On Dirac equation on a time scale

On Dirac equation on a time scaleGro Hovhannisyan Citation: Journal of Mathematical Physics 52, 102701 (2011); doi: 10.1063/1.3644343 View online: http://dx.doi.org/10.1063/1.3644343 View Table of Contents: http://scitation.aip.org/content/aip/journal/jmp/52/10?ver=pdfcov Published by the AIP Publishing Articles you may be interested in DiracKähler Theory and Massless Fields AIP Conf. Proc. 1205, 120 (2010); 10.1063/1.3382317 On nonautonomous Dirac equation J. Math. Phys. 50, 123507 (2009); 10.1063/1.3265922 Solution of the Dirac equation in the rotating Bertotti–Robinson spacetime J. Math. Phys. 49, 052501 (2008); 10.1063/1.2912725 Physical Fields Described By Maxwell’s Equations AIP Conf. Proc. 899, 706 (2007); 10.1063/1.2733447 New solutions of relativistic wave equations in magnetic fields and longitudinal fields J. Math. Phys. 43, 2284 (2002); 10.1063/1.1461428

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JOURNAL OF MATHEMATICAL PHYSICS 52, 102701 (2011)

On Dirac equation on a time scaleGro Hovhannisyana)

Kent State University at Stark, 6000 Frank Ave. NW, Canton, Ohio 44720-7599, USA

(Received 22 March 2011; accepted 7 September 2011; published online 4 October 2011)

We consider the non-autonomous linear Dirac equation on a time scale containingimportant discrete, continuous, and quantum time scales. A representation of thesolutions is established via an approximate solutions in terms of unknown phasefunctions with the error estimates. JWKB and other asymptotic representations arediscussed. The adiabatic invariants of the Dirac equation are described by using asmall parameter method. We also calculate the transition probabilities for the Diracequation. Using the asymptotic solutions we show that the electron-positron transitionprobability during a long period of time is about 1/3. Since this probability is high,there is a simple explanation of the stability of the revolution of an electron aboutthe proton only by the electromagnetic field. Indeed when the electron is far from theproton, it is attracted by the electromagnetic field of the proton. When the electronapproaches closer to the proton, it turns to the positron which is repelling fromthe proton by the same electromagnetic field. C© 2011 American Institute of Physics.[doi:10.1063/1.3644343]

I. INTRODUCTION

In this paper, we are describing the behavior of the solutions of the Dirac equation

iψ�(t) = D(τ )ψ(t), τ = tε, ε > 0, t ≥ t0, (1.1)

on a time scale (for a continuous time scale, see Refs. 5, 9, 12 and 13) with a small parameter ε, and

D(τ ) =3∑

k=1

αk(pk(τ ) − eAk(τ )) + α4m(τ ) + eϕ(τ ), (1.2)

where ψ�(t) is the delta derivative (see Refs. 1 and 3), {αk}4k=1 are Dirac matrices,

α1 =

⎛⎜⎜⎜⎜⎝

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

⎞⎟⎟⎟⎟⎠ , α2 =

⎛⎜⎜⎜⎜⎝

0 0 0 −i

0 0 i 0

0 −i 0 0

i 0 0 0

⎞⎟⎟⎟⎟⎠ ,

α3 =

⎛⎜⎜⎜⎜⎝

0 0 1 0

0 0 0 −1

1 0 0 0

0 −1 0 0

⎞⎟⎟⎟⎟⎠ , α4 =

⎛⎜⎜⎜⎜⎝

1 0 0 0

0 1 0 0

0 0 −1 0

0 0 0 −1

⎞⎟⎟⎟⎟⎠ , (1.3)

(A1(τ ), A2(τ ), A3(τ ), ϕ(τ )) is the (real) potential of the external electro-magnetic field, m(τ ) isthe mass of an electron, e is its charge, |ψ j (t, p1, p2, p3)|2 is the probability that electron has themomentum p = (p1, p2, p3) at the time t, and it is in jth state. θ1(t), θ3(t) are the (complex) phase

a)Electronic mail:[email protected].

