quantum entanglement for systems of identical bosons. ii ... - IOPscience

QUANTUM ENTANGLEMENT FOR SYSTEMS OF IDENTICAL BOSONS. II. SPIN SQUEEZING AND OTHER ENTANGLEMENT TESTS: APPENDICES B. J. Dalton1,2 , J. Goold1,3 , B. M. Garraway4 and M. D. Reid2 1

Physics Department, University College Cork, Cork City, Ireland Centre for Quantum and Optical Science ∗ , Swinburne University of Technology, Melbourne, Victoria 3122, Australia 3 The Abdus Salam International Centre for Theoretical Physics (ICTP), Trieste 34151, Italy 4 Department of Physics and Astronomy, University of Sussex, Falmer, Brighton BN19QH, United Kingdom 2

Email: [email protected] ∗Formerly Centre for Atom Optics and Ultrafast Spectroscopy

S1

Contents Appendix A Spin Operators - Multi-Mode Case

S4

Appendix B Alternative Spin Squeezing Criteria S6 B.1 Two Perpendicular Components . . . . . . . . . . . . . . . . . . S 6 B.2 Planar Spin Squeezing . . . . . . . . . . . . . . . . . . . . . . . . S 7 B.3 Spin Squeezing in Multi-Mode Cases . . . . . . . . . . . . . . . . S 7 Appendix C Significance of Spin Squeezing Test

S9

Appendix D Spin Squeezing and Entanglement - Multi-Mode Case S 15 D.1 Multi-Mode Separable States - Three Cases . . . . . . . . . . . . S 15 D.2 Spin Squeezing Test for Bipartite System (Case 1) . . . . . . . . S 16 D.2.1 Mode Expansions . . . . . . . . . . . . . . . . . . . . . . . S 18 D.2.2 Positive Definiteness . . . . . . . . . . . . . . . . . . . . . S 19 D.3 Spin Squeezing Tests for Single Mode Sub-Systems (Case 2) . . . S 21 D.4 Spin Squeezing Tests for Two Mode Sub-Systems (Case 3) . . . . S 24 Appendix E Hillery Spin Variance - Multi-Mode S 27 E.1 Hillery Spin Variance Test for Bipartite System (Case 1) . . . . . S 27 E.2 Hillery Spin Variance Test for Single Mode Sub-Systems (Case 2) S 28 E.3 Hillery Spin Variance Test for Two Mode Sub-Systems (Case 3) . S 29 Appendix F Raymer Entanglement Test

S 30

Appendix G Derivation of Sørensen et al Results

S 32

Appendix H Revising Sørensen Entanglement Test S 37 H.1 Revision Based on Localized Modes in Position or Momentum . . S 37 H.2 Revision Based on Separable State of Single Modes . . . . . . . . S 38 H.3 Revision Based on Separable State of Pairs of Modes with One Boson Occupancy . . . . . . . . . . . . . . . . . . . . . . . . . . . S 39 Appendix I

Benatti Entanglement Tests

S 42

Appendix J Heisenberg Uncertainty Principle Results S 44 J.1 Derivation of Inequalities . . . . . . . . . . . . . . . . . . . . . . S 44 J.2 Numerical Study of Inequalities . . . . . . . . . . . . . . . . . . . S 44 Appendix K “Separable but Non-Local” States

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Appendix L Quadrature Squeezing Entanglement Tests

S 52

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Appendix M Derivation of Interferometer Results M.1 General Theory - Two Mode Interferometer . . . M.2 Beam Splitter and Phase Changer . . . . . . . . M.3 Other Measurables . . . . . . . . . . . . . . . . . M.4 Squeezing in the xy Plane . . . . . . . . . . . . . M.5 General Theory - Multi-Mode Interferometer . .

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S 54 . S 54 . S 55 . S 55 . S 56 . S 57

Appendix N Limits on Interferometry Tests

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Appendix O Relative Phase State

S 61

References

S 63

S3

Appendix A

Spin Operators - Multi-Mode Case

As well as spin operators for the simple case of two modes we can also define spin operators in multimode cases involving two sub-systems A and B. For example, there may be two types of bosonic particle involved, each component distinguished from the other by having different hyperfine internal states |Ai , |Bi. Each component may be associated with a complete orthonormal set of spatial mode functions φai (r) and φbi (r), so there will be two sets of modes |φai i, |φbi i, where in the |ri representation we have hr |φai i = φai (r) |Ai and hr |φbi i = φbi (r) |Bi. Mode orthogonality between A and B modes arises from hA|Bi = 0 rather from the spatial mode functions being orthogonal. We can define spin operators for the combined multimode A and B subsystems [43] via 1R b † (r)Ψ b a (r) + Ψ b † (r)Ψ b b (r) dr Ψ Sbx = a b 2 R 1 b † (r)Ψ b a (r) − Ψ b † (r)Ψ b b (r) Sby = dr Ψ a b 2i R 1 b † (r)Ψ b b (r) − Ψ b †a (r)Ψ b a (r) Sbz = (193) dr Ψ b 2 where the field operators satisfy the non-zero commutation rules b c (r), Ψ b † (r0 )] = δ cd δ(r − r0 ) [Ψ d

c, d = a, b

(194)

It is then easy to show that the standard spin angular momentum commutation rules are satisfied. [Sbξ , Sbµ ] = iξµλ Sbλ . For convenience we can expand the field operators in terms of an orthonormal set of spatial mode functions φi (r). We can choose the spatial mode functions to be the same φai (r) = φbi (r) = φi (r) (these might be momentum eigenfunctions) and then the field annihilation operators for each component are b a (r) = P b b b (r) = P bbi φi (r) Ψ ai φi (r) Ψ (195) i

i

These expansions are consistent with the field operator commutation rules (194) based on the usual non-zero mode operator commutation rules [b ai , b a†j ] = [bbi , bb†j ] = δ ij . By substituting for the field operators we can then express the spin operators in terms of mode operators as 1 P b† 1 P b† 1 P b†b Sbx = bi b ai + b a†i bbi Sby = bi b ai − b a†i bbi Sbz = bi bi − b a†i b ai 2 i 2i i 2 i (196) and it is then easy to confirm that the standard spin angular momentum commutation rules are satisfied. [Sbξ , Sbµ ] = iξµλ Sbλ . We now have both field and mode expressions for spin operators in multimode cases involving two sub-systems A and B. S4

Finally, the total number of particles is given by R b = b † (r)Ψ b b (r) + Ψ b †a (r)Ψ b a (r) N dr Ψ b P b†b bi bi + b = a†i b ai i

=

bb + N ba N

(197)

in an obvious notation.

S5

Appendix B B.1

Alternative Spin Squeezing Criteria

Two Perpendicular Components

Other criteria for spin squeezing are also used, for example in the article by Wineland et al [17]. To focus on spin squeezing for Sbz compared to any orthogonal spin operators we can combine the second and third Heisenberg uncertainty principle relationships to give D E D E D E 1 D E D E | Sbx |2 + | Sby |2 ∆Sbz 2 ∆Sbx 2 + ∆Sby 2 ≥ (198) 4 Then we may define two new spin operators via Sb⊥ 1 = cos θ Sbx + sin θ Sby

Sb⊥ 2 = − sin θ Sbx + cos θ Sby

(199)

where θ corresponds to a rotation angle in the xy plane, and which satisfy the b b standard angular momentum commutation rules [Sb⊥ 1 , Sb⊥D2 ] = iSbEz , [S D⊥ 2 , SzE] = iSb⊥ 1 , [Sbz , Sb⊥ 1 ] = iSb⊥ 2 . It is straightforward to show that ∆Sbx 2 + ∆Sby 2 = D D E D E E 2 D E 2 D E 2 E 2 D ∆Sb⊥ 1 2 + ∆Sb⊥ 2 2 and Sb⊥ 1 + Sb⊥ 2 = Sbx + Sby so that E 2 D E 2 D E D E D E 1 D b b 2 2 2 b b b ∆Sz ∆S⊥ 1 + ∆S⊥ 2 ≥ S⊥ 1 + S⊥ 2 4

(200)

so that spin squeezing for Sbz compared to any two orthogonal spin operators such as Sb⊥ 1 or Sb⊥ 2 would be defined as s D E 2 D D E E 2 1 b b 2 b ∆Sz < S⊥ 1 + S⊥ 2 2

D E D E ∆Sb⊥ 1 2 + ∆Sb⊥ 2 2

and s D E 2 D E 2 1 b b > S⊥ 1 + S⊥ 2 2

(201)

D E For spin squeezing in ∆Sbz 2 we require the spin squeezing parameter ξ to satisfy an inequality D E ∆Sbz 2 1 1 ξ 2 = D ∼ (202) E 2 D E 2 < s D E 2 D E 2 N b b S⊥ 1 + S⊥ 2 2 Sb⊥ 1 + Sb⊥ 2 The last step is an approximation for an N particle state based on the assumption that the Bloch vector lies in the xy plane and close to the BlochD sphere, E this situation being the most conducive to detecting the fluctuation ∆Sbz 2 . S6

s D E 2 E 2 D b is approximately N/2. The conIn this situation S⊥ 1 + Sb⊥ 2 dition N ξ 2 < 1 is sometimes taken as the condition for spin squeezing [28], but it should be noted that this is approximate and Eq. (201) gives the correct expression.

