Peculiarities of squaring method applied to construct solutions of the Dirac, Majorana, and Weyl equations

0.21 MB·English

by  O. V. Veko

Peculiarities of squaring method applied to construct solutions
of the Dirac, Majorana, and Weyl equations
5 January 27, 2015
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0
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O.V. Veko (KalinkovichiGymnazium, Belarus, [email protected])

n
V.M. Red’kov (Institute of Physics, NAS of Belarus, [email protected])
a
J
Abstract

5
2 ItisshownthattheknownmethodtosolvetheDiracequationbymeansofthesquaringmethod,
whenrelyingonthescalarfunctionoftheformΦ=e−iǫteik1xeik2ysin(kz+α)leadstoa4-dimensional

] spaceoftheDiracsolutions. ItisshownthatsoconstructedbasisisequivalenttothespaceoftheDirac
h
p states relied on the use of quantum number k1,k2,±k and helicity operator; linear transformations
relatingthesetwospacesarefound. Applicationofthesquaringmethodsubstantiallydependsonthe
-
t choice of representation for the Dirac matrices, some features of this are considered. Peculiarities of
n
applying the squaring method in Majorana representation are investigated as well. The constructed
a
u bases are relevant to describe the Casimir effect for Dirac and Weyl fields in the domain restricted
q by two planes.
[
Pacs: 02.30.Gp, 02.40.Ky,03.65Ge, 04.62.+v

1 Keywords: Dirac, Majorana,Weyl fields, squaring method, Casimir effect, quantization
v
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7 1 Introduction
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In connection with the Casimir effect [1] for the Dirac field in the domain restricted by two planes,
0
. of special interest are solutions of the Dirac equation, which have vanishing the third projection of
1
the conserved current Jz on the plane boundaries. Such solutions are reachable when considering 4-
0
5 dimensional space
1
v: {Ψ}={Ψk1,k2,k3=k,σ ⊗Ψk1,k2,k3=−k,σ}, (1.1)
i
X that is the basisof four solutionswith opposite signs ofthe thirdprojectionof momentum+k and k ;
3 3
−
r σ is referred to polarization of the states.
a ItisshownthattheknownmethodtosolvetheDiracequationtroughsquaringmethod[2](elaboration
of such a method to the case of electromagnetic field see in [3, 4]), when relying onthe scalarfunction of
the following form
Φ=e−iǫteik1xeik2ysin(kz+α) = Ψ = Ψ ,Ψ ,Ψ ,Ψ (1.2)
1 2 3 4
⇒ { } { }
leads to a 4-dimensional space of the Dirac solutions. It is shown that so constructed basis (1.2) is
equivalent to the space of the Dirac states (1.1); linear transformations relating these two spaces are
found. Different values of parameter α in (1.2) determine only different bases in the same linear space
(1.2).
ApplicationofthesquaringmethodsubstantiallydependsonthechoiceofrepresentationfortheDirac
matrices, some features of this circumstance are considered.
1
Peculiarities of applying the squaring method in Majorana representation are investigated. It is
shownthatconstructedonthebaseofscalarfunctionscos(ǫt ~k~x)and isin(ǫt ~k~x),two4-dimensional
− − −
sets of real and imaginary solutions of the Majorana wave equation cannot be related by any linear
transformation.
General conditions for vanishing third projection of the current Jz at the boundaries of the domain
betweentwoparallelplanesareformulated: firstly,onthebasisofplanespinorwaves(1.1);andsecondly,
on the squared basis (1.2). In both cases, these conditions reduce to a linear homogeneous algebraic
systems which leads to a 4-th order algebraic equation, the roots of which are e2ika, where a is a half-
distancebetweentheplanes,andk standsforthethirdprojectionoftheDiracparticlemomentum. Each
solution of this equation,with representa complex number of the unit length, providesus with a certain
ruleforquantizationofthethirdprojectionofk. Explicitformsofthesealgebraicequationsaredifferent,
however their roots must be the same.
Conditions for vanishing the current Jz for Weyl neutrino field on the boundaries of the domain
between two planes are examined, the problem reduce to a 2-nd order algebraic equation.
Covariantizationof the conditions for vanishing Jz is performed.
