Detecting entanglement with non-hermitian operators
DDetecting entanglement with non-hermitianoperators
Mark Hillery (1) , Ho Trung Dung (1 , , and Julien Niset (1)(1) Department of Physics, Hunter College of CUNY695 Park AvenueNew York, NY 10065 (2)
Institute of Physics, Academy of Sciences and Technology1 Mac Dinh Chi Street, District 1Ho Chi Minh City, VietnamOctober 29, 2018
Abstract
We derive several entanglement conditions employing non-hermitianoperators. We start with two conditions that were derived previouslyfor field mode operators, and use them to derive conditions that can beused to show the existence of field-atom entanglement and entanglementbetween groups of atoms. The original conditions can be strengthened bymaking them invariant under certain sets of local unitary transformations,such as Gaussian operations. We then apply these conditions to severalexamples, such as the Dicke model. We conclude with a short discussionof how local uncertainty relations with non-hermitian operators can beused to derive entanglement conditions.
Entanglement has shown itself to be a valuable resource in quantum informationin applications ranging from communication protocols, such as dense coding,to quantum computing. This has led to a substantial effort to understand andcharacterize entanglement (see [1] and [2] for recent reviews). There is no simpleuniversal test that enables one to tell whether a given state is entangled, butthere are many sufficient conditions. These include, for example, entanglementwitnesses. An entanglement witness is an operator whose expectation value isnonnegative for separable states, but its expectation value can also be negative,and the states for which it is are entangled. For continuous-variable systems,there are entanglement criteria involving the expectation values of powers ofcreation and annihilation operators [3]-[8]. Here we would like to expand upon1 a r X i v : . [ qu a n t - ph ] D ec he work in [7] to find stronger entanglement conditions, and to find conditionsfor entanglement not just between field modes, but between atoms and fieldmodes or between groups of atoms.In [7] two conditions that enable one to determine whether the modes ina two-mode state are entangled were derived. These conditions are that themodes are entangled if either |(cid:104) A † B (cid:105)| > (cid:104) A † AB † B (cid:105) , (1)or |(cid:104) AB (cid:105)| > (cid:104) A † A (cid:105)(cid:104) B † B (cid:105) , (2)where A is any power of the annihilation operator for the first mode and B isany power of the annihilation operators for the second mode. These conditionsare sufficient, but not necessary, to demonstrate entanglement, that is if eitherone is satisfied, the state is entangled, but if neither one is satisfied, we cannotsay anything about the entanglement of the state.The derivations of these inequalities made no use of the special propertiesof annihilation operators. In fact, if we have a bipartite system described bya Hilbert space H = H a ⊗ H b , and A is any operator on H a and B is anyoperator on H b , then if a state satisfies either of the above conditions, thenthat state is entangled. If A or B is hermitian, these conditions are useless,because the Schwarz inequality guarantees that they will be violated for allstates. Consequently, we will be studying the detection of entanglement withnon-hermitian operators.There have been some previous studies of entanglement making use of non-hermitian operators. Toth, Simon and Cirac derived an inequality based onuncertainty relations with the number operator and the mode annihilation op-erator that can detect entanglement between modes [9]. As was mentioned inthe previous paragraph, Zubairy and one of us derived a set of entanglementconditions based on mode creation and annihilation operators, and a very gen-eral set of conditions based on the same operators was derived by Shchukin andVogel [8].We will also show how our original entanglement conditions can be strength-ened by making use of local unitary invariance. In the next section this will bestudied for some simple, low-dimesnional systems. We then go on to look atan atom, or atoms coupled to a single-mode field and develop entanglementconditions for the field-atoms system that are invariant under Gaussian trans-formations on the field mode. In the following section, we demonstrate that allof the entanglement conditions we find can be derived from the partial transposecondition, i.e. that if the partial transpose of a state is not positive, then theoriginal state is entangled. However, the conditions we derive are much easierto use than the partial transpose condition itself. We then go on to study theapplication of invariance under Gaussian operations to entanglement conditionsfor two modes. We then examine entanglement in two extended examples, theDicke model and light modes coupled by two beam splitters. In both cases wefind a connection between sub-Poissonian statistics of an input light mode and2ubsequent entanglement in the system. Finally, we show how a non-hermitianversion of local uncertainty relations can be used to derive further entanglementconditions between field modes, and between a collection of atoms and a fieldmode. As mentioned in the Introduction, another element we will be using in our studyof entanglement is its invariance under local unitary operations. Because of thisinvariance, it would be nice if our criteria for entanglement also possessed thisinvariance. As we shall see, we cannot always completely satisfy this condition,but we can often satisfy it for some subset of local unitary operators.Let us begin by considering the first inequality above. Let | α (cid:105) a and | α (cid:105) a be two orthogonal vectors in H a and | β (cid:105) b and | β (cid:105) b be two orthogonal vectorsin H b . Now let A = | α (cid:105)(cid:104) α | and B = | β (cid:105)(cid:104) β | , and consider the state | ψ (cid:105) = c | α (cid:105) a | β (cid:105) b + c | α (cid:105) a | β (cid:105) b , (3)where | c | + | c | = 1. We find that |(cid:104) A † B (cid:105)| = | c c | , (cid:104) A † AB † B (cid:105) = 0 , (4)so that Eq. (1) is a good way to detect the entanglement of | ψ (cid:105) .It will still detect it if we add noise to the state. Defining P α = (cid:88) j =1 | α j (cid:105)(cid:104) α j | , P β = (cid:88) j =1 | β j (cid:105)(cid:104) β j | , (5)and considering the density matrix ρ = s | ψ (cid:105)(cid:104) ψ | + 1 − s P α ⊗ P β , (6)where 0 ≤ s ≤
1, we find that Eq. (1) shows that the state is entangled if s > (1 + 16 | c c | ) / − | c c | . (7)If c = c = 1 / √ s > ( √ − / | c c | (cid:28)
1, then we have s > − | c c | . Another possible density matrix that can be detected by thiscondition is one of the form ρ = s | ψ (cid:105)(cid:104) ψ | + ρ , (8)where the vectors | α j (cid:105) a | β k (cid:105) b , for j, k = 1 , ρ , andTr( ρ ) = 1 − s . For this density matrix, Eq. (1) will show that it is entangled if s >
