Exact symmetry breaking ground states for quantum spin chains

0.13 MB·English

by  S. Alipour

Exact symmetry breaking ground states for quantum spin chains
Sahar Alipoura, Sima Baghbanzadeh,b and Vahid Karimipoura,c
8 a Department of Physics, Sharif University of Technology, P.O. Box 11365-9161, Tehran, Iran.
0 b Department of Physics, Iran University of Science and Technology, P.O. Box 16765-163, Tehran, Iran.
0 c Corresponding author,Email:[email protected]
2
Weintroduce afamily of spin-1/2 quantumchains, and show that their exact ground states break
n
a the rotational and translational symmetries of the original Hamiltonian. We also show how one
J can useprojection toconstruct a spin-3/2 quantumchain with nearest neighbor interaction, whose
6 exact ground states break the rotational symmetry of the Hamiltonian. Correlation functions of
1 bothmodelsaredeterminedinclosedform. Althoughweconfineourselvestoexamples,themethod
can easily beadapted to encompass more general models.

] PACS: 75.10.Jm
el Thequestionofexactgroundstateofagivenquantum tion to introduce another quantum spin chain, namely a
- manybodysystem,i.e. aquantumspinchain,hasalways spin 3/2 quantum chain with nearest neighbor interac-
r
t beenfascinatingandofgreatimportanceto physicistsin tion, with symmetry breaking ground states. We should
s
. all fields, from condensed matter physics and statistical stress that although we do this for a specific model with
t
a mechanics to mathematical physics and quantum field rotational and translational symmetry, our method can
m theory [1–3]. Besides their inherent interest as a the- be adapted for construction of a broader class of similar
- oretical laboratory, where the phenomenon of quantum models.
d
phase transition [4] can be studied in detail and in ana- Letusfirstfixournotation: weuse + , and S :=
n lytical form, quantum spin chains provide an important 1 (+, ,+ ) to denote respec|tiviely|−tihe sp|iniup,
o √2 | −i−|− i
c link with classical statistical mechanics on two dimen- spin down and a singlet state of two spins. Subscripts
[ sional lattice models, the latter is a subject with a rich indicate the sites of the lattice, thus S ,+ indicates
2 and fascinating variety of cooperative and critical phe- a singlet on sites 1 and 2 and a spin u|p12stat3eion site 3.
v nomena affording insights into many properties of real Consider the state
7 physical, chemical and biological systems.
4 As prototypes of correlated many-body systems, the |φ0i:=|S12,+3,S45,+6,···S3N−2,3N−1,+3Ni, (1)
2
Hilbert space dimension of quantum spin chains grows
1 shown in figure (1). We require that φ0 be the ex-
exponentially with system size, rendering any numerical | i
. act ground state of a spin chain Hamiltonian H =
1 treatmentoflargesystemsextremelydifficult. Therefore 3N
0 any exact solution of a new quantum spin chain, is not Pk=1hk,k+1,k+2,k+3, where we use periodic boundary
8 conditions. Tothisend,wedemandthatthelocalHamil-
onlyvaluableforitsownspecificcharacteristics,butmay
0 tonian h be a positive operator and annihilate φ0 . A
: provide us with a plethora of mathematical techniques | i
v localHamiltonianh finds four adjacentspins onlyin one
for constructing a whole new series of exactly solvable
i of the mixed states shownin figure (1). Thus, we should
X models.
have
r In this letter, we introduce a new model and tech-
a
nique for constructing exact ground states of quantum tr[h(S S + + I)]=0,
| ih |⊗| ih |⊗
spinchainswhichspontaneouslybreakthesymmetriesof h[+ S + ]=0,
| i| i| i
the parent Hamiltonian. The method is inspired by the tr[h(I + + S S )]=0. (2)
⊗| ih |⊗| ih |
original work of Affleck, Kennedy, Lieb and Tasaki who
firstintroduceditforconstructingexactinvariantground Since S is asinglet,the state + S + ,canbe written
| i | i| i| i
states for certain quantum chains [5]. Their method, de- as a linear combination of states with total spin 0 and
signed in the context of valence bond solids, led in the 1. Therefore a projector P2 which projects the product
course of time to the formalism of finitely correlated or of four adjacent spins on the the sector with total spin
matrix product states [6]. The model we introduce is 2, automatically satisfies equations (2). However there
a spin 1/2 quantum system on a ring, with interaction is one other projector which satisfies equations (2), and
rangeequalto4,havingbothtranslationalandrotational itsannihilationofthese statesisquitenontrivial. Thisis
symmetry, nevertheless we show that its exact ground the projector P0 := χ χ, where χ (or its embedding
| ih | | i
statesbreakboththesesymmetries. Wealsofindthe ex- onsites 1,2,3and4)is asingletoffourspins constructed
actmaximumenergystateandsomeoftheexactexcited as follows:
states, a progress which has not been reported for any
χ = S14,S23 + S13,S24 . (3)
of the models constructed in the AKLT or the matrix | i | i | i
product formalism [7–10]. Finally we use this construc- To show this we use the simple facts
FIG.1. Thethreeequationsin(2)correspond,respectively
from left to right, to the annihilation of these local mixed
states by h. For ease of comparison with the analysis in the
text we havelabeled thespins in all boxes as 1, 2, 3 and 4.