2 Squaring method
Let us start with a special solution of the Klein–Fock–Gordon equation with one formal change in
Φ=e−iǫt+~k~x: namely, we will change the factor e+ik3z by the real-valued factor
sin(kz+γ) sinϕ; (2.1)
≡
for brevity we use notation k =k . Thus, we start with
3
(iγa∂ M)(iγa∂ +M)=( ∂2+∂ ∂ M2),
a− a − t j j −
Φ=e−iǫteik1xeik2ysinϕ,ǫ2 k2 k2 k2 M2 =0. (2.2)
− 1 − 2− −
The 4 4- matrix of the four columns-solutions of the Dirac equation is constructed in accordance with
×
the following rule
Ψ ,Ψ ,Ψ ,Ψ =(iγa∂ +iγj∂ +M)Φ. (2.3)
1 2 3 4 t j
{ }
In spinor basis, at the choice (2.2), the matrix (2.3) will be
Msinϕ 0
Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) 0 (cid:12)(cid:12), Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) Msinϕ (cid:12)(cid:12),
1 (cid:12) (ǫsinϕ+ikcosϕ) (cid:12) 2 (cid:12) (k ik )sinϕ (cid:12)
(cid:12) (cid:12) (cid:12) − 1− 2 (cid:12)
(cid:12) (k1+ik2)sinϕ (cid:12) (cid:12) (ǫsinϕ ikcosϕ) (cid:12)
(cid:12) − (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(ǫsinϕ ikcosϕ) (k ik )sinϕ
1 2
− −
Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) (k1+ik2)sinϕ (cid:12)(cid:12), Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) (ǫsinϕ+ikcosϕ) (cid:12)(cid:12). (2.4)
3 (cid:12) Msinϕ (cid:12) 4 (cid:12) 0 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) Msinϕ (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
To decide on the linear(cid:12)dependence of Ψ (o(cid:12)r not), one should exam(cid:12)ine the following rela(cid:12)tion
j
a Ψ +a Ψ +a Ψ +a Ψ =0,
1 1 2 2 3 3 4 4
so we get the linear homogeneous system with respect to a ,a ,a ,a :
1 2 3 4
a Msinϕ+a (ǫsinϕ ikcosϕ)+a (k ik )sinϕ=0,
1 3 4 1 2
− −
a Msinϕ+a (k +ik )sinϕ+a (ǫsinϕ+ikcosϕ)=0,
2 3 1 2 4
2
a (ǫsinϕ+ikcosϕ) a (k ik )sinϕ+a Msinϕ=0,
1 2 1 2 3
− −
a (k +ik )sinϕ+a (ǫsinϕ ikcosϕ)+a Msinϕ=0. (2.5)
1 1 2 2 4
− −
Its determinant turns to be det Ψ =k4. This means that four solutions (2.4) of the Dirac equation (at
{ }
any α in the ϕ=kz+α) are linearly independent: they determine a 4-dimensional space.
Let us specify two possibilities for γ in (2.1):
γ =0, ϕ=kz, sinϕ=sinkz, cosϕ=coskz ,
Msinkz 0
Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) 0 (cid:12)(cid:12),Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) Msinkz (cid:12)(cid:12),
1 (cid:12)(cid:12) (ǫsinkz+ikcoskz) (cid:12)(cid:12) 2 (cid:12)(cid:12) −(k1−ik2)sinkz (cid:12)(cid:12)
(cid:12) (k +ik )sinkz (cid:12) (cid:12) (ǫsinkz ikcoskz) (cid:12)
(cid:12) − 1 2 (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (ǫsinkz ikcoskz) (cid:12) (cid:12) (k ik )sinkz (cid:12)
1 2
− −
Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) (k1+ik2)sinkz (cid:12)(cid:12), Ψ =e−iǫteik1xeik2y(cid:12)(cid:12) (ǫsinkz+ikcoskz) (cid:12)(cid:12), (2.6)
3 (cid:12) Msinkz (cid:12) 4 (cid:12) 0 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) Msinkz (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
and (cid:12) (cid:12) (cid:12) (cid:12)
π π
′ ′ ′
γ = , ϕ =kz , sinϕ = coskz, cosϕ =sinkz ,
2 − 2 −
Mcoskz 0
−
Ψ′ =e−iǫteik1xeik2y(cid:12)(cid:12) 0 (cid:12)(cid:12), Ψ′ =e−iǫteik1xeik2y(cid:12)(cid:12) −Mcoskz (cid:12)(cid:12),
1 (cid:12) ( ǫcoskz+iksinkz) (cid:12) (2) (cid:12) (k1 ik2)coskz (cid:12)
(cid:12) − (cid:12) (cid:12) − (cid:12)
(cid:12)(cid:12) (k1+ik2)coskz (cid:12)(cid:12) (cid:12)(cid:12) (−ǫcoskz−iksinkz) (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ( ǫcoskz iksinkz)(cid:12) (cid:12) (k ik )coskz (cid:12)
1 2
− − − −
