Table Of ContentRelativistic spinor equation of photon
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2 Xiang-YaoWua∗,HongLia,Xiao-JingLiua,Si-QiZhanga,
n JiMaa,Guang-HuaiWanga,Hai-XinGaob,Heng-MeiLicandHong-ChunYuanc
a
J a)InstituteofPhysics,JilinNormalUniversity,Siping136000China
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b)InstituteofPhysics,NortheastNormalUniversity,Changchun130024China
] c)CollegeofOptoelectronicEngineering,ChangzhouInstituteofTechnology,Changzhou213002,China
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- January 6, 2014
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g
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s Abstract
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h In this paper, we have proposed the spiron equation of free and non-free photon, and give
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the spin operator and spin wave function of photon. We calculate the helicity of photon and
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prove there are left-handed and right-handed photon. By the spironequation of non-free pho-
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v ton, we can studythe quantum propertyof photonin medium, which can be used in quantum
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optics,photoniccrystalsandsoon.
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0 Keywords: Photonspironequation;Spinoperator;Spinwavefunction
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1 PACS:03.65.Pm,42.50.-p,42.50.Ct
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i 1 Introduction
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r
a
Maxwell has unified the laws of electricity and magnetism in a consistent way into a set of four
equations,andhecalculatedthespeedofelectromagneticwavesandfoundthatitwas300,000km/sec,
whichwasthesameoflight. ThisforcedMaxwelltoponderaboutthenatureoflight: heconcluded
that light is a form of electromagnetic wave. For the explanation the spectra of black-body radi-
ation, Planck firstly proposed that emission of black-body was energy quantization with value of
~ω, and he introduced the Planck constant ~ [1]. In photoelectric effect experiment, Einstein only
considered the energy conservation and made use of Planck’s ~ω to propose a quanta concept
of light [2], and successfully explaining photoelectric effect phenomena, It was the beginning of
Quantum Physics.
Despite claims that massless particles cannot be localized in space-time [3,4], physicists have
pondered over the problem of photon localization and the related problem of the photon wave
∗E-mail: [email protected]
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functionfor almost80 yearsnow,beginningwith the work ofLandau andPeierls[5]. An extensive
review of the photon localization problem was presented by Keller [6]. The problem of photon lo-
calizationiscloselyrelatedtothewidelystudiedproblemofthephotonpositionoperator. Quantum
optics has long struggled to define a quantum wavefunction for photons that would be in agree-
ment with innumerable experiments revealing the possibility of photon localization. The photon
wave function appears in both ofthese descriptions and that it provides avery convenientconcept
to unify the two points of view. A very useful mathematical tool in this analysis is the Riemann-
Silbersteinvector [7–11]. Applicationsof the RS vector to manyphysical problemswerereviewed
in a verythoroughpaper by Keller[6].
After Diracdiscoveredthe relativistic equation for aparticlewith spin 1/2[12],muchworkwas
done to study spinor and vectors within the Lorentz group theory. Any quantity which transforms
linearlyunderLorentztransformationsis aspinor. For thatreason spinor quantities areconsidered
as fundamental in quantum field theory and basic equations for such quantities should be written
inaspinorform. AspinorformulationofMaxwellequationswasstudiedbyLaporteandUhlenbeck
[13],alsoseeRumer[14]. In1931,Oppenheimer[15]proposedtoconsidertheMaxwelltheoryof
electromagnetismasthe wavemechanicsofthe photon. Theyintroducedacomplex3-vectorwave
function satisfying the massless Dirac-like equations. In this paper, we shall give the relativistic
spinor equation for photon with spin 1 andmass 0.
2 Relativistic spinor equation of free photon
Dirac derived the Dirac equation by factorizing Einstein’s dispersion relation such that the field
equation becomes the first order in time derivative [16]. Namely, he factorized the relativistic
dispersion relation employingfour by four matrices
E2 c2~p 2 m2c4 = (E cp~ α~ m c2β)(E +cp~ α~ +m c2β), (1)
− − 0 − · − 0 · 0
where α~ and β are Dirac matrices. Fora photon, massm = 0, equation (1) becomes
0
E2 c2~p 2 = (E cα~ p~)(E +cα~ ~p), (2)
− − · ·
i.e.,
E cα~ ~p = 0, (3)
− ·
canonicalquantization equation (3),we have
∂
i~ ψ = ic~α~ ~ψ = Hψ, (4)
∂t − ·▽
wherethe photon’s Hamiltonian operatorH = ic~α~ ~,andψ isspiron wavefunctionofphoton.
