Equivalence between two-mode spin squeezed states and pure entangled states with equal spin

0.11 MB·English

by  Dominic W. Berry

Equivalence between two-mode spin squeezed states
and pure entangled states with equal spin
Dominic W. Berry1,2,3 and Barry C. Sanders2,3
1Department of Physics, The University of Queensland, St. Lucia, Queensland 4072, Australia
2Centre of Excellence for Quantum Computer Technology,
Macquarie University, Sydney, New South Wales 2109, Australia
3Institute for Quantum Information Science, University of Calgary, Alberta T2N 1N4, Canada
(Dated: February 1, 2008)
We prove that a pure entangled state of two subsystems with equal spin is equivalent to a two-
mode spin-squeezed state under local operations except for a set of bipartite states with measure
zero, and we provide a counterexample to the generalization of this result to two subsystems of
unequalspin.
5
0
PACSnumbers: 03.65.Ud,42.50.Dv
0
2

n I. INTRODUCTION
a
J
The uncertainty principle is a consequence of the noncommutativity of complementary variables. Specifically, two
3
1 complementary operators A and B have spreads ∆A and ∆B, given by roots-of-variances pV(A) and pV(B),
respectively, that satisfy the inequality ∆A∆B C /2 where iC = [A,B]. Often natural units can be employed
≥ |h i|
1 such that A and B are dimensionally equivalent and ∆A p C and ∆B p C for a typical state. Under
v ≥ |h i| ≥ |h i|
5 these conditions, p|hCi| represents a fundamental noise limit, which is known as the standard quantum limit. This
standard quantum limit is especially important in quantum metrology. One example is optical interferometry, for
6
0 whichsemiclassicalinputstates(whichcanbe expressedasa mixtureofGlauber-Sudarshancoherentstates)have,at
1 best, vacuum fluctuations; in this caseA and B are the canonicalharmonic oscillator operators,and eachof ∆A and
0 ∆B exceed 1. Spin systems, whose dynamical operators J , J , and J , satisfy iJ = [J ,J ], are another example:
2 x y z z x y
5
0 the standard quantum limit for ∆Jx and ∆Jy is p|hJzi|. An objective of quantum metrology is to prepare states
whose noise level is less than these standard quantum limits; such states are generically known as ‘squeezed’ because
/
h of the reduction, or squeezing, of the fluctuations below the standard quantum limit.
p
Spin squeezing,forwhichthe uncertaintyin onespin componentis reducedbelowthe standardquantumlimit, has
-
t been studied extensively [1, 2, 3, 4] and is especially important for applications to high-resolution spectroscopy [2].
n
Furthermore, spin squeezing has become important because spin squeezing implies entanglement within the spin
a
system[3]. Fortwospin-1/2systems,pure entangledstatesareequivalenttospin-squeezedstates underlocalunitary
u
q transformations [4]. For bipartite spin systems, pure two-mode spin-squeezed (TMSS) states [5, 6, 7, 8, 9], which
: exhibit strong correlations between spin components of the two subsystems, have been shown to be entangled [8, 9];
v
that is, two-mode spin squeezing is a sufficient condition for entanglement.
i
X Thus we know that two-mode spin squeezing implies entanglement for two-mode spin systems. The open question
r is whether entanglement is equivalent to two-mode spin squeezing up to local unitary transformations. Here we
a
show that, for pure states of two spin systems of equal dimension, two-mode spin squeezing after application of local
unitaries is a necessary condition for entanglement, except for a set of bipartite pure states of measure zero. Thus
two-mode spin squeezing can be considered to be approximately equivalent to entanglement, in the sense that the
exceptionsareofmeasurezero. Furthermoreweshowthattwo-modespinsqueezingisnotequivalenttoentanglement
for (a) mixed states, (b) two spin systems of unequal dimension, or (c) the restriction of the local unitary operations
to rotations. For these cases, the set of exceptions has nonzero measure.
II. EQUIVALENCE
We apply the superscript (1) or (2) to the spin operators J to indicate the subsystem. We take these subsystems
k
toeachbeoffixedtotalspin;thatis,wedonotconsidersuperpositionsofdifferenttotalspins. Inthissectionwetake
(±) (1) (2)
bothsubsystemstohavethesamespin,j. SumsanddifferencesofthespinoperatorsaredenotedbyJ =J J ,
k k ± k
and the usual criterion for two-mode spin squeezing may then be expressed as
V(J(+))+V(J(−))< J(+) . (1)
y x h z i
2
j
Any pure entangled state may be expressed, via a Schmidt decomposition, as Ψ = ψ ϕ χ , where
| i Pm=−j m| mi| mi
ψ are the Schmidt coefficients, and ϕ and χ are orthonormal bases for the two spin systems. The Schmidt
m m m
| i | i
coefficients are nonnegative real numbers, and we label them such that they are in nondescending order.