0022-2488/2011/52(10)/102701/16/$30.00 C©2011 American Institute of Physics52, 102701-1


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mailto: [email protected]

102701-2 Gro Hovhannisyan J. Math. Phys. 52, 102701 (2011)

functions of an electron with the spin up and down accordingly, θ2(t), θ4(t) are phase functions ofa positron with the spin up and down accordingly. Numerical solutions of the Dirac equation viadiscretisation had been investigated in Ref. 2.

A time scale T is an arbitrary nonempty closed subset of the real numbers (see Refs. 1 and 3).We consider the time scales that are unbounded above, using the notation T0 := T ∩ [t0,∞). Definethe forward jump operator σ (t) and the graininess μ(t)

σ (t) = inf{s ∈ T , s > t}, μ(t) = σ (t) − t. (1.4)

If T has a left-scattered minimum m, then T κ = T − {m}, otherwise T κ = T . The set of functionssuch that their nth delta derivative exists and is rd-continuous on T (see Refs. 1 and 3) is denotedby Cn

rd . The function f (t) is called regressive on T if 1 + μ(t) f (t) �= 0 for all t ∈ T . The setof regressive functions in Cn

rd is denoted by Rn. For any positive ε define auxiliary time scalesTε = {εt = τ, t ∈ T } with forward jump operator and graininess function

σ1(τ ) = inf{sε ∈ Tε, sε > τ }, μ1(τ ) = εμ(t), σ1(τ ) = εσ (t), τ = tε. (1.5)

Frequently, we suppress dependence on τ = tε or t . To distinguish the differentiation by t or τ weshow the argument of differentiation in parenthesizes: f �(t) = f �t (t) or f �(τ ) = f �τ (τ ).

II. FUNDAMENTAL SOLUTION: ADIABATIC INVARIANTS

Consider Dirac equation (1.1) written in the form

ψ�(t) = S(τ )ψ(t), τ = tε, t ∈ T κ , (2.1)

where S(t) is anti-hermitian matrix (with the entries sk j = −s jk)

S(t) = −i D(t) =

⎛⎜⎜⎜⎜⎝

s11 s12 s13 s14

s21 s22 s23 s24

s31 s32 s33 s34

s41 s42 s43 s44

⎞⎟⎟⎟⎟⎠ , (2.2)

s11 = s22 = −i(m + eϕ)(τ ), s33 = s44 = i(m − eϕ)(τ ), s12 = s21 = s34 = s43 = 0,

s13 = i(eA3 − p3)(τ ), s14 = s32 = [eA2 − p2 − i(p1 − eA1)](τ ), s24 = −s13. (2.3)

Our construction of the fundamental solution is valid under the conditions

p1(τ ) − eA1(τ ) + i(p2(τ ) + ieA2(τ )) �= 0, τ ∈ Tε, (2.4)

W [τ, p j (τ ) − eA j (τ ), p3(τ ) − eA3(τ )] ≡ 0, j = 1, 2, τ ∈ Tε, (2.5)

where W [τ, a(τ ), b(τ )] ≡ ε(a(τ )b�(τ ) − a�(τ )b(τ )) is a Wronskian.Note that if condition (2.5) does not satisfied, the structure of fundamental matrix will be

different from (2.18) below, and it would contain the expression (2.5) in the denominator. It meansthat in general it is necessary to consider both cases when (2.5) is satisfied, and when it is not. Herewe consider only the first case, since it contains important autonomous case. Introduce the auxiliaryequations

L j u(t) = u��(t) + Pj (t)u�(t) + Q j (t)u(t) = 0, j = 1, 2, (2.6)

where

P1,2(t) = i[2eϕ(t) + μ(t)(±m + eϕ)�(t)]+

1 + iμ(±m − eϕ)

E2 − m2

⎛⎝iW [t, p1 − eA1, p2 − eA2] −

3∑j=1

(p j − eA j )�(t)(p j − eA j )

⎞⎠ , (2.7)


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102701-3 On Dirac equation on a time scale J. Math. Phys. 52, 102701 (2011)

Q1,2(t) = E2 − e2ϕ2 + i(±m + eϕ)�(t) + i(±m + eϕ) − μ(E2 − e2ϕ2)