− ·▽
FortheproperLorentzgroupL ,theirreducibilityrepresentationsofspin1photonisD10,D01 and
p
2
11
D22, respectively, and the dimension of irreducibility representations are corresponding to three,
three and four. We choose photon’s spiron wave function as the basis vector of three dimension
irreducibility representations,it is
ψ
1
ψ = ψ , (5)
2
ψ
3
andα~ matricesare taken as
0 0 0 0 0 i~ 0 i~ 0
−
α = 0 0 i~ ,α = 0 0 0 ,α = i~ 0 0 , (6)
x y z
−
0 i~ 0 i~ 0 0 0 0 0
−
they areHermitian matrices
α~† = α~, (7)
andphoton’s Hamiltonian operator is Hermitian
H† = H. (8)
3 The spin operators of photon
In the following, we shall prove the selection of α~ matrices are reasonable, they confirm that
equation (4) is photon’s spiron equation of spin 1. equation (4) canbe written as
∂
i~ ψ = c(p~ α~)= Hψ, (9)
∂t ·
where
H = cp~ α~, (10)
·
The orbital angularmomentumof photon satisfy
d 1
L = [L ,H]
dt x i~ x
= c(α p α p )
y z z y
−
= c(α~ p~) , (11)
x
×
i.e.,
[L~,H] = i~c(α~ ~p), (12)
×
theequation(12)isshownthattheorbitalangularmomentumofphotonisn’tconservation,butthe
totalangularmomentumoffreephotonshouldbeconservative. So,photonshouldhaveanintrinsic
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angularmomentum,i.e., spin angularmomentum~s, the total angularmomentumofphoton J~is
J~= L~ +~s, (13)
andJ~is conservative
[J~,H] = 0, (14)
with equations (12) and (14),we have
[~s,H] = [L~,H] = i~c(α~ p~), (15)
− − ×
i.e.,
[s ,H] = [s ,cα~ p~] = i~c(α~ p~)
x x x
· − ×
= i~c(α p α p ), (16)
z y y z
−
or
[s ,cα p +cα p +cα p ]= c[s ,α ]p +c[s ,α ]p +c[s ,α ]p
x x x y y z z x x x x y y x z z
= i~c(α p α p ), (17)
z y y z
−
there are commutationrelations
[s ,α ] = 0, [s ,α ] = i~α , [s ,α ]= i~α , (18)
x x x y z x z y
−
it is easy to summarizethat
[s ,α ] = i~ǫ α , (19)
i j ijk k
with equation (18),we can calculatethe s matrices,letthe s matrix is
x
a a a
11 12 13
s = a a a , (20)
x 21 22 23
a a a
31 32 33
with the commutation relation
[s ,α ]= i~α , [s ,α ] = i~α , (21)
x y z x z y
−
we get
a 0 0
s = 0 a i~ , (22)
x
−
0 i~ a
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letthe s matrix is
y
b b b
11 12 13
s = b b b , (23)
y 21 22 23
b b b
31 32 33
with the commutation relation
[s ,α ]= i~α , [s ,α ]= i~α , (24)
y x z y z x
−
we get
b 0 i~
s = 0 b 0 , (25)
y
i~ 0 b
−
letthe s matrix is
z
c c c
11 12 13
s = c c c , (26)
z 21 22 23
c c c
31 32 33
with the commutation relation
[s ,α ] = i~α , [s ,α ]= i~α , (27)
z x y z y x
−
we get
c i~ 0
−
s = i~ c 0 . (28)
z
0 0 c
In the following,we should calculatethe eigenvaluesofs , s , s .
