Table Of ContentEntanglement versus observables
An Min WANG∗
Quantum Theory Group, Department of Modern Physics,
University of Science and Technology of China, Hefei 230026, People’s Republic of China
A general scheme to seek for the relations between entanglement and observables is proposed
in principle. In two-qubit systems with enough general Hamiltonian, we find the entanglement to
be the functions of observables for six kinds of chosen state sets and verify how these functions be
invariantwithtimeevolution. Moreover,wedemonstrateandillustratethecaseswithentanglement
versus a set of commutable observables under eight kinds of given initial states. Our conclusions
show how entanglement become observable even measurable by experiment, and they are helpful
for understandingof thenatureof entanglement in physics.
7 PACSnumbers: 03.67.Mn,03.65.-w
0
0
2 Quantum entanglement lies at the heart of quantum Thus, we can build the relation
n mechanicsandis viewedasa useful resourceinquantum
a information and quantum computation [1]. Recently, Eq =fhgij(cid:16)nOα,{λαk}o(cid:17)i. (5)
J
the relationsbetweenquantumentanglementandenergy
8 [2, 3, 4, 5] as well as physical phenomena [6] have at- This is a generalscheme to seek for the relationbetween
1
tracted a lot of attentions from many physicists working entanglement and observables in principle. It is clear
2 in this area. Obviously, these studies and their conclu- that only if this relation is true for a set or a kind of
v sionsarehelpfultoclearlyunderstandthenatureofquan- states, it will be really useful and significant because, in
9 tum entanglement and effectively use it in the practical the subspace of this set or this kind of states, quantum
1 tasks of quantum information processing. entanglementbecomesobservable. Particularly,itwillbe
1
Entanglement indeed relates closely with energy, in more interesting and important if this relation is invari-
1
0 particular,for spinHamiltoniansystems. However,their antwith time evolutionandthe involvedobservablesare
7 relation is not one to one from our point of view. We commutable each other.
0 prefertothinkthatthisfactindicatesquantumentangle- Now, let we give out six sets of states of two qubits
h/ ment versus quantumobsevables. This is the motivation with the above features and the following structures
p and propose of our study in this letter.
a 0 0 e−iα1v
t- In fact, for a given state in a Hilbert space H spanned 01 0 0 0 1
n byall{|ii,i=1,2,···},its entanglementis afunctionof ρ = , (6)
a 1 0 0 0 0
u (perhaps partial) density matrix elements ρij as follows eiα1v1 0 0 1−a1
q
: Eq =f({ρij}). (1) 0 0 0 0
v 0 b e−iα2v 0
Xi Whenthereexistsasetofobservables{Oα,α=1,2,···} ρ2 = 0 e2iα2v2 1−b22 0 , (7)
r intheH,andcorrespondingeigenvectorsandeigenvalues 0 0 0 0
a of any observable Oα are, respectively, |vαi and λα (i=
i i 0 0 0 0
1,2,···). Thus,withoutlossofgenerality,thegivenstate 0 1−c −d e−iα3v 0
can be expanded as ρ = 3 3 3 , (8)
3 0 eiα3v3 c3 0
ρ= ρ |iihj|= aα ({ρ }))|vαi vα . (2) 0 0 0 d3
Xi,j ij Xi,j ij kl i (cid:10) j(cid:12)(cid:12) a4 0 0 0
0 b e−iα4v 0
ρ = 4 4 , (9)
Hence 4 0 eiα4v4 1−a4−b4 0
0 0 0 0
Oα =Tr(Oαρ)= aα({ρ }))λα. (3)
X ii kl i 1−c −d 0 0 e−iα5v
i 5 5 5
0 0 0 0
ρ = , (10)
If there are such a set of observables giving an equation 5 0 0 c5 0
systemmade ofthe aboveformequationswith variousα eiα5v5 0 0 d5
that we can solve this equation system to obtain all ρij 1−b −d 0 0 e−iα6v
6 6 6
appearing at Eq.(1), that is,
0 b 0 0
ρ = 6 , (11)
6 0 0 0 0
ρij =gij(cid:16)nOα,{λαk}o(cid:17). (4) eiα6v6 0 0 d6
2
where we have used the Hermi and unit trace proper- with time can not be well determined and described by
ties of the density matrix. As to the positive conditions only using energy from our point of view.