Ψ′ =e−iǫteik1xeik2y(cid:12)(cid:12) −(k1+ik2)coskz (cid:12)(cid:12), Ψ′ =e−iǫteik1xeik2y(cid:12)(cid:12) (−ǫcoskz+iksinkz) (cid:12)(cid:12). (2.7)
3 (cid:12) Mcoskz (cid:12) 4 (cid:12) 0 (cid:12)
(cid:12) − (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) Mcoskz (cid:12)
(cid:12) (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Wereadilyfindexpre(cid:12)ssionsforthefollowingc(cid:12)ombinations(thetotalf(cid:12)actore−iǫteik1xeik2y isom(cid:12) itted):
M 0
Ψ′ +iΨ =eikz(cid:12)(cid:12) 0 (cid:12)(cid:12), Ψ′ +iΨ =eikz(cid:12)(cid:12) M (cid:12)(cid:12),
− 1 1 (cid:12)(cid:12) ǫ−k (cid:12)(cid:12) − 2 2 (cid:12)(cid:12) −(k1−ik2) (cid:12)(cid:12)
(cid:12) (k1+ik2) (cid:12) (cid:12) ǫ+k (cid:12)
(cid:12) − (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ǫ+k (cid:12) (cid:12) (k1 ik2) (cid:12)
−
−Ψ′3+iΨ3 =eikz(cid:12)(cid:12)(cid:12) (k1+Mik2) (cid:12)(cid:12)(cid:12), −Ψ′4+iΨ4 =eikz(cid:12)(cid:12)(cid:12) ǫ−0k (cid:12)(cid:12)(cid:12). (2.8)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) M (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Notethatsolutions(2.8)exactlycoincidewiththoseobtainedbyapplyingthesquaringmethodwhen
one starts with a scalar function as the ordinary plane wave:
M 0 (ǫ+k) (k ik )
1 2
−
U =e−iǫt+ik1x+ik2y+ikz(cid:12)(cid:12) 0 M (k1+ik2) (ǫ−k) (cid:12)(cid:12). (2.9)
(cid:12) (ǫ k) (k ik ) M 0 (cid:12)
(cid:12) − − 1− 2 (cid:12)
(cid:12) (k1+ik2) (ǫ+k) 0 M (cid:12)
(cid:12) − (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
The rank of the matrix in (2.9) equals to 2. Therefore, among four solutions of the Dirac equations
givenby(2.9)onlytwoofthemarelinearlyindependent. Fordefiniteness,letuschosesolutionsU and
(1)
3
U :
(2)
M 0
U =e−iǫt+ik1x+ik2y+ikz(cid:12)(cid:12) 0 (cid:12)(cid:12), U =e−iǫt+ik1x+ik2y+ikz(cid:12)(cid:12) M (cid:12)(cid:12). (2.10)
1 (cid:12)(cid:12) ǫ−k (cid:12)(cid:12) 2 (cid:12)(cid:12) −(k1−ik2) (cid:12)(cid:12)
(cid:12) (k +ik ) (cid:12) (cid:12) ǫ+k (cid:12)
(cid:12) − 1 2 (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
It is readily checked that remain(cid:12)ing solutions U(cid:12),U are expressed through(cid:12)U ,U as follow(cid:12)s
3 4 1 2
1 1
U = [(ǫ+k)U +(k +ik )U ] , U = [(k ik )U +(ǫ k)U ] . (2.11)
3 1 1 2 2 4 1 2 1 2
M M − −
Also, instead of (2.8), one can construct other combinations
M 0
Ψ′ iΨ =e−ikz(cid:12)(cid:12) 0 (cid:12)(cid:12), Ψ′ iΨ =e−ikz(cid:12)(cid:12) M (cid:12)(cid:12),
− 1− 1 (cid:12)(cid:12) ǫ+k (cid:12)(cid:12) − 2− 2 (cid:12)(cid:12) −(k1−ik2) (cid:12)(cid:12)
(cid:12) (k +ik ) (cid:12) (cid:12) ǫ k (cid:12)
(cid:12) − 1 2 (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ǫ k (cid:12) (cid:12) (k1 ik2) (cid:12)
− −
Ψ′ iΨ =e−ikz(cid:12)(cid:12) (k1+ik2) (cid:12)(cid:12), Ψ′ iΨ =e−ikz(cid:12)(cid:12) ǫ+k (cid:12)(cid:12). (2.12)
− 3− 3 (cid:12) M (cid:12) − 4− 4 (cid:12) 0 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) M (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
These coincidewith thoseobtainedby(cid:12) applyingth(cid:12)e squaringmethodto a(cid:12)scalarfunct(cid:12)ionas the ordinary
plane wave with the only change k into k:
−
M 0 (ǫ k) (k ik )
1 2
− −
U′ =e−iǫt+ik1x+ik2y−ikz(cid:12)(cid:12) 0 M (k1+ik2) (ǫ+k) (cid:12)(cid:12). (2.13)
(cid:12) (ǫ+k) (k1 ik2) M 0 (cid:12)
(cid:12) − − (cid:12)
(cid:12) (k +ik ) (ǫ k) 0 M (cid:12)
(cid:12) − 1 2 − (cid:12)
(cid:12) (cid:12)
Again, the rank of the matrix in (2(cid:12).13) equals 2, so only two solutions are independent:(cid:12)
M 0
U′ =e−iǫt+ik1x+ik2y−ikz(cid:12)(cid:12) 0 (cid:12)(cid:12), U′ =e−iǫt+ik1x+ik2y−ikz(cid:12)(cid:12) M (cid:12)(cid:12). (2.14)
