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Two-photon spin states and entanglement states 4 1 Xiao-Jing Liua,b, Xiang-Yao Wub∗, Jing-Bin Lua, Si-Qi Zhangb, Hong-Lib 0 2 Ji Mab, Hong-Chun Yuanc, Heng-Mei Lic, Hai-Xin Gaod and Jing-Wu Lie g a)InstituteofPhysics,JilinUniversity,Changchun130012China u A b)InstituteofPhysics,JilinNormalUniversity,Siping136000China 8 c)CollegeofOptoelectronicEngineering,ChangzhouInstituteofTechnology,Changzhou213002,China d)InstituteofPhysics,NortheastNormalUniversity,Changchun130024China ] h e)InstituteofPhysics,XuzhouNormalUniversity,Xuzhou221000China p - t n August 11, 2014 a u q [ 2 Abstract v 2 In this paper, we have given the spin states of two-photon, which are expressed by the 5 1 quadraticcombinationoftwosingle photonspinstates,theyareakindofquantumexpression. 2 Otherwise,wegive allentanglementstatesoftwo-photon,whicharedifferentfromthexanzsd . 1 classicalpolarizationvectorexpressionoftwo-photonentanglementstate. 0 PACS:42.50.-p,42.50.Dv,03.65.Ud 4 1 Keywords: two-photon;spinstates;entanglementstates : v i X 1 Introduction r a Quantumentanglement,andits inherentnonlocalproperties, areamongthe mostfascinating and challengingfeaturesofthequantumworld. Inaddition, entanglementplaysacentralroleinquan- tum information [1–5]. Since its first description in the decade of 1930 [6], and in spite of the decisive contribution of Bell [7] and the subsequent experimental studies [8], entanglementstays evennowasarathermysteriousandpuzzlingpropertyofbipartitequantumobjects. Entanglement is not only a fundamental concept in Quantum Mechanics with profound implications, but also a basic ingredient of many recent technological applications that has been put forward in quantum communicationsandquantumcomputing[9,10]. Entanglementisaveryspecialtypeofcorrelation between particles that can exist in spite of how distant they are. Nevertheless, the term entanglement is sometimes also used to refer to certain correlations existing between different degrees of freedom of a single particle [11]. By and large, the most common method to generate photonic entanglement, that is entanglement between photons, is the process of spontaneous parametric ∗E-mail: [email protected] 1 down-conversion (SPDC) [12]. In SPDC, two lower-frequency photons are generated when an intense higher-frequency pump beam interacts with the atoms of a non-centrosymetric nonlinear crystal. Entanglementcan reside in any ofthe degreesoffreedomsthat characterizelight: angular momentum(polarizationandorbitalangularmomentum),momentumandfrequency,orinseveral ofthem,whatis knownashyper-entanglement. Undoubtedly,polarizationis the mostwidely used resource to generate entanglement between photons thanks to the existence of many optical