x y z
The s eigenvalueproblems ψ = λ ψ is
x x 1
a 0 0 ψ ψ
1 1
0 a i~ ψ = λ ψ , (29)
2 1 2
−
0 i~ a ψ ψ
3 3
thereforethe characteristic equation is
a λ 0 0
1
−
0 a λ i~ = 0, (30)
(cid:12) 1 (cid:12)
(cid:12) − − (cid:12)
(cid:12) 0 i~ a λ (cid:12)
(cid:12) − 1 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
i.e., (cid:12) (cid:12)
(a λ )[(a λ )2 ~2]= 0, (31)
1 1
− − −
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when a = 0, the rootsλ are
1
λ = ~, (32)
1
±
the s eigenvalueproblems ψ = λ ψ is
y y 2
b 0 i~ ψ ψ
1 1
0 b 0 ψ = λ ψ , (33)
2 2 2
i~ 0 b ψ ψ
− 3 3
thereforethe characteristic equation is
b λ 0 i~
2
−
0 b λ 0 = 0, (34)
(cid:12) 2 (cid:12)
(cid:12) − (cid:12)
(cid:12) i~ 0 b λ (cid:12)
(cid:12) − − 2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
i.e., (cid:12) (cid:12)
(b λ )[(b λ )2 ~2]= 0, (35)
2 2
− − −
when b = 0, the rootsλ are
2
λ = ~, (36)
2
±
the s eigenvalueproblems ψ =λ ψ is
z z 3
c i~ 0 ψ ψ
1 1
−
i~ c 0 ψ = λ ψ , (37)
2 3 2
0 0 c ψ ψ
3 3
thereforethe characteristic equation is
c λ i~ 0
3
− −
i~ c λ 0 = 0, (38)
(cid:12) 3 (cid:12)
(cid:12) − (cid:12)
(cid:12) 0 0 c λ (cid:12)
(cid:12) − 3 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
i.e., (cid:12) (cid:12)
(c λ )[(c λ )2 ~2] = 0, (39)
3 3
− − −
when c =0, the roots λ are
3
λ = ~. (40)
3
±
By calculation,we obtain the spin matricesofphoton, they are
0 0 0 0 0 i~ 0 i~ 0
−
s = 0 0 i~ ,s = 0 0 0 ,s = i~ 0 0 . (41)
x y z
−
0 i~ 0 i~ 0 0 0 0 0
−
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Obviously, these matrices are photon’s spin matrices, which describe the spin s = 1. On the one
hand, their eigenvaluesare ~. On the other hand,the matricessquare is
±
1 0 0 1 0 0
~s 2 = s2 +s2 +s2 = 2 0 1 0 ~2 = s(s+1)~2 0 1 0 , (42)
x y z
0 0 1 0 0 1
i.e., the spin sis
s = 1. (43)
Comparing (6) with (41),we find
s = α , s = α , s = α . (44)
x x y y z z
4 The helicity of photon
By studying the helicity of photon, we can obtain the photon left-handed and right-handed. The
helicity is definedas the projection ofspin in the momentumdirection,it is
~s p~ α~ p~
h = · = · , (45)
p~ p~
| | | |
and
0 ip ip
z y
−
α~ ~p= α p +α p +α p = ip 0 ip , (46)
x x y y z z z x
· −
ip ip 0
− y x
the α~ ~p eigenvalueproblemα~ p~ψ =λψ is
· ·
0 ip ip ψ ψ
z y 1 1
−
ip 0 ip ψ = λ ψ , (47)
z x 2 2
−
ip ip 0 ψ ψ
− y x 3 3
thereforethe characteristic equation is
λ ip ip
z y
− −
ip λ ip = 0, (48)
(cid:12) z x (cid:12)
(cid:12) − − (cid:12)
(cid:12) ip ip λ (cid:12)
(cid:12) − y x − (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
i.e., (cid:12) (cid:12)
λ(p¯2 λ2) =0, (49)
−
the rootsλ are
λ = p~, p~, (50)
| | −| |
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andthe helicity hare
~= +1, 1. (51)
−
When λ = +1 the photon is called right-handed photon, and when λ = 1 the photon is called
−
left-handedphoton.
5 The probability conservation equation of photon
In the following, we should give the probability density and probability conservation equation of
photon.
