibility of the Hamiltonian matrix at U = + , and
F (a=b) j ∞
B = |Kab| 6 (12) application of the Perron-Frobenius theorem were dis-
Kab (cid:26)KaFa (a=b), cussed earlier by Tasaki12,13 in the context of Nagaoka’s
ferromagnetism. Nagaoka’s ferromagnetism is a mech-
wherethediagonaltermscorrespondto andthe off-
Hint anism of ferromagnetism in the Hubbard model with a
diagonal terms correspond to . With finite U ’s, one
Ht j single hole with Uj = + , and can be understood as a
sitecanbeoccupiedbyonespin-upparticleandonespin- ∞
consequence of the Perron-Frobenius theorem. For that,
down particle. Thus spin-up particles can move as spin-
the irreducibility of the Hamiltonian matrix in a certain
less particles for any given configuration of spin-down
basisisrequired. InRef.13,asufficientconditionforthe
particles,andvice versa. Ofcourse,the interactionterm
irreducibility was presented: if the entire lattice is con-
, which is diagonal in this basis, is affected by the
Hint nected by exchange bonds, then the Hamiltonian matrix
presence of particles with opposite spins. However, as
in the occupation number basis is irreducible. Here “ex-
far as the irreducibility (connectivity) of Hamiltonian is
change bond” is defined by a pair of sites which belongs
concerned,onecanregardthesystemastwoindependent
to a loop of length three or four, and the whole lattice
systems of hard-core particles. As a consequence, when
remains connected via nonvanishing hopping amplitudes
the lattice Λ is connected, any pair of basis states Φa
| iB even when the two sites are removed. Thus we obtain
and Φb are connected to each other by a successive
B
appli|catiions of the hopping term in B. Together with Theorem 3. (Natural inequality for spinful case below
K
the property KaBb ≥ 0, KB satisfies the condition of the half filling)
Perron-Frobenius theorem. When the number of parti- When U ’s are either + or finite, if all the hop-
j
clesM 3,thereareatleasttwoparticleswiththesame pingamplitudest arereala∞ndnonnegative,theinequal-
≥ jk
spin. The conditionM 2N 3guaranteesthere areat ity (1) holds for any given number of particles M N
least two spaces which≤can a−ccommodate two particles on a finitelattice Λ with N sites. Furthermore, if the≤en-
with the same spin. Thus, when the number of parti- tire lattice Λ is connected by exchange bonds, and if the
cles falls in the range3 M 2N 3, we can exchange number of particles satisfies 3 M N 1, the strict
twoidenticalparticlesan≤dret≤urnba−cktothesamestate, inequality holds. ≤ ≤ −
based on the branch structure as in Fig. 1. Therefore,
when U ’s are finite, the lattice is connected and has a With infinite on-site repulsion, the maximum number
j
branch structure, and 3 M 2N 3, two-particle ex- of particles is N. The condition M 3 is to guarantee
≤ ≤ − ≥
change always happens. As in the proof of Theorem 1 there are at least two particles with the same spin such
5
that they can be exchanged. For a lattice connected by for fermions, leading to the nonvanishing energy differ-
exchange bonds, two particles on an exchange bond can ence between fermions and bosons. The exception oc-
be exchanged without changing the configuration out- curs when the on-site attractive interaction between up
side, by hopping a hole around the loop on which both and down spin particles is infinite (U = ). Then
j
−∞
the exchange bond and the hole lie13. Hence, when the the pairsare completely robust,and no virtualexchange
number of particle M satisfies 3 M N 1, two ofconstituent particles occurs;the ground-stateenergies
≤ ≤ −
particles with the same spin can be exchanged on an for fermions and bosons become identical in this limit.
exchange-bond lattice by successive particle hoppings. On the other hand, with the infinite attraction, the irre-
The property that the entire lattice is connected by ducibilitycannotbesatisfied,becausebreakingaCooper
exchangebonds canbe verified13 invariouscommonlat- pair (or bosonic molecule) costsan infinite energy andis
tices,suchastriangular,square,simple cubic,fcc,orbcc thus prohibited. This implies that the bosonic molecules
lattices, in which nearest neighbor sites are connected arecompletelylocalizedinthemodel(9). Thusthenatu-
by nonvanishing hopping amplitudes. Thus, the above ral inequality is reduced to the trivial equality EB =EF
0 0
theorem holds for these lattices. in the limit Uj .