ele- mentstocontrolthepolarizationoflightandtotheeasinessofitsmanipulationwhencomparedto other characteristics of alightbeam,e.g., its spatial shape or bandwidth. Entangledstates of photons are the basic resourcein the successful implementation of quantum information processing applications, namely optical quantum computing [13,14], and quantum cryptography,orquantumkeydistribution[15,16]. Also,intheexperimentalstudyoffundamental problems in Quantum Mechanics, as loophole free tests of the violation of the Bell’s inequalities [17],thedelayed-choicequantumeraser[18],quantumteleportationandentanglementswapping [19], generation of states with a large number of particles and generalizedtypes of entanglement [20], etc. In this paper, we have given the spin states of two-photon, which are expressed by the quadratic combination of two single photon spin states, they are a kind of quantum expression. Otherwise, we give all entanglement states of two-photon, which are different from the xanzsd classical polarization vector expression oftwo-photon entanglementstate. 2 The spin state of two-photon The following will give some comments on the real meaning of the state of photon. In quantum electrodynamics,the spin vectoroperator ofthe photon is [21,22](in natural unit system ~ = c = 1) 0 0 0 0 0 i 0 i 0 − s = 0 0 i ,s = 0 0 0 ,s = i 0 0 , (1) x   y   z   − 0 i 0 i 0 0 0 0 0    −          1 0 0 S~2 = s2 +s2 +s2 = 2 0 1 0 , (2) x y z   0 0 1     the spin vector χ ofS2 ands satisfy the following equations µ z S~2χ =2χ , (3) µ µ s χ = µχ , (4) z µ µ 2 these are 0 1 1 1 1 χ0 =  0 ,χ1 =  i ,χ−1 =  i , (5) −√2 √2 − 1 0 0             the total spin S~ is the sum oftwo photons spin s and s 1 2 S~ =~s +~s , (6) 1 2 andS~2 eigenvalueis S~2 = S(S +1), (7) andquantum numberS are S = 0,1,2. (8) For the given S, s eigenvalues are: µ = S, S +1, S, i.e., there are 2S +1 eigenvalues, and z − − ··· 2S + 1 eigenfunctions of spin. For S = 0, there is one eigenfunction, for S = 1, there are three eigenfunctions,andforS = 2,therearefiveeigenfunctions,i.e.,thereareninespinwave-functions χ for two-photon, which are expressed by the quadratic combination of two single photon spin Sµ wave functionsχ and χ . We try to write the ninespin wave functionsof two-photon, they are µ1 µ2 (1) S = 2 spin wave functions (1) χ = χ (s ,s ) = χ (s )χ (s ), (S = 2,µ = 2) (9) S 22 1z 2z 1 1z 1 2z 1 (2) χ =χ (s ,s ) = [χ (s )χ (s )+χ (s )χ (s )], (S = 2,µ = 1) (10) S 21 1z 2z √2 0 1z 1 2z 0 2z 1 1z 1 (3) χS = χ20(s1z,s2z)= √2[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)], (S = 2,µ = 0) (11) 1 (4) χS = χ2−1(s1z,s2z) = √2[χ0(s1z)χ−1(s2z)+χ0(s2z)χ−1(s1z)], (S = 2,µ = −1) (12) (5) χS = χ2−2(s1z,s2z) = χ−1(s1z)χ−1(s2z), (S = 2,µ = −2) (13) (2) S = 0 spin wave function (6) χ = χ (s ,s ) = χ (s )χ (s ), (S = 0,µ = 0) (14) S 00 1z 2z 0 1z 