B.2

Planar Spin Squeezing

A special case of recent interest is that referred to as planar squeezing [18] in which the Bloch vector for a suitable choice of spin operators lies in a plane and along one of the axes. D E If this plane D isE chosen to be the xy plane and the b x axis is chosen then Sz = 0 and Sby = 0, resulting in only one Heisenberg uncertainty principle where the right side is non-zero, namely D relationship D ED E D E 2 ED E ∆Sby 2 ∆Sbz 2 ≥ 41 Sbx . Combining this with ∆Sbx 2 ∆Sby 2 ≥ 0 D E 2 D E D E D E gives ∆Sby 2 + ∆Sbx 2 ∆Sbz 2 ≥ 41 Sbx . So the total spin fluctuD E D E D E ation in the xy plane defined as ∆Sbpara 2 = ∆Sby 2 + ∆Sbx 2 will be squeezed compared D E D to the E spin fluctuation perpendicular to the xy plane given 2 2 b b by ∆Jperp = ∆Sz if D E 1 D E E 1 D E (203) ∆Sbpara 2 < Sbx and ∆Sbperp 2 > Sbx 2 2 D E D E D E By minimizing ∆Sbpara 2 whilst satisfying the constraints Sbz = Sby = 0 D E a spin squeezed state is found that satisfies (203) with ∆Sbpara 2 ˜ J 2/3 , D E D E ∆Sbperp 2 ∼ J 4/3 , Sbx ∼ J for large J = N/2 [18]. The Bloch vector is on the Bloch sphere. D

B.3

Spin Squeezing in Multi-Mode Cases

Since the multi-mode spin operators defined in Eq. (193) satisfy the standard angular momentum operator commutation rules, the usual Heisenberg Uncertainty rules analogous to (5) apply, so that spin squeezing can also exist in the multi-mode case as well. Thus D E ∆Sbx 2 D E ∆Sby 2

Sby 2 2

for Sbx being squeezed compared to Sby , and so on. S7

(204)

Similar alternative criteria to (201) can also be obtained, for example for Sbz being squeezed compared to any two orthogonal spin operators such as Sb⊥ 1 or Sb⊥ 2 defined similarly to (199).

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Appendix C

Significance of Spin Squeezing Test

The spin squeezing test for two mode systems was based on the general form (15) for all separable states together with the requirement that the sub-system density operators b ρA ρB R and b R were compliant with the local particle number SSR. From the point of view of a supporter for applying the local particle number SSR if the result of an experiment is that spin squeezing has occurred, the immediate conclusion is that the state is entangled. On the other hand from the point of view of a sceptic about being required to apply the local particle number SSR for the sub-system states, such a sceptic would draw different conclusions from an experiment that demonstrated spin squeezing. They would immediately point out that in this case spin squeezing is not a test for entanglement. However, as we will now see these conclusions are still of some interest. To discuss this it is convenient to divide possible mathematical forms for the density operator into categories. Considering all general two mode quantum states that are compliant with the global particle number SSR, we may first divide such quantum states into three categories, as set out in Table 1. REGION

OVERALL QUANTUM STATE

A

b ρ=

X

ρB PR b ρA R R ⊗b

SUB-SYSTEM QUANTUM STATE

CATEGORY

ρB Both b ρA R are local R and b

* Separable

R

particle number SSR compliant B

b ρ=

X


ρB Neither b ρA R is local R nor b

* Separable but

particle number SSR compliant

non-local [4]; * Entangled [2]

N/A

* Entangled

R

C

b ρ 6=

X


R

Table I. Categories of two mode quantum states. The regions referred to are shown in Figure 2. All authors would regard the quantum states in Region A as being separable and those in Region C as being entangled - it is only those in Region B where the category is disputed. Those such as [2] (local SSR supporters) who require local particle number SSR compliance for each sub-system state would classify the overall state as entangled, those who do not require this (local SSR sceptics) such as [4] would classify the overall state as separable but non-local. Note that no further subclassification is needed. In Appendix N of paper I we show that if states of the form (44) are globally SSR compliant, then both the sub-system states b ρA ρB R and b R are local particle S9

number SSR compliant in general. However, we point out that there are special matched choices for both b ρA ρB ρA R and b R along with the PR , where neither b R nor B b ρR is local particle number SSR compliant even though b ρ is global particle number SSR compliant. But the case where just one of b ρA ρB R or b R is non SSR compliant does not occur, so Region B does not need to be sub-divided along these lines.

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Figure 2. Categories of two mode quantum states that are global particle number SSR compliant. The regions A, B and C are described in Table I and represent the sets of separable, separable but non-local and entangled states respectively.

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D E D E Now let us consider quantum states for which ∆Sbx 2 ≥ 21 | Sbz | and ρ ρ D E D E 1 2 b b ∆Sy ≥ 2 | Sz |. Such states are clearly not spin squeezed. Firstly, we ρ

ρ

know that all states in Region A satisfy these inequalities. However, some states in Region B and some states in Region C may also satisfy these inequalities. In Figure 3 the quantum states in Region B that satisfy these inequalities are depicted as lying in Region D, those in Region C that do so are depicted as lying in Region F. Hence, if we find that the quantum state is such that spin squeezing does occur (as in the test of (47)) we can definitely say that it does not lie in Regions A, D or F. It must therefore be located in Regions E or L. The question is - Does this determine whether the state is entangled or not according to the supporters of applying the local SSR as in the definition of entanglement used in the present paper? The answer is that it does. This is because the quantum state must be located within either of Regions B or C, since these regions include E and L respectively. In both cases it would be entangled according to the definition used here [2] (see Table 1).

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of two D Figure D E D mode D Estates satisfying inequalities E 3. Categories E quantum ∆Sbx 2 ≥ 12 | Sbz | and ∆Sby 2 ≥ 21 | Sbz | Such states are represented ρ

ρ

ρ

ρ

as being in regions are A, D and F, where A includes all the separable states, D includes some of the separable but non-local states and F includes some of the entangled states (apart from those in D). The region A ⊕ D ⊕ F includes all the unsqueezed states. Referring to Figure 2, B = D ⊕ E and C = F ⊕ L.

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However, the sceptics of applying the local SSR would draw a different conclusion from the experiment that demonstrated spin squeezing (as in the test of (47)). They would agree that the mathematics shows that a state in Region A could not demonstrate spin squeezing. Nor by assumption could states in Regions D or F. This means that the state must lie in either Region L or Region E. So from the point of view of the sceptic, either the state is entangled (if it lies in Region L) or the sub-system states in all separable states (Region E) do not comply with the local particle number SSR. The sceptic’s conclusion is clearly interesting - in the first case the quantum state is entangled, and hence may demonstrate other non-classical features, and in the second case the possibility exists of finding sub-systems in states that have the unexpected feature in non-relativistic many body physics of having coherences between states with differing particle number. If there was a second experimental test that could show that the state was not entangled, then this would demonstrate the existence of quantum states (sub-systems are themselves possible quantum systems) in which the particle number SSR breaks down. The second experiment would seem to require a test for entanglement which is necessary as well as being sufficient - the latter alone being usually the case for entanglement tests. Such criteria and measurements are a challenge, but not impossible even though we have not met this challenge in these two papers. Thus, in principle there could be a pair of experiments that give evidence of entanglement, or failure of the super-selection rule. For such investigations to be possible, the use of entanglement criteria that do invoke the local super-selection rules is also required. Such tests are the focus of these two papers, though here our primary reason is because we consider applying the local particle number SSR is required by the physics of non-relativistic quantum many body systems involving massive particles.

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Appendix D

Spin Squeezing and Entanglement - Multi-Mode Case

In this Appendix we discuss three cases of separable states for multimode systems and then examine whether spin squeezing tests demonstrate entanglement for each of these cases.

D.1

Multi-Mode Separable States - Three Cases

As we have seen the multi-mode case involves a set of n modes with annihilation operators b ai for bosons with hyperfine component A, and another set of n modes with annihilation operators bbi for bosons with hyperfine component B. Since entanglement implies a clear choice of what sub-systems are to be entangled, there are numerous choices possible here for the present multi-mode case. Case 1 involves two sub-systems, one consisting of all the b ai modes as sub-system A and the other consisting of all the bbi modes as sub-system B. Case 2 involves 2n sub-systems, the Ai th containing the mode b ai and the Bi th containing the mode bbi . Case 3 involves n sub-systems, the ith containing the two modes b ai and bbi . These three cases relate to entanglement causing interactions in differing circumstances. Case 1 might apply to cases where separable states can b be created with all the b ai modes coupled together to produce states b ρA R and the bi B modes coupled together to produce states b ρR . Case 2 might apply to cases where separable states can be created with the b ai and all the bbi modes independent of a(i) b(i) each together to produce states b ρR ⊗ b ρR . Case 3 might apply to cases where separable states can be created with the b ai and the matching bbi modes coupled ab(i) together to produce states b ρR . Cases 2 and 3 will be discussed further in SubSection 4.4 dealing with the entanglement test introduced by Sørensen et al [14]. The density operators for separable states in the three cases will be of the form X b ρsep = PR b ρR , R

b ρR = b ρA ρB R ⊗b R b ρR = b ρR =

a(1) a(i) a(n) b(1) b ρR ⊗ .. ⊗ b ρR .. ⊗ b ρR ⊗ b ρR ⊗ .. ab(1) ab(2) ab(i) ab(n) b ρR ⊗ b ρR ⊗ .. ⊗ b ρR .. ⊗ b ρR

b(n) ⊗b ρR

Case 1

(205)

Case 2

(206)

Case 3

(207) (208)

Discussion of whether there is a spin squeezing test for Case 1 in the multimode case involves a generalization of the theory set out inD SubSection 3.1. E D E b b There is a Bloch vector entanglement test, in that if either of S x or S y is non-zero, then the state is entangled. We also find that spin squeezing in any spin component requires the state to be entangled, thus generalizing the spin