of them are also required. In order to clearly determine Itis easy to verifythe sufficientconditions to keepthe
their entanglement by negativity measure [7, 8], we as- form invariance of time evolution of the above relations
sume all diagonal elements and v in these density ma- (15,16)underthe followingtwokinds formofHamiltoni-
i
trices are not negative (negative v can be rewritten by ans
i
absorbing its minus sign into e±iαi). It is easy to calcu-
H[1] = h s +h s +f (s −s )
late their negativity 30 30 03 03 1 12 21
+g (s +s )+h s , (22)
1 11 22 33 33
N =|v |, N =|v |,
1 1 2 2 H[2] = h s +h s +f (s +s )
30 30 03 03 2 12 21
N3 =pd23+v32−d3, N4 =pa24+v42−a4, (12) +g2(s11−s22)+h33s33. (23)
N5 =pc25+v32−c5, N6 =pb26+v42−b6. In fact, the verification process is standard and simple.
We first expand the density matrix by using the basis
Introduce the spin tensor for two qubit systems
madeoftheeigenvectorsofchosenHamiltonianandtheir
conjugate vectors. Then, acting the time evolution op-
s =σ ⊗σ (µ,ν =0,1,2,3), (13)
µν µ ν
erator (independent of time) on it, we can obtain the
where σ is 2×2 identity matrix and σ (i=1,2,3) are final density matrix at a given time. It is key matter
0 i
the Pauli matrices. Thus, the total spin vector and its to make the final state has the same structure as some
square read ρi. Finally,wecalculateitsnegativity. Itisclearthatwe
chosesuchHamiltonians(22,23)andfollowingtheirsim-
3 plifications (26)–(31) in order to guarantee the required
1
Si = 2(si0+s0i), S2 =XSiSi. (14) structure of final state. This is arrived at because these
i=1 chosen Hamiltonians have appropriate eigenvectors and
Here,~istakenas1forsimplicity. Denotingtheexpected eigenvalues. Thedetailsisomittedinordertosavespace.
Similarly, we can verify that the relation (17,18) and
value of an operator O in the state ρ by hOi , we can
i i
(19,20) have the form invariance of time evolution, re-
expresstheentanglementbythesomecomponentsofspin
spectively, under H[1] and H[2].
tensor, that is
It is more significant if, in the above the relations be-
1
N = (hs i2+(hs i2, (15) tween entanglement and observables, the involved ob-
1 2q 11 1 12 1
servables are commutable each other. Actually, this can
1
N2 = 2qhs11i22+hs12i22, (16) be obtained by taking particular initial states. Firstly,
let us set the initial state as
1
N3 = 2qhs11i23+hs12i23+hSzi23+hSzi3, (17) |ψi=sinθ |00i+e−iα1cosθ |11i. (24)
1 1
1
N4 = 2qhs11i24+hs12i24+hSzi24−hSzi4, (18) Obviously, it is entangled and will evolute to the first
kind ρ . Its negativity reads
1 1
N5 = 2qhs11i25+hs12i25+(2−hS2i5)2+hS2i5−2, (19) 1
N (t)= 1−hS i (t). (25)
1 ψ 2p z 2
N6 = 2qhs11i26+hs12i25+(2−hS2i6)2+hS2i6−2. (20) It is both formally invariant and numerically conversed
Therefore, the entanglement, for the given six kinds of forthetimeevolutionwithH[1],thatis,Nψ(0)=Nψ(t).