0. 3ow let us go to a more complicated situation. Let S be the linear spanof the vectors {| α (cid:105) , | α (cid:105)} , and S be the linear span of the vectors {| α (cid:105) , | α (cid:105)} ,where the vectors {| α j (cid:105) ∈ H a | j = 1 , . . . , } form an orthonormal set. We wantan entanglement condition that will detect entanglement in vectors of the form | ψ (cid:105) = 1 √ | v (cid:105) a | β (cid:105) b + | v (cid:105) a | β (cid:105) b ) , (9)where | v (cid:105) ∈ S and | v (cid:105) ∈ S . We would also like the condition to be inde-pendent of the specific vectors | v (cid:105) a and | v (cid:105) a . One way to approach this is tomake use of the fact that entanglement is invariant under local unitary trans-formations. Suppose that U a is a unitary operator on H a that leaves S and S invariant. For any two vectors in S , | v (cid:105) a and | v (cid:48) (cid:105) a , and two vectors in S , | v (cid:105) a and | v (cid:48) (cid:105) a , we can find a U a such that U a | v (cid:105) a = | v (cid:48) (cid:105) a and U a | v (cid:105) a = | v (cid:48) (cid:105) a .Therefore, if we have an entanglement condition that is invariant under thetransformations, U a , we will have one that is independent of the vectors | v (cid:105) a and | v (cid:105) a in the equation for | ψ (cid:105) above. We can find such a condition by notingthat if we have an operator of the form | α (cid:105) a (cid:104) α | that maps S to S , then U a | α (cid:105) a (cid:104) α | U − a will be a linear combination of the operators | α j (cid:105) a (cid:104) α k | , where j = 3 , k = 1 ,
2. Therefore, let us choose A = z | α (cid:105) a (cid:104) α | + z | α (cid:105) a (cid:104) α | + z | α (cid:105) a (cid:104) α | + z | α (cid:105) a (cid:104) α | , (10)where the complex numbers z , . . . , z are arbitrary. The operator B is as before.We then have that |(cid:104) A † B (cid:105)| − (cid:104) A † AB † B (cid:105) = (cid:88) j,k =1 z ∗ j M jk z k , (11)where the 4 × M depends on the state being considered. If M has apositive eigenvalue, then by choosing the z j , j = 1 , . . . , A that shows thatthe state we are considering is entangled.Let us carry this out explicitly for the density matrix ρ = s | ψ (cid:105)(cid:104) ψ | + 1 − s P α ⊗ P β , (12)where P α is the projection onto S ∪ S and P β is as before. We then find that M jk = s η j η ∗ k − − s δ jk , (13)where η = (cid:104) v | α (cid:105)(cid:104) α | v (cid:105) , η = (cid:104) v | α (cid:105)(cid:104) α | v (cid:105) ,η = (cid:104) v | α (cid:105)(cid:104) α | v (cid:105) , η = (cid:104) v | α (cid:105)(cid:104) α | v (cid:105) . (14)4e can then express M as M = s | η (cid:105)(cid:104) η | − − s , (15)where the vector | η (cid:105) has components given by η j , j = 1 , . . . ,
4. It is then clearthat M has three negative eigenvalues, corresponding to directions orthogonalto | η (cid:105) , with the remaining eigenvalue equal to (2 s + s − /
8. This is positiveif s > /
2, and therefore, the state is entangled if s > /
2. Note that thiscondition is independent of | v (cid:105) a and | v (cid:105) a . The above entanglement condition can be applied to a two-level atom coupledto a single-mode field. The atom can either absorb a photon and go from itslower to its upper state, or emit a photon and go from its upper to its lowerstate. Let the lower state of the atom be | g (cid:105) and the excited state be | e (cid:105) , andlet us consider photon states with 0, 1, 2, and 3 photons. If we start the atomin its excited state and a superposition of 0, and 2 photons, as the systemevolves there will be some amplitude to be in these states, but there will alsobe amplitudes to be in the states | g (cid:105)| (cid:105) and | g (cid:105)| (cid:105) . The upper state of the atomwill be correlated with photon subspace spanned by | (cid:105) and | (cid:105) , and the lowerstate will be correlated with the subspace spanned by | (cid:105) and | (cid:105) . Therefore,the entanglement conditions developed in the previous section could be used totest for entanglement in this system.In considering an atom interacting with a single-mode field, we do not usuallyconfine our attention to states of three photons or fewer, so a different set ofentanglement conditions could be more useful. Instead of limiting the photonnumber, we will consider all possible photon states. In that case we will haveto give up invariance of the conditions under all unitary transformations of thephoton Hilbert space. We can, however, derive some simple conditions if werestrict the set of unitary transformations to those corresponding to Gaussianoperations. These operations consist of translations, rotations and squeezingtransformations. In particular, if a and a † are the annihilation and creationoperators for the mode, the translations are given by the operators D ( α ) =exp( αa † − α ∗ a ), where α is an arbitrary complex number, rotations are givenby R ( θ ) = exp( iθa † a ), where 0 ≤ θ < π , and the squeezing transformationis given by S ( z ) = exp[( z ∗ a − z ( a † ) ) / z = r exp( iφ ) is a complexnumber. These transformations act as follows: D ( α ) † aD ( α ) = a + α,R ( θ ) † aR ( θ ) = e iθ a,S ( z ) † aS ( z ) = a cosh r − a † e − iφ sinh r. (16)Note that these transformations send the creation and annihilation operatorsinto a linear combinations of the annihilation operator, the creation operator,and a constant. 5n order to find an entanglement condition that is invariant under Gaussiantransformations of the field mode, we set A = σ − = | g (cid:105)(cid:104) e | ,B = z ( a − (cid:104) a (cid:105) ) + z ( a † − (cid:104) a † (cid:105) ) , (17)and substitute these into Eq. (1). The result is (cid:88) j,k =1 z ∗ j M jk z k > , (18)where now M = (cid:18) |(cid:104) σ + ∆ a (cid:105)| − (cid:104) P e (∆ a ) † ∆ a (cid:105) (cid:104) σ − (∆ a ) † (cid:105)(cid:104) σ + (∆ a ) † (cid:105) − (cid:104) P e ((∆ a ) † ) (cid:105)(cid:104) σ − ∆ a (cid:105)(cid:104) σ + ∆ a (cid:105) − (cid:104) P e (∆ a ) (cid:105) |(cid:104) σ − ∆ a (cid:105)| − (cid:104) P e ∆ a (∆ a ) † (cid:105) (cid:19) , (19)where ∆ a = a − (cid:104) a (cid:105) , σ + = | e (cid:105)(cid:104) g | and P e = | e (cid:105)(cid:104) e | . If we transform the state ofthe field mode with a Gaussian transformation, the effect in Eq. (18) is just tochange the values of z and z .Now if we can find a value of z and z so that Eq. (18) is satisfied, thenthe state is entangled. This will be possible if the matrix M has a positiveeigenvalue. Therefore, our entanglement condition, which is now invariant underGaussian transformations of the field mode, is that the state is entangled if M has a positive eigenvalue. That means, for example, that it can detectentanglement in states of the form I at ⊗ S ( z ) D ( α )( | e (cid:105)| (cid:105) + | g (cid:105)| (cid:105) ) / √
2, where I at is the identity operator for the atomic system, for any value of α and z .Let us now look at two examples of atom-field entanglement. Let us firstconsider a two-level atom in the rotating wave approximation interacting witha single field mode, which is initially in a thermal state. On resonance, theHamiltonian for this system is H = ωa † a + ω σ z + κ ( σ + a + σ − a † ) . (20)The thermal state density matrix is given by ρ therm = 1(1 + ¯ n ) ∞ (cid:88) n =0 (cid:18) ¯ n n (cid:19) n | n (cid:105)(cid:104) n | . (21)With this initial state we find that the off-diagonal elements of M are initiallyzero, and remain zero for all times. The element in the lower right-hand corner( M ) is always negative, so only the element in the upper left-hand corner( M ) can possibly become positive. We plot M for several different values of¯ n in Figure 1. As can be seen, for sufficiently small values of ¯ n , our conditionshows that there are times at which the atom and the field mode are entangled.Let us now look at two two-level atoms interacting on resonance with a singlemode field. The Hamiltonian of the system is now H = ωa † a + ω σ z + σ z ) + κ (cid:0) ( aσ +1 + a † σ − ) + ( aσ +2 + a † σ − ) (cid:1) . (22)6 ! t -0.00100.0010 0.1 0.2M n - Figure 1: The eigenvalue M as a function of time for average photon numbers¯ n = 0 .