FIG.2. The states |φ0i, |φ1i and |φ2i.
1 1
+1 S12 = 2 , 1 S12 = − +2 ,
h | i √2|− i h− | i √2| i in this way. Direct calculation from (3) is notoriously
tedious. We can circumvent this problem by calculating
to arrive at the identity
the other projector P = χ χ and subtract it from
0′ ⊥ ⊥
| ih |
1 0 to obtain P0. It is straightforward to check from (4)
S23 S12,S34 = − S14 , (4) P
h | i 2 | i that the other singlet which should be perpendicular to
χ is the following
where 1,2,3 and 4 are any four different indices. From | i
this identity one can easily derive other identities sim-
χ⊥ = S12,S34 . (9)
ply by permuting indices in an appropriate way, namely | i | i
hS34|S14,S23i = −21|S12i, or hS34|S13,S24i = 21|S12i. This readily gives
These two identities, immediately imply that S34 χ =
0. SimilarlyonefindshS12|χi=0,hencethefirshtan|dithe P0 = 0 χ⊥ χ⊥ = 0 (1 s1 s2)(1 s3 s4).
lastequationsof(2)aresatisfied. Forthesecondrelation P −| ih i P − 4 − · 4 − ·
of equation (2) we note that
Putting all these together, and discarding overal con-
3 stants, we find
χS23 = S14,S23 S23 + S13,S24 S23 = S14 , (5)
h | i h | i h | i 2h |
3N
hence H =X si si+1(1 2si+2 si+3)
· − ·
i=1
3
4+J 1+J
χ+1,S23,+4 = S14 +1,+4 =0. (6) S S + (S S )2, (10)
h | i 2h | i i i i i
− 3 · 6 ·
This proves the assertion that P0 annihilates all three wheres isthespinoperatorofonesiteandS isthespin
i i
kinds oflocalstates. Thus,we findthat the localHamil-
operatorofablockoffourspinsatsitesi,i+1,i+2,i+3.
tonian can be of the form h = 2JP2+P0, where J is a Note that the relative strength of consecutive couplings
positivebutarbitrarycouplingconstant,thefactorof2is
are approximately 1: 0.48: 0.06.
for convenience and one of the couplings, the coefficient
of P0, has been re-scaled to unity. The final form of the The state φ0 is the exact ground state of H which
Hamiltonian in terms of spin operators can be found as | i
breaks translational and rotational symmetry of the
follows. First we note that 21⊗4 =2⊕13⊕02 where the Hamiltonian. Clearly, it is not invariant under rotation,
exponents indicate the multiplicities and P0 is the pro- itis the top state ofthe totalspin-N sector,other states
2
jector to one of the above two zeros. Let us denote the
which are degenerate with it in energy, are obtained by
sum of spin-j projectors in the above decomposition by 3N
acting with total L = Pk=1σk− operator, which does
. Then, from S S = j(j +1) , where S is the −
Pj · Pj Pj not affects the singlets but flips the up spins. None
spin operator of four consecutive sites, we have
of these N + 1 states, are invariant under translation.
Thus the ground state has a 3(N +1) fold degeneracy,
I = 0+ 1+ 2,
P P P generatedby the brokensymmetries ofthe Hamiltonian,
S S =2 1+6 2,
(S S·)2 =4P1+3P6 2, (7) namely the continuous so(3) symmetry and the discrete
· P P translational symmetry. To find translational invariant
from which we obtain states, we form the states Ψj , j =0,1,2 as follows:
| i
P0 =1− 23S·S+ 112(S·S)2, |Ψji:= √13(|φ0i+ωj |φ1i+ω2j |φ2i), (11)
1 1
2 = S S+ (S S)2. (8)
P −12 · 24 · whereω3 =1,and φi =T φi 1 ,whereT istheone-site
| i | − i
translation operator, figure (2). These states are eigen-
This readily gives P2 = ( 2) in terms of spin operators.