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squeezing test to the multi-mode case, for two sub-systems consisting of all the modes b ai and all the modes bbi The details are covered in SubSection D.2. For Case 2 a spin squeezing test for entanglement also be obtained. The test is again that spin squeezing in any spin component Sbx , Sby or Sbz confirms entanglement of the 2n sub-systems consisting of single modes b ai and bbi . Furthermore, D E Dthere E is also a Bloch vector entanglement test, in that if either of Sb x or Sb y is non-zero, then the state is entangled. As these systems can have quantum states with large numbers N of bosonic particles, it can be said that entanglement in an N particle system has occurred if spin squeezing is found. The proof of these tests is set out in SubSection D.3. For Case 3 there is also a spin squeezing test for entanglement, but it is restricted. Here the test is that spin squeezing in Sbz confirms entanglement of the n sub-systems consisting of pairs of modes b ai and bbi , but the test is restricted to the situation where exactly one boson occupies each mode pair. No spin squeezing test was found for the other spin operators, nor was a Bloch vector entanglement test obtained. The proof of this result is set out in SubSection D.4. That no general spin squeezing test for entanglement exists can be shown by a counter-example. If all the N bosons occupied one mode pair b ai and bbi , ab(i) and the quantum state b ρR for this pair corresponded to the relative phase eigenstate with phase θp = 0 (see SubSection 3.7) then although the D overall E b b state is separable, spin squeezing in Sy compared to Sz occurs (with ∆Sby2 = D E D E 1 1 π2 π 1 2 b b2 4 + 8 ln N , ∆Sz = 6 − 64 N and S x = N 8 . Thus there is a situation where a non-entangled state for sub-systems consisting of mode pairs is spin squeezed, so spin squeezing does not always confirm entanglement. As in the previous two mode cases, having established in multi-mode cases that spin squeezing requires entanglement a further question then is: Does entanglement automatically lead to spin squeezing? The answer is no, since cases where the quantum state is entangled but not spin squeezed can be found - an example is given in the previous paragraph. Thus in general, spin squeezing and entanglement are not equivalent - they do not occur together for all states. Spin squeezing is a sufficient condition for entanglement, it is not a necessary condition.

D.2

Spin Squeezing Test for Bipartite System (Case 1)

We now consider spin squeezing for the multi-mode spin operators given in Eqs. (193) and (196) in Appendix A. We consider separable states for Case 1, the density operator being given in Eq. (205). In this bipartite case the two subsystems consist of all modes b a and all modes bbi . The development involves exD EC i D EC b c (r) b c (r)b b † (r) b † (r)b pressions such as Ψ = T rC (Ψ ρC ), Ψ = T rC (Ψ ρC R c c R ) and R R D EC D EC b †c (r)Ψ b †c (r0 ) b †c (r)Ψ b †c (r0 )b b b 0 b c (r)Ψ b c (r0 )b Ψ = T rC (Ψ ρC = T rC (Ψ ρC R ), Ψc (r)Ψc (r ) R ), R R D EC b †c (r)Ψ b c (r0 ) b †c (r)Ψ b c (r0 )b Ψ = T rC (Ψ ρC R ), where C = A, B. R

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Firstly, we have D EB EA D EA D D E EB D 1R † † b b b b b Ψb (r) =0 Ψa (r) + Ψa (r) Sx = dr Ψb (r) 2 R R R R R

(209)

since from the local particle number SSR for sub-systems A and B we have D EB D EA D E b † (r) b a (r) Ψ = Ψ = 0. A similar result applies to Sb y so it then b R R R follows that D E D E Sb x = Sb y = 0 (210) This immediately yields the Bloch vector entanglement test. It also leads to the spin squeezing in Sb z entanglement test, namely if Sb z is squeezed with respect to Sb x or Sb y (or vice versa), then the state must be entangled. The question then is: Does spin squeezing in Sb x with respect to Sb y (or vice versa) require the state to be entangled for the two n mode sub-systems A and B? To obtain an inequality for the variance in Sb x , we see that D E D EB D EA 1 RR b † (r)Ψ b † (r0 ) b a (r)Ψ b a (r0 ) Sb 2x = dr dr0 × { Ψ Ψ b b 4 R R R D EB D EA D EB D EA † b (r)Ψ b b (r0 ) b a (r)Ψ b † (r0 ) + Ψ b b (r)Ψ b † (r0 ) b † (r)Ψ b a (r0 ) + Ψ Ψ Ψ a a b b R R R R D EB D EA b b (r)Ψ b b (r0 ) b †a (r)Ψ b †a (r0 ) } + Ψ Ψ (211) R

R

D EB b † (r)Ψ b † (r0 ) From the local particle number SSR for sub-systems A and B we have Ψ = b b R EA D b a (r)Ψ b a (r0 ) Ψ = 0, so the first and fourth terms are zero. Using the field opR erator commutation rules we then obtain EA EB D D D E 1 RR b †a (r0 )Ψ b a (r) b † (r)Ψ b b (r0 ) Ψ dr dr0 Ψ Sb 2x = b 2 R R R D EB D EA 1R † b (r)Ψ b b (r) b † (r)Ψ b a (r) } + dr { Ψ + Ψ (212) a b 4 R R so that D E ∆Sb 2x

=

R

D EB D EA 1 RR b † (r)Ψ b b (r0 ) b †a (r0 )Ψ b a (r) dr dr0 Ψ Ψ b 2 R R D EB D EA R 1 † † b (r)Ψ b b (r) b a (r)Ψ b a (r) } + dr { Ψ + Ψ b 4 R R

(213)

Hence from (20) D E ∆Sb 2x ≥

D EB D EA 1 RR b † (r)Ψ b b (r0 ) b †a (r0 )Ψ b a (r) dr dr0 Ψ Ψ b 2 R R R D E D E B A R 1 b † (r)Ψ b a (r) } b † (r)Ψ b b (r) + dr { Ψ + Ψ a b 4 R R P

PR {

S 17

(214)

E D The same result applies to ∆Sb 2y . Now we can easily show that D E P D EB D EA 1R b † (r)Ψ b b (r) b †a (r)Ψ b a (r) } Sb z = PR dr { Ψ − Ψ b 2 R R R

(215)

so that EA D EB D P 1R 1 Db E b †a (r)Ψ b a (r) } b † (r)Ψ b b (r) (216) + Ψ dr { Ψ | S z | ≤ PR b 2 4 R R R D EA D EB b †a (r)Ψ b a (r) b † (r)Ψ b b (r) are real and positive. and Ψ as Ψ b R R Hence we find that D E 1 D E ∆Sb 2x − | Sb z | 2 D EB D EA P 1 RR b † (r)Ψ b b (r0 ) b † (r0 )Ψ b a (r) ≥ PR dr dr0 Ψ Ψ (217) a b 2 R R R P 1 RR b b (r0 ) b b† b b† 0 dr dr0 T rB {Ψ ρB ρA = PR R Ψb (r)} T rA {Ψa (r) b R Ψa (r )} (218) 2 R n o 1 RR b a (r) Ψ b b (r0 ) b b † (r0 ) Ψ b † (r) = dr dr0 T r Ψ ρsep Ψ (219) a b 2 E D D E giving three forms that the inequality for ∆Sb 2x − 21 | Sb z | has to satisfy in the case of a separable state. The last form involves a double space integral of a quantum correlation function. Note the order of r and r0 . It is straightforward to show that the right side of the inequality is real, but to achieve an entanglement test involving spin squeezing for Sb x we D needE to show D that E it is non-negative. 1 2 b b Identical inequalities can be found for ∆S y − 2 | S z |. D.2.1

Mode Expansions

If we use Eq. (195) to expand the field operators then using Eq. (218) we have D E 1 D E ∆Sb 2x − | Sb z | 2 P 1 P P RR ≥ PR dr dr0 2 ij kl R × φi (r)φ∗j (r0 )φk (r0 )φ∗l (r) n o b† × T rA {b ai b ρA a†j } T rB {bbk b ρB R b R bl } o P 1 Pn b† = PR T rA {b ai b ρA a†j } T rB {bbj b ρB R b R bi } 2 ij R P 1P R R R = PR (Aij Bji + Bij AR (220) ji ) 4 ij R P 1 = PR T r{AR B R + B R AR } (221) 4 R S 18

where mode orthogonality has been used and we have introduced matrices AR and B R whose elements are AR ai b ρA a†j } ij = T rA {b R b

R b† Bji = T rB {bbj b ρB R bi }

(222)

R ∗ R R ∗ It is easy to show that AR the matrices ij = (Aji ) and Bij = (Bji ) showing thatP R R R R R R R A and B are Hermitian, as is A B + B A . The quantity (AR ij Bji + ij

R Bij AR ji ) is real. The question is: Is it also positive ? For the simple case where there is only one spatialn mode for each component o P b† ab ρA a† } T rB {bb b ρB the right side of the inequality is just equal to PR 21 T rA {b R b R b } = R P PR 21 NRA NRB , where NRA and NRB give the mean numbers of bosons in subR

systems A and B for the states b ρA ρB R and b R . The right side of the inequality is positive, showing that the separable state is not spin squeezed for Sb x with respect to Sb y (or vice versa), leading as before to the test that such spin squeezing requires entanglement. D.2.2

Positive Definiteness

For the multi-mode case we now take into account that the sub-system density AR operators b ρA ρB and π BR are R and b R are positive-definite. Their eigenvalues π λ µ real and non-negative as well as summing to unity, and we can write the density operators in terms of their orthonormal eigenvectors |AR, λi and |BR, µi as P BR P AR π µ |BR, µi hBR, µ| (223) π λ |AR, λi hAR, λ| b ρB b ρA R = R = µ

λ

Then from (222) X AR π AR a†j b ai |AR, λi ij = λ hAR, λ| b

R Bji =

X

b†b π BR µ hBR, µ| bi bj |BR, µi

µ

λ

(224) Consider a 1 × n row matrix ξ † = {ξ ∗1 , ξ ∗2 , .., ξ ∗n } X ξ † AR ξ = ξ ∗i AR ij ξ j ij