→−∞
We alsonote that, Nagaoka’sferromagnetismonly ap-
plies to the system with single hole with respect to half
filling. However,thisrestrictionisonlynecessarytoguar-
antee that all the matrix elements are nonnegative. The
irreducibilityofthe Hamiltonianmatrixdoesnotrequire
that there is only one hole. In fact, the breakdown of
the positivity in the presence of more than one holes in
the Hubbard model with U = + is precisely due to FIG.2. Four-sitebranch latticewith four spinsat halffilling
j
the Fermi statistics of the electron∞s. If we consider the and Sz =0.
“Bose-Hubbard model” with spin-1/2 bosons instead of
electrons, all the matrix elements are nonnegative in the
0.0
occupationnumber basis,for any number of holes. Thus
the Bose-Hubbard model with spin-1/2 bosons exhibit -0.1
ferromagnetism for any filling fraction.14 This nonnega-
-0.2
tivity of the matrix elements for bosons is also essential
for Theorem 3, which holds for any filling fraction. (cid:144)Et -0.3
D
The proofs of Theorems 1 and 2 are insensitive to the
-0.4
signs of the interaction terms V and U . Namely the
jk j
natural inequality holds no matter the interaction is re- -0.5
pulsive or attractive. The interesting aspect of the at-
-0.6
tractive interaction is that it will induce Cooper pair of -20 -10 0 10 20
fermions. In the case of spinless fermions, orbital part U(cid:144)t
of the Cooper pair wavefunction must be antisymmetric
withrespecttotheexchangeoftwofermions. Thisresults FIG. 3. Difference of ground-state energy (∆E = E0B−E0F)
in an extra cost in the kinetic energy. Such a fermionic between hard-core bosons and fermions on the 4-site lattice
BECstatethushasahigherground-stateenergythanits with a branch, in Sz = 0 sector with 4 spins. The absolute
value of energy difference decreases down to 10−8t around
bosoniccounterpart,infullagreementofthe Theorem1. ∼
U /t=100.
Incontrast,inthecaseofspinfulfermions,withattrac- | |
tive interaction, fermions could pair up in the nodeless
Inthefollowing,wenumericallydemonstrateaboveob-
s-channel. Inthiscase,thereisnoobviousreasonwhythe
servations in spinful Bose-Hubbard and Fermi-Hubbard
fermions havea higher ground-stateenergythanbosons.
models on a 4-site cluster as shown in Fig. 2. Here the
Nevertheless, according to Theorem 2, spinful fermions
bosons in the “Bose-Hubbard”model still obey a partic-
still have strictly higher ground-state energy than corre-
ular hard-core condition n = 0,1. The Hamiltonian is
spondingbosons,evenwhenthepairingisinthenodeless jσ
given by
s-channel.
Thiscanbeinterpretedphysicallyinthefollowingway. = t (c† c +H.c.)+U n n , (13)
If the paring of two particles is completely robust, the H − hXi,jiXσ iσ jσ Xj j↑ j↓
problem is reduced to the identical problem of bosonic
“molecules”, whether the original particles are fermions wheret>0,and i,j denotesapairofneighboringsites.
h i
or bosons. Then the ground-stateenergiesshould be the We consider the spin-1/2 bosons and fermions at half
same for fermions and bosons. However, in general, the filling (the total number of particles per site ν = 1) and
pairing is not completely robust, and two pairs can (vir- Sz =0. That is, on this 4-site cluster, there are two up-
tually) exchange each one of their constituent particles. spin particles and two down-spin particles. The energy
The amplitude for such a process has negative sign only difference between spinful bosons and spinful fermions
6
(∆E = EB EF) is shown as a function of U = U in quantum particles, the direction of the spins in the xy-
0 − 0 j
Fig. 3. plane corresponds to the quantal phase of particles at
ConformingtoTheorem2,EB EFholdsforallrange eachsite. If all the hopping amplitudes are nonnegative,
0 ≤ 0
of U, independent of the signs of U. Moreover, ∆E(U) everyhopping term can be simultaneously minimized by
is symmetric along U =0 due to particle-hole symmetry choosingauniformphasethroughoutthesystem. Inthis
of Hubbard model at half filling ν =115. sense, bosons with nonnegative hopping amplitudes are
WhenU isfinite,fermionshavestrictlyhigherground- unfrustrated with respect to their quantal phase.