P states of T, with eigenvalues 1,ω and ω2 respectively.
However,theindividualprojectorP0cannotbeobtained
One can readily show that for i=j,
6
1
φ φ =( +,S S,+ )N =(− )N,
i j
h | i h | i 2
from whichwe find the states Ψj are normalizedin the
| i
thermodynamic limit. In this same limit, the correlation
functions are easy to find. First the magnetization turns
out to be
1
Ψj Sz Ψj = , j. (12)
h | | i 6 ∀
and the two-point correlation function becomes
FIG.3. The dispersion relation for the holes of model (10).
1
hΨj|S1zSrz|Ψji= −12(δr,2−Xδr,3n+1),
n=1
1
hΨj|S1xSrx|Ψji= −12δr,2, ∀j. (13)
We now ask if it is possible to find some of the exact
excited states of the Hamiltonian. We will show that
this is indeed possible, although the states thus found
are not the low-lying states of the energy spectrum. To FIG. 4. The states in each bulb, are projected to a spin
proceed, we note that the state 3/2 state, toform anewquantumspinchain with spin equal
to 3/2 and with nearest neighbor interactions.
Ω := +,+,+, ,+,+ , (14)
| i | ··· i
3N
is an exact energy eigenstate, indeed the maximum en- ψr = Xein23πNr n , (16)
ergy eigenstate with Emax = 6NJ (here we are working | i | i
n=1
with h = 2JP2+P0 and not with (10), in which multi-
plicative and additive constants have been ignored). To with
simeeatlihtyisonfottheethenaetrPgy0|Ωisifo=un0danbdytPa2k|Ωinig=J|vΩeir.yTlahregemaanxd- E =2J(3N 4+2cos3 2πr +2cos2 2πr). (17)
r
− 3N 3N
noting the continuity of the energy with respect to J.
Consider now the one ”hole” state of the form Note that E is the exact ground state energy of
max
−
H,inwhichcase E givesthe energyofthe low-lying
r
n = +,+, , , ,+,+ , e−xcitationsorspinw−avesof H. Thedispersionrelation
| i | ··· − ··· i −
is plotted in figure (3). Note that if it were not for the
where the state in the n th position has been flipped.
− P0 projector, then a Majumdar-Ghosh state, a singlet,
A general one-hole state is of the form
would have been the ground state of H. The role of P0
is to elevate the energy of this state and hence break the
3N
|ψri= Xψnr|ni. (15) soynmemcaentroyb.taFiinnaalnlyexwaectngortoeutnhdatstfartoemfotrhaesapbionv3e mmooddeell
n=1 2
with nearest neighbor interaction. Let us project three
ItisobviousthatP0 n =0,foranyembeddingofprojec- spins into a spin 3 site as shownin figure(4). The prod-
| i 2
tion P0 on four consecutive sites. To see this, note that uctoftwo adjacentspin 3’s onthe lowerchain,being ef-
2
the total z component of each four consecutive spins is fectivelythe productoffourspin 1’s onthe upper chain,
2
either 2 or 1. Also from the explicit form of the spin-2 can combine to give only spins 2, 1 and 0. Therefore a
states, one can check that P2 does not take a state |ni projector P3 on spin 3 sector, can annihilate the local
outside the one-hole sector. In fact, a simple calculation state of any two adjacent spins in the lower chain. This
shows that means that the state constructed as such, will be an ex-
act groundstate of a Hamiltonianwith nearestneighbor
3
J
H n =6J(N 1)n + X(4 k)(n+k + n k ). interaction H = Pk(P3)k,k+1. The projector is found in
| i − | i 2 − | i | − i the same way as in (7) and (8) and we arrive at
k=1
InsertingthisintheeigenvalueequationH ψr =Er ψr , N 81 29 2
gives | i | i H =C+X Si Si+1+ (Si Si+1)2+ (Si Si+1)3,
8 · 6 · 3 ·
i=1
whereC = 165N. Inthisway,weconstructarotationally and
32
invariant Hamiltonian whose exact ground state breaks 1 1
ξ = , ξ = . (20)
this rotational symmetry. The remarkable point is that l ln(2+√3) t ln(1+√3)
we can also obtain the explicit form of this state, be-
The correlation functions for other states in the ground
yond the constructionshown in figure (4). With this ex-
state multiplet can be determined by the application of
plicit expressionone canthen determine the correlations
rotational symmetry. In summary, we have introduced
in this state which are no longer restricted to nearest
a one-parameter family of isotropic spin 1/2 quantum
neighbor sites. To achieve this, let us first construct a
chains,whoseexactgroundstatesbreakthetranslational
Matrix Product (MP) representationfor the parent (up-
and rotationalsymmetry of the Hamiltonian. The total-
per) dimer state in figure (4). In view of the MP repre-
ity of degenerate vacua of these models are generated
sentation [10] of dimerized Majumdar-Ghosh state [11],
by the application of rotation and translation operators.
