Table Of ContentAll Possible Coupling Schemes in XY Spin Chains for Perfect State Transfer
Yaoxiong Wang and Feng Shuang
Institute of Intelligent Machines, Chinese Academy of Sciences, Hefei, China 230031
Herschel Rabitz
Department of Chemistry, Princeton University, Princeton, NJ 08544
We investigate quantum state transfer in XY spin chains and propose a recursive procedure to
construct the nonuniform couplings of these chains with arbitrary length to achieve perfect state
transfer(PST). We show that this method is capable of finding all possible coupling schemes for
PST. These schemes, without external control fields, only involve preengineered couplings but not
dynamical control of them, so they can besimply realized experimentally. Theanalytical solutions
1 provideall information for coupling design.
1
0 PACSnumbers: 03.67.Hk,05.50.+q
2
n
Quantum information and quantum computation can "verifiable" but not "constructive" way. Thus, we have
a
J process lots of tasks which are intractable with classi- not yet got all possible coupling schemes for PST.
6 cal technologies. Although many schemes such as quan- In this Letter, we start from the necessary and suf-
tum dots[1], ion trap[2], NMR[3] have been discussed
ficient conditions of PST. After preselecting the eigen-
] extensively, a macroscopic scalable quantum computer
h values of a XY spin chain Hamiltonian, we propose two
p still seems to need a channel, often known as quantum recursiveformulasofthecouplingsforbothevenandodd
- wire, to transmit or exchange quantum states between N casesandprovethembymathematicalinduction. Fur-
t
n inner parts of the quantum computer. These architec- therdiscussionsdemonstratethatthismethodiscapable
a tures require to implement a transmission process for
of finding all possible coupling schemes for PST in XY
u
an unknown quantum state from one place to another
q chain with arbitrary length. Experimentally, our PST
whichisoftencalledquantumstatetransfer. Inaseminal
[ schemes can be realized, for example, by superconduct-
paper[4],Boseproposedaspinchainmodel,whoseevolu-
ingcircuitsandquantumbus[14],nanoelectromechanical
1
tionwasgovernedbyareasonableHamiltonian,andcon-
resonator arrays[15] or cold-atom optical lattice[16].
v
sideredthefidelityofstatetransferinthismodel. Similar
6 Next, we first review some basic concepts of state
5 results were also derived by studying dynamical proper-
transfer protocol using spin chain as the channel[4, 17].
1 ties of entanglement transition in Heisenberg XY spin
An unknown qubit, as encoded in site 1, is attached to
1 chain[5]. This model, in which two processors are con-
. one end of a spin chain when the chain is initialized to
1 nectedthroughaspinchainasquantumwire,isusefulfor
the allspin-downgroundstate (state initializationis not
0 quantumcomputationbasedonHeisenberginteraction[6]
necessary[18],andourresultscanbegeneralizedtothese
1 or measurements[7].
1 cases). Due to the coupling between site 1 and 2, free
Although some important and significant results have
: evolutionofthe system causesthe unknown state to dis-
v been found, see for example[4, 5], all of the available
i tribute among the chain. After a specific interval, we
X results are just concerned with uniform interaction, i.e.
want to recover this unknown state at the opposite end
the couplingsbetweenanytwonearest-neighborsitesare
r of the chain to achieve state transfer.
a the same. For this case, however, it is shown that when
N ≥4, where N is the number of the sites in XY chain, A reasonable Hamiltonian for this task is XY type
PST is impossible[8]. This drawback of uniform inter- Hamiltonian
action motivates people to find some modified models
to achieve "long" distance PST. Some works considered 1N−1 1 N
long-rangeinteractions[9],andsomeconcentratedonnu- H = 2 X Ji(σixσix+1+σiyσiy+1)−2XBi(σiz−1), (1)
mericalsimulations[10]. Onefeasiblechoiceistopreengi- i=1 i=1
neerthecouplings[8],i.e. choosespecialnonuniformcou-
plings to achieve PST, and some specific analytical cou- whereJiisthecouplingstrengthbetweensitesiandi+1,
pling schemes were found[8, 11, 12]. The necessary and andBi istheexternalstaticpotential,orcontrolfield,at
sufficient conditions for the couplings of PST, which can site i. σx,σy,σz are the three Pauli matrices. One im-
provideacriteriontoverifythepreengineeredschemesas portant observation is that Hamiltonian(1) commutates
well as to find new analytical ones, were derived from a withthetotalz-spinoperator N σz. Thus, N σz is
Pi=1 i Pi=1 i
more systematical treatment of this problem[13] by mir- a conservation, and the evolution of the system in these
ror inversion[11] and quantum computation. However, statetransfercaseswilljustinvolvethesubspacespanned
all these preengineered schemes are obtained through a by ground state and N one-site excited states. By the
2
Jordan-Wigner transform, which maps(1) to For even N cases, we assume the eigenvalues of H
N
are {±Λ ,...,±Λ } where n = N/2 , Λ ∈ N and Λ >
1 n i 1
Λ > ... > Λ > 0 (if none of {J } is zero, then the
N−1 N 2 n i
H = X Ji(a†iai+1+a†i+1ai)+XBia†iai, (2) eigenvalues oτf HN are nondegenerate[19]), and omit the
scalefactor . {J }and{Λ } areconnectedthroughthe
i=1 i=1 π i i
characteristic polynomial of the Hamiltonian(4):
XY model can be solved exactly. Hamiltonian(2) de-
scribes an N-site hopping model subjects to nonuniform n
external fields. Let |ii denotes the single excited state Det(H −λI)= (λ2−Λ2) (5)
N Y i
at site i, Hamiltonian(2) in a 2N-dimensional space will i=1
reduce to an N-dimensional subspace spanned by {|ii}.