1 (cid:12) ǫ+k (cid:12) (2) (cid:12) (k1 ik2) (cid:12)
(cid:12) (cid:12) (cid:12) − − (cid:12)
(cid:12)(cid:12) −(k1+ik2) (cid:12)(cid:12) (cid:12)(cid:12) ǫ−k (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Solutions U and U are cons(cid:12)tructed throug(cid:12)h U ,U in accordance with(cid:12) the rules (cid:12)
(3) (4) (1) (2)
1 1
′ ′ ′ ′ ′
U = (ǫ k)U +(k +ik )U , U = (k ik )U +(ǫ+k)U . (2.15)
(3) M h − (1) 1 2 (2)i (4) M h 1− 2 (1) (2)i
3. Solutions in the basis of momentum 4-vector and helicity
Solutions of the Dirac equation in Cartesian coordinates can be searched in the form
f
1
Ψǫ,k1,k2 =e−iǫt eik1x eik2y eik3z(cid:12)(cid:12)(cid:12) ff2 (cid:12)(cid:12)(cid:12) . (2.16)
(cid:12) 3 (cid:12)
(cid:12) f4 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
With the use of spinor basis for Dirac matrices, one gets the l(cid:12)inear(cid:12)system for f :
i
ǫf +k f ik f +k f Mf =0,
3 1 4 2 4 3 3 1
− −
ǫf +k f +ik f k f Mf =0,
4 1 3 2 3 3 4 2
− −
ǫf k f +ik f k f Mf =0,
1 1 2 2 2 3 1 3
− − −
ǫf k f ik f +k f Mf =0. (2.17)
2 1 1 2 1 3 2 4
− − −
4
Let us diagonalize the known helicity operator Σ = σ p . With the substitution (2.16), from eigenvalue
j j
equation ΣΨ=pΨ we arrive at
k f ik f +k f =pf ,
1 2 2 2 3 1 1
−
k f + ik f k f =pf ,
1 1 2 1 3 2 2
−
k f ik f +k f =pf ,
1 4 2 4 3 3 3
−
k f +ik f k f =pf . (2.18)
1 3 2 3 3 4 4
−
Considering eqs. (2.17) and (2.18) jointly, we obtain the system (note that p2 =ǫ2 M2)
−
ǫf +pf Mf =0, ǫf +pf M f =0,
3 3 1 4 4 2
− −
ǫf pf Mf =0, ǫf pf M f =0.
1 1 3 2 2 4
− − − −
This system results in two values for p and corresponding restrictions on f :
i
ǫ p ǫ p
f = − f , f = − f . (2.19)
3 1 4 2
M M
Allowing for (2.19), from (2.18) one gets the system for f and f :
1 2
(k p)f +(k ik )f =0,
3 1 1 2 2
− −
(k +p)f (k +ik )f =0. (2.20)
3 2 1 2 1
−
Further we obtain
i k +ik
1 2
f = (k p)f = f .
2 3 1 1
−ik +k − k +p
1 2 3
Thus, two independent solutions (we will mark them by α and β) at the fixed~k are (further let f = 1
1
and p=+√ǫ2 M2)
−
ǫ p k +ik
1 2
(α), − =α, f =1, f = =s,
1 2
M k +p
3
ǫ+p k +ik
1 2
(β), =β , f =1, f = =t; (2.21)
1 2
M k p
3
−
these may be presented in a shorter form
1 1
Ψ =e−iǫteik1xeik2yeik3z(cid:12)(cid:12) s (cid:12)(cid:12) , Ψ =e−iǫteik1xeik2yeik3z(cid:12)(cid:12) t (cid:12)(cid:12) , (2.22)
(α) (cid:12) α (cid:12) (β) (cid:12) β (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) αs (cid:12) (cid:12) βt (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
3 Relationships between two bases: those obtained from squar-
ing method and the momentum-helicity solutions
FourtypesofthedifferentsolutionsoftheDiracequationcanbeconstructedbethemethodofseparation
of variables (we take k =k and k = k)
3 3
−
1 1
Φ =Ψ (k)=eikz(cid:12)(cid:12) k1k++ipk2 (cid:12)(cid:12),Φ =Ψ ( k)=e−ikz(cid:12)(cid:12) k−1+k+ikp2 (cid:12)(cid:12),
1 (α) (cid:12) α (cid:12) 2 (α) − (cid:12) α (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) αk1+ik2 (cid:12) (cid:12) αk1+ik2 (cid:12)
(cid:12) k+p (cid:12) (cid:12) −k+p (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
5
1 1
Φ =Ψ (k)=eikz(cid:12)(cid:12) k1k+−ipk2 (cid:12)(cid:12),Φ =Ψ ( k)=e−ikz(cid:12)(cid:12) k−1+k−ikp2 (cid:12)(cid:12). (3.1)
3 (β) (cid:12) β (cid:12) 4 (β) − (cid:12) β (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) βk1+ik2 (cid:12) (cid:12) βk1+ik2 (cid:12)
(cid:12) k−p (cid:12) (cid:12) −k−p (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
and they must be related to four squaring solutions in Section 2.