E2 − m2×

⎛⎝iW [t, p1 − eA1, p2 − eA2] −

3∑j=1

(p j − eA j )�(t)(p j − eA j )

⎞⎠ , (2.8)

E(t) =√√√√m2(t) +

3∑j=k

(pk(t) − eAk(t))2. (2.9)

If

(p1 − eA1)(t)(p2 − eA2)(t)(p3 − eA3)(t) �= 0, t ∈ T , (2.10)

then formulas (2.7), (2.8) are simplified,

P1,2(t) = 2ieσ ϕσ (t) − (p3 − eA3)�(t) + iμ(t)W [t, p3 − eA3, eϕ ∓ m]

p3(t) − eA3(t), (2.11)

Q1,2(t) = (p3 − eA3)σ (t)(E2(t) − e2ϕ2(t)) + iW [t, p3 − eA3, eϕ ± m]

p3(t) − eA3(t). (2.12)

Using the auxiliary functions {Pj (t)}2j=1 :

Pj (t) = 3

⎧⎨⎩

12 Pj (t), μ(t) ≡ 0,∫ t

t0e−2/μ(t, s)Pj (s) �s

μ(s) , μ(t) �= 0,, Pj (t) = Pσ

j (t) + Pj (t), (2.13)

define

I j (t) = Q j (t) − P�j (t) − P2

j (t), j = 1, 2, (2.14)

which are the invariants of transformation u(t) → v(t)w(t) on a continuous time scale. Assumingθs ∈ R1, s = 1, 2, 3, 4, j = 1, 2, introduce characteristic equations

C L j (t) = C L(θs) = L j [es(t)]

es(t)= θ�

s (t) + θσs (t)θs(t) + Pj (t)θs(t) + Q j (t) = 0, (2.15)

where C L(θs) are characteristic (Riccati) functionals of (2.1),

j = s

2+ 1 + (−1)s−1

4, s = 1, 2, 3, 4, (2.16)

and {es(t)}4s=1 are the exponential functions on a time scale (see Ref. 3)

es(t) ≡ eθs (t, t0) = exp∫ t

t0

limh↘μ(y)

Log(1 + hθs(y))�y

h, s = 1, . . . , 4.

Define the phase functions {θs}4s=1 as the linearly independent asymptotic solutions of the

characteristic Eqs. (2.15). If conditions (2.4), (2.5) are satisfied then any solution ψ(t) of Diracsystem (2.1) may be represented in the form

ψ(t) = �(t)C, (2.17)

where the fundamental matrix

�(t) =

⎛⎜⎜⎜⎜⎝

e1(t) e2(t) 0 0

− s13(τ )e1(t)s14(τ ) − s13(τ )e2(t)

s14(τ )(θ3(t)−s33(τ ))e3(t)

s14(τ )(θ4(t)−s33(τ ))e4(t)

s14(τ )

0 0 e3(t) e4(t)(θ1(t)−s11(τ ))e1(t)

s14(τ )(θ2(t)−s11(τ ))e2(t)

s14(τ ) − s13(τ )e3(t)s14(τ ) − s13(τ )e4(t)

s14(τ )

⎞⎟⎟⎟⎟⎠ . (2.18)


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Remark 2.1: For discrete time scale T = hZ h > 0

es(t, 0) = (1 + hθs(t))t/h, s = 1, 2, 3, 4.

For quantum time scale T = qN0

es(t, 1) =∏

r∈T∩(0,t)

[1 + (q − 1)rθs(r ))], s = 1, 2, 3, 4.