=

X

=

X

π AR λ

X

ξ ∗i hAR, λ| b a†j b ai |AR, λi ξ j

ij

λ

b† Ω b π AR hAR, λ| Ω λ A A |AR, λi

(225)

λ

b A = P ξ ∗i b where we have introduced the operator Ω ai . Since ξ † AR ξ is always i R non-negative for all ξ, this shows that A is a positive definite matrix. Similarly, considering a 1×n row matrix η † = {η ∗1 , η ∗2 , .., η ∗n } and introducing the operator b B = P η i bbi we find that Ω i X b† Ω b η† B R η = π BR hBR, µ| Ω (226) µ B B |BR, µi µ

S 19

which is also always non-negative, showing that B R is also a positive definite matrix. We can then express the positive definite Hermitian matrices AR and B R B in terms of their normalized column eigenvectors θA α and ζ β respectively, where the corresponding real, positive eigenvalues are ν α and σ β Thus we have (for ease of notation R will be left understood) X A † † A AR θ A = ν α θA (θA AR = ν α θA α α α ) θ γ = δ αγ α (θ α ) α

B

R

ζB β

=

† B (ζ B β ) ζ

σβ ζ B β

= δ β

B

R

=

X

B † σβ ζ B (227) β (ζ β )

β

Then T r{AR B R + B R AR } XX A † B B † = T r{ ν α σβ θA α (θ α ) ζ β (ζ β ) } α

β

XX B † A A † +T r{ ν α σβ ζ B β (ζ β ) θ α (θ α ) } α

=

XX α

+ 2

B † A † B ν α σ β [(θA α ) ζ β ] [(ζ β ) θ α ]

β

XX α

=

β

β

XX α

A † B † A ν α σ β [(ζ B β ) θ α ] [(θ α ) ζ β ]

† B 2 ν α σ β | [(θA α ) ζβ ] |

(228)

β

Hence we have using (221) D E 1 D E ∆Sb 2x − | Sb z | 2 P 1 XX † B 2 ≥ PR ν α σ β | [(θA α ) ζβ ] | 2 R α

(229)

β

where of the inequality is non-negative. The same result applies D theE right DsideE 1 b 2 b to ∆S y − 2 S z . Thus separable states are not spin squeezed in Sb x or in Sb y . Thus we have established the spin squeezing test for the multi–mode Case 1 - states that are spin squeezed in Sb x compared to Sb y (or vice versa) must be entangled states for the two subsystems consisting of all modes b ai and all modes bbi . D E For the other spin components, the Bloch vector result in (210) that Sb x = D E Sb y = 0 for separable states enables us to show that if Sb z is squeezed compared to Sb x (or vice versa) or if Sb z is squeezed compared to Sb y (or vice versa) then the state must be entangled. Thus spin squeezing in any spin component requires the state to be entangled, just as for the two mode case. S 20

D.3

Spin Squeezing Tests for Single Mode Sub-Systems (Case 2)

We now consider spin squeezing for the multi-mode spin operators given in Eqs. (193) and (196) in Appendix A. We consider separable states for Case 2, the density operator being given in Eq. (206). In this single mode sub-system case there are 2n subsystems consist of all modes b ai and all modes bbi . This case is that involved in the modified approach E et al and we Eto Sørensen D D will see that it leads to a useful inequality for ∆Sb 2x or ∆Sb 2y that applies when non-entangled states are those when all the separate modes b ai and bbi are the sub-systems. We will follow the approach used for the simple two mode case in Section 3. b in a mixed state Firstly, the variance for a Hermitian operator Ω X b ρ= PR b ρR (230) R

is always greater than or equal to the the average of the variances for the separate components D E X D E b2 ≥ b2 ∆Ω PR ∆ Ω (231) R

R

D E D E D E b 2 = T r(b b 2 ) with ∆Ω b = Ω− b Ω b and ∆Ω b2 b R 2) where ∆Ω ρ ∆Ω = T r(b ρ R ∆Ω R D E bR = Ω b− Ω b . The proof is straight-forward and given in Ref. [19]. with ∆Ω R D E D E D E D E D E Next we calculate ∆Sb 2x , ∆Sb 2y and Sbx , Sby , Sbz for the R R R R R case where X b1 (232) ρbR3 ⊗ .... ρaR3 ⊗ b ρbR2 ⊗ b ⊗ b ρaR2 ⊗ b ρR b ρ= PR b ρaR1 ⊗ b R

as is required for a general non-entangled state all 2n modes. This situation is that of Case 2 for the sub-systems, as described in SubSection D.1. As the density operators for the individual modes must represent possible physical states for such modes, so the super-selection rule for atom number applies and we have D E h(b ai )p ia i = T r(b ρaRi (b ai )p ) = 0 (b a†i )p = T r(b ρaRi (b a†i )p ) = 0 ai D E D E (bbi )m = T r(b ρbRi (bbi )m ) = 0 (bb†i )m = T r(b ρbRi (bb†i )m ) = 0 bi

bi

(233)

S 21

The Schwinger spin operators are X † X Sbx = (bbi b ai + b a†i bbi )/2 = Sbxi i

Sby

i

=

X

=

X

ai − b a†i bbi )/2i = (bb†i b

X i

i

Sbz

Sbyi

(bb†i bbi − b a†i b ai )/2 =

X

i

Sbzi

(234)

i

where b ai , bbi and b a†i , bb†i respectively are mode annihilation, creation operators. Note that this expression for the spin operators is the same as (196) for the multi-mode case treated in Appendix A. From Eqs. (234) we find that P P bi bj Sb 2x = (Sbxi )2 + Sx Sx i

(235)

i6=j

so that on taking the trace with b ρR and using Eqs. (232) we get after applying † b the commutation rules [b e, eb ] = 1 (b e=b a or bb) D E P D bi 2 E P D bi E D bj E Sb 2x = (Sx ) + Sx Sx (236) R

As we also have D E P D bi E Sb x = Sx R

R

i6=j

R

D E2 P D bi E2 P D bi E D bj E Sb x = Sx + Sx Sx

R

i

R

i

R

R

i

R

i6=j

(237) R

E D is using Eqs. (232) and we see finally that the variance ∆Sb 2x R

D

∆Sb 2x

E

= R

PD

(Sbxi )2

i

E R

P D bi E2 − Sx i

(238)

R

all the terms with i 6= j cancelling out. and therefore from Eq. (231) D E X D E2 P D bi 2 E 2 b ∆S x ≥ PR (Sx ) − Sbxi R

R

i

(239)

R

D E An analogous result applies for ∆Sb 2y . But using (233) (Sbxi )2 D E (Sbxi )2 R

and

= =

1 b† b† (b b ai b b ai + bb†i b ai b a†i bbi + b a†i bbibb†i b ai + b a†i bbi b a†i bbi ) 4 i i

† 1 D b†b E 1 † D b†b E ( (b b)i + (b a b a)i R ) + ( (b a b a)i R (b b)i ) (240) 4 2 R R D E Sbxi =0 R

S 22

(241)

It then follows that D E X D E Sb x = PR Sb x =0

D

R

R

E X D E Sb y = PR Sb y =0 R

(242)

R

so that D E X

† P 1 D b†b E 1 † D b†b E 2 b ∆S x ≥ PR ( (b b)i a b + (b a b a)i R ) + ( (b a)i R (b b)i ) 4 2 R R i R (243) D E 2 b The same result applies for ∆S y . Now using (233) D

D

Sb z

E

Sbzi

=

E

= R

X

† 1 D b†b E ( (b b)i − (b a b a)i R )) 2 R PR

R

1 D b E Sz 2

(244)

P D bi E Sz

R

i

P 1 D † E

† b b = PR ( (b b)i − (b a b a)i R )) 2 2 R i R X

† 1 P D b†b E ≤ PR − (b a b a)i R )) ( (b b)i 4 i R R X

† 1 P D b†b E ≤ ( (b b)i PR + (b a b a)i R )) 4 i R 1X

(245)

R

and thus D E 1 D E ∆Sb 2x − Sb z 2 X

† P 1 D b†b E 1 † D b†b E ≥ PR ( (b b)i + (b a b a)i R ) + ( (b a b a)i R (b b)i ) 4 2 R R i R D E X

1 P b†b − ( (b b)i PR + (b a† b a)i R )) 4 i R R X 1 P † D b†b E ( (b a b a)i R (b b)i ) = PR 2 i R R

≥ 0

(246) D E D E A similar proof shows that ∆Sb 2y − 21 Sb z ≥ 0 for the non-entangled state of all 2n modes. b This shows that for the general non-entangled state with all modes ai and D b E bi as the sub-systems, the variances for two of the spin fluctuations ∆Sb 2x and D E D E ∆Sb 2y are both greater than 21 Sb z , and hence there is no spin squeezing S 23

s D E D E 2 E 2 D b for Sbx or Sby . Note that as Sb y = 0, the quantity S⊥ 1 + Sb⊥ 2 D E is the same as Sb z , so the alternative criterion in Eq. (201) is the same as that in Eq. (6) which is used here. Hence we have shown that for a non-entangled physical state for all the 2n modes b ai and bbi E 1 D E D E 1 D E D ∆Sb 2x ≥ | Sb z | and ∆Sb 2y ≥ | Sb z | (247) 2 2 b b requires entanglement. so that spin squeezing in either D S Ex orDSy E From (242) we see that Sb x = Sb y = 0 for the general separable state, D E showing there is a Bloch vector test for entanglement such that if either Sb x D E or Sb y is non-zero, then the state must be entangled. Finally, if there is spin squeezing in Sbz with respect to Sbx or vice versa, D or E spin squeezing in Sbz with respect to Sby or vice versa, it follows that one of Sb x D E or Sb y is non-zero. But as both these quantities are zero for a non-entangled state, if follows that spin squeezing in Sbz also requires entanglement. Thus, spin squeezing in any spin operator Sbx , Sby or Sbz is a sufficiency test for entanglement of all the separate mode sub-systems.