03 (solid line), ¯ n = 0 .
02 (dashed line), and ¯ n = 0 .
01 (dotted line). Theinset shows M as a function of ¯ n at κt = 2 . H = ⊕H ( n ) . For n > H ( n ) is spanned by the vectors | n, g, g (cid:105) , | n − , g, e (cid:105) , | n − , e, g (cid:105) , and | n − , e, e (cid:105) , where the first slot contains the photon number, and the remainingtwo slots contain the states of the atoms. For n = 1 it is spanned by | , g, g (cid:105) , | , g, e (cid:105) , and | , e, g (cid:105) . If we start in the state | n, g, g (cid:105) at time t = 0, we find that M = M = 0 for all times, and M ≤
0. Thus our entanglement conditionreduces to M >
0. Finding an explicit expression for M (see the Appendix)this condition becomes n n − (Ω t )[cos(Ω t ) + 2( n − − ( n − (cid:104) n − t ) − + (2 n −
1) sin (Ω t ) (cid:105) > , (23)where Ω = κ (cid:112) n − n = 1 and n = 2, the condition reads assin (Ω t ) cos (Ω t ) > , (24) | cos(Ω t ) + 2 | > √ , (25)respectively. For the single-photon case, it is not known which of the two atomsabsorbs the photon, so that the atoms are always entangled. The conditionbecomes more restrictive as n increases. For times in the neighborhood of 2 kπ/ Ω t = (cid:15) + k π Ω , (26)where (cid:15) is small but nonvanishing and k is an integer, one can make an (cid:15) -7xpansion of the left-hand side of Eq. (23) around 0 to obtain(2 n − (cid:15) −
16 (3 n + n + 4) (cid:15) + O ( (cid:15) ) > . (27)This can be satisfied for an arbitrarily large n , provided (cid:15) is small enough, thatis there always exists a time interval about 2 kπ/ Ω during which entanglementis detected. Obviously, for an increasing n , this time interval is reduced.We can also study the entanglement between both atoms and the field. Nowwe choose A = a and B = J − with J − = σ − + σ − in Eq. (1). The conditionfor entanglement becomes |(cid:104) a † J − (cid:105)| − (cid:104) a † aJ + J − (cid:105) > , (28)or, by using the state at time t as given in the Appendix, n n − (Ω t )[cos(Ω t ) + 2( n − − ( n − (cid:104) ( n − t ) − + (2 n −
1) sin (Ω t ) (cid:105) > . (29)For n = 1 and n = 2 the above equation is the same as Eq. (23). In general,this condition is easier to satisfy than the condition (23) due to the fact thatits second term is smaller than its counterpart in Eq. (23). As a consequence,there are moments when we detect entanglement between the field and bothatoms, while no entanglement is detected between the field and a single atom.For times about 2 kπ/ Ω, see Eq. (26), one can expand (29) with respect to (cid:15) around 0 (2 n − (cid:15) −
112 (3 n + 11 n + 2) (cid:15) + O ( (cid:15) ) > . (30)Again for large n one can detect entanglement during time windows centeredaround 2 kπ/ Ω. The derivations of the entanglement conditions, Eqs. (1) and (2), which werepresented in [7], did not make use of the Positive Partial Transpose (PPT)condition. This condition states that the partial transpose of a separable densitymatrix is a positive operator, which implies that if the partial transpose ofa density matrix is not positive, the original density matrix was entangled.However, it was subsequently shown that in the case that A and B are powersof mode creation and annihilation operators, the conditions are a consequenceof the PPT condition [10]. We will now show that this is true in general, i.e.that no matter what the choice of A and B , the entanglement conditions in Eqs.(1) and (2) are consequences of the PPT condition.Let us begin by discussing one of these conditions. Our system is dividedinto subsystems a and b , and let A be an operator acting on subsystem a and B
8n operator acting on subsystem b . A separable state must satisfy the condition[7] |(cid:104) A † B (cid:105)| ≤ (cid:104) A † AB † B (cid:105) . (31)If this condition is violated, the state is entangled. Now let us see if we canderive this condition from the PPT condition. We start with the condition |(cid:104) AB (cid:105)| ≤ (cid:104) A † AB † B (cid:105) , (32)which follows from the Schwarz inequality. These expectation values are takenwith respect to a density matrix ρ , and the matrix element of this density matrixare ρ m,µ ; n,ν = ( a (cid:104) m | b (cid:104) µ | ) ρ ( | n (cid:105) a | ν (cid:105) b ) , (33)where {| m (cid:105) a } is an orthonormal basis for a and {| µ (cid:105) b } is an orthonormal basisfor b . Now define the partial transpose of ρ with respect to subsystem a , ρ T a ,to be the operator with matrix elements ρ T a m,µ ; n,ν = ρ n,µ ; m,ν . (34)For a separable density matrix, ρ T a should also be a valid density matrix, thatis it should be positive and have a trace equal to one. That implies that theinequality in Eq. (32) will hold if the expectation values are taken with respectto ρ T a . We have thatTr( ABρ T a ) = (cid:88) m,n (cid:88) µ,ν A nm B νµ ρ T a mµ ; nν = (cid:88) m,n (cid:88) µ,ν A nm B νµ ρ n,µ ; m,ν = (cid:88) m,n (cid:88) µ,ν ( A † mn ) ∗ B νµ ρ n,µ ; m,ν . (35)If A † mn is real, then we see that Tr( ABρ T a ) = Tr( A † Bρ ). This would be thecase, for example, if A were a product of mode creation and annihilation opera-tors and the basis is the number-state basis. Similarly, we find that if ( A † A ) mn is real, then Tr( A † AB † Bρ T a ) = Tr( A † AB † Bρ ). If the density matrix is sepa-rable and all of the reality conditions are satisfied, then we have that Eq. (32),with the expectation values taken with respect to ρ T a , implies Eq. (31), withthe expectation values taken with respect to ρ . Under these conditions, theinequality in Eq. (31) follows from the PPT condition.As we can see, the above derivation depends on the fact that certain matrixelements are real. The condition can be relaxed a bit; as long as the matrixelements A † mn all have the same phase, the derivation will work. However, theoriginal derivation of the Eq. (31) did not require these conditions on the matrixelements.It is the none the less the case that Eq. (31) can be derived from the PPTcondition with no additional assumptions. Consider two operators F acting on9 a and B acting on H b . We have that, as in the previous paragraph, for aseparable state |(cid:104) F B (cid:105)| ≤ (cid:104) F † F B † B (cid:105) . (36)Now, from above, we see thatTr( F Bρ T a ) = (cid:88) m,n (cid:88) µ,ν ( F † mn ) ∗ B νµ ρ n,µ ; m,ν , (37)Tr( F † F B † Bρ T a ) = (cid:88) m,n (cid:88) µ,ν [( F † F ) mn ] ∗ ( B † B ) νµ ρ n,µ ; m,ν . (38)Define the operator G , acting on H a , by G = (cid:88) m,n F ∗ mn | m (cid:105)(cid:104) n | . (39)The quantities involving the partially transposed density matrix can then beexpressed as Tr( F Bρ T a ) = (cid:104) G † B (cid:105) , (40)Tr( F † F B † Bρ T a ) = (cid:104) G † GB † B (cid:105) . (41)For any separable density matrix, we must have |(cid:104) G † B (cid:105)| ≤ (cid:104) G † GB † B (cid:105) . (42)In order to recover our original inequality, choose F = (cid:88) m,n ( A mn ) ∗ | m (cid:105)(cid:104) n | , (43)which implies that G = A , and this completes the derivation of the inequalityfrom the PPT condition. The derivation, for separable states, of the condition |(cid:104) AB (cid:105)| ≤ (cid:104) A † A (cid:105)(cid:104) B † B (cid:105) , (44)using the PPT condition, is similar. Let us now look at the case of two modes with annihilation operators a and b .We will first find an entanglement condition that is invariant under Gaussiantransformations of one of the modes. Let us set A = z (∆ a ) † + z ∆ a and B = b .Then the condition |(cid:104) A † B (cid:105)| > (cid:104) A † AB † B (cid:105) can be written as (cid:104) v | M | v (cid:105) > , (45)where | v (cid:105) is the two-component vector | v (cid:105) = (cid:18) z z (cid:19) , (46)10nd M is the 2 × M = (cid:18) |(cid:104) ∆ ab (cid:105)| − (cid:104) ∆ a (∆ a ) † b † b (cid:105) (cid:104) ∆ ab (cid:105)(cid:104) (∆ a ) † b (cid:105) ∗ − (cid:104) (∆ a ) b † b (cid:105)(cid:104) (∆ a ) † b (cid:105)(cid:104) ∆ ab (cid:105) ∗ − (cid:104) (∆ a † ) b † b (cid:105) |(cid:104) (∆ a † b (cid:105)| − (cid:104) (∆ a ) † ∆ ab † b (cid:105) (cid:19) . (47)In this form, we can see that if M has at least one positive eigenvalue, we canfind a vector | v (cid:105) so that the entanglement condition (cid:104) v | M | v (cid:105) > M that results from ithas a positive eigenvalue. As before, this condition is invariant under Gaussianoperations, and, consequently, so is our entanglement condition. For this simple2 × M by M jk , where j, k = 1 ,
2, the larger of the two eigenvalues is λ = 12 { ( M + M ) + [( M − M ) + 4 | M | ] / } , (48)and the condition that λ > M + M ) >
0, or that | M | >M M .As a short example of how our new condition is stronger than the one orig-inally proved in [7], consider the state | ψ (cid:48) (cid:105) = 1 √ S a ( z ) ⊗ I b ( | (cid:105) a | (cid:105) b + | (cid:105) a | (cid:105) b ) . (49)This state is obtained by applying a Gaussian operation to the state | ψ (cid:105) =( | (cid:105) a | (cid:105) b + | (cid:105) a | (cid:105) b ) / √ | ψ (cid:105) , the matrix in the previous para-graph is diagonal, and the lower right-hand element is positive, so that the stateis entangled. Since | ψ (cid:48) (cid:105) is obtained from | ψ (cid:105) by a Gaussian operation, the cri-terion in the previous paragraph will also show that | ψ (cid:48) (cid:105) is entangled for any z .However, if we apply the criterion |(cid:104) a † b (cid:105)| > (cid:104) a † ab † b (cid:105) , (50)presented in [7], we find that it only shows that the state is entangled if tanh | z | < / √
2. Therefore, the new condition is an improvement on the old one.Now suppose we set A = z ( a − (cid:104) a (cid:105) ) + z ( a † − (cid:104) a † (cid:105) ) ,B = w ( b − (cid:104) b (cid:105) ) + w ( b † − (cid:104) b † (cid:105) ) . (51)Let H and H be two two-dimensional Hilbert spaces with | u (cid:105) = (cid:18) z z (cid:19) ∈ H , (52)and | v (cid:105) = (cid:18) w w (cid:19) ∈ H . (53)11he condition |(cid:104) A † B (cid:105)| > (cid:104) A † AB † B (cid:105) can be written as ( (cid:104) u | ⊗ (cid:104) v | ) X ( | u (cid:105) ⊗ | v (cid:105) ) >
0, where X is a linear hermitian operator on H ⊗H , and this operator dependson the state being considered. Therefore, a state is entangled if there exists aproduct state, | u (cid:105) ⊗ | v (cid:105) in H ⊗ H , such that( (cid:104) u | ⊗ (cid:104) v | ) X ( | u (cid:105) ⊗ | v (cid:105) ) > . (54)One way of determining whether the above condition can be satisfied is byexamining the operator X v = (cid:104) v | X | v (cid:105) , where | v (cid:105) is a vector in H and X v isan operator in H . If for any vector | v (cid:105) ∈ H the operator X v has a positiveeigenvalue, then the condition in Eq. (54) can be satisfied. This can be somewhattedious since we have to consider all possible vectors | v (cid:105) , so having some simplercriteria is also useful.The first criterion involves the eigenvalues of the reduced matrix X =Tr ( X ). We will show that if X has a positive eigenvalue, then a productvector satisfying the above equation exists. Suppose that X has a positiveeigenvalue, λ , and that the corresponding eigenstate is | u (cid:105) . In addition, let {| v j (cid:105)| j = 1 , } be an orthonormal basis for H . We then have that (cid:104) u | X | u (cid:105) = (cid:88) j =1 ( (cid:104) u | ⊗ (cid:104) v j | ) X ( | u (cid:105) ⊗ | v j (cid:105) ) = λ > . (55)Since the sum in the above equation is positive, at least one of the terms mustbe positive, i.e. ( (cid:104) u | ⊗ (cid:104) v j | ) X ( | u (cid:105) ⊗ | v j (cid:105) ) > j . Therefore, if X hasa positive eigenvalue, the state is entangled. This is, however, only a sufficientcondition for a product vector satisfying Eq. (54) to exist, not a necessary one.That is, it is possible for X to have only negative or zero eigenvalues and stillbe able to find a product vector satisfying Eq. (54). An example is provided bythe operator X = ( | (cid:105) + | (cid:105) )( (cid:104) | + (cid:104) | ) − | (cid:105) − | (cid:105) )( (cid:104) | − (cid:104) | ) , (56)for which X = − I , but ( (cid:104) + x |(cid:104) + x | ) X ( | + x (cid:105)| + x (cid:105) ) > X itself. If X has two or morepositive eigenvalues, then we can find a product vector satisfying Eq. (54). Thiscan be seen by showing that a product vector can be found in the subspacespanned by the eigenvectors corresponding to the two positive eigenvalues. Let | x (cid:105) and | x (cid:105) be the two eigenvectors with positive eigenvalues, and let theSchmidt decomposition of | x (cid:105) be given by | x (cid:105) = (cid:88) j =1 κ j | ξ j (cid:105)| ζ j (cid:105) . (57)We can expand | x (cid:105) in the Schmidt basis of | x (cid:105) as | x (cid:105) = (cid:88) j =1 2 (cid:88) k =1 d jk | ξ j (cid:105)| ζ k (cid:105) . (58)12he vector α | x (cid:105) + β | x (cid:105) is then given by α | x (cid:105) + β | x (cid:105) = (cid:88) j =1 2 (cid:88) k =1 c jk | ξ j (cid:105)| ζ k (cid:105) , (59)where c jk = ακ δ j δ k + ακ δ j δ k + βd jk . (60)Now, α | x (cid:105) + β | x (cid:105) will be a product state if ( c /c ) = ( c /c ) , and makinguse of the above equation, this condition can be expressed as yκ + d d = d yκ + d , (61)where y = α/β . This leads to a quadratic equation for y , which can be solved.Therefore, there is a product vector in the span of | x (cid:105) and | x (cid:105) .A simple example of this procedure is given by examining the entanglementof the state ρ = s | ψ (cid:105)(cid:104) ψ | + 1 − s P ( a )01 ⊗ P ( b )01 , (62)where | ψ (cid:105) = ( | (cid:105) a | (cid:105) b + | (cid:105) a | (cid:105) b ) / √