We will use more general approximate representation,

ψ(t) = �(t)(C + δ(t)), (2.19)

where δ(t) is the error 4-vector-function. Expressions |Cs + δs |2 (that are changing slowly with timefor small δs(t)) are called the adiabatic invariants. From (2.19)

Cs + δs = (�−1(t)ψ(t))

s , s = 1, 2, 3, 4. (2.20)

Define adiabatic functionals by formulas Js(t, ψ) = |Cs + δs |2, or

Js(t, ψ) = | (�−1ψ)

s |2 = Es(t, ψ)

|es(t)(θs − θ4 j−s−1)|2 , s = 1, 2, 3, 4,

J5(t, ψ) =√

E1 E2

|θ1 − θ2|2|e1(t)e2(t)| , J6(t, ψ) =√

E3 E4

|θ3 − θ4|2|e3(t)e4(t)| , (2.21)

where

Es(t, ψ) = |ψ�s − θ4 j−1−sψs |2, s = 1, 2, 3, 4. (2.22)

If {θs}4s=1 are solutions of (2.15), then we get the laws of conservation of energy:

Js(t, ψ(t)) = const, s = 1, . . . , 6, t ∈ T . (2.23)

Theorem 2.1: Assume that D ∈ C1rd , θs ∈ R1, and conditions (2.4), (2.5),∫ ∞

tε

∣∣∣∣∣ es(τ )C L(θs)�τ

eσp (τ )(θs − θ4 j−s−1)σ (τ )

∣∣∣∣∣ ≤ C0εm+1, s = 1, 2, 3, 4, p = s, 4 j − s − 1 (2.24)

are satisfied for some positive numbers C0, ε, and for some natural number m, where j = j(s) isdefined in (2.16). Assume also that parameter ε is so small that

0 < C0εm < 1. (2.25)

Then for any solution ψ(t) of (1.1), and for all t1, t2 ∈ T exists c0 > 0 such that

Js(ψ, ε) ≡ |Js(t1, ψ, ε) − Js(t2, ψ, ε)| ≤ c0εm, s = 1, 2, . . . , 6. (2.26)

To check (2.24) we seek the solutions θ (t) of (2.15) in the form of JWKB series (see Refs. 7, 8 and11),

θs(t) =m∑

k=0

εkζks(t), s = 1, 2, 3, 4, (2.27)

where auxiliary functions ζ0s are defined as

ζos(t) = iνs − Pj

2, νs(t) = (−1)s−1

√Q j − P2

j /4, s = 1, 2, 3, 4, (2.28)

and ζks for k = 1, 2, . . . m, s = 1, 2, 3, 4 are defined by recurrent relations

ζks(t) = i

2νs

⎡⎣ζ�

k−1,s(τ )(1 + μζ0s) +k−1∑p=1

ζps(ζk−p,s + μζ�k−1−p,s(τ ))

⎤⎦ . (2.29)


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Denote for s = 1, 2, 3, 4

Zs(t) = ζ�ms(τ )(1 + μζ0s) +

2m+1∑k=m+1

εk−m−1

⎛⎝ k−1∑

p=k−m−1

ζps[ζk−p,s + μζ�k−p−1,s(τ )]

⎞⎠ . (2.30)

Theorem 2.2: Assume that D ∈ C1rd , θs ∈ R1, and conditions (2.4), (2.5) and

∫ ∞

tε

∣∣∣∣∣ es(t)Zs(τ )

eσp (t)(θs(t) − θ4 j−s−1(t))

∣∣∣∣∣�τ ≤ C, s = 1, . . . , 4, p = s, 4 j − s − 1 (2.31)

are satisfied. Then for for any solution ψ(t) of (1.1) the estimate (2.26) is true for some sufficientlysmall parameter ε > 0.

In the case m = 1, assuming

|μν�s (τ )|

|νs | ≤ C,|P�

j (τ )(2 − μPj )||νs |2 ≤ C, t ∈ T0, (2.32)

condition (2.31), and formula (2.21) of the adiabatic invariants are simplified,∫ ∞

tεmax

s=1,...,4,

p=s,4 j−s−1

∣∣∣∣∣ζ�1s (τ )(1 + 2μζ0s + εμζ1s) + ζ1s(ζ1s + μζ�

0s (τ ))

eσp (t)/es(t)

∣∣∣∣∣�τ ≤ C, (2.33)

Js(t) = Es(t, ψ)

4|ν2s e2

s (t)| = |ψ�s − θ4 j−s−1ψs |2

4|ν2s e2

s (t)| , s = 1, 2, 3, 4. (2.34)

Corollary 2.3: Assume that D ∈ C1rd , θs ∈ R1, and conditions (2.4), (2.5), (2.32), (2.33) are

satisfied. Then estimate (2.26) is true for adiabatic invariants (2.34).