which, by expanding it with respect to λ2, is equivalent
Explicitly,
to a set of equations:
B J 0 ··· 0
1 1 N−1 n
J1 B2 J2 ··· 0 J2 = Λ2 (6a)
H = 0 J2 B3 ··· 0 (3) X i X i
N i=1 i=1
... ... ... ... JN−1 ...
0 0 0 JN−1 BN
J2 ···J2 = Λ2 ···Λ2 , (6b)
X k1 kn X k1 kn
in {|ii} basis. The fidelity of this transfer procedure can ki+1−ki≥2 ki+1>ki
be expressed as hN|e−iHNτ |1i, where τ is the time in- .
.
terval of the free evolution. The equivalent conditions .
for PST, i.e. (cid:12)hN|e−iHNτ|1i(cid:12) = 1, are: (a) the reflec- J2J2...J2 = n Λ2 (6c)
tion symmetry(cid:12)Bi = BN+1−i(cid:12)and Ji = JN−i. (b) after Y 1 3 N−1 Y i
τ i=1
sorting the eigenvalues of H in decreasing order, the
N
π and we want to derive {J } from {Λ }. This is often
differencebetweenanytwoadjacenteigenvaluesisanodd i i
called an inverse problem. Notice that we still use J
number[13]. All schemes discovered before required (a) N−i
rather than J when i ≤ N/2 despite they are equal
aspartoftheirprotocolsanddesignedtheeigenvaluesof i
τ just for elegance of the expressions. We first introduce
HNπ to be {−k,−k+1,···k−1,k} for 2k ∈ N[8, 11], some notations for convenience. Denote jn[N] = Jn[N]
{q(k2 +k)+(2p+1)k} for k = 0,...,N[11] or {−k+ and j[N] = (J[N])2 for 1 ≤ i ≤ n − 1 whose mean-
12−n,...,−k−23,−k−12,k+21,k+23,...,k−12+n}[12] ing wiill becomie clear soon. Here, the superscript N
whichallsatisfy(b). Allthese coupling schemesarespe- denotes the dimension of the matrix H and we im-
N
cialsolutionsforPST, andourmainresultinthis Letter ply the eigenvalues of H are {±Λ ,...,±Λ } and
D 1 D/2
is to show how to get all possible couplings for PST in its couplings are {J[D]}. The main idea is to obtain
theabsenceofexternalfields,i.e. B =0. Becauseofthe i
i [D] [D−2]
{j } from {j } when we require the Hamiltonians
perfecttransfercondition(a)andthepostulationBi =0, i i
construct by them respectively share the same eigen-
Hamiltonian(3) becomes
values {±Λ ,...,±Λ }. Further, denote ΓN =
1 (D−2)/2 k
J01 J01 J02 ······ 00 00 jn[N−j−1n[N2]]jjn[n[NN−−−2]12·]····j·k[jN+k[N−]1−12] for 1 ≤ k < n, where j0[D] ≡ 0, and
0 J2 0 ··· 0 0 ΓN = j[N−2]. Denote ∆N = jk[N−2]jk[N+−22]jk[N+−42]··· and
HN = ... ... ... ... J2 0 (4) ∆nN =nj−[1N−2] , ∆N = 1k where tjkh[N+e]2jpk[N+ro]4jdk[Nu+]c6·t·s· in the
0 0 0 J 0 J n−1 n−1 n
2 1 numerators and denominators involve terms only if the
0 0 0 0 J1 0 indices of them are not larger than n−1 and n respec-
tively. With these notations, we will show the following
whose eigenvalues are symmetric about zero. Owing to
equation permits us to get {J } from {Λ } directly:
this symmetry, there are only [N/2] independent cou- i i
plings and [N/2] independent eigenvalues in(4) (0 is al-
ways an eigenvalue when N is odd). Our purpose is to
j[N] = ΓN −(−1)iΛ ∆N i=1,··· ,n. (7)
construct the couplings {J } from a set of preselected i i n i
i
eigenvalues {Λ } satisfy (b). We will first consider even
i Eq.(7) allows to construct j[N] from {j[N−2],...,j[N−2]}
N cases and show how to derive {J } effectively. Then, i n−1 i−1
i
we generalize these results to odd N cases, and finally ,{jn[N],··· ,ji[+N1]}andΛn. Thus,whenweknow{ji[N−2]},
show the completeness of this method, i.e. it can get all by adding one more parameter Λn, we can derive
possible coupling schemes for PST. j[N],j[N] ,···j[N] one by one explicitly. Now, we need
n n−1 1
3
to proveEq.(7)is consistentwith Eq.(6). Direct calcula- 1,...,n are all eigenvalues of (4). Fig.(1) shows idea of
tion shows Λ satisfies a continued fraction: the proof.