this state has the following representation:
Besidesthegroundstate,themaximumenergystateand
φ0 =tr(A A B A A B )i1,i2,i3,i4,i5,i6, some of the lower energy states have also been derived
| i i1 i2 i3 i4 i5 i6 ··· | ···i
in closed analytical form. A related spin 3/2 quantum
where chain with nearest neighbor interactions has also been
constructed, the exact ground state of which has an MP
A = + 0 + 0 ,
+ | ih | | ih−| representation. The method presented in this letter can
A = 0 0 +,
− |−ih |−| ih | be pursued further, to make other spin models whose
B = 0 0, B =0, (18)
+ | ih | − groundstatebreaktherotationalsymmetryoftheHamil-
tonian. Thereisalsoaninterestingun-answeredquestion
and + , and 0 arethreeortho-normalstates. Using
| i |−i | i which deserve further investigation: Is there a singlet
theClebsh-Gordoncoefficientstodecomposetheproduct
state which is invariant under rotation and has a higher
of three states in a bulb of the upper chain in figure (4)
energy than the degenerate ground states [13]? V. K.
and noting that B = 0, we find the following matrices
− wouldlike to thank LalehMemarzadehfor very valuable
corresponding to the states of a single spin 3/2 site on
and constructive discussions. S. A. and S.B. would also
the lower chain:
thank Ali Rezakhani for valuable discussions.
1
A =A B A , A = (A B A +A B A ),
3/2 + + + 1/2 √3 − + + + + −
1
A =0, A = A B A ,
−3/2 −1/2 √3 − + −
leading to
[1] R. J. Baxter, Exactly Solved Models in Statistical Me-
A =σ+, A = −1σ , A = −1σ , A =0. chanics, Academic Press, London, 1982.
3/2 1/2 √3 z −1/2 √3 − −3/2 [2] E.H.Lieb,andD.C.Mattis(Eds.)MathematicalPhysics
in One Dimension, Academic Press, N.Y. 1966.
The state of the lower chain is now given by a matrix
[3] E.H.Lieb,T.Schultz,andD.Mattis,AnnalsofPhysics,
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(1987).
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Sz = 1, Sx = Sy =0, and [6] M.Fannes,B.Nachtergaele,andR.F.Werner,Commun.
h i 2 h i h i Math. Phys. 144, 443 (1992); A. Klu¨mper, A. Schad-
u schneider, and J. Zittartz, J. Phys. A, Math. Gen. 24,
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h i − h i h i 2
[7] A. Klu¨mper, A. Schadschneider, and J. Zittartz, Euro-
where u := 1 . With the notation Ga(1,r) := phys. Lett. 24, 293 (1993); M. A. Ahrens, A. Schad-
SaSa Sa 1S+a√3, schneider, and J. Zittartz, Europhys. Lett. 59 (6), 889
h 1 ri−h 1ih ri (2002);E.Bartel,A.SchadschneiderandJ.Zittartz,Eur.
Gz(1,r)=Al( 1)r−1 e−ξrl , Gt(1,r)=At( 1)r−1 e−ξrt Phys. Jour. B, 31, 2, 209-216 (2003).
− − [8] M.Asoudeh,V.Karimipour,andA.Sadrolashrafi,Phys.
Rev. B 75, 224427 (2007); Phys. Rev. A 76, 012320
where
(2007); Phys.Rev.B 76, 064433 (2007).
12+6√3 12+7√3 [9] S. Alipour, V. Karimipour and L. Memarzadeh, Phys.
Al := , At := , (19) Rev. A 75, 052322 (2007).
4 4
[10] V. Karimipour and L. Memarzadeh,The Matrix Product
Formalism and the Generalization of Majumdar-Ghosh
Model to Arbitrary Spins, e-print,arXiv:0712.2018.
[11] C. K. Majumdar and D. P. Ghosh, J. Math. Phys. 10,
1388 (1969); J. Math. Phys.10, 1399 (1969).
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[13] Direct and lengthy calculation shows that for a ring of
size 4, |αi:=|S ,S i−|S ,S i and |βi:=|S ,S i
14 23 12 34 13 24
are eigenstates with energies 1 and 3 resp. and for a
ring of size 6, |γi = |S ,S ,S i−2|S ,S ,S i+
16 25 34 12 36 45
|S ,S ,S i is an eigenstate with energy 2.
14 23 56