state energy than bosons, again in agreement with the Let us now consider the case of fermions. Since Fermi
latter half of Theorem2. When U =+ , the present4- statistics brings in negative signs even if all the hop-
∞
siteclusterdoesnotcontainanyexchangebond,andthus pings tjk are nonnegative, it would be natural to expect
the strict inequality cannot be proven. In fact, in this that Fermi statistics leads to some kind of frustration.
limit, it is easy to see that the particles are completely However, it is difficult to formulate this based on the
immobile and no particle-exchange occurs. The ground- above mapping to an S = 1/2 spin system. To un-
state energy is indeed exactly the same for fermions and derstand the frustration induced by Fermi statistics in
forbosonsinthislimit. Likewise,inthelimitofU = , many-particlesystems,weintroduceanalternativemap-
−∞
either bosons or fermions form completely robust (and pingofthemany-bodyHamiltonianintoasingle-particle
immobile) pairs, and the ground-state energies are ex- tight-binding model. That is, we identify each of the
actly the same. In the present case, this can also be many-body basis states Φa with a site on a fictitious
| i
understoodasaconsequenceofthe particle-holesymme- lattice. If two basis states Φa and Φb are connected
try15 at half filling, which maps U U. by Hamiltonian, Φb Φa |= 0i, there| isia link connect-
→− h |H| i 6
ing sites a and b in the fictitious lattice. If we can start
from an initial state, and return back to the same state
III. UNIFIED UNDERSTANDING OF by successive applications of the Hamiltonian (2), there
FRUSTRATION AND DIAMAGNETIC is a loop in the fictitious lattice. For bosons, there is no
INEQUALITY extra phase in the loop. In other words, the fictitious
lattice for hard-core bosons is flux free. Therefore, there
is no frustration for bosons because there is a construc-
Theroleplayedbyfrustration isofcentralimportance
tive interference among all the paths. In contrast, for
in the proofsof the theorems. The terminology“frustra-
fermions,inthe originalmany-bodyproblem,if twopar-
tion” is often used for antiferromagnetically interacting
ticles are exchanged and the system returns back to the
spin system,ongeometricallyfrustratedlattices, suchas
initialstate,thesystemacquiresanextraπ phase. Upon
triangular, kagome and pyrochlore lattices. When there
themappingtothesingle-particleproblem,thisisequiv-
is no global state of the system that minimizes every
alent to the presence of a π-flux in the corresponding
antiferromagnetic interaction, there is some frustration.
loop in the fictitious lattice. This can be interpreted as
Moregenerally,frustrationmaybeapplicabletoasystem
frustration,whichcausesdestructiveinterferencesamong
withcompetinginteractions,whenthegroundstatedoes
not minimize individual interaction simultaneously16. different paths.
For a single-particletight-binding model, introduction
To see that the sign of hopping amplitudes t in a
jk
of a flux always raises or does not change the ground-
many-boson system is related to frustration, it is illumi-
stateenergy,whichisknownasdiamagneticinequality18.
nating to map the hard-core boson problem to a spin-
1/2 quantum spin system17. The mapping is based on The first half of Theorem 1, which states the non-strict
inequality, may be then regarded as a corollary of the
the equivalence between hard-core boson operators and
diamagnetic inequality. On the other hand, the latter
spin-1/2 operators:
half of the Theorem 1 concerning the strict inequality
1 does not, to our knowledge, follow from known results
Sj+ ∼b†j, Sj− ∼bj, Sjz ∼b†jbj − 2. (14) on the diamagnetic inequality. In fact, the arguments in
the proofofTheorem1canbe appliedto astrictversion
It is then easy to see that a hopping term for hard-core of the diamagnetic inequality on general lattices. The
bosons maps to an in-plane exchange interaction: general result can be summarized as follows.