Assuming

ν2s (t) = Q j (t) − P2

j (t)

4≥ 0, s = 1, 2, 3, 4 j = 1, 2, (2.35)

for continuous time scale we get

|es(t)|2 = exp

(∫ t

t0

2 [θs(y)]dy

)= νs(t0)/νs(t)

exp(∫ t

t0 [Pj (y)]dy

) = C

|p3(t) − eA3(t)|νs(t),

and from (2.34) we get adiabatic invariant formula with the frequencies νs(t) :

Js(t) = |p3(t) − eA3(t)||ψ ′s(t) − θ4 j−s−1ψs(t)|2

4|νs(t)| , s = 1, 2, 3, 4. (2.36)

Note that (2.36) is close to the well-known adiabatic invariant formula for Lorentz’s pendulums(see Ref. 10) with the additional factor |p3(t) − eA3(t)|.

In continuous time scale (μ(t) ≡ 0) Eqs. (2.6) are much more simpler since

P1(t) = P3(t) = 2ieϕ(t) − (p3 − eA3)′(t)(p3 − eA3)(t)

,

Q1,3(t) = E2(t) − e2ϕ2(t) + i(eϕ(t) ± m)′(t) − (ieϕ(t) ± im)(p3 − eA3)′(t)(p3 − eA3)(t)

. (2.37)


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III. ASYMPTOTIC SOLUTIONS

In this section, we study the asymptotic behavior of the solutions of (2.6) for t → ∞ withoutsmall parameter method, that is, we assume that ε ≡ 1, τ ≡ t . See also Refs. 5 and 6

Theorem 3.1: Let (u1(t), u2(t)), (u3(t), u4(t)), be the complex-valued pairs of functions{us}4

s=1 ∈ C2rd such that

W [us, u4 j−s−1] = us(t)u�4 j−s−1(t) − u�

4 j−s−1(t)us(t) �= 0, s = 1, 3, t ∈ T0, (3.1)∫ ∞

t0

M0(t)�t < ∞, (3.2)

where for s = 1, 2, 3, 4

M0(t) = maxs,p

|Msp(t)|, Msp(t) ≡ uσp(t)L j [us(t)]

W [t, us, u4 j−s−1]σ, p = s, 4 j − s − 1. (3.3)

Then for arbitrary constants {Cs}4s=1 there exist the solutions u(t) of (2.6j) that may be written in the

form (j=j(s) is defined in (2.16))

u(t) = [Cs + δs(t)] us(t) + [C4 j−s−1 + δ4 j−s−1(t)]

u4 j−s−1(t), s = 1, 3,

u�(t) = [Cs + δs(t)] u�s (t) + [C4 j−s−1 + δ4 j−s−1(t)

]u�

4 j−s−1(t), s = 1, 3, (3.4)

and the error 2-vector-functions δs(t) = (δs(t), δ4 j−s−1), s = 1, 3 are estimated as

‖δs(t)‖ ≤ ‖C‖(

−1 + exp∫ ∞

tM0(s)�s

), ‖δs(t)‖ =

√δ2

s (t) + δ24 j−s−1(t). (3.5)

The estimate (3.5) means that if quotients Msp(t) are absolutely integrable on T0, then the errorδs(t) becomes small for sufficiently large time t.

Remark 3.1: If we seek solutions us(t) of (2.6) in the Euler form

us(t) = es(t) = eθs (t, t0), s = 1, 2, 3, 4, (3.6)

then in view of L j [es(t)] = C L j (t)es(t) formula (3.3) turns to

M0(t) = maxp,s

∣∣∣∣∣ eσp (t)C L j (t)es(t)/eσ

s (t)

eσ4 j−s−1(t)[θσ

s (t) − θσ4 j−s−1(t)]

∣∣∣∣∣ , p = s, 4 j − s − 1, s = 1, . . . , 4. (3.7)