Using local unitary operations, we map the bases ϕ and χ to the bases of J eigenstates, thus obtaining
m m z
| i | i
j
ψ = X ψm m,m z (2)
| i | i
m=−j
with m ,m m m . The state (2) satisfies certain symmetry conditions. The first is that the expectation
1 2 z 1 z 2 z
values| of theix-≡an|d yi-sp|in icomponents are zero. To see this, note that J(−) ψ =0 so
z
| i
J(+,−) = e−iπJz(1)J(+,−)eiπJz(1) = J(+,−) (3)
h x,y i −h x,y i −h x,y i
where we have used the subscripts x,y and superscripts (+, ) to indicate that the same derivation holds for any of
these operators. Thus we have J(+) = J(−) =0; hence −
y x
h i h i
V(J(+))= (J(+))2 , V(J(−))= (J(−))2 . (4)
y h y i x h x i
We may show that these variances are equal using a similar method:
h(Jy(+))2i=he−iπ2Jz(−)(Jy(+))2eiπ2Jz(−)i=h(Jx(−))2i. (5)
Therefore, in order to show that the state ψ is a TMSS state, it is sufficient to establish that
| i
(J(−))2 J(+)/2 <0. (6)
h x − z i
We can show this result by simply evaluating the left-hand side. Expanding Eq. (6) yields
(J(−))2 J(+)/2 =2 (J(1))2 2 J(1)J(2) J(+)/2 (7)
h x − z i h x i− h x x i−h z i
where we have used (J(1))2 = (J(2))2 , which follows from symmetry. From the calculations in Appendix A, we
x x
h i h i
obtain
j−1
h(Jx(−))2−Jz(+)/2i= X (ψm−ψm+1)ψm[j(j+1)−m(m+1)]. (8)
m=−j
Because the ψ are nonnegative and in nondescending order, (ψ ψ )ψ 0. In addition, if the Schmidt
m m m+1 m
− ≤
coefficients take more than one nonzero value, there must be a value of m such that ψ and ψ are not equal and
m m+1
both nonzero, which implies that (ψ ψ )ψ <0.
m m+1 m
Therefore, providedthe nonzero Sch−midt coefficients of ψ are not all equal, (J(−))2 J(+)/2 is stricly less than
x z
| i h − i
zero,and ψ is a TMSS state. Thus we haveshownthat all states with Schmidt coefficients that take more than one
| i
nonzero value are equivalent to TMSS states under local unitary operations.
III. CASE OF EQUAL SCHMIDT COEFFICIENTS
ForthecasewherethenonzeroSchmidtcoefficientsofthestateareallequal,theproofgivenintheprevioussection
doesnotapply. Thiscaseincludes(i)unentangledstates,(ii)maximallyentangledstateswithall Schmidtcoefficients
equal, and (iii) states with some of the Schmidt coefficients zero, and the remainder equal. For the third case, if we
restrict to the subspaces for the two subsystems that are spanned by the states in the Schmidt decomposition, the
state is maximally entangled. For case (i), the states are unentangled, so it is clear that they can not be equivalent
to TMSS states [8, 9]. Cases (ii) and (iii) comprise a set of bipartite pure spin states of measure zero so, although
these states can be exceptions to the principle of equivalence between TMSS and entanglement, they are rare in that
the probability for such states is zero when selected according to the Haar measure.
It is easily seen that, for case (ii), the states are counterexamples to the principle of equivalence between TMSS
and entanglement. For maximally entangled states, the reduced density operator for each subsystem is the identity,
so J(+) = 0, and the state clearly cannot be TMSS under any local unitaries. For case (iii), the state ψ in the
z
h i | i
form (2) is not TMSS. This does not eliminate the possibility that there are local unitary operations that bring the
states to TMSS form. However, numerical searches have failed to find such operations.