2, and P ( a )01 and P ( b )01 project onto thezero and one-photon states of the a and b modes, respectively, i.e. P ( a )01 = | (cid:105) a (cid:104) | + | (cid:105) a (cid:104) | , and similarly for P ( b )01 . In Ref. [7] it was found that using thecondition in Eq. (1) with the choice A = a and B = b shows that this state isentangled for s > ( √ − /
2. With the choice of A and B given above, we findthat the eigenvalues of X ( v ) are given by the solutions of the equation λ − (cid:18) s − (cid:19) λ + 116 [1 − s ( | w | − | w | ) − s ( | w | + | w | )] = 0 , (63)where we have set | w | + | w | = 1, because only the direction of | v (cid:105) is impor-tant. One of the eigenvalues will be positive if the last term in this equation isnegative, and this gives us the condition1 < s ( | w | − | w | ) + 2 s ( | w | + | w | ) . (64)From this we find that the state is entangled if s > . √ − / . We would now like to use the simplest versions of our entanglement conditionsto study entanglement between different subsystems in the Dicke model. Theentanglement of the ground state in the Dicke model was studied in [11], butwe will concentrate on entanglement generated by the dynamics. This model13onsists of N two-level atoms, which are contained in a volume whose dimensionsare small compared to an optical wavelength, coupled to a single mode of theradiation field. Assuming again that the interaction is on resonance, and thatthere is no atom-atom interaction, the Hamiltonian of the system can be writtenas H = ωa † a + ω N (cid:88) i =1 σ zi + κ N (cid:88) i =1 ( aσ + i + a † σ − i )= ωa † a + ωS z + κ ( aS + + a † S − ) , (65)after the introduction of the collective spin operators S + = N (cid:88) i =1 σ + i , S − = N (cid:88) i =1 σ − i ,S z = 12 [ S + , S − ] . (66)The equations of motion resulting from this Hamiltonian are difficult to solve, sowe will introduce an approximation based on the Holstein-Primakoff represen-tation of the spin operators. The Holstein-Primakoff transformation expressesthe collective spin operators in terms of the bosonic operators ξ † and ξ . If wechoose the ground state of these new operators to correspond to the atomic statein which all of the atoms are in their lower energy level, the transformation isgiven by S + = ξ † (cid:112) S − ξ † ξ, S − = (cid:112) S − ξ † ξξ (67)with S = N/
2. From these definitions, it follows that S z = − S + ξ † ξ. (68)If the number of excitations of the atomic system remains small with respectto the total number of atoms, i.e. (cid:104) ξ † ξ (cid:105) (cid:28) N , then we can expand the squareroots keeping only the lowest order terms S + ≈ √ N ξ † , S − ≈ √ N ξ, (69)and the Hamiltonian of the system simplifies to H = ωa † a + ωξ † ξ + κ √ N ( aξ † + a † ξ ) . (70)So far, we have grouped the atoms into a single subsystem. However, wewould like to study the entanglement between groups of atoms, so we will splitthem into two groups, each group having its own collective spin operators. Letus divide the atoms in two groups, one consisting of k atoms and the otherconsisting of N − k atoms, S +1 = (cid:80) ki =1 σ + i ≈ √ kξ † , S +2 = (cid:80) Ni = k +1 σ + i ≈ √ N − kξ † ,S − = (cid:80) ki =1 σ − i ≈ √ kξ , S − = (cid:80) Ni = k +1 σ − i ≈ √ N − kξ . (71)14n the Holstein-Primakoff representation with only the lowest order terms re-tained, the Hamiltonian of this tripartite system is H = ω ( a † a + ξ † ξ + ξ † ξ ) + κ √ k ( aξ † + a † ξ ) + κ √ N − k ( aξ † + a † ξ ) . (72)We would now like to diagonalize this Hamiltonian. In order to do so, wefirst express it as H = v † M v (73)with v = ( a, ξ , ξ ) T and M = ω κ √ k κ √ N − kκ √ k ω κ √ N − k ω . (74)The matrix M can be diagonalized by the introduction of the new modes b = 1 √ a − (cid:114) k N ξ − (cid:114) N − k N ξ ,b = (cid:114) N − kN ξ − (cid:114) kN ξ ,b = 1 √ a + (cid:114) k N ξ + (cid:114) N − k N ξ , (75)to which correspond the eigenvalues λ = ω − κ √ Nλ = ω,λ = ω + κ √ N , (76)respectively. Setting Ω = κ √ N , the Hamiltonian can now be expressed as H = ( ω − Ω) b † b + ωb † b + ( ω + Ω) b † b . (77)Note that Eq. (75) can be inverted to give a = 1 √ (cid:0) b + b (cid:1) ,ξ = (cid:114) k N (cid:0) b − b (cid:1) + (cid:114) N − kN b ,ξ = (cid:114) N − k N (cid:0) b − b (cid:1) − (cid:114) kN b . (78)From the Hamiltonian, we easily find the time evolution of the operators b , b and b in the Heisenberg picture, i.e. b ( t ) = e − i ( ω − Ω) t b (0) ,b ( t ) = e − iωt b (0) ,b ( t ) = e − i ( ω +Ω) t b (0) . (79)15ombining these relations with (75) and (78), we obtain the time evolution ofthe field and atomic operators: a ( t ) = e − iωt (cid:104) cos(Ω t ) a (0) − i (cid:114) kN sin(Ω t ) ξ (0) − i (cid:114) N − kN sin(Ω t ) ξ (0) (cid:105) ,ξ ( t ) = e − iωt (cid:104) − i (cid:114) kN sin(Ω t ) a (0) + (cid:0) N − kN + kN cos(Ω t ) (cid:1) ξ (0)+ (cid:112) k ( N − k ) N (cid:0) cos(Ω t ) − (cid:1) ξ (0) (cid:105) ,ξ ( t ) = e − iωt (cid:104) − i (cid:114) N − kN sin(Ω t ) a (0) + (cid:112) k ( N − k ) N (cid:0) cos(Ω t ) − (cid:1) ξ (0)+ (cid:0) kN + N − kN cos(Ω t ) (cid:1) ξ (0) (cid:105) . (80)Let us first consider the entanglement condition in Eq. (2), and choose A = ξ and B = ξ . The resulting inequality will tell us if the two groups of atoms areentangled. If the initial state is of the form | Ψ (cid:105) = | ψ (cid:105) | , (cid:105) , (81)then the quantities at time t appearing in the entanglement condition can beeasily calculated, since only the terms involving the field operator at t = 0 canbe non-zero. In particular, we obtain (cid:104) ξ ( t ) ξ ( t ) (cid:105) = − e − iωt (cid:112) k ( N − k ) N sin (Ω t ) (cid:104) ψ | ( a (0)) | ψ (cid:105) , (cid:104) ξ † ( t ) ξ ( t ) (cid:105) = kN sin (Ω t ) (cid:104) ψ | a † (0) a (0) | ψ (cid:105) , (cid:104) ξ † ( t ) ξ ( t ) (cid:105) = N − kN sin (Ω t ) (cid:104) ψ | a † (0) a (0) | ψ (cid:105) , (82)hence for t >