Theorem 3.2: Assume P1,2 ∈ C1rd , Q1,2 ∈ Crd ,− [P1,2] ∈ R, and there exists non-zero real

number m such that conditions

eim−Pj(t)e−im−Pj

(t) �= 0, t ∈ T0, j = 1, 2, (3.8)

and (3.2) are satisfied, where {I j }2j=1 are defined in (2.14), and

M0(t) = maxj=1,2

|e±1q (t)(I j (t) − m2)|/m√

(1 − [Pj ]μ(t))2 + (m ± �[Pj ])2μ2(t), q(t) = −2im

1 + (im − Pj )μ(t). (3.9)

Then for arbitrary constants Cs, s = 1, 2, 3, 4, there exist the solutions of (2.6), that may bewritten in form (3.4), with error estimate (3.5), where

us(t) = e−Pj(t) sinρ(t), u4 j−s−1(t) = e−Pj

(t) cosρ(t), ρ(t) = m

1 − μPj (t). (3.10)

sinρ(t) = eiρ(t) − e−iρ(t)

2i, cosρ(t) = eiρ(t) + e−iρ(t)

2. (3.11)


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Note that (3.11) is the definition of trigonometric functions on a time scale (see Ref. 1).

Remark 3.2: If condition (2.10) is satisfied then P1,2 are defined from (2.11). Assume that

�[P1,2(t)] = 2(eϕ)σ (t) − μ(t)W [t, p3 − eA3, eϕ ∓ m]

p3(t) − eA3(t)≡ 0, t ∈ T0,

then |eq | = 1, P1(t) = P2(t), and formula (3.9) is simplified,

M0(t) = maxj=1,2

|I j (t) − m2|m√

(1 − μ(t)P1(t))2 + m2μ2(t).

Using JWKB approximation (see Ref. 8) we get the following theorem.

Theorem 3.3: Assume P1,2 ∈ C3rd , Q1,2 ∈ C2

rd , conditions (3.2), and

eθ1 (t)eθ2 (t) �= 0, eθ3 (t)eθ4 (t) �= 0, t ∈ T0, (3.12)

are satisfied, where

θs(t) = (−1)s−1i − p j (t)

w2j (t)

∈ R1, w j (t) = I −1/4j (t), s = 1, 2, 3, 4, (3.13)

p j (t) = − (1 − μPj )(w2j )

�(t)

2, j = 1, 2, M0(t) = max

j=1,2|M j (t)e

±q j

(t)|, (3.14)

M j (t) = p2j (t)

w2j (t)

+ μ(w2j )

�(t)

w4j (t)

+ p�j (t)

(1 − Pjμ(t) + (±i − p j )μ(t)

w2j (t)

), (3.15)

q j (t) = −2i

w2j (t) + μ(t)(i − p j (t))

, j = 1, 2. (3.16)

Then for arbitrary constants Cs, s = 1, 2, 3, 4, there exist the solutions of (2.6) that may be writtenin form (3.4), (3.6), (3.13) with error estimate (3.5). Here I 1/4

j (t) is the principal value of the fourthroot of the complex valued function I j (t).

Note that for a continuous time scale M j (t) = [w′j (t)]

2 − [w′j (t)w j (t)]′ = −w′′

j (t)w j (t).

IV. TRANSITION PROBABILITIES

Assuming that initially electron has the spin oriented up, we have

ψ1(0) = 1, ψ2(0) = 0, ψ3(0) = 0, ψ4(0) = 0. (4.1)

Define for i p ∈ R slightly different from (3.11) trigonometric functions on T

sintp(t) = 1

2i

(eip(t) − 1

eip(t)

), costp(t) = 1

2

(eip(t) + 1

eip(t)

). (4.2)

Trigonometric functions (4.2) on a continuous time scale are the same as (3.11). Definition (4.2)

simplifies the transition probability formulas below. In view of e−i p(t) − 1eip(t) = ep2μ

(t)−1

eip(t) , we haveconversion formulas

sinp(t) = sintp(t) + 1 − ep2μ(t)

2ieip(t), cosp(t) = costp(t) − 1 − ep2μ(t)

2eip(t). (4.3)