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IV. EXTENSIONS TO OTHER CASES
The results that we have presented have three requirements; that the spins are equal, the states are pure, and
arbitrary local unitary operations are allowed. Below we present examples demonstrating that if any of these three
requirementsarelifted, byallowingunequalspin,mixedstates,orrestrictingtolocalrotations,thenequivalencedoes
not hold for a set of states with nonzero measure.
A. Unequal spin
The approachwe use is to show that there is a state such that the strict inequality
V(J(−))+V(J(+))> J(+) , (9)
x y h z i
holds under any combination of local unitary operations. From the continuity of fidelity and the expectation values,
theremustbeafiniteregionofstatesnearthisstatethatarealsonotTMSSunderanylocalunitaries. Hence,finding
a state satisfying (9) is sufficient to show that the equivalence between entanglement and two-mode spin squeezing
fails to hold for a region of states with nonzero measure.
For unequal spin, we consider the state for j = 1 and j =1:
1 2 2
1
ψ = ( 1,1 + 1,0 ). (10)
| i √2 |2 iz |−2 iz
Because [J(−),J(+)]=iJ(−), the variances for J(−) and J(+) satisfy
x y z x y
V(J(−))+V(J(+)) J(−) . (11)
x y ≥|h z i|
From the derivation of this inequality, for equality it would be necessary that V(J(−)) = V(J(+)) = J(−) /2. The
x y z
inequality V(J(−))V(J(+)) J(−) 2/4 follows from the generalised uncertainty principle, and it|his knio|wn that
x y z
equality is only possible if (J≥(−|)h λJi|(+))φ for some value of λ. If, in addition, V(J(−)) = V(J(+)), then it would
x y x y
follow that (J(−) J(+))φ = 0.−Howeve|r,ifor j = 1 and j = 1, the determinant of (J(−) J(+)) is nonzero, so
x − y | i 1 2 2 x − y
there is no state such that (J(−) J(+))φ =0. Hence it is not possible for equality to be obtained in Eq. (11).
x y
− | i
Forthespecificexampleofthestate ψ giveninEq.(10),thereduceddensitymatrixforsubsystem1istheidentity,
so J(1) = 0 for ψ and all states r|elaited to ψ by local unitaries. Thus J(−) = J(+) , and we obtain the
z z z
|h i| | i | i |h i| |h i|
strict inequality (9). Hence equivalence between TMSS and entanglement fails for a set of states whose measure is
not zero.
B. Mixed states
For the case of mixed states, we consider the example of the Werner state
1 α
ρα =α|ΦihΦ|+ (2J−+1)211, (12)
with Φ a maximally entangled state. This state is entangled for α > 1/(2J + 2) [10]. In addition, the local
| i
reduced density matrices are proportionalto the identity, and therefore J(+) must be equal to zero under any local
z
operations. In addition, because this state is not maximally entangledh, it cain not satisfy V(J(−)) = V(J(+)) = 0
x y
(see Appendix B). Thus, the strict inequality (9) is satisfied, and there is a set of nonzero measure such that the
equivalence between TMSS and entanglement fails.
C. Local rotations
We canuse a similarmethod for the casewhere the localoperationsare restrictedto rotations. For j =1,consider
the state
ψ =(1,1 + 1, 1 )/√2, (13)
z z
| i | i |− − i
4
whichsatisfies J(1,2) =0,wherek x,y,z . Therefore,itwillbethecasethat J(+) =0underanycombinationof
localrotations.hInk adidition,thisstat∈e{isnotm}aximallyentangled,andthereforechanznoitsatisfyV(J(−))=V(J(+))=
x y
0. Thus, this state must satisfy the strict inequality (9) under any combination of rotations. This means that, in
additiontothestate ψ ,thereisafiniteregionofstatesthatarenotequivalenttoTMSSstatesunderlocalrotations.
| i
V. CONCLUSIONS
Wehaveshownthatanyentangledpurebipartitestateoftwosubsystemswithequalspinisequivalenttoatwo-mode
spin-squeezed state under local unitary operations except for a set of states with measure zero. This equivalence be-
tweentwo-modespinsqueezingandentanglementestablishesaprofoundconnectionbetweencorrelationsthatexceed
the standard quantum limit, which is so important for quantum metrology, especially high-resolution spectroscopy,
and entanglement, which provides a resource for quantum information processing. Aside from the fundamental im-
portance of this equivalence principle, a practical merit of this result is that two-mode spin squeezing provides a
macroscopic signature of underlying entanglement, which will be useful in identifying the presence of entanglement
in physical systems.