state sets, can be expressed as the functions of some ob- While under H[2], it only keeps invariance in form, but
servables. the conversationis broken in general. In order to under-
It is more interesting how these relations evolute with standhowentanglementevolute,wechosetwosimplified
time. Withoutlossofgenerality,thegeneralHamiltonian forms of H[2]
of two-qubit systems can be written as 1
H[2,1] = ω (σ ⊗σ −σ ⊗σ )
2 3 0 0 3
2
3 3
H = X hµνσµ⊗σν = X hµνsµν. (21) +g2(σ1⊗σ1−σ2⊗σ2)+h2σ3⊗σ3,(26)
µ,ν=0 µ,ν=0 1
H[2,2] = ω (σ ⊗σ −σ ⊗σ )
2 3 0 0 3
2
where h are real. Based on the facts that Hamilto-
µν +f (σ ⊗σ +σ ⊗σ )+h σ ⊗σ .(27)
2 1 2 2 1 2 3 3
nian is made of the spin tensor and that entanglement
can expressed as the spin tensor, it is not surprised that Then, let us fix α , for example, α = 0, we can draw
1 1
there exists the relation between entanglement and en- Fig.1 and Fig.2 to show the surface graphics determined
ergy. However, entanglement quantity and its evolution by θ, hS i(t) and N(t).
z
3
N0.4 1 00..44
0.2 0.5 00 20000..22N
0
00 0
11 Sz 11 1.5
-0.5
ΘΘ 22 ΘΘ 22 1 S2
33 -1 0.5
33
0
FIG. 1: Under H[2,1], hSziψ(t) = −cos(2T)cos(2θ1),
Nψ(t)=p1−cos2(2T)cos2(2θ1)/2 (α1 =0, T =2g2). FIG. 3: Under H[1,2], hS2iφ(+)(t) = 1 + sin2(2T +2θ2),
N (t)=psin2(2T +2θ )/2 (α =0, T =2f ).
φ(+) 2 2 1
0.4
N 1
0.2 00..44
0.5 00..22
000 0 00 200 N
Sz
11 1.5
-0.5 11
ΘΘ 22 1
33 -1 ΘΘ 22 S2
0.5
FIG. 2: Under H[2,2], hSziψ(t)=−cos(2T −2θ1), Nψ(t)= 33
psin2(2T −2θ )/2 (α =0, T =2f ). 0
1 1 2
FIG. 4: Under H[1,2], hS2iφ(−)(t) = 1 + sin2(2T −2θ2),
Nφ(−)(t)=psin2(2T −2θ2)/2 (α2 =π, T =2f1).
When the initial state is taken as
|φi=sinθ |01i±cosθ |10i, (28) ing six kinds of mixed states at the initial time
2 2
ρM(0) = sin2θ|00ih00|+cos2θ|11ih11|, (32)
its negativity reads 1
ρM(0) = sin2θ|01ih01|+cos2θ|10ih10|, (33)
2
1 ρM(0) = sin2θ|10ih10|+cos2θ|11ih11|, (34)
Nφ(t)= 2q(hS2i(t)−1)2. (29) ρM3 (0) = sin2θ|01ih01|+cos2θ|11ih11|, (35)
4
ρM(0) = sin2θ|00ih00|+cos2θ|10ih10|, (36)
It is both formally invariant and numerical conserved 5
ρM(0) = sin2θ|01ih00|+cos2θ|01ih01|. (37)
only under H[2](h03 =h30), or only invariant in form 6
under H[1,2]. Here, we have defined that
Theyareseparable,buttheywillbecomeentangledafter
the finite time evolution. Actually, these mixed states
1
H[1,1] = ω (σ ⊗σ +σ ⊗σ ) can be thought of as the reduced density matrix from
1 3 0 0 3
2
the compound systems initially with entanglement. For
+g (σ ⊗σ +σ ⊗σ )+h σ ⊗σ ,(30)
1 1 1 2 2 1 3 3 example,thestate(sinθ|00i+cosθ|11i)⊗|10ihasentan-
1
H[1,2] = ω (σ ⊗σ +σ ⊗σ ) glementbetweenthefirstandsecondqubits. Topartially
2 3 0 0 3
2 trace qubits 2 and 4, we will get ρM(0) made of qubits
1
+f (σ ⊗σ −σ ⊗σ )+h σ ⊗σ .(31)
1 1 2 2 1 1 3 3 1 and 3, or partially trace qubits 1 and 3, we will get
ρM(0) made of qubits 2 and 4, and so on. This implies
3
Fig.3 andFig.4show the surface graphicsdetermined by that our methods and following discussions can be simi-
θ, hS2i(t) and Nφ(±)(t) under H[1,2]. larlyusedtothe casesofentanglementtransferandwith
Similarly,wecanobtaintheinvarianceofthe relations the environment-system interactions [3].