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


By using fundamental matrix (2.18) we calculate the transitions probabilities

|ψ1(t)|2 =∣∣∣∣2(m(0) + eϕ(0) − iθ2(0))

θ1(0) − θ2(0)eη2 (t)

[eη1 (t)sintq1 (t) + eη2 (t)

]∣∣∣∣2

,

|ψ2(t)|2 = ∣∣R1eη1 (t)eη2 (t)sintq1 (t) − R2eη3 (t)eη4 (t)sintq2 (t) + eη2 (t)eη3 (t)R3sintq3 (t) + R4e3(t)∣∣2 ,

|ψ3(t)|2 = 4

∣∣∣∣ (p3(0) − eA3(0))

θ3(0) − θ4(0)eη3 (t)eη4 (t)sintq2 (t)

∣∣∣∣2

,

|ψ4(t)|2 = 1 − |ψ1(t)|2 − |ψ2(t)|2 − |ψ3(t)|2, (4.4)

where

R1(t) = 2(p3 − eA3)(t)(iθ2(0) − eϕ(0) − m(0))

(p1 − eA1 − i p2 + ieA2)(t)(θ1(0) − θ2(0)),

R2(t) = 2(p3 − eA3)(0)(iθ4 − eϕ + m)(t)

(θ3 − θ4)(0)(p1 − eA1 − i p2 + ieA2)(t), R3(t) = −2i(p3 − eA3)(t)

(p1 − eA1 − i p2 + ieA2)(t),

R4(t) = (p3 − eA3)(0)

(p1 − eA1 − i p2 + ieA2)(t)

(θ3(t) − θ4(t)

θ3(0) − θ4(0)− (p3 − eA3)(t)

(p3 − eA3)(0)

), (4.5)

ηs(t) =⎧⎨⎩

θs (t)2 , μ(t) ≡ 0,

−1+√

1+μ(t)θ j (t)

μ(t) μ(t) �= 0,s = 1, 2, 3, 4, (4.6)

q1 = θ1 − θ2

i(√

1 + μθ1 + √1 + μθ2)

√1 + μθ2

,

q2 = θ3 − θ4

i(√

1 + μθ3 + √1 + μθ4)

√1 + μθ4

, q3 = θ2 − θ3

i(√

1 + μθ3 + √1 + μθ2)

√1 + μθ3

.

(4.7)

For the special choice of electromagnetic field: p3(t) − eA3(t) ≡ 0 we get

|ψ1(t)|2 =∣∣∣∣2(m(0) + eϕ(0) − iθ2(0))

θ1(0) − θ2(0)eη2 (t)

[eη1 (t)sintq1 (t) + eη2 (t)

]∣∣∣∣2

,

|ψ2(t)|2 = |ψ3(t)|2 = 0, |ψ4(t)|2 = 1 − |ψ1(t)|2. (4.8)

These formulas mean that for the special case p3(t) − eA3(t) ≡ 0, the only transition of an electronto the positron with the opposite spin orientation is possible.

For continuous time scale the first transition probability formula (4.4) turns to

|ψ1(t)|2 =∣∣∣∣2(m + eϕ − iθ2)(0)

θ1(0) − θ2(0)e

12

∫ t0 (θ1+θ2)dy sin

(1

2i

∫ t

0(θ1 − θ2)dy

)+ e

∫ t0 θ2dy

∣∣∣∣2

. (4.9)

In the simple case of continuous time scale and

(pk − eAk)�(t) = 0, (eϕ)�(t) = 0, m�(t) = 0, k = 1, 2, 3, (4.10)


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


from (2.7), (2.8), (2.15) we get

Pj = 2ieϕ, Q j = E2 − e2ϕ2, j = 1, 2,

θ�(t) + θσ (t)θ (t) + 2ieϕθ (t) + E2 − e2ϕ2 = 0, (4.11)

with the exact solutions θs, that are the eigenvalues of S = −i D (see (2.2))

θ1 = θ3 = −ieϕ − i E, θ2 = θ4 = i E − ieϕ. (4.12)