Exceptions to this equivalence principle are maximally entangled states, either in the full Hilbert space or in
a restricted Hilbert space. Such states comprise a set of measure zero, so the equivalence principle holds in an
approximatesense. Wehaveshown,byprovidingcounterexamples,thatthisequivalenceprinciplecannotbeextended
to the general cases of hybrid-spin systems or mixed states.
APPENDIX A
In order to establish TMSS, we must evaluate the three expectation values in Eq. (7) for the state (2).
j j
h(Jx(1))2i= X ψm1ψm2zhm1,m1|(Jx(1))2|m2,m2iz = X ψm2 zhm|(Jx(1))2|miz
m1,m2=−j m=−j
j
= X ψm2 zhm|Jx(1)(cid:2)αm−1|m−1iz +αm|m+1iz(cid:3), (A1)
m=−j
with αm =pj(j+1) m(m+1)/2. We allow the states j 1 and j+1 , provided they are multiplied by zero,
− |− − i | i
which is the case here because α−j−1 =αj =0. This expression simplifies to
j j
1
h(Jx(1))2i= X ψm2 (α2m−1+α2m)= 2 X ψm2 [j(j+1)−m2]. (A2)
m=−j m=−j
Similarly,
j
hJx(1)Jx(2)i= X ψm1ψm2zhm1,m1|(cid:2)αm2−1|m2−1iz +αm2|m2+1iz(cid:3)(cid:2)αm2−1|m2−1iz +αm2|m2+1iz(cid:3)
m1,m2=−j
j
= X ψm1ψm2zhm1,m1|(cid:2)α2m2−1|m2−1,m2−1iz +α2m2|m2+1,m2+1iz(cid:3)
m1,m2=−j
j j−1 j−1
= X ψm−1ψmα2m−1+ X ψm+1ψmα2m =2 X ψm+1ψmα2m. (A3)
m=1−j m=−j m=−j
Lastly, evaluating J(+)/2 gives
z
h i
j j
hJz(+)/2i= X ψm1ψm2zhm1,m1|m2|m2,m2iz = X mψm2 . (A4)
m1,m2=−j m=−j
5
APPENDIX B
Consider a state that satisfies
V(J(+))=V(J(−))=0. (B1)
y x
This state must be an eigenstate of both J(+) and J(−). In addition
y x
h(Jy(+))ni=heiπ2Jx(−)(Jy(+))ne−iπ2Jx(−)i=h(Jz(−))ni, (B2)
where n=1 or 2. Therefore V(J(−))=0, so the state is also an eigenstate of J(−). Similarly we find
z z
hJx(−)i=he−iπJy(+)Jx(−)eiπJy(+)i=−hJx(−)i,
hJy(+)i=he−iπJx(−)Jy(+)eiπJx(−)i=−hJy(+)i,
hJz(−)i=he−iπJy(+)Jz(−)eiπJy(+)i=−hJz(−)i. (B3)
Thus the state must be an eigenstate with eigenvalue zero of each of J(−), J(+) and J(−). If the state is pure, it may
x y z
be written in any of the three forms
jmin jmin jmin
′ ′′
|ψi= X ψm|m,mix = X ψm|m,−miy = X ψm|m,miz. (B4)
m=−jmin m=−jmin m=−jmin
For generality we have allowed the possibility of unequal spins in the two subsystems, and j = min j ,j . Note
min 1 2
that the spins must be both integer or both half-odd integer, because otherwise J(−), J(+) and J(−) c{ould n}ot have
x y z
zero eigenvalues.
The reduceddensitymatricesfor subsystems1 and2,ρ andρ ,commute with eachofthe localspinoperatorsJ ,
1 2 x
J , and J , and therefore are multiples of the identity. This implies that the spins for the two subsystems are equal,
y z
and ψ is maximally entangled.
| i
We may show the corresponding result for a general mixed state ρ from the fact that the variance is a concave
function of the state. A general mixed state ρ can be written in the form ρ= p ψ ψ . If ρ satisfies Eq. (B1),
Pk k| kih k|
then the individual pure states ψ must satisfy Eq. (B1) also. However, as shown above the only pure state that
k
| i
satisfies Eq. (B1) is the maximally entangled state. Therefore this is the only solution if we allow mixed states also.
ACKNOWLEDGMENTS
This research has been supported by iCORE and the Australian Research Council. DWB is grateful for valuable
discussions with Robert W. Spekkens and Ernesto F. Galv˜ao.
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