Theorem 4. (General diamagnetic inequality and its
t b†b +b†b J⊥ SxSx+SySy , (15)
− jk(cid:16) j k k j(cid:17)∼ jk(cid:0) j k j k(cid:1) strict version)
Let us consider a single particle on a finite lattice Ξ,
where Jj⊥k = −2tjk. Thus the nonnegative tjk corre- with the eigenequation
sponds to ferromagneticinteraction,in terms of the spin
system. When all the exchange couplings are ferromag- τ ψ =Eψ . (16)
αβ β α
netic, there is no frustration. Namely, every in-plane −βX∈Ξ
exchange interaction energy can be minimized simulta-
neously by aligning all the spins to the same direction In general, τ is complex, with τ =τ∗ . The ground-
αβ αβ βα
in the xy-plane. Going back to the original problem of state energy E for a given set of the hopping amplitudes
0
7
τ satisfies The non-strict versionis the standard diamagnetic in-
αβ
{ } equality18. However, the strict inequality obtained here
E ( τ′ τ ) E ( τ ). (17)
0 { αβ ≡| αβ|} ≤ 0 { αβ} appears new, also in the general context of diamagnetic
Furthermore, the strict inequality, inequality.
E ( τ′ τ )<E ( τ ), (18)
0 { αβ ≡| αβ|} 0 { αβ}
Mapping of the original quantum many-particle prob-
holds, provided that the lattice Ξ is connected and there
lem to the single-particle problem on a fictitious lattice
is at least one loop which contains a nonvanishing flux.
providesaunifiedunderstandingoffrustrationofquantal
A sequence of sites α ,α ,α ,...,α , which satisfies
{ 0 1 2 n} phase. When there is a nonvanishing flux in the original
α =α , τ =0 and α =α is called a loop. The
l 6 l+1 αlαl+1 6 n 0 many-particle problem, we observed that there is a frus-
loop contains a nonvanishing flux when the product
tration among local quantal phases, which we may call
τ τ τ ...τ (19) hopping frustration. On the other hand, when the par-
α0α1 α1α2 α2α3 αn−1αn
ticles in the original problem are fermions, there is also
is not positive (either negative or not real). a frustrationamong quantal phases, which we name sta-
Proof. TheproofissimilartothatofTheorem1. Wecan tistical frustration. In the original many-particle prob-
define the matrices , ′ by lem, the statistical frustration appears rather different
K K from the hopping frustration. However, upon mapping
αβ ταβ +Cδαβ, (20) to the single-particle problem on the fictitious lattice,
K ≡
′ τ′ +Cδ , (21) both hopping frustration and statistical frustration are
Kαβ ≡ αβ αβ
represented by a nonvanishing flux in the fictitious lat-
with a sufficiently large constant C so that and ′ is
positive definite. We then define n anKd ′ K′n, tice. This provides a unified understanding of hopping
L ≡ K L ≡ K and statistical frustrations.
forthelengthnoftheloopwithanonvanishingflux. The
positive definiteness of and ′ implies that and ′
K K L L
arealsopositivedefinite,andthusallthediagonalmatrix A system of many bosons with only nonnegative hop-
elements and ′ are strictly positive. Similarly to ping amplitudes t represents a frustration-free system.
the proofLoαfαTheoreLmαα1, ′ holds for any α,β. Introduction of anjyk frustration into such a system is ex-
Lαβ ≥|Lαβ|
In particular, the diagonal matrix elements of ′ and pected not to decrease the ground-state energy. For ex-
L L
are expanded as ample, introduction of magnetic flux (hopping frustra-
tion) does not decrease the ground-state energy. This
L′α0α0 = Kα′0α1Kα′1α2...Kα′n−1α0, (22) is a lattice versionof Simon’s universal diamagnetism in
α1,·X··,αn−1 bosonic systems19.
= ... . (23)
Lα0α0 Kα0α1Kα1α2 Kαn−1α0
α1,·X··,αn−1 On the other hand, when one type of frustration al-
Each term in the expansion satisfies ready exists, the effect of introducing another type of
frustration is a nontrivial problem. For example, in
Kα′0α1Kα′1α2...Kα′n−1α0 ≥ Kα0α1Kα1α2...Kαn−1α0 , a system of fermions, the statistical frustration exists.