D.4

Spin Squeezing Tests for Two Mode Sub-Systems (Case 3)

We now consider spin squeezing for the multi-mode spin operators given in Eqs. (193) and (196) in Appendix A. We consider separable states for Case 3, the density operator being given in Eq. (207). In this mode pair sub-system case there are n subsystems consist of all pairs of modes b ai and bbi . This case is also involved D in aEmodified approach to Sørensen et al and we show a useful inequality for ∆Sb 2z applies when non-entangled states are those when the pairs of modes b ai and bbi are the separate sub-systems, but only in restricted situations. The pairs of modes corresponding to localized modes on different lattice sites or pairs of modes with the same momenta does represent the closest way of simulating the approach used by Sørensen et al where identical particles i were regarded as the sub-systems. Now the general non-entangled state will be X b ρ= PR b ρ1R ⊗ b ρ2R ⊗ b ρ3R ⊗ ... (248) R

where the b ρiR are now the density operators for sub-system i consisting of the pair of modes b ai and bbi (which are of the form given in Eq. (303)) and the conditions S 24

in Eq. (233) no longer apply. The Fock states are of the form |Nia i⊗|Nib i for the pair of modes b ai and bbi , and for this Fock state the total occupancy of the pair of modes is Ni = Nia + Nib . From the super-selection rule the density operator b ρiR for the ith pair of modes b ai and bbi is diagonal in the total occupancy. For Ni = 0 there is one non zero matrix element (h0|ia ⊗ h0|ib ) b ρiR (|0iia ⊗ |0iib ). For Ni = 1 there are four non zero matrix elements, which may be written (h1|ia ⊗ h0|ib ) b ρiR (|1iia ⊗ |0iib )

=

ρiaa

(h1|ia ⊗ h0|ib ) b ρiR (|0iia ⊗ |1iib )

=

ρiab


= ρiba


= ρibb

(249)

For Ni = 2 there are nine non zero matrix element (h2|ia ⊗h0|ib ) b ρiR (|2iia ⊗|0iib ), ..., (h0|ia ⊗ h2|ib ) b ρiR (|0iia ⊗ |2iib ) and the number increases with Ni . If we restrict ourselves to general entangled states for one particle, where Ni = 1 for all pairs of modes, then the density operator b ρiR is of then form b ρiR

= ρiaa (|1iia h1|ia ⊗ |0iib h0|ib ) + ρiab (|1iia h0|ia ⊗ |0iib h1|ib ) +ρiba (|0iia h1|ia ⊗ |1iib h0|ib ) + ρibb (|0iia h0|ia ⊗ |1iib h1|ib ) (250)

In addition Hermitiancy, positivity, unit trace T r(b ρiR ) = 1 and T r(b ρiR )2 ≤ 1 can be used as in Eq (295) to parameterize the matrix elements in (249). ρiaa

=

ρiab

=

sin2 αi ρibb = cos2 αi q sin2 αi cos2 αi sin2 β i exp(+iφi )

ρiba =

q

sin2 αi cos2 αi sin2 β i exp(−iφi ) (251)

The expectation values for the spin operators Sbxi , Sbyi and Sbzi associated with the ith pair of modes are then D E 1 Sbxi = T r(b ρiR (bb†i b ai + b a†i bbi ) 2 R 1 i ρab + ρiba = 2 D E 1 i i b ρab − ρiba Sy = 2i R D E 1 i i b Sz = ρbb − ρiaa (252) 2 R which are of exactly the same form as in Eq. (294) as in the Appendix G derivation of the original Sørensen et al [14] results based on treating identical particles as the sub-systems. The proof however is now different and rests on restricting the states b ρiR to each containing exactly one boson. The remainder of the proof is exactly the same as in Appendix G and we find that D E 1 D b E2 D b E2 2 b ∆S z ≥ Sx + Sy (253) N S 25

for non-entangled pairs of modes b ai and bbi . Thus when the interpretation is changed so that are the separate sub-systems are these pairs of modes, it follows that spin squeezing in Sb z with respect to Sb x or Sb y requires entanglement of all the mode pairs, but only if there is one particle in each mode pair. In general, spin squeezing in either Sb x or Sb y is not linked to entanglement for Case 3 sub-systems, as has been pointed out in SubSection D.1 by a counter-example involving the relative phase state. Also there is no Bloch vector entanglement test. For we have in general D

E Sbxi R D E Sbyi R

= =

1 ai + b a†i bbi ) T r(b ρiR (bb†i b 2 1 ai − b a†i bbi ) T r(b ρiR (bb†i b 2i

(254)

and the local particle number SSR does not require these quantities to D beE zero for sub-systems consisting of pairs of modes b ai and bbi . Thus in general Sbx and D E Sby can be non-zero for a separable state, so the Bloch vector entanglement test does not apply.

S 26

Appendix E

Hillery Spin Variance - Multi-Mode

It turns out that the Hillery spin variance test can also be applied in multi-mode situations, where the spin operators are defined as in Appendix A. As explained in Appendix D three cases occur in regard to specifying the sub-systems. For Case 1, where there are two sub-systems each consisting of all the modes b ai or all the modes bbi . the Hillery spin variance test as in (84) applies. The proof set out below again does not require the sub-system density operators to be local SSR compliant. Also, for Case 2 where there are 2n subsystems consisting of all modes b ai and all modes bbi the Hillery spin variance test as in (84) applies. The proof is set out below again does not require the sub-system density operators to be local SSR compliant. However, for Case 3 where there are n sub-systems consisting of all mode pairs b ai and bbi the Hillery spin variance test does not ab(i) apply. Basically, this is because specific sub-system density operators b ρR (see (207)) could be entangledDstates ai and bbi all of which do satisfy E of the modes b b the Hillery test involving Ni for this ith sub-system. If we choose a special R

separable state of the form (207) with just one term (no sum over R), it is easy to see that the Hillery test will be satisfied for the full system. However, the full system state involving these sub-systems is still a separable state, showing that satisfying the Hillery spin variance test does not always require the state to be entangled.

E.1

Hillery Spin Variance Test for Bipartite System (Case 1)

We first consider Case 1 where there are two sub-systems each consisting of all the modes b ai or all the modes bbi . We use the results from (214) to find that for a separable state D E D E ∆Sb 2x + ∆Sb 2y EA EB D D RR P b †a (r0 )Ψ b a (r) b † (r)Ψ b b (r0 ) ≥ PR { dr dr0 Ψ Ψ b R

R

D EB D EA 1R b † (r)Ψ b b (r) b †a (r)Ψ b a (r) } dr { Ψ + Ψ + b 2 R R

R

(255)

The same result would have occurred if the local sub-system SSR had been EA D EB D RR b † (r)Ψ b † (r0 ) b a (r)Ψ b a (r0 ) disregarded, the terms such as 41 dr dr0 × { Ψ Ψ b b R R cancelling out. The mean number of bosons is obtained from (197) and hence D EB D EA R 1 D bE 1 X b † (r)Ψ b b (r) b †a (r)Ψ b a (r) N = PR dr Ψ + Ψ (256) b 2 2 R R R

S 27

Thus we have

≥

D E D E 1D E b ∆Sb 2x + ∆Sb 2y − N 2 D EB D EA RR P b † (r)Ψ b b (r0 ) b †a (r0 )Ψ b a (r) PR dr dr0 Ψ Ψ b R

R

(257)

R

Using the mode expansion (195) we then get E 1D E E D D b N ∆Sb 2x + ∆Sb 2y − 2 D EB D EA X X RR P ≥ PR dr dr0 φ∗i (r)φj (r0 )φ∗k (r0 )φl (r) bb†i bbj b a†k b al =

R

ij

P

X

R

=

P R

PR

R

kl

b† T rA {b ai b ρA a†j }T rB {bbj b ρB R b R bi }

R

(258)

ij

1 PR T r(AR B R + B R AR ) 2

(259)

R after orthogonality is used and the matrix elements AR ij and Bji are introduced from (222). Since we have shown in Appendix D.2 that the right side of the last inequality is always non-negative, the Hillery spin variance entanglement test follows that if D E D E 1D E b ∆Sb 2x + ∆Sb 2y < N (260) 2 then the quantum state must be an entangled state for the case of two subsystems each consisting of all the modes b ai or all the modes bbi .

E.2

Hillery Spin Variance Test for Single Mode Sub-Systems (Case 2)

We now consider separable states for Case 2, the density operator being given in Eq. (206). In this single mode sub-system case there are 2n subsystems consisting of all modes b ai and all modes bbi . We use the results from (243) to find that for a separable state D E D E ∆Sb 2x + ∆Sb 2y X

†

† D † E P 1 D b†b E b b ≥ PR ( (b b)i + (b a b a)i R ) + ( (b a b a)i R (b b)i ) (261) 2 R R i R

The same result would have occurred D E if the local sub-system SSR had been 1 b†b† disregarded, the terms such as 4 bi bi hb ai b ai iR cancelling out. Ri

The mean number of bosons is obtained from (197) E

† PD 1 D bE 1 X N = PR ( (bb†bb)i + (b a b a)i R ) 2 2 R i R

S 28

(262)

Thus we have D E D E 1D E b ∆Sb 2x + ∆Sb 2y − N 2 X P † D b†b E ≥ PR (b a b a)i R (b b)i

(263)

R

i

R

which is always non-negative. The Hillery spin variance entanglement test follows that if the inequality in (260) occurs then the quantum state must be an entangled state for the case of 2n sub-systems consisting of all the modes b ai and all the modes bbi .