By physical reason, we should expect existence of the following linear expansions
U =aΦ +bΦ , U =cΦ +dΦ . (3.2)
1 1 3 2 1 3
Evidently, if a,b,c,d are known, one can derive
1
U = [(ǫ+k)U +(k +ik )U ]
(3) M (1) 1 2 (2)
1
= [(ǫ+k)(aΦ +bΦ )+(k +ik )(cΦ +dΦ )]; (3.3)
1 3 1 2 1 3
M
1
U = [(k ik )U +(ǫ k);U ]
(4) M 1− 2 (1) − (2)
1
= [(k ik )(aΦ +bΦ )+(ǫ k)(cΦ +dΦ )]. (3.4)
1 2 1 3 1 3
M − −
Analogously, there must exist expansions
′ ′ ′ ′ ′ ′
U =aΦ +bΦ , U =cΦ +dΦ , (3.5)
1 2 4 2 2 4
with the help of which one can derive
1
′ ′ ′
U = [(ǫ k)(aΦ +bΦ )+(k +ik )U ], (3.6)
(3) M − 2 4 1 2 (2)
1
′ ′ ′ ′
U = [(k ik )(aΦ +bΦ )+(ǫ+k)(cΦ +dΦ )]. (3.7)
(4) M 1− 2 2 4 2 4
In turn, after that, one can derive the next expansions by the formulas
1 1
′ ′ ′
Ψ = (U U ), Ψ = (U +U ). (3.8)
j 2i j − j j −2 j j
Let us consider the first relation in (3.2), U =aΦ +bΦ ; explicitly it reads
1 1 3
M 1 1
(cid:12)(cid:12) 0 (cid:12)(cid:12)=a(cid:12)(cid:12) k1k++ipk2 (cid:12)(cid:12)+b(cid:12)(cid:12) k1k+−ipk2 (cid:12)(cid:12); (3.9)
(cid:12) ǫ k (cid:12) (cid:12) α (cid:12) (cid:12) β (cid:12)
(cid:12) − (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)(cid:12) −(k1+ik2) (cid:12)(cid:12) (cid:12)(cid:12) αk1k++ipk2 (cid:12)(cid:12) (cid:12)(cid:12) βk1k+−ipk2 (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
from whence it follows
a b aα bβ
M =a+b, 0= + , ǫ k =aα+bβ, 1= + . (3.10)
k+p k p − − k+p k p
− −
From the first and second equations we get
M M
a= (k+p), b= (k p); (3.11)
2p −2p −
two remaining equations turn to be identities.
6
Now, let us consider the second equation in (3.2): U =cΦ +dΦ ; it reads explicitly
2 1 3
0 1 1
(cid:12)(cid:12) M (cid:12)(cid:12)=c(cid:12)(cid:12) k1k++ipk2 (cid:12)(cid:12)+d(cid:12)(cid:12) k1k+−ipk2 (cid:12)(cid:12); (3.12)
(cid:12) (k ik ) (cid:12) (cid:12) α (cid:12) (cid:12) β (cid:12)
(cid:12) − 1− 2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ǫ+k (cid:12) (cid:12) αk1+ik2 (cid:12) (cid:12) βk1+ik2 (cid:12)
(cid:12) (cid:12) (cid:12) k+p (cid:12) (cid:12) k−p (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
or (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
k +ik k +ik
1 2 1 2
0=c+d, M =c +d ,
k+p k p
−
cα(k +ik ) dβ(k +ik )
1 2 1 2
(k ik )=cα+dβ , ǫ+k = + .
1 2
− − k+p k p
−
From the first and second equations, it follows
M
d= c, c= (k ik ); (3.13)
1 2
− 2p −
two remaining ones are identities. Thus, we have obtained
M M
U = [(p+k)Φ +(p k)Φ ] , U = [(k ik )Φ (k ik )Φ ]; (3.14)
1 1 3 2 1 2 1 1 2 3
2p − 2p − − −
and further
M p+k p k M k ik k ik
1 2 1 2
U = [ Φ + − Φ ], U = [ − Φ − Φ ]; (3.15)
3 1 3 4 1 3
2p α β 2p α − β
the identities ǫ p=M/β, ǫ+p=M/α should be remembered.