betweentheentanglementandobservablesforthefollow- Obviously, NρM1 under both H[2,1] and H[2,2], NρM2
4
under H[1,2] are, respectively, 22 SS22
11..55
11
1
N (t) = hS i(0)2−hS i(t)2, (38) 00..55
ρM1 2p z z
00
1 1
NρM2 (t) = 2q(hS2i(t)−1)2. (39) 0.75
While, under H[1,2] 0.5N
0.25
NρM(t) = qhSzi2i(t)+(hS2ii(t)+hSzii(t)−1)2 00..7711055
i
+hS i (t), (40) 00..55
z i 00..2255 SSzz
NρM(t) = qhSzi2j(t)+(hS2ij(t)−hSzij(t)−1)2 00
j
−hSzij(t), (41) FIG. 6: Under H[1,2], hSzij(t) = sin2θj, hS2ij(t) =
and under both H[2,1] and H[2,2], ˆ3−cos(2θj)−(−1)isin(2T)cos2θj˜/2, NρMj (t) =
psin4θj+sin2(2T)cos4θj−sin2θj (j =5,6, T =2f2t).
NρM(t)=q(2−hS2ik(t))2+(hS2ik(t)−1)2−hSzi2k(t)
k
+hS2i (t)−2, (42)
i
where, i = 3,4, j = 5,6 and k = 3,4,5,6. The surface 1
figures determined by S , S2 and N are Figs. 5 - 8 00..7755
z
00..55
N
00..2255
--11 200
--00..55 1.75
00 1.5S2
1 SSzz 00..55 1.25
11 1
0.75
N0.5 2 FIG. 7: Under H[2,1], hSzii(t)=−cos(2T)sin2θi, hS2ii(t)=
0.25 1.5 [3+cos(2θi)]/2, NρM(t) = psin4θi+sin2(2T)cos4θi −
i
----00011..7755 1 S2 sin2θi (i=3,4, T =2f1t).
--00..55 0.5
SSzz --00..2255
important and interesting thing in such a claim is that
00 0
wedemonstrateandillustratethattheentanglementcan
FIG. 5: Under H[1,2], hSzii(t) = −cos2θi, be expressedas the functions of a set ofcommutable ob-
hS2ii(t) = ˆ3+cos(2θi)−(−1)isin(2T)sin2θi˜/2, servablesS2 andS for eightkinds ofinitialstate sets in
z
NρM(t) = pcos4θi+sin2(2T)sin4θi − cos2θi (i = 3,4 two-qubit systems. As well-known, the involved observ-
i
and T =2f2). able should be measurable at the same time. Actually,
thismeansthatentanglementcanbemeasuredbyexper-
In summary, we propose a general scheme to seek for imentinthesestatesets. Weexpectitwillbetrueinthe
the relations between entanglement and observables in near future.
principle, and we find and verify such relations for the We are grateful all the collaborators of our quantum
given state sets in two-qubits system, that is, entangle- theorygroupinouruniversity. This workwassupported
ment is expressed as the function of observables. In ad- by the National Natural Science Foundation of China
dition, our methods canbe similarly used to the casesof under Grant No. 60573008.
entanglement transfer as well as with the environment-
system interaction. Because, Hamiltonian can be ex-
pressedbytheinvolvedobservables,ourconclusionshave
implied the relation between entanglement and energy.
∗
Electronic address: [email protected]
Moreover,ourconclusionsaremoredeterminedrelations
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FIG. 8: Under H[2,2], hSzij(t)= cos(2T)sin2θj, hS2ij(t) =
[3−cos(2θj)]/2, NρM(t) = pcos4θj+sin2(2T)sin4θj −
j
cos2θj (j =5,6, T =2f2t).