E.3

Hillery Spin Variance Test for Two Mode Sub-Systems (Case 3)

We now consider separable states for Case 3, the density operators being given in Eq. (207). In this two mode sub-system case there are n subsystems consisting of all mode pairs b ai and bbi . We consider a special separable state with just one term where b ρsep = b ρab(1) ⊗ b ρab(2) ⊗ .. ⊗ b ρab(i) .. ⊗ b ρab(n) (264) We use the results from (238) to find that D E D E ∆Sb 2x + ∆Sb 2y P D bi 2 E D bi 2 E = (∆Sx ) + (∆Sy )

(265)

i

D E where ∆Sbαi = Sbαi − Sbαi for α = x, y. This result did not depend on applying R the local SSR. Now suppose each of the two mode states b ρab(i) is an entangled state of the modes b ai and bbi in which the Hillery spin variance test is satisfied. Then D E D E 1 ni i (∆Sbxi )2 + (∆Sbyi )2 < hb 2 Hence

< =

D E D E ∆Sb 2x + ∆Sb 2y P1DbE Ni i 2 1 D bE N 2

(266)

b†b a† b b =P N b b where N i ai i i is the total number operator and Ni = bi bi + b Thus the Hillery spin variance test is satisfied even though the state (264) is separable, shoeing that the test cannot be applied for multi-mode Case 3. S 29

Appendix F

Raymer Entanglement Test

In this Appendix the proof of the Raymer entanglement test in SubSection 4.3 is presented. b A, Λ b A and Ω bB, Λ b B for the two sub-systems we With Hermitian operators Ω consider b = αΩ b A + βΩ bB b A − βΛ bB U Vb = αΛ (267) P A B where α, β are real. Then with b ρ = R PR b ρR and b ρR = b ρR ⊗ b ρR and using (20) it can first be shown that E E X D E D E X D D b2 b2 ≥ (268) PR ∆VbR2 ∆Vb 2 ≥ PR ∆ U ∆U R R

R

D E D E D E D E bR = U b− U b , ∆VbR = Vb − Vb b bb where ∆U with U = T r(U ρR ), Vb = R

R

R

R

T r(Vb b ρR ). b and Vb from (267) and using b ρB Substituting for U ρR = b ρA R we can then R ⊗b evaluate the various terms as follows. D ER D ER D E D ER D ER b2 b 2A b 2B bA bB U = α2 Ω + β2 Ω Ω + 2αβ Ω R A B A B D E D ER D ER b bA bB U = α Ω +β Ω R

D E 2 b U R

D E bR2 ∆U

A

B

D E 2 D E 2 D ER D ER R R bA bB bA bB Ω Ω Ω = α2 + β2 + 2αβ Ω A B A B ! ! D ER D ER 2 D ER D ER 2 2 2 2 2 b b b b = α ΩA ΩA − +β ΩB ΩB − A

A

B

B

(269) D E D ER b2 = with a similar result for ∆VbR2 . Here for sub-system A we define Ω A A D ER D ER D ER A A b2 b b bA b b b2 b2 b bA b = T r(Λ ρA = T r(Ω ρA = T r(Λ T r(Ω R) R ) and ΛA A ρR ), ΛA A ρR ), ΩA A A A with analogous expressions for sub-system B. We thus have D E D ER D ER X X b2 b 2AR b 2BR ∆U ≥ α2 PR ∆ Ω + β2 PR ∆ Ω A

R

D E ∆Vb 2

≥ α2

X

D

b2 PR ∆ Λ AR

R

ER A

R

+ β2

X R

B

D ER b2 PR ∆ Λ BR

(270)

B

D ER D ER D ER b AR = Ω bA − Ω b A , ∆Ω b BR = Ω bB − Ω b B , ∆Λ b AR = Λ bA − Λ bA where ∆Ω B A D ER A b b b and ∆ΛBR = ΛB − ΛB . B

S 30

Adding the two results gives E E D D b 2 + ∆Vb 2 ∆U D ER D ER X 2 2 2 b b ≥ α + ∆ΛAR PR ∆ΩAR A

R

+β 2

X

PR

D

b 2BR ∆Ω

R

A

ER B

D ER b 2BR + ∆Λ

(271)

B

a general variance inequality for separable states. This last result can be developed further based on the commutation rules b A, Λ b A ] = iΘ bA [Ω

bB, Λ b B ] = iΘ bB [Ω

(272)

The Schwarz inequalities - valid for all real λA and λB D ER b AR − iλA ∆Λ b AR ) b b b (∆Ω ρA R (∆ΩAR + iλA ∆ΛAR ) A ER D b b b BR − iλB ∆Λ b BR ) b (∆Ω ρB R (∆ΩBR + iλB ∆ΛBR )

≥

0

≥

0

(273)

B

lead to the following inequalities D ER D ER D ER b2 bA b2 ∆Ω + λA Θ + λ2A ∆Λ AR AR A A A D ER D ER D ER b 2BR bB b 2BR ∆Ω + λB Θ + λ2B ∆Λ B

B

0

≥

0

(274)

B

so by taking λA,B = 1 or −1 we have D ER D ER b 2AR b 2AR ∆Ω + ∆Λ A A D ER D ER 2 2 b b ∆Ω + ∆Λ BR BR B

≥

B

D ER bA | | Θ A D ER bB ≥ | Θ |

≥

(275)

B

D ER D ER D ER b2 b2 b A |2 /4 The Heisenberg Uncertainty principle results ∆Ω ∆ Λ ≥ | Θ AR AR A

A

A

etc also follow from (274). D ER D ER D E D ER X X X bA | ≥ | bA | = | Θ b A | and bB Noting that PR | Θ PR Θ PR | Θ | ≥ A

|

X R

PR

D

A

R R ER R D E bB b B | since the modulus of a sum is never greater than Θ | =| Θ A

the sum of the moduli, we finally arrive at the final inequality for separable states D E D E D E D E b A + βΩ b B )2 + ∆(αΛ b A − βΛ b B )2 ≥ α2 | Θ b A | + β2| Θ b B | (276) ∆(αΩ This leads to the following test for entanglement D E D E D E D E b A + βΩ b B )2 + ∆(αΛ b A − βΛ b B )2 < α2 | Θ b A | + β2| Θ b B | (277) ∆(αΩ which is usually based on choices where α2 = β 2 = 1. S 31

B

Appendix G

Derivation of Sørensen et al Results

Sørensen et al [14] derive a number of inequalities from which they deduce a further inequality for the spin squeezing parameter in the case of a non-entangled state. From this result they conclude that spin squeezing implies entanglement. The final inequality they obtain for a non-entangled state is E D 1 D b E2 D b E2 2 b ∆S z ≥ Sx + Sy (278) N Their approach is based on writing the density operator for a non-entangled state of N identical particles as in Eq. (105) X X PR b ρR (279) ρ3R ⊗ ... = ρ2R ⊗ b b ρ= PR b ρ1R ⊗ b R

R

The spin operators are defined as P bi X Sbx = Sx = (|φb (i)i hφa (i)| + |φa (i)i hφb (i)|)/2 i

i

P bi X Sy = (|φb (i)i hφa (i)| − |φa (i)i hφb (i)|)/2i

Sby

=

Sbz

P bi X = Sz = (|φb (i)i hφb (i)| − |φa (i)i hφa (i)|)/2

i

i

i

(280)

i

where the sum i is over the identical atoms and each atom is associated with two states |φa i and |φb i. Clearly, the spin operators satisfy the standard commutation rules for angular momentum operators. Sørensen et al [14] state that the variance for Sbz satisfies the result D E2 D E2 D E N P P D bi E2 P − PR Sz + PR Sbz ∆Sb 2z = − Sbz 4 R R i R R

(281)

To prove this we have D E X P P bi bj Sb 2z = PR T r(b ρR Sz Sz ) i

R

=

X

PR

j

Sbzi

2

i

R

=

P

N X + PR 4 R

P D bi E D bj E + Sz Sz R

i6=j

R

P D bi E D bj E Sz Sz i6=j

S 32

R

R

!