(cid:12) ((cid:12)24) What happens if one further introduces hopping frus-
(cid:12) (cid:12)
tration (magnetic flux)? In such a case, there cannot
thanks to ′ . By assumption, there is a non-
Kαβ ≥ |Kαβ| be a general statement: the ground-state energy may
vanishing contribution to from the loop of length
Lα0α0 or may not decrease, depending on the system in the
n,
question. Thatis,diamagnetismis notuniversalinspin-
... =τ τ ...τ , (25) less fermion systems. Correspondingly, the orbital mag-
Kα0α1Kα1α2 Kαn−1α0 α0α1 α1α2 αn−1α0
netism of fermions can be either paramagnetic or dia-
which is not positive. Here we usedthe fact that the off-
magnetic, depending on the model.
diagonal elements of and τ are identical. Combining
K
with the contribution from its reverse loop
This means that, in some cases, the hopping frustra-
... , (26)
Kα0αn−1Kαn−1αn−2 Kα1α0 tionmay(partially)canceltheeffectofstatisticalfrustra-
which is the complex conjugate of Eq. (25), we find the tion, so that the introduction of the hopping frustration
strict inequality actuallydecreasestheground-stateenergy. Thepossibil-
ity of partial cancellation between the two kinds of frus-
′ ′ ... ′ +c.c. trations can be again naturally understood by the map-
Kα0α1Kα1α2 Kαn−1α0
ping to the single-particleproblemona fictitious lattice.
> ... +c.c.. (27)
Kα0α1Kα1α2 Kαn−1α0 Each of the frustrations introduces a particular pattern
Thus ′ > >0. InvokingthePerron-Frobenius of magnetic flux in the fictitious lattice. It is certainly
Lα0α0 Lα0α0
theorem again, the strict diamagnetic inequality (18) is possible that these two magnetic flux (partially) cancel
proved. with each other.
8
IV. VIOLATION OF THE NATURAL The difference of ground-state energy is exactly calcu-
INEQUALITY lated as
EPBC EAPBC 2[1 cos(π/N)]
Inthefollowing,wediscusshowthenaturalinequality 0 0 = − >0, (30)
canbeviolated. Theorems1and2leavethepossibilityof N − N Nsin(π/N)
violationoftheinequalitybyintroducingahoppingfrus-
for any N > 1. The antiperiodic boundary condition
tration,thatis,bychoosingnegativeorcomplexhopping
gives the lower ground-state energy. The leading order
amplitudes t . However, the hopping frustration is a
jk
of difference can be extracted in the limit of large N as,
necessary but not sufficient condition to reverse the nat-
ural inequality. We will demonstrate that the violation
EPBC 2 2π 2π3 1
ofnaturalinequalityindeedhappensinseveralfrustrated 0 = + + +O( ) (31)
systems. For simplicity, we limit ourselves to the com- N −π 3N2 45N4 N6
parison between spinless fermions and hard-core bosons, EAPBC 2 π 7π3 1
0 = +O( ), (32)
with no interaction other than the hard-core constraint. N −π − 3N2 − 180N4 N6
The case with other interactions will be discussed at the
end of this section. forthe periodic (PBC)andantiperiodic(APBC)bound-
ary conditions. The leading term of O(1/N2) is also de-
termined by conformal field theory22,23. It can be seen
that the noninteracting fermions on a ring have a lower
A. Particles on a ring
ground-stateenergywiththeantiperiodicboundarycon-
dition.
We startwith the best understoodandsolvablemodel
As a result, with periodic boundary condition, hard-
in one dimension:
core bosons have a lower ground-state energy than
fermions, in full agreement with Theorem 1. On the
N
= (c†c +H.c.). (28) other hand, the ground-state energy of hard-core bosons
H −Xj=1 j j+1 ishigherthanthatoffermionswithanti-periodicbound-
arycondition. Theanti-periodicboundaryconditioncan
The hard-core boson version of this model, which is be understood as a result of insertion of π-flux inside
equivalent to the spin-1/2 XY chain, can be mapped the ring. This hopping frustration cancels the statistical
to free fermions on a ring by Jordan-Wigner transfor- frustration so that the natural inequality is violated.