− ′ ′ ′
Now, let us consider the equation U =aΦ +bΦ ; it reads explicitly as
1 2 4
M 1 1
(cid:12)(cid:12) 0 (cid:12)(cid:12)=a′(cid:12)(cid:12) k−1+k+ikp2 (cid:12)(cid:12)+b′(cid:12)(cid:12) k−1+k−ikp2 (cid:12)(cid:12); (3.16)
(cid:12) ǫ+k (cid:12) (cid:12) α (cid:12) (cid:12) β (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)(cid:12) −(k1+ik2) (cid:12)(cid:12) (cid:12)(cid:12) αk−1+k+ikp2 (cid:12)(cid:12) (cid:12)(cid:12) βk−1+k−ikp2 (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
what differs formally from (3.9) only in the change k k a presence of primes at variables, so the
→ −
result can be written at once
M M
′ ′
a = ( k+p), b = ( k p).
2p − −2p − −
′ ′ ′
Consider the equation U =cΦ +dΦ ; it reads
2 2 4
0 1 1
(cid:12)(cid:12) M (cid:12)(cid:12)=c′(cid:12)(cid:12) k−1+k+ikp2 (cid:12)(cid:12)+d′(cid:12)(cid:12) k−1+k−ikp2 (cid:12)(cid:12); (3.17)
(cid:12) (k ik ) (cid:12) (cid:12) α (cid:12) (cid:12) β (cid:12)
(cid:12) − 1− 2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ǫ k (cid:12) (cid:12) αk1+ik2 (cid:12) (cid:12) βk1+ik2 (cid:12)
(cid:12) − (cid:12) (cid:12) −k+p (cid:12) (cid:12) −k−p (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
which differs from (3.12) only in notation, so its solution looks d′ = c′, c′ = M(k ik ).
− 2p 1− 2
Thus, we have derived decompositions
M M
′ ′
U = [(p k)Φ +(p+k)Φ ] , U = [(k ik )Φ (k ik )Φ ].
1 2p − 2 4 2 2p 1− 2 2− 1− 2 4
M p k p+k M k ik k ik
′ ′ 1 2 1 2
U = − Φ + Φ , U = − Φ − Φ . (3.18)
3 2p (cid:18) α 2 β 4(cid:19) 4 2p (cid:18) α 2− β 4(cid:19)
7
and repeat (3.14), (3.15):
M M
U = [(p+k)Φ +(p k)Φ ] , U = [(k ik )Φ (k ik )Φ ],
1 1 3 2 1 2 1 1 2 3
2p − 2p − − −
M p+k p k M k ik k ik
1 2 1 2
U = Φ + − Φ , U = − Φ − Φ . (3.19)
3 1 3 4 1 3
2p (cid:18) α β (cid:19) 2p (cid:18) α − β (cid:19)
Next, relying on f the formulas
1 1 1 1
′ ′ ′ ′ ′ ′
Ψ = (U U ), Ψ = (U +U ), Ψ = (U U ), Ψ = (U +U ),
1 2i 1− 1 1 −2 1 1 2 2i 2− 2 2 −2 2 1
1 1 1 1
′ ′ ′ ′ ′ ′
Ψ = (U U ), Ψ = (U +U ), Ψ = (U U ), Ψ = (U +U ), (3.20)
3 2i 3− 3 3 −2 3 3 4 2i 4− 4 4 −2 4 4
we derive decompositions
′ ′
Ψ =a Φ , Ψ =a Φ ,
l ln n l ln n
where the involved matrices are given by
(p+k) (p k) (p k) (p+k)
− − − −
iaij = M4p (cid:12)(cid:12)(cid:12) α(−k11(−p+ikk2)) −α(−k11(−p ikk2)) −β(−k11(−p ikk2)) β(−k11(−p+ikk2)) (cid:12)(cid:12)(cid:12), (3.21)
(cid:12)(cid:12) α−1(k ik ) α−−1(k −ik ) β−1(k −ik ) β−−1(k ik ) (cid:12)(cid:12)
(cid:12) 1− 2 − 1− 2 − 1− 2 1− 2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(p+k) (p k) (p k) (p+k)
− −
−a′ij = M4p (cid:12)(cid:12)(cid:12) α(−k11(−p+ikk2)) α(−k11(−p ikk2)) −β(−k11(−p ikk2)) −β(−k11(−p+ikk2)) (cid:12)(cid:12)(cid:12). (3.22)
(cid:12)(cid:12) α−1(k1 ik2) α−1(k1 −ik2) β−1(k1 −ik2) β−1(k1 ik2) (cid:12)(cid:12)
(cid:12) − − − − − − (cid:12)
(cid:12) (cid:12)
From those we ca(cid:12)n construct a new matrix (cid:12)
(p+k) 0 (p k) 0
−
Sij =−a′ij +iaij = M2p (cid:12)(cid:12)(cid:12) α(−k11(−p+ikk2)) 00 −β(−k11(−p ikk2)) 00 (cid:12)(cid:12)(cid:12); (3.23)
(cid:12)(cid:12) α−1(k1 ik2) 0 β−1(k1 −ik2) 0 (cid:12)(cid:12)
(cid:12) − − − (cid:12)
(cid:12) (cid:12)
(itcorrespondstotheuseinthefactorei(cid:12)kz intheinitialscalarsubstitutionforΦ)T(cid:12)herankofthismatrix
equals 2, and it is responsible for transformation
S
U ,U ,U ,U = Φ ,Φ .