The hermitian conjugateof(4) is
∂ψ†
i~ = i~c~ψ† α~, (52)
− ∂t ∇ ·
multiplying(52) by ψ, there is
∂ψ†
i~ ψ = i~c~ψ† α~ψ, (53)
− ∂t ∇ ·
multiplying(4) by ψ+,there is
∂ψ
i~ψ†( ) = i~cψ†α~ ~ψ, (54)
∂t − ·∇
taking the difference,we get
∂ψ ∂ψ†
i~(ψ† + ψ)+i~cψ†α~ (~ψ)+i~c~ψ† α~ψ = 0, (55)
∂t ∂t · ∇ ∇ ·
or
1 ∂
(ψ†ψ)+ψ†α~ (~ψ)+(~ψ†) α~ψ = 0, (56)
c∂t · ∇ ∇ ·
i.e.,
∂ρ
+ J~= 0, (57)
∂t ∇·
where
ρ = ψ†ψ, J~= cψ†α~ψ, (58)
are the probability and probability currentdensity ofphoton, respectively.
6 The plane wave of free photon
We have the spiron equation of freephoton
∂
i~ ψ = Hψ, (59)
∂t
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where
ψ
1
H = cα~ p~,ψ = ψ , (60)
2
·
ψ
3
since ∂H = 0 and [p~,H] = 0, the energy E and momentum ~p of photon are conserved quantity.
∂t
Their commoneigenstate is
ψ (~r,t) = u(p~)ei(p~·~r−Et)/~, (61)
E,p~
where
u (p~)
1
u(p~) = u (p~) , (62)
2
u (p~)
3
substituting equations(61) and(62) into (59),we have
cα~ p~ u(p~)= Eu(p~), (63)
·
i.e.,
u u
1 1
(cα p +cα p +cα p ) u = E u , (64)
x x y y z z 2 2
u u
3 3
andexpanding (64),we get
0 icp icp u u
z y 1 1
−
icp 0 icp u = E u , (65)
z x 2 2
−
icp icp 0 u u
− y x 3 3
or
Eu +icp u icp u = 0, (66)
1 z 2 y 3
−
icp u Eu icp u = 0, (67)
z 1 2 x 3
− −
icp u icp u +Eu = 0, (68)
y 1 x 2 3
−
thereforethe characteristic equation is
E icp icp
z y
−
icp E icp = 0, (69)
(cid:12) z x (cid:12)
(cid:12) − − (cid:12)
(cid:12) icp icp E (cid:12)
(cid:12) y − x (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
the rootsE are (cid:12) (cid:12)
E = +cp~, E = cp~. (70)
1 2
| | − | |
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Fromequations (67) and (68),we have
icp p u Ep u icp p u = 0, (71)
z y 1 y 2 x y 3
− −
icp p u icp p u +Ep u = 0, (72)
y z 1 x z 2 z 3
−
taking the differenceof equations(71) and(72),we get
(Ep icp p )u +(Ep +icp p )u = 0, (73)
y x z 2 z x y 3
−
i.e.,
u Ep +icp p
2 z x y
= , (74)
u −Ep icp p
3 y x z
−
substituting (74) into (66),there is
u ic(p2 +p2)
1 y z
= , (75)
u Ep icp p
3 y x z
−
by (74) and (75),we have
u ic(p2 +p2)
1 y z
= , (76)
u −Ep +icp p
2 z x y
the u(p)can be written as
u ic(p2 +p2)
1 y z
u(p)= u = N (Ep +icp p ) , (77)
2 z x y
−
u Ep icp p
3 y − x z
where N is normalizationconstant, it can be obtainedby the normalization
ic(p2 +p2)
y z
u†(p)u(p) = N2 ic(p2 +p2) (Ep icp p ) (Ep +icp p ) (Ep +icp p )
− y z − z − x y y x z − z x y
(cid:16) (cid:17) Ep icp p
y − x z
= 2E2N2(p2 +p2) = 1, (78)
y z
i.e.,
1
N = , (79)
2E2(p2 +p2)
s y z
then
ic(p2 +p2)
y z
1
u(p) = (Ep +icp p ) , (80)
s2E2(p2y +p2z) − z x y
Ep icp p
y − x z
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