0 2z (3) S = 1 spin wave functions 1 (1) χ =χ (s ,s ) = [χ (s )χ (s ) χ (s )χ (s )], (S = 1,µ = 1) (15) A 11 1z 2z √2 0 1z 1 2z − 0 2z 1 1z 1 (2) χA = χ10(s1z,s2z)= √2[χ1(s1z)χ−1(s2z)−χ1(s2z)χ−1(s1z)], (S = 1,µ = 0) (16) 3 1 (3) χA = χ1−1(s1z,s2z) = √2[χ0(s1z)χ−1(s2z)−χ0(s2z)χ−1(s1z)], (S = 1,µ = −1) (17) In the following, we shall check the above spin wave functions, the two-photon total spin square S~2 andits z components are z S~2 = (s~ +s~ )2 =~s 2+~s 2+2(s s +s s +s s ) 1 2 1 2 1x 2x 1y 2y 1z 2z = 4+2(s s +s s +s s ), (18) 1x 2x 1y 2y 1z 2z s = s +s , (19) z 1z 2z by spin wave functions(5),we have 0 0 0 0 1 sxχ0 =  0 0 i  0  = (χ1+χ−1), (20) − √2 0 i 0 1       0 0 0 1 1 1 s χ = 0 0 i i = χ , (21) x 1    0 −√2 − √2 0 i 0 0       0 0 0 1 1 1 sxχ−1 =  0 0 i  i  = χ0, (22) √2 − − √2 0 i 0 0       i syχ0 = (χ1 χ−1), (23) −√2 − i s χ = χ , (24) y 1 0 √2 i syχ−1 = χ0, (25) −√2 s χ = 0 χ , (26) z 0 0 · s χ = 1 χ , (27) z 1 1 · szχ−1 = 1 χ−1, (28) − · with equations (20)-(28), we cancheckspin wave functions. (1) checkingS = 2spin wave functions (1) (a) checkingspin wave functionχ : S (1) s χ = (s +s )χ (s )χ (s ) z S 1z 2z 1 1z 1 2z = (s χ (s ))χ (s )+χ (s )(s χ (s )) 1z 1 1z 1 2z 1 1z 2z 1 2z = 2χ (s )χ (s ), (29) 1 1z 1 2z 4 i.e., µ = 2, (30) S~2χ(1) = [4+2(s s +s s +s s )]χ (s )χ (s ) S 1x 2x 1y 2y 1z 2z 1 1z 1 2z 1 1 i i = 4χ (s )χ (s )+2[ χ (s ) χ (s )+ χ (s ) χ (s )+χ (s )χ (s )] 1 1z 1 2z 0 1z 0 2z 0 1z 0 2z 1 1z 1 2z √2 √2 √2 √2 = 6χ (s )χ (s ), (31) 1 1z 1 2z i.e., S = 2, (32) (1) we have checkedχ is the spin wave function of two-photon correspondingto S = 2and µ = 2. S (2) (b) checkingspin wave function χ : S 1 (2) s χ = (s +s ) [χ (s )χ (s )+χ (s )χ (s )] z S 1z 2z √2 0 1z 1 2z 0 2z 1 1z 1 = [χ (s )χ (s )+χ (s )χ (s )], (33) 0 2z 1 1z 0 1z 1 2z √2 i.e., µ = 1, (34) 1 S~2χ(2) = [4+2(s s +s s +s s )] [χ (s )χ (s )+χ (s )χ (s )] S 1x 2x 1y 2y 1z 2z √2 0 1z 1 2z 0 2z 1 1z = 2√2[χ (s )χ (s )+χ (s )χ (s )] 0 1z 1 2z 0 2z 1 1z 1 1 +√2[ (χ1(s1z)χ0(s2z)+χ−1(s1z)χ0(s2z))+ (χ1(s2z)χ0(s1z)+χ−1(s2z)χ0(s1z))] 2 2 1 1 +√2[ (χ1(s1z)χ0(s2z) χ−1(s1z)χ0(s2z))+ (χ1(s2z)χ0(s1z) χ−1(s2z)χ0(s1z))] 2 − 2 − 1 = 6 (χ (s )χ (s )+χ (s )χ (s )), (35) 0 1z 1 2z 0 2z 1 1z · √2 i.e., S = 2, (36) (2) we have checkedχ is the spin wave function of two-photon correspondingto S = 2and µ = 1. S 5 (3) (c)checkingspin wave function χ : S 1 (3) szχS = (s1z +s2z)√2[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] 1 = [(s1zχ1(s1z))χ−1(s2z)+χ1(s2z)(s1zχ−1(s1z)] √2 1 + [χ1(s1z)(s2zχ−1(s2z))+(s2zχ1(s2z))χ−1(s1z)] √2 = 0, (37) i.e., µ = 0, (38) 1 S~2χ(S3) = [4+2(s1xs2x+s1ys2y +s1zs2z)]√2[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] = 2√2[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] 1 1 1 1 +√2[ χ (s ) χ (s ))+ χ (s ) χ (s ))] 0 1z 0 1z 0 2z 0 1z √2 √2 √2 √2 i 1 i 1 +√2[ χ (s )( )χ (s ))+ χ (s )( )χ (s ))] 0 1z 0 2z 0 2z 0 1z √2 −√2 √2 −√2 +√2[ χ1(s1z)χ−1(s2z)) χ1(s2z)χ−1(s1z))] − − 1 = 2 [χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] · √2 1 +2 [χ (s )χ (s )+χ (s )χ (s )], (39) 0 1z 0 2z 0 2z 0 1z · √2 we find that χ(3) is not the common eigenstate of S~2,s , corresponding to S = 2 and µ = 0. S { z} (3) (3) Their common eigenstate χ can be obtained by the linear superposition of χ and χ , the χ S 20 00 S can bewritten as (3) χS = aχ0(s1z)χ0(s2z)+b[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)], (40) by the eigenequations S~2χ(3) = 6χ(3) S S , (41) (3) ( szχS = 0 we can calculatethe superposition coefficientsa andb (3) szχS = a(s1zχ0(s1z))χ0(s2z)+b[(s1zχ1(s1z))χ−1(s2z)+(s1zχ−1(s1z))χ1(s2z)] +a(s2zχ0(s2z))χ0(s1z)+b[(s2zχ−1(s2z))χ1(s1z)+(s2zχ1(s2z))χ−1(s1z)] = b[χ1(s1z)χ−1(s2z) χ1(s2z)χ−1(s1z)]+b[ χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] − − = 0, (42) 6 i.e., µ 0, (43) ≡ S~2χ(S3) = 4[aχ0(s1z)χ0(s2z)+b(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))] +2[a(s1xχ0(s1z))(s2xχ0(s2z))+b((s1xχ1(s1z))(s2xχ−1(s2z))+(s1xχ−1(s1z))(s2xχ1(s2z)))] +2[a(s1yχ0(s1z))(s2yχ0(s2z))+b((s1yχ1(s1z))(s2yχ−1(s2z))+(s1yχ−1(s1z))(s2yχ1(s2z)))] +2[a(s1zχ0(s1z))(s2zχ0(s2z))+b((s1zχ1(s1z))(s2zχ−1(s2z))+(s1zχ−1(s1z))(s2zχ1(s2z)))] +2[b( χ1(s1z))χ−1(s2z) (χ−1(s1z))χ1(s2z)] − − = 4[aχ0(s1z)χ0(s2z)+b(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))]+2[a(χ1(s1z)χ−1(s2z) +χ1(s2z)χ−1(s1z))+2bχ0(s1z)χ0(s2z) b(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))] − = 4(a+b)χ0(s1z)χ0(s2z)+2(a+b)(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)) = 6[aχ0(s1z)χ0(s2z)+b(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))], (44) comparingthe coefficientsof the same terms, we obtain 4(a+b) = 6a , (45) ( 2(a+b) = 6b i.e., a = 2b, (46) then (3) χS =2bχ0(s1z)χ0(s2z)+b[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)], (47) by condition ofnormalization (χ(3))+χ(3) = 1, (48) S S we have 1 b = , (49) √6 (3) we getthe eigenstate χ ,it is S 2 1 (3) χS = √6χ0(s1z)χ0(s2z)+ √6[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)], (50) 7 (4) (d) checkingspin wave function χ : S 1 (4) szχS = √2[(s1zχ0(s1z))χ−1(s2z)+χ0(s2z)(s1zχ−1(s1z))] 1 + [χ0(s1z)(s2zχ−1(s2z))+(s2zχ0(s2z))χ−1(s1z)] √2 1 = [χ0(s1z)χ−1(s2z)+χ0(s2z)χ−1(s1z)], (51) −√2 i.e., µ = 1, (52) − 1 S~2χ(S4) = 4χ(S4) +2√2[(s1xχ0(s1z))(s2xχ−1(s2z))+(s2xχ0(s2z))(s1xχ−1(s1z))] 1 +2 [(s1yχ0(s1z))(s2yχ−1(s2z))+(s2yχ0(s2z))(s1yχ−1(s1z))] √2 1 +2 [(s1zχ0(s1z))(s2zχ−1(s2z))+(s2zχ0(s2z))(s1zχ−1(s1z))] √2 1 (4) = 4χS +2√2[χ−1(s1z)χ0(s2z)+χ−1(s2z)χ0(s1z)] (4) = 6χ , (53) S i.e., S = 2, (54) (4) wehavecheckedχ isthe spinwavefunctionoftwo-photon correspondingto S = 2andµ = 1. S − (5) (e)checkingspin wave function χ : S (5) szχS = (s1zχ−1(s1z))χ−1(s2z)+χ−1(s1z)(s2zχ−1(s2z)) = χ1(s1z)χ−1(s2z) χ−1(s1z)χ−1(s2z) − − = 2[χ1(s1z)χ−1(s2z)], (55) − i.e., µ = 2, (56) − S~2χ(S5) = 4χ−1(s1z)χ−1(s2z)+2[(s1xχ−1(s1z))(s2xχ−1(s2z)) +(s1yχ−1(s1z))(s2yχ−1(s2z))+(s1zχ−1(s1z))(s2zχ−1(s2z))] 1 1 i i = 4χ−1(s1z)χ−1(s2z)+2[ χ0(s1z) χ0(s2z)+ − χ0(s1z)− χ0(s2z) √2 √2 √2 √2 +χ−1(s1z)χ−1(s2z)] = 6χ−1(s1z)χ−1(s2z), (57) 8 i.e., S = 2, (58) (5) wehavecheckedχ isthe spinwavefunctionoftwo-photon correspondingto S = 2andµ = 2. S − (6) (2) checkingS = 0spin wave functionsχ : S s χ(6) = (s χ (s ))χ (s )+χ (s )(s χ (s )) = 0, (59) z s 1z 0 1z 0 2z 0 1z 2z 0 2z i.e., µ = 0, (60) S~2χ(6) = 4χ (s )χ (s )+2[(s χ (s ))(s χ (s )) S 0 1z 0 2z 1x 0 1z 2x 0 2z +(s χ (s ))(s χ (s ))+(s χ (s ))(s χ (s ))] 1y 0 1z 2y 0 2z 1z 0 1z 2z 0 2z 1 1 = 4χ0(s1z)χ0(s2z)+2[ (χ1(s1z)+χ−1(s1z)) (χ1(s2z)+χ−1(s2z)) √2 √2 i i + − (χ1(s1z) χ−1(s1z))− (χ1(s2z) χ−1(s2z))] √2 − √2 − = 4χ0(s1z)χ0(s2z)+2[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)] = 0, (61) 6 we find χ(6) is not the common eigenstate of S~2,s , corresponding to S = 0 and µ = 0. Their S { z} (6) commoneigenstate χ can be written as equation (40), we have S (6) s χ = 0, (62) z S i.e., µ 0, (63) ≡ and S~2χ(S6) = 4[aχ0(s1z)χ0(s2z)+b(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))] +4bχ0(s1z)χ0(s2z)+2(a b)(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)) − = 4(a+b)χ0(s1z)χ0(s2z)+2(a+b)(χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z))) = 0, (64) we have 4(a+b) = 0 , (65) ( 2(a+b) = 0 i.e., a = b, (66) − 9 then (6) χS = aχ0(s1z)χ0(s2z)−a[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)], (67) by condition ofnormalization (χ(6))+χ(6) = 1, (68) S S 1 a = , (69) √3 (6) we getthe eigenstate χ ,it is S 1 1 (6) χS = √3χ0(s1z)χ0(s2z)− √3[χ1(s1z)χ−1(s2z)+χ1(s2z)χ−1(s1z)]. (70) (3) S = 1 spin wave functions (1) (a) checkingspin wave functionχ : A 1 (1) s X = (s +s ) [χ (s )χ (s ) χ (s )χ (s )] z A 1z 2z √2 0 1z 1 2z − 0 2z 1 1z 1 = [(s χ (s ))χ (s ) χ (s )(s χ (s ))] 1z 0 1z 1 2z 0 2z 1z 1 1z √2 − 1 + [χ (s )(s χ (s )) (s χ (s ))χ (s )] 0 1z 2z 1 2z 2z 0 2z 1 1z √2 − 1 = [χ (s )χ (s ) χ (s )χ (s )], (71) 0 1z 1 2z 0 2z 1 1z √2 − i.e., µ = 1, (72) S~2χ(1) = 4χ(1) +√2[(s χ (s ))(s χ (s )) (s χ (s ))(s χ (s ))] A A 1x 0 1z 2x 1 2z − 2x 0 2z 1x 1 1z +√2[(s χ (s ))(s χ (s )) (s χ (s ))(s χ (s ))] 1y 0 1z 2y 1 2z 2y 0 2z 1y 1 1z − +√2[(s χ (s ))(s χ (s )) (s χ (s ))(s χ (s ))] 1z 0 1z 2z 1 2z 2z 0 2z 1z 1 1z − 1 1 1 1 = 4χ(A1) +√2[√2(χ1(s1z)+χ−1(s1z))√2χ0(s2z)− √2(χ1(s2z)+χ−1(s2z))√2χ0(s1z)] i i i i +√2[ (χ1(s1z) χ−1(s1z)) χ0(s2z)+ (χ1(s2z) χ−1(s2z)) χ0(s1z)] −√2 − √2 √2 − √2 = 4χ(1) +√2[χ (s )χ (s ) χ (s )χ (s )] A 1 1z 0 2z − 1 2z 0 1z (1) (1) (1) = 4χ 2χ = 2χ , (73) A − A A i.e., S = 1, (74) (1) we have checkedχ is the spin wave function of two-photon correspondingto S = 1and µ = 1. A 10