1 2 3 4 1 3
⇐
Analogously,we have anothercombination(it correspondsto the use in the factor e−ikz in the initial
scalar substitution for Φ)
0 (p k) 0 (p+k)
−
Si′j =−a′ij −iaij = M2p (cid:12)(cid:12)(cid:12) 00 α(−k11(−p ikk2)) 00 −β(−k11(−p+ikk2)) (cid:12)(cid:12)(cid:12); (3.24)
(cid:12)(cid:12) 0 α−1(k −ik ) 0 β−1(k ik ) (cid:12)(cid:12)
(cid:12) 1− 2 − 1− 2 (cid:12)
(cid:12) (cid:12)
the rank of this matrix equals 2, and it c(cid:12)orresponds to the transformation (cid:12)
′ ′ ′ ′ S
U ,U ,U ,U = Φ ,Φ .
1 2 3 4 ⇐ 2 4
Because the determinants do not vanish
Mα2k2
det(a )=+ (k ik )2 α2+β2 2 ,
ij 1 2
ip − −
(cid:0) (cid:1)
Mk2
det(a′ )= (k ik )2 α2+β2 2 , (3.25)
ij − ip 1− 2 −
(cid:0) (cid:1)
8
inverse matrices exist and have the form
α α(k−p) 1 (k−p)
k1−ik2 − −k1−ik2
[a−1]= 2ip =(cid:12)(cid:12) α −αk1(k−+ikp2) −1 k(1k−+ipk)2 (cid:12)(cid:12), (3.26)
ij kM(α−β) (cid:12)(cid:12)(cid:12)(cid:12) ββ ββk1((kk−−+ikpp2)) −11 −k((1kk−−+ippk))2 (cid:12)(cid:12)(cid:12)(cid:12)
(cid:12) −k1−ik2 − k1−ik2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
α α(k−p) 1 (k−p)
− −k1−ik2 k1−ik2
[(a′)−1]= 2ip =(cid:12)(cid:12) α −αk1(k−+ikp2) −1 k(1k−+ipk)2 (cid:12)(cid:12). (3.27)
ij kM(α−β) (cid:12)(cid:12)(cid:12)(cid:12) −ββ −ββk1((kk−−+ikpp2)) 11 k((1kk−−+ippk))2 (cid:12)(cid:12)(cid:12)(cid:12)
(cid:12) −k1−ik2 − k1−ik2 (cid:12)
(cid:12) (cid:12)
(cid:12) ′ (cid:12)
Evidently, we have inverse transformations relating Ψ and Ψ :
j j
Ψ′ =a′ Φ =[a′ (a−1) ]Ψ ,
j jk k jk kl k
(p+k)α (p−k)2α (p k) p2−k2
− k1−ik2 − − k1−ik2
a′jk(a−1)kl = 2k(αi−β)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (k(1k(1k−−+αipikk)2β2)β)α −((k(p(1kp2−−−+ikkk)22pβ))β)αα (kk−11−−p−βiikkk22) −−((k(kp1p−−++kikk)2p2))β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12), (3.28)
(cid:12) α α β β (cid:12)
(cid:12) (cid:12)
(cid:12)Ψ =a Φ =[a (a′−1) ]Ψ′ , (cid:12)
j jk k jk kl k
(p+k)α (p−k)2α (p k) p2−k2
− − k1−ik2 − k1−ik2
[ajk(a′−1)kl]= k(α1−β)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (k(−1k1(−−kα+iikpk2)2)ββ)α (((kkp12(−+p−−ikkpk22))))ββαα (kk11p−−−βikikk22) −((kk(1p+p−+−ikkpk)2)2)β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12). (3.29)
(cid:12) α − α β − β (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
The following conclusion can be given: the choice of an initial phase γ in the function sin(kz +γ)
does notinfluence on the whole structure of the spaceof solutions – it only determines a specific basis in
the same space:
sin(kz+γ)=cosγ [sinkz] sinγ [ coskx],
− −
sin(kz+γ) = Ψγ =cosγ Ψ sinγ Ψ′ . (3.30)
⇒ j j − j
4 Dependence of squared solutions on the choice of Dirac ma-
trices, standard basis
Let us follows two representations for Dirac matrices
γa , Γa =SγaS−1.