n
For odd N cases, we assume Λ >Λ >...>Λ >
j[N] n n−1 1
jn[N−]1 n+(−1)n−1Λ =1. (8) n0.−D1e,finne=j(n[NN]−=12)/(J2,n[Na]n)d2 danefidnjei[NΓ]N=an(Jdi[N∆]N)2afosrth1e≤saim≤e
. n k k
.. as even N cases. The corresponding recursive formula
jΛ1[Nn]−Λn+Λn for odd N is
Eq.(8) is equivalentto Det(H −Λ I)=0. Actually, by j[N] = Λ2∆N −ΓN (10a)
N n k n k k
expanding Det(Hi −ΛI) in terms of order i−1 deter- j[N] = ∆N −ΓN (10b)
minants, we will find the original continued fraction for k−1 k−1 k−1
Det(HN −ΛnI)=0 is where k = n,n−2,n−4,··· till we get j[N]. The dif-
1
(J[N])2 ference between n and k in Eq.(10a) is an even number
(J.N[N−]1)2 N +(−1)N−1Λn+(−1)NΛn =0. (9) awlhteicrhnaitmelpyl.ieJsusΛt2nlikaeppeveaenrsNincathsees,riwgehtcahnanddireocftlEyqc.h(1ec0k)
.
. Λ is an eigenvalue of (4) by the continued fraction rep-
(J1[Λ(NJn2[]N)2])−2Λn+Λn−Λn fraencsetonrta2tiaopnpweahresnin{Jtih}eadreefienxitpiroenssoefdjbn[Ny]Eaqls.o(1c0o)m, aensdfrtohme
DuetothesymmetrybetweenJ andJ ,wecanmove thecontinuedfractionstructure. Althoughthemainidea
i N−i
upper half of the continued fraction to the right hand of is the same,there arestill somedifferences betweeneven
the equal sign. After taking a square root on both sides, andoddcases. First,{j[N]}arenolongerunchangedun-
i
we obtain Eq.(8), which means Λn is actually an eigen- der the permutation of Λn and −Λn−1 when N is odd.
value of (4). The square root operation is exactly the Instead, the permutations of {Λn,Λn−2,Λn−4,...} and
origin of why we denote jn[N] = Jn[N] but ji[N] = (Ji[N])2 {Λn−1,Λn−3,Λn−5,...} form two groups keep {ji[N]} in-
for i 6= n before. Next, we will prove Eq.(7) is correct variant respectively (if we consider all the eigenvalues
for arbitrary N by mathematical induction. We assume {±Λi}, then both even and odd cases have two groups
the permutations of {Λ ,−Λ ,Λ ,...,(−1)nΛ } form formedby interlacedeigenvaluesrespectively whichkeep
1 2 3 n−1
a groupkeeps {j[N−2]} unchangedwhichis actually true {j[N]} unchanged, see Fig.(1)). Second, it’s interesting
i i
for {j[4]}. The following step is to prove {j[N]} are also to see j3[N] and j2[N−2] have the same
winivtahritaihnetuasnsduemrpthtieonpearmbouvtea,tidoinreocftlΛyninadnudc−eisΛ{n±−1Λwi}hicfohr, Λjn1[N−]j12[N−Λ]n−1+Λn−1 jΛ1[Nn−−22]−Λn−2
i = 1,...,n are the eigenvalues of H when {J[N]} are structure when a similar replacement as in even N cases
N i beenmade. With the helpofthis property,we canprove
[N]
constructed from Eq.(7). Obviously, j is unchanged
n Λ isalsoaneigenvaluesofH . Combiningitwiththe
under the permutation of Λn and −Λn−1. For jn[N−]1, fanc−t1that Λn is an eigenvalue oNf HN and the symmetry
we can expand it using Λ , Λ and {j[N−4]} in which property between Λ and Λ , by means of mathemat-
n n−1 i n n−2
{ji[N−4]} are irrelevantto Λn and Λn−1. Notice that this ical induction, we assert the eigenvalues of HN whose
expression has similar form with j[N−2] when j[N−2] is off-diagonal elements are constructed from Eq.(10) are
n−5 n−5
expandedbyΛ ,Λ and{j[N−6]},andifwereplace
n−1 n−2 i
Λ , Λ and {j[N−4]} in j[N] by −Λ , −Λ and
n n−1 i n−1 n−1 n−2
{j[N−6]} respectively, we will find they are indeed the -Λ -Λ-Λ -Λ ...-Λ -Λ 0 Λ Λ ... Λ Λ Λ Λ
i 1 2 3 4 n-1 n n n-1 4 3 2 1
same one. Owing to the assumption that the permu-
tation of Λ and −Λ keeps j[N−2] unchanged, we (a)
n−1 n−2 n−5
concludethatj[N] isalsounchangedunderthepermuta-