mation20,21. Thus energy eigenvalue problem of hard- This example of tight-binding model may look triv-
core bosons and fermions on a ring are almost the same, ial, and indeed the calculation itself has been known for
except for the subtle difference in the boundary condi- years. Nevertheless, it is very useful in highlighting the
tion. For the periodic or antiperiodic boundary condi- central physics of the problem, that is, the effect of the
tions c c , the Jordan-Wigner fermions f˜ obey statistical frustration of fermions can be canceled by the
N+1 1 j
≡ ±
the boundary condition f˜ = eiπMf˜, where M is flux orhopping frustration. The presentfinding canalso
N+1 1
the number of Jordan-Wigner fer∓mions (equals to the be applied to construction of more nontrivial examples,
number of bosons). If M is assumed as even, it implies as we will discuss in the Sec. IVB.
that hard-core bosons with the periodic (antiperiodic)
boundary condition is mapped to free fermions with the
antiperiodic (periodic, respectively)boundary condition. B. Coupled rings
Now let us discuss the dependence ofthe ground-state
energy on the boundary condition. Assuming M =N/2 Since hard-core bosons have a higher ground-stateen-
is even, the ground-state energy density (ground-state ergy than fermions on a ring containing π flux inside
energy per site) is given as the ring as proved in Sec. IVA, we can construct a se-
ries of systems where EB > EF, by taking many such
E 2 0 0
ǫ = 0 = cosk, (29) small rings and connecting them with weak hoppings. If
0
N −N Xk the inter-ring hoppings are weak enough, they are ex-
pected not to revert the inequality and EB > EF would
0 0
where k is taken over all the momenta in the Fermi sea, be kept24.
π/2 k < π/2. For the periodic boundary condition, Weproverigorouslythat,thereversednaturalinequal-
− ≤
the wavenumber k is quantized as k = 2πn/N, while ity is indeed still kept in coupled π-flux rings, connected
k =π(2n+1)/N fortheantiperiodicboundarycondition, byweakhoppings,eveninthethermodynamiclimit. The
where n ( N/4 n<N/4) is an integer. first example is π-flux octagon-square model. The lat-
− ≤
The ground-state energy density asymptotically con- tice structure is shown in Fig. 4 (a), where one unit
verges,inthe thermodynamiclimitN ,tothesame cell is shown in green with basis vectors~a = (3,0) and
1
→∞
integral for either boundary condition. Nevertheless, it ~a =(0,3). Thehoppingamplitudesonthickandbroken
2
does depend on the boundary condition for a finite N. lines are denoted by t and t′, respectively. The Hamilto-
9
nian is given by
= t eiπ/4c†c t′ c†c +H.c., π π
H − i j− i j
X X
hi,ji∈thick,oriented hi,ji∈broken
(33)
where “thick, oriented” and “broken” refer respectively
tothelinksdrawnwitharrowsandthosedrawnasbroken π
lines in Fig. 4(a). We also assume t>t′ >0.
By the choice of eiπ/4 hopping phase on the oriented
thick lines, there is a π flux in every square. Therefore,
it can be regarded as a model of coupled π-flux rings by (a) (b)
weak hopping t′. In order to prove EB > EF rigorously
0 0
in the coupled rings, we seek a lower bound for EB and FIG. 4. (a) π-flux octagon-square lattice, in which a unit
0
an upper bound for EF. If the former is higher than the cell is shown in green. (b) The lowest two bands of Hamilto-
latter,thedesiredineq0ualityisproved. Weintroducethe nian (33) with t=1, t′=0.1.
positive semi-definite operators,
m ǫF✸(m) ǫB✸(m)
A=t′ (c†i +c†j)(ci+cj)≥0, (34) 1 √2t+t′ √2t t′
hi,ji∈XBroken 2 −2√2t+2t′ −2t −2t′
B =t′ (c† c†)(c c ) 0, (35) 3 −√2t+3t′ −√2t− 3t′
X i − j i− j ≥ 4 − 4t′ − 4−t′
hi,ji∈Broken −
where A 0 means ΦAΦ 0 for any state Φ .