0, the entanglement condition becomes | (cid:104) ψ | ( a (0)) | ψ (cid:105) | > (cid:104) ψ | a † (0) a (0) | ψ (cid:105) . (83)We first note that a coherent state will not satisfy this condition. That is notsurprising, because under the action of our Hamiltonian, an initial state withthe field in a coherent state and the atoms in their ground states will evolveinto a product of coherent states in all three modes. Such a state is clearly notentangled. If the field is initially in a squeezed vacuum state | ψ (cid:105) = S ( z ) | (cid:105) thesituation is very different. The entanglement condition reduces tocosh r > sinh r (84)which is satisfied for any r (cid:54) = 0. Therefore, squeezing in the initial field will leadto entanglement between the two groups of atoms.Let us now consider the entanglement condition in Eq (1) and again choose A = ξ and B = ξ . If the initial state at t = 0 is of the same form as before,16 Ψ (cid:105) = | ψ (cid:105) | , (cid:105) , then the condition can be again greatly simplified as most ofthe terms on both sides of the equation are zero. We find that (cid:104) ξ † ( t ) ξ ( t ) (cid:105) = (cid:112) k ( N − k ) N sin (Ω t ) (cid:104) ψ | a † (0) a (0) | ψ (cid:105) , (cid:104) ξ † ( t ) ξ ( t ) ξ † ( t ) ξ ( t ) (cid:105) = k ( N − k ) N sin (Ω t ) (cid:104) (cid:104) ψ | ( a † (0) a (0)) | ψ (cid:105)− (cid:104) ψ | a † (0) a (0) | ψ (cid:105) (cid:105) , (85)and the entanglement condition becomes (cid:104) ψ | a † (0) a (0) | ψ (cid:105) > (cid:104) ψ | ( a † (0) a (0)) | ψ (cid:105) − (cid:104) ψ | a † (0) a (0) | ψ (cid:105) . (86)Expressed in terms of the photon number operator, n = a † a , this is just∆ n (0) < (cid:104) n (0) (cid:105) , where ∆ n = (cid:104) n (cid:105) − (cid:104) n (cid:105) is the variance of n . Therefore,we see that there will be entanglement between the two groups of atoms pro-vided that the field is initially in a state with sub-Poissonian photon statistics.Connections between nonclassical states and entanglement have been noted be-fore. For example, for the output state of a beam splitter to be entangled, theinput state must be nonclassical [12]. The amount of two-mode entanglementthat can be produced by a single-mode field using linear optics, auxiliary clas-sical fields and ideal photodetectors has even been proposed as a measure ofthe nonclassicality of a state [13]. Here we see yet another connection betweennonclassical states an entanglement in the dynamics of the Dicke model.Light has, in fact, been used to entangle atomic ensembles, though thisis often done by letting the light interact with both sets of atoms and thenmeasuring it [14]-[17]. The system we have considered here is a cartoon versionof the one in the experiment [17], but it does provide useful information in thatit shows that certain kinds of nonclassical light can lead to atomic entanglement. Let us now consider a system of field modes that is closely related, at least inits description, to the atom-field system we have just studied. The system isdepicted in Fig. 2. It consists of three field modes and two beam splitters. Onemode, mode a , passes through both beam splitters, while the other two, modes b and c , pass through just one. Our goal is to entangle modes b and c by makingthem interact with mode a through the array of two beam splitters. Becausethis system closely resembles our previous example where a light mode wasused to entangle two groups of atoms, it is not surprising that the entanglementcondition Eq. (1) again provides a usefull tool to derive simple conditions onthe input state of the ancillary mode a that will result in modes b and c beingentangled at the output. This condition was applied to determine when theoutput of a single beam splitter is entangled in [18].The system we are examining here was first studied in [19]. In that paper aseries of beam splitters was used to model a reservoir, and, conditions for the17 ,t r ,t a b c Figure 2: Entangling modes b and c using the auxiliary mode a .entanglement of reservoir modes, which would correspond to our modes b and c , were found. However, in that paper only Gaussian states were considered.We recall that a beam splitter acts on two input modes a and b as a out = ta in + rb in ,b out = − ra in + tb in , (87)where r and t are positive, and r + t = 1. For beam splitters with transmittanceand reflectance ( t , r ) and ( t , r ) respectively, the relationship between theoutput modes and the input modes is given by a out = t ( t a in + r b in ) + r c in ,b out = − r a in + t b in ,c out = − r ( t a in + r b in ) + t c in . (88)Let us suppose for simplicity that modes b and c are initially in the vacuum,i.e. | Ψ in (cid:105) = | ψ (cid:105) | (cid:105) | (cid:105) . The entanglement condition in Eq. (1) for A = b out and B = c out is |(cid:104) b † out c out (cid:105)| > (cid:104) b † out b out c † out c out (cid:105) . (89)Replacing b out and c out by their expression Eq. (88) we find (cid:104) b † out c out (cid:105) = t r r (cid:104) ψ | a † in a in | ψ (cid:105) , (cid:104) b † out b out c † out c out (cid:105) = ( r t r ) (cid:104) (cid:104) ψ | ( a † in a in ) | ψ (cid:105) − (cid:104) ψ | a † in a in | ψ (cid:105) (cid:105) , (90)and the entanglement condition becomes, assuming that r t r (cid:54) = 0, (cid:104) (cid:104) ( a † in a in ) (cid:105) − (cid:104) a † in a in (cid:105) (cid:105) < (cid:104) a † in a in (cid:105) . (91)i.e., b and c will be entangled provided that mode a is initially in a state withsub-Poissonian statistics. 