tion of Λn andn−−Λ1n−1. This method, demonstrating the -Λn -Λn-1-Λn-.2..0 Λ1 Λ2 Λ3 Λ4 Λ5 ... Λn-2 Λn-1 Λn
invariance by replacement, is applicable for other ji[N], (b)
and we can further prove all {j[N]} are invariant under
i
the permutation of Λ and −Λ . Figure 1: (a) and (b) are recursive procedures for even and
Combining this pronof and then−fa1ct that {Λ ,−Λ } ac- oddN casesrespectively. Permutationsof{Λi}withthesame
1 2 colour form one group. Although 0 is always an eigenvalue
tually form a group for {j1[4],j2[4]}, we can prove the in (b), the permutation group does not contain it. Arrows
permutationsof{Λ1,−Λ2,...,(−1)n+1Λn}formagroup onthestraightlinesindicatetherecursivedirections,e.g. the
keeps {j[N]} unchanged. Furthermore, if Λ is an eigen- rightestarrowin(b)impliesifΛnisaneigenvalueof (4)when
value ofi(4), then, according to Eq.(8), {±nΛ } for i = {Ji} are constructed from Eq.(10) then so does Λn−2.
i
4
{0,±Λ } for i=1,...,n indeed. chains can not afford PST. We also prove these formu-
i
The completeness of this method comes from the fact las are complete. Although this method is numerically
that (3) is uniquely determined by its eigenvalues when effective, there are still some interesting issues. We set
{J } are all positive[20]. This also implies all real cou- thediagonalelementstobezeros,i.e. thereisnoexternal
i
pling schemes for (4) are uniquely determined by its controlfieldinspinchainorthelaserresonanceswithany
eigenvalues. Since the completeness is available only if twoadjacentlevelsinanN-levelsystem. Thisisnotnec-
all {J } are real, we need to prove the positivity of {J2} essary for PST or perfect population transfer. Non-zero
i i
for Eq.(7) and Eq.(10). This is more apparent when diagonalelementsbreakthesymmetryofthespectrumof
we factor out the common factors of each equation in theHamiltonian,andtheeigenvaluesnolongerappearin
Eq.(7) and Eq.(10). After factorization, we will find pairs. Nevertheless, the continued fraction is also avail-
each expression contains two factors, one is positive and able when we replace Λ by B −Λ. We expect similar
i
the other is monotone with respect to Λ . Considering formula for cases involve control fields which, of course,
n
Λ > Λ > 0 and Λ > Λ in even and odd cases will contain N recursive equations but not [N/2] for an
n−1 n n n−1
respectively, we assert {j[N]} are all positive and {J } N-sitespinchain. Anotherquestioniswhether thereare
n i
are all real which satisfy the completeness condition. In other simple coupling schemes for special selected eigen-
a word, Eq.(7) and Eq.(10) are complete for all possible values. We have tested some simple sets of eigenvalues,
coupling schemes. but the couplings still seem complicated.
Upto now,we havesolvedbothevenandodd N cases
This work was supported in part by Foundation of
in the absence of external control fields {B }. This con-
i President of Hefei Institutes of Physical Science CAS,
structive method allows us to calculate the couplings
one of us (F.S.) was also partly supported by National
from a set of preselected eigenvalues. We have chosen
Natural Science Foundation of China (No. 61074052).
several sets of 50 numbers whose interval between any
two adjacent ones in each set is a random odd number
in the domain [1,100]. In general, we got the couplings
within 10 seconds. This numerical calculationshows our
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