Table Of ContentTwo-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
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t
n August 11, 2014
a
u
q
[
2 Abstract
v
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In this paper, we have given the spin states of two-photon, which are expressed by the
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1 quadraticcombinationoftwosingle photonspinstates,theyareakindofquantumexpression.
2 Otherwise,wegive allentanglementstatesoftwo-photon,whicharedifferentfromthexanzsd
.
1 classicalpolarizationvectorexpressionoftwo-photonentanglementstate.
0
PACS:42.50.-p,42.50.Dv,03.65.Ud
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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 entangle-
ment 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