18ow let us see what happens when we apply a more powerful entanglementcondition. Motivated by considerations similar to those in Section V, we set A = z b † out + z b out and B = c out in Eq. (1). We then find that a state isentangled if the 2 × M , where M = |(cid:104) b out c out (cid:105)| − (cid:104) b out b † out c † out c out (cid:105) ,M = M ∗ = (cid:104) b out c out (cid:105)(cid:104) b out c † out (cid:105) − (cid:104) b out c † out c out (cid:105) ,M = |(cid:104) b † out c out (cid:105)| − (cid:104) b † out b out c † out c out (cid:105) , (92)has a positive eigenvalue, which will be the case if | M | > M M . When weexpress this condition in terms of the input operators, and assume that the b and c modes are initially in the vacuum state, we find that the two-mode outputstate is entangled if |(cid:104) a in (cid:105)(cid:104) n (cid:105) − (cid:104) na in (cid:105)| > [ (cid:104) n (cid:105) − (∆ n )] (cid:20) |(cid:104) a in (cid:105)| − (cid:104) n (cid:105) + (cid:18) − r (cid:19) (cid:104) n (cid:105) (cid:21) , (93)where n = a † in a in . Note that the Schwarz inequality implies that the secondfactor on the right-hand side is negative, so that if the input field has sub-Poissonian statistics, then this inequality is satisfied. Therefore, this conditionincludes the one in Eq. (91). There are, however, states that satisfy the newcondition but do not satisfy the old one. For example, for the state | ψ (cid:105) = (cid:34) − (cid:18) √ (cid:15) (cid:19) (cid:35) / | (cid:105) + (cid:18) √ (cid:15) (cid:19) | (cid:105) , (94)we find that (∆ n ) − (cid:104) n (cid:105) = − √ (cid:15) , to lowest order in (cid:15) . On the other hand,Eq. (93) becomes, again keeping only lowest order terms in (cid:15) > −√ (cid:15). (95)So, if (cid:15) < So far we have been concentrating on the consequences of the inequalities inEqs. (1) and (2). Needless to say, it is possible to derive other entanglementconditions involving non-hermitian operators. One way, which we will brieflyexplore here is to apply a variant of the local uncertainty entanglement conditiondue to Hofmann and Takeuchi [20]. Let us start by considering two modes. Nowsuppose that A is an operator on mode a and B is an operator on mode b . Wewant to first find an expression for (cid:104) ( A † + B † )( A + B ) (cid:105)−|(cid:104) A + B (cid:105)| for a separablestate ρ = (cid:88) k p k ρ ( a ) k ⊗ ρ ( b ) k . (96)19e first note that for an operator D and a density matrix ρ = (cid:88) k p k ρ k , (97)we have that (cid:104) D † D (cid:105) − |(cid:104) D (cid:105)| = (cid:88) k p k [ (cid:104) D † D (cid:105) k − (cid:104) D † (cid:105) k (cid:104) D (cid:105) k + ( (cid:104) D † (cid:105) k − (cid:104) D † (cid:105) )( (cid:104) D (cid:105) k − (cid:104) D (cid:105) )] ≥ (cid:88) k p k ( (cid:104) D † D (cid:105) k − (cid:104) D † (cid:105) k (cid:104) D (cid:105) k ) , (98)where expectation values with the subscript k denote the expectation value withrespect to ρ k . If we now apply this to D = A + B with the separable densitymatrix above, we find that (cid:104) ( A † + B † )( A + B ) (cid:105) − |(cid:104) A + B (cid:105)| ≥ (cid:88) k p k ( (cid:104) A † A (cid:105) k − |(cid:104) A (cid:105) k | + (cid:104) B † B (cid:105) k − |(cid:104) B (cid:105) k | ) . (99)This condition can easily be extended to the case in which we have more thanone operator for each subsystem. If we have M operators for mode a , A j and M for mode b , B j , then the above inequality generalizes to M (cid:88) j =1 (cid:104) ( A † j + B † j )( A j + B j ) (cid:105) − |(cid:104) A j + B j (cid:105)| ≥ (cid:88) k p k M (cid:88) j =1 ( (cid:104) A † j A j (cid:105) k − |(cid:104) A j (cid:105) k | + (cid:104) B † j B j (cid:105) k − |(cid:104) B j (cid:105) k | ) . (100)These two inequalities will hold for all separable density matrices, so if a stateviolates them, it must be entangled.As a simple example of an entanglement condition that can be derived fromEq. (99), let us set A = a and B = b † . We then have that (cid:104) ( a † + b )( a + b † ) (cid:105) − |(cid:104) a + b † (cid:105)| ≥ (cid:88) k p k ( (cid:104) a † a (cid:105) k − |(cid:104) a (cid:105) k | + (cid:104) bb † (cid:105) k − |(cid:104) b (cid:105) k | ) ≥ . (101)If this inequality is violated by a state, that state is entangled. A two-modesqueezed vacuum state, | ψ sq (cid:105) = e r ( a † b † − ab ) | (cid:105) , (102)will violate this inequality. We find that for this state (cid:104) ( a † + b )( a + b † ) (cid:105) − |(cid:104) a + b † (cid:105)| = e − r , (103)so that for r >
0, the condition is violated, and we can conclude that the stateis entangled. It should be noted that this condition can also be derived fromthe analysis given by Shchukin and Vogel [8].20 very similar condition can be derived for atom-field entanglement. If wehave N two-level atoms represented by collective spin operators S + , S − , and S z , then setting A = a † and B = J + we find that for a separable state (cid:104) ( a + S − )( a † + S + ) (cid:105) − |(cid:104) a + S − (cid:105)| ≥ , (104)so that any state that violates this inequality must be entangled. A very simpleexample is the state, for N = 1 | ψ (cid:105) = cos θ | e (cid:105)| (cid:105) + sin θ | g (cid:105)| (cid:105) , (105)for 0 < θ < π/ We have presented a number of entanglement conditions that employ non-hermitian operators. While these conditions follow from the partial transposecondition, they are much simpler to use. It is only necessary to compute a smallnumber of correlation functions rather than to have to diagonalize the partiallytransposed density matrix for the entire system. We used these conditions toshow the presence of entanglement in a number of systems of interest in quantumoptics, including Jaynes-Cummings model and the Dicke model. The conditionsare sufficiently flexible that we can study entanglement between modes, entan-glement between a field mode and an atom, and entanglement between groupsof atoms. Most of these conditions are invariant under some set of local unitarytransformations. For field modes we considered local Gaussian operations. Bymaking the conditions invariant under a set of local operations, we are able tomake them able to detect entanglement in a wider class of states.
Acknowledgments
Julien Niset acknowledges support from the Belgian American EducationalFoundation, and Ho Trung Dung acknowledges support from the Vietnam Ed-ucation Foundation. This research was partially supported by the NationalScience Foundation under grant PHY-0903660.