≥ h | | i ≥ | i TABLE I. The ground-state energies of fermions and hard-
Therefore, the Hamiltonian for fermions and bosons can
corebosons on athick-linesquare,where mis thenumberof
be written as
particles on a π-fluxsquare.
F = ˜F A= hF A, (36)
✸
H H − −
X✸
By exact diagonalization, we obtain the ground-state
HB =H˜B+B = hB✸+B, (37) energies ǫB✸,F(m) in given m particles sectors, shown in
X✸
Table I.
wanhderehBhF✸== −ttP44i=(1e(ieπi/π4/c4†cc†ici+1++HH.c.c.).)+tt′′P44i=1cc††icci, resTuhltesonfuemxbaecrtdoifaguonnitalcizealltsioins,aasslouwmeerdboausnNd,fofrrobmosothnes
✸ − i=1 i i+1 − i=1 i i is given by EB 2N(t+t′) when t′/t 2 √2, or
the cluster HamPiltonians defined on a solid-liPne square 0 ≥ − ≤ −
EB N(√2t + 3t′) when 2 √2 < t′/t < 1. An
for fermions and bosons, respectively. Noticing h✸ com- 0 ≥ − −
mutes with each other, the ground-state energy of ˜ is upper bound for fermions is given by the E˜0F, which is
simply given by the summation25: H dependent on the density pattern on the whole lattice.
Athalffilling,anupperboundoffermionsisobtainedas
E˜0 =X✸i ǫ✸i, (38) rEa0Fng≤e t−′/2tN≤(√(√2t2−−t1′))./2T,hwues,hwavheenE0tBh≥e rEat0Fio. falls in this
Instead of searching an upper bound of fermions, the
wthhaetreofE˜h0✸anodnǫi✸-tihaπre-fltuhxesgqruoaurned,-rsetsapteecetniveerlgyy. of H˜ and gartocuenrtda-isntafitelliennge.rgFyorofcfoenrvmeinoinesnccea,ntbiesesxeatcetqlyuaclaltcoul1a.teInd
i
Because the operators B is positive semi-definite, the the single particle sector, the exact dispersion relations
ground-state energy of bosons satisfies are obtained by Fourier transformation:
E0B =hΦ|HB|Φi≥hΦ|H˜B|Φi≥E˜0B =X✸ ǫB✸i, (39) E±(1) =±r(t′)2+2−2t′q1−sin(3kx)sin(3ky),
i
where |Φi is assumed as the ground state of HB. E±(2) =±r(t′)2+2+2t′q1−sin(3kx)sin(3ky),
Ontheotherhand,anupperboundoffermionscanbe
derived as, where (k ,k ) is the wavenumber which belongs to the
x y
reduced Brillouinzone π/3 k <π/3. The ground-
x,y
EF = Ψ F Ψ Ψ˜ F Ψ˜ Ψ˜ ˜F Ψ˜ =E˜F = ǫF , state energy of fermion−s at µ≤= 0, which corresponds to
0 h |H | i≤h |H | i≤h |H | i 0 ✸i
X✸ the half filling, is given as
i
(40)
where Ψ and Ψ˜ are the ground states of F and ˜F, EF = E(1)(k ,k )+E(2)(k ,k ) . (41)
| i | i H H 0 − x y − x y
respectively. kXx,ky(cid:2) (cid:3)
10
m ǫF△(m) ǫB△(m)
π π 1 t+2t′ t 2t′
− ′ − − ′
2 2t+4t t 4t
3 − 6t′ − −6t′
−
π π
TABLE II. The ground-state energies of fermions and hard-
corebosonsonathick-lineuptriangle,wheremisthenumber
of particles on a triangle.