Having applied the squaring method in both bases, we will obtain two sets of solutions:
(iγa∂ +M)Φ= Ψ ,Ψ ,Ψ ,Ψ , (iΓa∂ +M)Φ= ϕ ,ϕ ,ϕ ,ϕ . (4.1)
a 1 2 3 4 a 1 2 3 4
{ } { }
These two sets, or two matrices, are linked to each other according to the formula
S Ψ ,Ψ ,Ψ ,Ψ S−1 = ϕ ,ϕ ,ϕ ,ϕ . (4.2)
1 2 3 4 1 2 3 4
{ } { }
9
This relation tells that if at a fixed initial scalar function Φ we have a matrix of squared solutions
with rank 2, the corresponding matrix of solutions in any other Dirac basis will have also the rank 2.
Analogously,ifwehavethe matrixofsolutionsoftherank4inoneDiracbasis,wewillhaveinanyother
Dirac basis a matrix of solutions of the same rank 4.
Forinstance,letus specify squaredsolutionsinthe commonly usedstandardrepresentationfor Dirac
matrices:
I 0 0 σi
γ0 = , γi = .
(cid:12) 0 I (cid:12) (cid:12) σi 0 (cid:12)
(cid:12) − (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Taking into account the relation (cid:12) (cid:12) (cid:12) (cid:12)
i∂ +M 0 i∂ i∂ +∂
t 3 1 2
(iγa∂a+M)=(cid:12)(cid:12)(cid:12)(cid:12) −i0∂3 −i∂i∂t1+−M∂2 −i∂i∂1t−+∂M2 −i0∂3 (cid:12)(cid:12)(cid:12)(cid:12),
(cid:12) i∂1+∂2 i∂3 0 i∂t+M (cid:12)
(cid:12) − − (cid:12)
(cid:12) (cid:12)
and choosing (cid:12) (cid:12)
Φ=e−iǫteik1xeik2ysinϕ, sinϕ=kz+γ ,
we get an explicit form for the matrix of solutions
(ǫ+M)sinϕ 0 +ikcosϕ ( k +ik )sinϕ
1 2
−
[W]=(cid:12)(cid:12) 0 (ǫ+M)sinϕ (−k1−ik2)sinϕ −ikcosϕ (cid:12)(cid:12). (4.3)
(cid:12) ikcosϕ (k ik )sinϕ ( ǫ+M)sinϕ 0 (cid:12)
(cid:12) − 1− 2 − (cid:12)
(cid:12) (k1+ik2)sinϕ +ikcosϕ 0 ( ǫ+M)sinϕ (cid:12)
(cid:12) − (cid:12)
(cid:12) (cid:12)
Thus, we constr(cid:12)uct four different solutions of the Dirac equation: (cid:12)
(ǫ+M)sinϕ 0
(cid:12) 0 (cid:12) (cid:12) (ǫ+M)sinϕ (cid:12)
W =(cid:12) (cid:12), W =(cid:12) (cid:12),
1 (cid:12) ikcosϕ (cid:12) 2 (cid:12) (k ik )sinϕ (cid:12)
(cid:12) − (cid:12) (cid:12) 1− 2 (cid:12)
(cid:12) (k1+ik2)sinϕ (cid:12) (cid:12) +ikcosϕ (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) +ikcosϕ (cid:12) (cid:12)( k1+ik2)sinϕ (cid:12)
−
(cid:12) ( k ik )sinϕ (cid:12) (cid:12) ikcosϕ (cid:12)
W3 =(cid:12)(cid:12) −( ǫ1+−M2)sinϕ (cid:12)(cid:12), W4 =(cid:12)(cid:12) − 0 (cid:12)(cid:12). (4.4)
(cid:12) − (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) ( ǫ+M)sinϕ (cid:12)
(cid:12) (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Due to statement after (4.2), they are linearly independent.
Two choices of the phase γ lead to two sets of solutions:
ϕ=kz, sinϕ=sinkz, cosϕ=coskz ,
(ǫ+M)sinkz 0
(cid:12) 0 (cid:12) (cid:12) (ǫ+M)sinkz (cid:12)
W =(cid:12) (cid:12), W =(cid:12) (cid:12),
1 (cid:12) ikcoskz (cid:12) 2 (cid:12) (k1 ik2)sinkz (cid:12)
(cid:12) − (cid:12) (cid:12) − (cid:12)
(cid:12)(cid:12) (k1+ik2)sinkz (cid:12)(cid:12) (cid:12)(cid:12) +ikcoskz (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) +ikcoskz (cid:12) (cid:12)( k1+ik2)sinkz (cid:12)
−
W3 =(cid:12)(cid:12)(cid:12) (−( kǫ1+−Mik2))sisninkkzz (cid:12)(cid:12)(cid:12), W4 =(cid:12)(cid:12)(cid:12) −ikc0oskz (cid:12)(cid:12)(cid:12); (4.5)
(cid:12) − (cid:12) (cid:12) (cid:12)
(cid:12) 0 (cid:12) (cid:12) ( ǫ+M)sinkz (cid:12)
(cid:12) (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) π (cid:12) (cid:12) (cid:12)
′ ′ ′
ϕ =kz , sinϕ = coskz, cosϕ =sinkz ,
− 2 −
10