Appendix
Here we present the details needed for the calculations involving the Tavis-Cummings model. As we noted, the Hilbert space splits up into a direct sumof invariant subspaces, H ( n ) ,which are characterized by the total excitationnumber, n . H (0) is spanned by {| , g, g (cid:105)} so the ground state of the system is (cid:12)(cid:12)(cid:12) ψ (0)0 (cid:69) = | , g, g (cid:105) (106)21ith eigenvalue E (0)0 = (cid:104) , g, g | H | , g, g (cid:105) = − ω. (107) H (1) is spanned by {| , g, g (cid:105) , | , e, g (cid:105) , | , g, e (cid:105)} , and the Hamiltonian can bewritten in this basis as H (1) = κ κκ κ . (108)This Hamiltonian has three eigenvalues: E (1)0 = − κ √ ,E (1)1 = 0 ,E (1)2 = κ √ , (109)to which we associate the three eigenvectors (cid:12)(cid:12)(cid:12) ψ (1)0 (cid:69) = 1 √ | , g, g (cid:105) − √ (cid:2) | , e, g (cid:105) + | , g, e (cid:105)√ (cid:3) , (cid:12)(cid:12)(cid:12) ψ (1)1 (cid:69) = 1 √ (cid:2) | , e, g (cid:105) − | , g, e (cid:105) (cid:3) , (cid:12)(cid:12)(cid:12) ψ (1)2 (cid:69) = 1 √ | , g, g (cid:105) + 1 √ (cid:2) | , e, g (cid:105) + | , g, e (cid:105)√ (cid:3) . (110) H ( n ) is spanned by {| n, g, g (cid:105) , | n − , e, g (cid:105) , | n − , g, e (cid:105) , | n − , e, e (cid:105)} , and theHamiltonian can be written in this basis as H ( n ) = ω ( n − κ √ n κ √ n κ √ n ω ( n −
1) 0 κ √ n − κ √ n ω ( n − κ √ n − κ √ n − κ √ n − ω ( n − . (111)This Hamiltonian has four eigenvalues: E ( n )0 = ω ( n − − κ (cid:112) n − ,E ( n )1 , = ω ( n − ,E ( n )3 = ω ( n −
1) + κ (cid:112) n − , (112)to which we can associate the four eigenvectors (cid:12)(cid:12)(cid:12) ψ ( n )0 (cid:69) = (cid:114) n n − | n, g, g (cid:105) − √ (cid:2) | n − , e, g (cid:105) + | n − , g, e (cid:105)√ (cid:3) , + (cid:115) n − n − | n − , e, e (cid:105) , (cid:12)(cid:12)(cid:12) ψ ( n )1 (cid:69) = 1 √ (cid:2) | n − , e, g (cid:105) − | n − , g, e (cid:105) (cid:3) , (cid:12)(cid:12) ψ ( n )2 (cid:69) = (cid:114) n − n − | n, g, g (cid:105) − (cid:114) n n − | n − , e, e (cid:105) , (cid:12)(cid:12)(cid:12) ψ ( n )3 (cid:69) = (cid:114) n n − | n, g, g (cid:105) + 1 √ (cid:2) | n − , e, g (cid:105) + | n − , g, e (cid:105)√ (cid:3) + (cid:115) n − n − | n − , e, e (cid:105) . (113)Suppose that at time t = 0, the field contains n excitations while both atomsare in their ground state, i.e. | Ψ(0) (cid:105) = | n, g, g (cid:105) (114)= (cid:114) n n − (cid:12)(cid:12)(cid:12) ψ ( n )0 (cid:69) + (cid:114) n − n − (cid:12)(cid:12)(cid:12) ψ ( n )2 (cid:69) + (cid:114) n n − (cid:12)(cid:12)(cid:12) ψ ( n )3 (cid:69) . Note that this expression is valid for any n >
0, provided that the appropriateeigenvectors are chosen. In particular, when n = 1, (cid:12)(cid:12)(cid:12) ψ ( n )3 (cid:69) corresponds to (cid:12)(cid:12)(cid:12) ψ (1)2 (cid:69) as defined in Eq. (110).After a time t , the state of the system will have evolved according to | Ψ( t ) (cid:105) = e − iH ( n ) t | Ψ(0) (cid:105) = (cid:114) n n − e − iE ( n )0 t (cid:12)(cid:12)(cid:12) ψ ( n )0 (cid:69) + (cid:114) n − n − e − iE ( n )2 t (cid:12)(cid:12)(cid:12) ψ ( n )2 (cid:69) + (cid:114) n n − e − iE ( n )3 t (cid:12)(cid:12)(cid:12) ψ ( n )3 (cid:69) = e − iω ( n − t (cid:104) n − (cid:0) n cos(Ω t ) + n − (cid:1) | n, g, g (cid:105) + (cid:112) n ( n − n − (cid:0) cos(Ω t ) − (cid:1) | n − , e, e (cid:105)− i (cid:114) n n − t ) (cid:2) | n − , e, g (cid:105) + | n − , g, e (cid:105)√ (cid:3)(cid:105) , (115)where Ω = κ (cid:112) n − References [1] R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, Rev. Mod.Phys. , 865 (2009).[2] O. G¨uhne and G. Toth, Physics Reports , 1 (2009).[3] R. Simon, Phys. Rev. Lett. , 2726 (2000).[4] L. -M. Duan, G. Giedke, J. I. Cirac, and P. Zoller, Phys. Rev. Lett. ,2722 (2000). 235] S. Mancini, V. Giovannetti, D. Vitali, and P. Tombesi, Phys. Rev. Lett. , 120401 (2002).[6] G. S. Agarwal and A. Biswas, New Journal of Physics , 211 (2005).[7] Mark Hillery and M. Suhail Zubairy, Phys. Rev. Lett. , 050503 (2006).[8] E. Shchukin and W. Vogel, Phys. Rev. Lett. , 230502 (2005).[9] Geza Toth, Christoph Simon, and Juan Ignacio Cirac, Phys. Rev. A ,062310 (2003).[10] Hyunchul Nha and Jaewan Kim, Phys. Rev. A , 012317 (2006).[11] V. Buˇzek, M. Orszag, and M. Rosko, Phys. Rev. Lett. , 163601 (2005).[12] M. S. Kim, W. Son, V. Buˇzek, and P. L. Knight, Phys. Rev. A , 173602(2005).[14] L. -M. Duan, J. I. Cirac, P. Zoller, and E. S. Polzik, Phys. Rev. Lett. ,5643 (2000).[15] A. Kuzmich and E. Polzik, Phys. Rev. Lett. , 5639 (2000).[16] For a review see K. Hammerer, A. S. Sorensen, and E. S. Polzik, quant-ph/0807.3358.[17] B. Julsgaard, A. Kozhekin, and E. Polzik, Nature , 6854 (2001).[18] Mark Hillery and M. Suhail Zubairy, Phys. Rev. A , 032333 (2006).[19] Daniel Nagaj, Peter ˇStelmachoviˇc, Vladimir Buˇzek, and Myungshik Kim,Phys. Rev. A , 062307 (2002).[20] Holger Hofmann and Shigeki Takeuchi, Phys. Rev. A68