(a) (b)
with π-flux, coupled by weak hopping t′. To obtain a
lower bound for the ground-state energy of bosons and
an upper bound for that of fermions, the Hamiltonians
are written as F = hF A and B = hB +B
H △ △− H △ △
with the same definiPtions of A and B in EqPs. (34)(35),
where hF = t 3 (eiπ/3c†c +H.c.)+2t′ 3 c†c
△ − i=1 i i+1 i=1 i i
and hB = t P3 (eiπ/3c†c +H.c.) 2t′ P3 c†c ,
△ − i=1 i i+1 − i=1 i i
the cluster HaPmiltonians defined on a solid-lPine point-
ing up triangle. Therefore, we have EB ǫB ,
0 ≥ △i △i
EF ǫF . The ground-state energies inPgiven m-
(c) pa0rt≤iclePs△ecito△rsi are demonstrated in Table II. Also the
number of unit cells is assumed as N, a lower bound for
FIG. 5. (a) π-flux hexagon-triangle lattice, in which a unit bosons is given by EB N(t+4t′) when t′/t 1/2 or
cellisshowningreen. (b)ThefirstBrillouin zone. Thebasis EB 6Nt′ when 10/2≥<−t′/t < 1. An upper b≤ound for
vectorsaredenotedby~b1and~b2. (c)Dispersionsofthelowest fe0rm≥ion−s is givenby E˜F, which also depends on the den-
′ 0
two bandswith t=1, t =0.2.
sity pattern on the whole lattice. At 2/3 filling, we find
EF 2N(t 2t′)N. According to the results of exact
di0ag≤on−alizatio−n on a cluster, we find when t′/t 1/8,
Under the assumption that the lattice is of size 9L2, the EB N(t+4t′) 2N(t 2t′) EF. ≤
numberofunitcellsN equalsL2. Inthethermodynamic 0 ≥− ≥− − ≥ 0
The second approach for the ground-state energy of
limit L , the ground-state energy of fermions per
→ ∞ fermions is to calculate the dispersion. The dispersion
unitcellathalffillingisgivenbytheintegralofthelowest
relations are (t=1 is assumed):
two bands (shown in Fig. 4 (b)) in the reduced Brillouin
zone, 1
E(1) = (1 t′ 9(t′)2+6t′+9+8t′Λ(~k)),
2 − −q
E0F= πdk˜x πk˜y (t′)2+2+2t′ 1 sink˜ sink˜ E(2) =t′ 1,
N −Z−π2π Z−π2πhr q − x y 1−
E(3) = (1 t′+ 9(t′)2+6t′+9+8t′Λ(~k)),
2 − q
+ (t′)2+2 2t′ 1 sink˜ sink˜ . (42)
r − q − x yi where Λ(~k)=cosk +cosk cosk , k =~k ~a and
1 2 3 1,2 1,2
− ·
Itiseasilyverifiedthatthereversednaturalinequality k3 =k1 k2. Theground-stateenergyoffermionsat2/3
−
holdswithsmallratiooft′/t,bycomparisonofthe lower filling is given by the integral of the lowest two bands in
bound of bosons and numerical integral of Eq. (42) with the Brillouin zone, which is shown in Fig. 5 (c),
given value of t′. For example when t = 1 and t′ = 0.1,
EE00BB ≥≥−−22.8.2NN>>E0FE0=F −=2.−8825.8937119N67.NTh.eWreshuelntsto′b=tai0n.e4d, E0F =kXx,ky(cid:2)E(1)(kx,ky)+E(2)(cid:3)
by two approaches are consistent with each other. √3N dk dk
Thesecondexampleistheπ-fluxhexagon-trianglelat- = x y E(1)(kx,ky)+E(2) ,(44)
2 ZZ 2π 2π
tice, which is shown in Fig. 5 (a). One unit cell is shown BZ (cid:2) (cid:3)
in green in Fig. 5 (a), with basis vectors~a = (0,1) and where k BZ as shown in Fig. 5 (b). The basis vec-
1 x,y
∈
~a2 =(1/2,√3/2). The Hamiltonian is defined as tors~b1 and~b2 are chosen accordingly as 2π(1, 1/√3)
−
and 2π(0, 2/√3), respectively. The reversed natural in-
= t eiπ/3c†c t′ c†c +H.c., equality holds when t′ t. For example, when t = 1
H − hi,ji∈thiXck,oriented i j− hi,jiX∈broken i j (43) ta′n=d t0′.=2, 0E.1B, E0B ≥1.8−N1.≪4>NE>F E=0F =2.−0127.000347N34,9Nin;wwhhiechn
where “thick, oriented” and “broken” links are specified case the ine0qu≥al−ity EB E0F is−not reverted by weak
in Fig. 5(a). This model can be regarded as triangles inter-ring coupling t′ a0s e≥xpec0ted.