R

! (282)

where we have used 2 1 Sbzi = (|φ (i)i hφb (i)| − |φa (i)i hφa (i)|)2 4 b 1 = (|φ (i)i hφb (i)|φb (i)i hφb (i)| − (|φb (i)i hφb (i)|φa (i)i hφa (i)|) 4 b 1 + (−(|φa (i)i hφa (i)|φb (i)i hφb (i)| + (|φa (i)i hφa (i)|φa (i)i hφa (i)|) 4 1 = ((|φb (i)i hφb (i)| + (|φa (i)i hφa (i)|) 4 1b 1i (283) = 4 a result based on the orthogonality, normalization and completeness of the states |φa (i)i , |φb (i)i. Also D E P Sb z = T r(b ρR Sbzi ) R i P D bi E = Sz R i ! D E D E2 D E D E 2 P P P P PR Sbz = PR Sbzi + Sbzi Sbzj (284) R

R

P

so eliminating the term

R

R

i

R

P D bi E D bj E Sz Sz

PR

R

i6=j

R

i6=j

R

! gives the required expresR

E D E D E2 D sion for ∆Sb 2z = Sb 2z − Sbz . Next, Sørensen et al [14] state that D E2 D E2 X P P D bi E2 P D bi E 2 Sbx ≤ N PR Sx Sb y ≤ N PR | Sy | R

i

R

(285)

R

i

R

To prove this we have D E Sb x

=

X

=

X

≤

X

PR T r(b ρR

P bi Sx ) i

R

P D bi E PR Sx i

R

D E | Sb x |

R

P D bi E PR | Sx |

(286)

R

i

R

since the modulus of a sum is less than or equal to the sum of the moduli. Now

D

Sb x

E2

=

D E | Sb x |2 ≤

X

!2 P D bi E PR | Sx | i

R

≤

X R

PR

2 P D bi E | Sx | i

S 33

R

R

(287)

2

√

P

≤

using the general result that PR CR R D E √ P bi here CR = | Sx |. Next consider

P

PR CR , where

R

P

PR = 1 with

R

R

i

y

= N

P D bi E 2 | Sx | R

i

2 2 P D bi E P D bi E | Sx | = | Sx | R R i i D E P D bi E 2 j (| Sx | − | Sbx |) ≥ 0

z

=

y−z

=

R

i 0 applies for the case of indistinguishable particles. For the (N − 1) ∆Jbn3 D E D E − − − case where n1 = → x , n2 = → y and n3 = → z, Sbx2 + Sby2 − N/2 − (N − D E2 D E E D E D E D E D = − N/2 − (N ) Sbz2 + (N − 1) Sbz Sbx2 + Sby2 + Sbz2 1) ∆Sbz2 = D E D E2

P N (N + 2)/4 − N/2 − N N 2 /4 − R PR ( b a† b a R bb†bb ) + (N − 1) Sbz . R D E2 D E D E D E 2 2 2 b b b As Sz ≤ N /4 we see that Sx + Sy − N/2 − (N − 1) ∆Sbz2 ≤

† D † E P N R PR ( b a b a R bb bb ), which is certainly ≥ 0 and not ≤ 0 as required. R However, perhaps an entanglement test such that if is could be shown that D E D E D E X

† D † E Sbx2 + Sby2 − N/2 − (N − 1) ∆Sbz2 > N PR ( b a b a R bb bb ) (320) D

R

R

always applies then it could be included that the state is entangled. Unfortunately the right side could be too large for the left side toD always E exceed the

† right side for some separable states. Noting that b a b a R + bb†bb = N for the R N bosons states D Ebeing considered we find that the right side is maximized when

† b a b a R = bb†bb = N/2 for all PR , giving a maximum for the right side of R

N 3 /2 - and this can occur for some separable states. To show that the state is entangled the left side must exceed this value, otherwise the state might be one of the separable states. However, the left side is at most of order N 2 from the first two terms and the negative terms only make the left side smaller. Hence there is no entanglement test of the form (320).

S 43

Appendix J J.1

Heisenberg Uncertainty Principle Results

Derivation of Inequalities

Here D we derive to inequalities for the variE the results in SubSection 4.6 D leading E 2 b b ance ∆J x considered as a function of | J z | for states where the spin opD E D E erators are chosen such that Jb x = Jb y = 0. D E2 D E From the Schwarz inequality Jb z ≤ Jb 2z so that D E D E D E2 D E D E D E Jb 2x + Jb 2y + Jb z ≤ Jb 2x + Jb 2y + Jb 2z = J(J + 1)

(321)

D E2 D E2 giving Eq. (113). Subtracting Jb x = Jb y = 0 from each side gives D E D E D E2 ∆Jb 2x + ∆Jb 2y + Jb z ≤ J(J + 1) (322) E D Substituting for ∆Jb 2y from the Heisenberg uncertainty principle result in Eq. (114) gives D D E2 D E2 E 1 D E2 2 b ∆J x − J(J + 1) − Jb z ∆Jb 2x + ξ Jb z ≤ 0 (323) 4 E D The left side is a parabolic function of ∆Jb 2x and for this to be negative D E requires ∆Jb 2x to lie between the two roots of this function, giving D

D

∆Jb 2x

∆Jb 2x

E

E

≥

≤

  s D E 2 2 D E2  D E2 1 J(J + 1) − Jb z − ξ Jb z J(J + 1) − Jb z −  2 (324)   s 2   D E D E D E 2 2 2 1 J(J + 1) − Jb z + J(J + 1) − Jb z − ξ Jb z  2 (325)

which are the required inequalities in Eq. (115) and (116).

J.2

Numerical Study of Inequalities

D E Here we consider the question: Is it possible to find values for ∆Sb 2x and D E | Sb z | in which all three inequalities (115), (116) and (117) are satisfied? S 44

E D E D Results showing the regions in the ∆Sb 2x versus | Sb z | plane corresponding to the three inequalities are shown in Figures 4 and 5 for the cases where J = 1000 and with ξ = 1.0 and ξ = 10.0 respectively. The for Dwhich D quantities E E the 2 b b regions are shown are the scaled variance and mean ∆S x /J and | S z |/J, D E D E with ∆Sb 2x given as a function of | Sb z | via (115), (116) and (117). That D E 2 2 D E2 regions exist where the quantity J(J + 1) − Sb z −ξ Sb z then becomes negative is seen in Figure 4. The spin squeezing region is always consistent with the second Heisenberg inequality (116) and for large J = 1000 there is a large region of overlap with the first inequality (115). For small J and large ξ the region of overlap becomes much smaller, as the result in Figure 6 for J = 1 and with ξ = 10.0 shows.

S 45

Figure 4. Regions in the < ∆Sbx2 > versus | < Sbz > | plane (shown shaded) for states that satisfy (a) the spin squeezing inequality Eq. (117) (b) the smaller Heisenberg uncertainty principle inequality Eq. (115) and (c) the larger HUP inequality Eq. (116). The case shown is for J = 1000 and HUP factor ξ = 1. Both < ∆Sbx2 > and | < Sbz > | are in units of J. The spin operators are chosen so that < Sbx >=< Sby >= 0.

S 46

Figure 5. As in Figure 4, but with J = 1000 and HUP factor ξ = 10.0.

S 47

Figure 6. As in Figure 4, but with J = 1 and HUP factor ξ = 10.0. Inspection of the three figures shows that there are regions where all three inequalities are satisfied.

S 48

Appendix K

“Separable but Non-Local” States

It is instructive to apply the various entanglement tests to the so-called separable but non-local states considered in Refs. [4], [44], for which the sub-system states are definitely not SSR compliant. These states should not pass the Hillery tests [22], [32] for SSR neglected entanglement, but they may pass the entanglement tests in this paper and in Ref. [2] since these states would be regarded as SSR compliant entangled. Note that these states are consistent with the global particle number SSR, so there is no dispute about whether they are possible two mode quantum states. The issue is rather whether they should be categorized as separable or entangled, and that depends on how separable (and hence entangled) states are first defined. As discussed previously, the interferometric measurements discussed here do not enable us to choose one definition over the other - that is an issue involved what types of quantum states would be allowed in the separate sub-systems. The first example of such states is the mixture of two mode coherent states is represented by the two mode density operator R dθ |α, αi hα, α| 2π R dθ = (|αi hα|)a ⊗ (|αi hα|)b 2π

b ρ =

(326)

where |αiC is a one mode coherent state for mode c = a, b with α = |α| exp(−iθ), and modes a, b are associated with bosonic annihilation operators b a, bb. The magnitude |α| is fixed. This state globally but not locally SSR compliant. Now D E R dθ † b a†bb = Tr b a bb (|αi hα|)a ⊗ (|αi hα|)b 2π R dθ = Tr |αi hα| b a† a ⊗ bb |αi hα| 2π b = |α|2 (327) But R dθ b a† b a |αi hα| a ⊗ bb†bb |αi hα| 2π b R dθ † †b b = hα| b a b a |αi a ⊗ hα| b b |αi 2π b = |α|4 (328) D E D E D E a†bb |2 = b a† b a bb†bb . This shows the state Hence we have | b a†bb |2 > 0 and | b is SSR compliant entangled. However, it fails the Hillery test for SSR neglected entanglement which is consistent with being a separable state if the local particle SSR is neglected [4, 44]. D E b a† b a bb†bb

= Tr

S 49

The second example of such states has an overall density operator which is a statistical mixture given by 1 1 (|ψ i hψ 1 |)a ⊗ |ψ 1 i hψ 1 |)b + (|ψ i i hψ i |)a ⊗ |ψ i i hψ i |)b 4 1 4

1 1 + ( ψ −1 ψ −1 )a ⊗ ψ −1 ψ −1 )b + ( ψ −i ψ −i )a ⊗ ψ −i ψ −i )b 4 4 (329) √ where |ψ ω i = (|0i + ω |1i)/ 2, with ω = 1, i, −, −i. The |ψ ω i are superpositions of zero and one boson states and consequently the local particle number SSR is violated by each of the sub-system density operators |ψ ω i hψ ω |)a and |ψ ω i hψ ω |)b . √ √ a† = (h0| ω ∗ )/ 2 and |ω|2 = 1 Now using bb |ψ ω i = (ω |0i)/ 2, hψ ω | b b ρ =

D E b a†bb

1X † (b a |ψ ω i hψ ω |a ) ⊗ (bb |ψ ω i hψ ω |b ) 4 ω 1X hψ ω | b a† |ψ ω ia hψ ω | bb |ψ ω ib 4 ω 1 X 1 ∗1 ω ω 4 ω 2 2

= Tr = = =

1 4

=

Tr

(330)

But D E b a† b a bb†bb

= = =

1X † (b a b a |ψ ω i hψ ω |a ) ⊗ (bb†bb |ψ ω i hψ ω |b ) 4 ω 1X hψ ω | b a† b a |ψ ω ia hψ ω | bb†bb |ψ ω ib 4 ω 1 X 1 21 2 |ω| |ω| 4 ω 2 2 1 4

(331)

D E D E D E a†bb |2 < b a† b a bb†bb . This shows the state Hence we have | b a†bb |2 > 0 and | b is SSR compliant entangled. However, it again fails the Hillery test for SSR neglected entanglement which is consistent with being a separable state if the local particle SSR is neglected [4, 44]. It should be noted, however, that the density operator can also be written as b ρ =

1 1 (|0i h0|)A ⊗ |0i h0|)B + (|1i h1|)A ⊗ |1i h1|)B 4 4 1 + (|Ψ+ i hΨ+ |)AB 2 S 50

(332)

√ where |Ψ+ iAB = (|0iA |1iB + |1iA |0iB )/ 2. In this form the terms correspond to a statistical mixture of states with 0, 1, 2 bosons. The first two terms correspond to separable states, in which the sub-system density operators are SSR compliant. The final term however is a one boson Bell state which is generally regarded as the paradigm of a two mode entangled state. Hence regarding the overall state as separable is highly questionable.

S 51

Appendix L

Quadrature Squeezing Entanglement Tests

D E D E b 2 (+) ≥ 1/2 and ∆Pb2 (+) ≥ We can show that for separable states both ∆X θ θ bθ (+) or Pbθ (+) is a test for 1/2, so two mode quadrature squeezing in either X two mode entanglement. Firstly, for SSR compliant sub-system states D E D E D E X bθ (+) = √1 baθ bbθ ) = 0 X PR ( X + X R R 2 R

(333)

D E sincehb aiR = bb = 0. Secondly, R

E D bθ2 (+) X D E D E

†

† 1X PR ( b ab a R+ b a b a R + bbbb† + bb†bb ) = 4 R R R X 1 1 = PR ( + (hb na iR + hb nb iR ) 2 2 R 1 1 D bE = + N 2 2 1 (334) ≥ 2

2 D 2E D E where for local SSR compliant states other terms involving b a R , bb , b abb = R R D E D E D E etc. are all zero. Hence abb† = hb aiR bb† hb aiR bb , b R

R

R

D E D E D E2 bθ2 (+) = X bθ2 (+) − X bθ (+) ≥ 1 ∆X 2

(335)

D E bθ+π/2 (+) we also have ∆Pb2 (+) = which establishes the result. Since Pbθ (+) = X θ D E 1 1 1 b + 2 2 N ≥ 2 for a separable state. Hence the two mode quadrature squeezing test. If D E D E 1 bθ2 (+) < 1 ∆X or ∆Pbθ2 (+) < (336) 2 2 bθ (+) and Pbθ (+) cannot both be then the state is entangled. Obviously X squeezed for the same state. D E b 2 (−) = 1 + A similar proof shows that for separable states both ∆X θ 2 D E D E D E 1 b ≥ 1/2 and ∆Pb2 (−) = 1 + 1 N b ≥ 1/2, so if N θ 2 2 2 D E bθ2 (−) < 1 ∆X 2

or

S 52

D E 1 ∆Pbθ2 (−) < 2

(337)

bθ (+), Pbθ (+), X bθ (−), Pbθ (−) being then the state is entangled. Hence any one of X squeezed will demonstrate two mode entanglement. The question then arises - Can two of the four two mode quadrature operators be squeezed? For simplicity we only discuss θ = 0 cases in detail. b0 (+), Pb0 (+) or X b0 (−), Pb0 (−) cannot. Next, we conObviously pairs such as X b0 (+) and Pb0 (−). We note that for all global SSR compliant states sider X D E D E D E

2 2 b b 2 (+) + ∆Pb2 (−) = 1 ∆(b ∆X x + x b ) + ∆(b p − p b ) = 1 + N A B A B 0 0 2 D E D E b0 (+) is squeezed ∆X b 2 (+) < 1 then ∆Pb2 (−) > using (140), so that if X 0 0 2 D E 1 b b b 2 + N , showing that both X0 (+) and P0 (−) cannot both be squeezed - in b0 (−) and spite of the operators commuting. A similar conclusion applies to X D E D E b0 (+) and X b0 (−) we have ∆X b 2 (+) + ∆X b 2 (−) = Pb0 (+). For the pair X 0 0 D E

1 b using (138) and (142), so the ∆(b xA + x bB )2 + ∆(b xA − x bB )2 = 1 + N 2 b0 (+) and Pb0 (−) applies, and thus X b0 (+) and X b0 (−) same situation as for X b b cannot both be squeezed. A similar conclusion applies to P0 (−) and P0 (+). In bθ (+), Pbθ (+), X bθ (−), Pbθ (−) can be squeezed. general, only one of X

S 53

Appendix M M.1

Derivation of Interferometer Results

General Theory - Two Mode Interferometer

Introducing the free and interaction evolution operators via b U b0 U

b0 U bint = U =

b 0 t/~) exp(−iH

(338)

it is straightforward to show that for c = 1 (bb†bb − b M a† b a) 2

(339)

we have D

E c M D E c2 ∆M

cH b = T r(M ρ) D Eo2 n cH cH − M b ρ) = T r( M

(340)

giving the mean and variance in terms of the input density operator and interaction picture Heisenberg operators cH M bbH

1 b† b (b bH − b a†H b aH ) 2 H b −1 bb U bint b −1 b b = U b aH = U int int a Uint

=

(341)

b −1 bb U b0 = exp(−iω b t) bb and U b −1 b b where we have used the results U a. 0 0 a U0 = exp(−iω a t) b The interaction picture Heisenberg operators satisfy i~

∂b bH = [bbH , VbH ] ∂t

i~

∂ b aH = [b aH , VbH ] ∂t

(342)

where VbH

= A(t) exp(−iω 0 t) exp(iφ) bb†H b aH exp(+iω ba t) +A(t) exp(+iω 0 t) exp(−iφ) b a† bbH exp(−iω ba t) H

= A(t) exp(iφ) bb†H b aH + A(t) exp(−iφ) b a†H bbH

(343)

for resonance. We then find that the Heisenberg picture operators satisfy coupled linear equations i~

∂b bH = A(t) exp(+iφ) b aH ∂t

i~

∂ b aH = A(t) exp(−iφ) bbH ∂t

(344)

which after replacing the time t by the area variable s then involve time independent coefficients i

∂b bH (s) = exp(+iφ) b aH (s) ∂s

i

∂ b aH (s) = exp(−iφ) bbH (s) ∂s

S 54

(345)

The equations can then be solved via Laplace transforms giving bbH (s, φ) = cos s bb − i exp(iφ) sin s b a

b aH (s, φ) = −i exp(−iφ) sin s bb + cos s b a (346) where now 2s is the area for the classical pulse. Hence we have in general cH (2s, φ) M

1 b† (b (s, φ)bbH (s, φ) − b a†H (s, φ)b aH (s, φ)) 2 H = sin 2s (sin φ Sbx + cos φ Sby ) + cos 2s Sbz =

(347)

The versatility of the measurement follows from the range of possible choices of the pulse area 2s and the phase φ. E D2s = θE we can then substitute into Eq. (340) to obtain results for D Writing c and ∆M c2 . These are set out in SubSection 7.3 in Eqs. (170) and M (171) in terms of the mean values of the spin operators and the elements of the covariance matrix for the spin operators, evaluated for the input quantum state b ρ.

M.2

Beam Splitter and Phase Changer

For the beam splitter we have 2s = π/2 and φ (variable) so that cH ( π , φ) = sin φ Sbx + cos φ Sby M 2

(348)

whilst for the phase changer we have 2s = π and φ (arbitrary) so that cH (π, φ) = −Sbz M

M.3

(349)

Other Measurables

We can also consider other choices for the measurable, which then enable us to determine other moments of the spin operators. A case of particular interest is the square of the population difference c2 = M

1 b†b (b b − b a† b a) 2

2 (350)

It is then straightforward to show for the beam splitter case with 2s = π/2 and φ (variable) c2H ( π , φ) M 2

2

=

=

sin2 φ (Sbx )2 + cos2 φ (Sby )2 + sin φ cos φ (Sbx Sby + Sby Sbx )

sin φ Sbx + cos φ Sby

(351)

S 55

Hence E E D E D E D D c2 = sin2 φ (Sbx )2 + cos2 φ (Sby )2 + sin φ cos φ (Sbx Sby + Sby Sbx ) (352) M c2 is a sinusoidal function of the showing that the mean for the new observable M BS interferometer variable φ with coefficients that depend on the means of Sbx2 , Sby2 and Sbx Sby + Sby Sbx .

M.4

Squeezing in the xy Plane

b# 3π The question is does squeezing in either Sbx# ( 3π 2 + φ) or Sy ( 2 + φ) demonstrate entanglement of the modes b a and bb? For a separable state we have 3π 3π Sbx# ( (353) + φ) = Sby# ( + φ) = 0 2 2 ρ ρ D E D E # 3π ( 2 + φ) are just linear combinations of the zero Sbx,y . as before, since Sbx,y ρ

ρ

Since [ Sbx# ( 3π + φ), Sby# ( 3π + φ)] = iSbz the Heisenberg uncertainty principle D 2 E2 D E D E 3π # 2 2 shows that ∆Sbx ( 2 + φ) ∆Sby# ( 3π ≥ 14 | Sbz |2 so spin squeez2 + φ) ρ

ing in Sbx# ( 3π 2 + φ) with that 3π ∆Sbx# ( + φ)2 < 2 ρ

ρ

ρ

respect to Sby# ( 3π 2 + φ) or vice versa requires us to show

1 Db E | Sz | 2 ρ

or

∆Sby# (

3π + φ)2 2

< ρ

1 Db E | Sz | 2 ρ (354) E

D E D 2 2 + φ) Since we measure ∆Sbx# ( 3π the spin squeezing test is ∆Sbx# ( 3π 2 2 + φ) ρ D E 1 b |. 2 | Sz