In this case from (4.4) we get

|ψ1|2(t) = cos2(Et) + m2

E2sin2(Et), |ψ2(t)|2 = 0,

|ψ3|2(t) = (p3 − eA3)2

E2sin2(Et), |ψ4(t)|2 = 1 − |ψ1(t)|2 − |ψ2(t)|2 − |ψ3(t)|2. (4.13)

Formula |ψ2|2(t) = 0 means that transition of the electron with the spin up to the positron withthe same orientation of the spin is impossible. From these formulas electron-positron transitionprobability is given by the formula

P Rep(t) = |ψ2(t)|2 + |ψ4(t)|2 = (p1 − eA1)2 + (p2 − eA2)2

E2sin2(t E). (4.14)

Calculating average of this probability during a long period of time we get

APep = limT →∞

∫ T0 P Rep(t)dt

T= (p1 − eA1)2 + (p2 − eA2)2

2[m2 + (p1 − eA1)2 + (p2 − eA2)2 + (p3 − eA3)2]. (4.15)

If the ratio m(p1−eA1)2+(p2−eA2)2+(p3−eA3)2 is small, and (p j − eA j )2 are approximately same in all

directions, then electron-positron transition probability during a long period of time is APep = 13 .

Since the probability of the positron-electron transition is pretty high, there is a simple explanationof the stability of the revolution of an electron about the proton only by the electromagnetic field.Indeed when the electron is far from the proton, it is attracted by the electromagnetic field of theproton. When the electron approaches closer to the proton, it turns to the positron which is repellingfrom the proton by the same electromagnetic field.

V. CONSTRUCTION OF THE FUNDAMENTAL SOLUTION

To construct the fundamental solution of the Dirac equation (2.1) we solve

ψ�1 (t) − s11ψ1(t) = s13ψ3(t) + s14ψ4(t), ψ�

2 (t) − s11ψ3(t) = −s14ψ3(t) − s13ψ4(t), (5.1)

ψ�3 (t) − s33ψ3(t) = s13ψ1(t) + s14ψ2(t), ψ�

4 (t) − s33ψ4(t) = −s14ψ1(t) − s13ψ2(t), (5.2)

as linear algebraic equations for ψ1, ψ2, ψ3, ψ4

ψ1 = s13(ψ�3 − s33ψ3) + s14(ψ�

4 − s33ψ4)

m2 − E2, ψ2 = s13(ψ�

4 − s33ψ4) + s14(ψ�3 − s33ψ3)

E2 − m2,

ψ3 = s13(ψ�1 − s11ψ1) + s14(ψ�

2 − s11ψ2)

m2 − E2, ψ4 = s13(ψ�

2 − s11ψ2) + s14(ψ�1 − s11ψ1)

E2 − m2. (5.3)

From (5.1), (5.2) we get

ψ4(t) = ψ�1 (t) − s11ψ1(t) − s13ψ3(t)

s14, ψ2(t) = ψ�

3 (t) − s13ψ1(t) − s33ψ3(t)

s14. (5.4)


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


By differentiation of (5.1), in view of the rule ( f g)�(t) = g f �(t) + f σ g�(t), we get

ψ��1 (t) = sσ

11ψ�1 (t) + sσ

13ψ�3 (t) + sσ

14ψ�4 (t) + ψ1s�

11(t) + ψ3s�13(t) + ψ4s�

14(t),

and excluding ψ�1 (t), ψ�

3 (t), ψ�4 (t) by using (5.1), (5.2) we have

ψ��1 (t) = sσ

11(s11ψ1 + s13ψ3 + s14ψ4) + sσ13(s13ψ1 + s14ψ2 + s33ψ3)+

sσ14(−s14ψ1 − s13ψ2 + s33ψ4) + s�

11ψ1 + s�13ψ3 + s�

14ψ4,

Further excluding ψ2, ψ3, ψ4 we get

ψ��1 (t) + P1(t)ψ�

1 (t) + Q1(t)ψ1(t) = B1(t)(ψ�2 (t) − s11ψ2(t))+

[s13(ψ�4 − s33ψ4) + s14(ψ�

3 − s33ψ3)]μW [s14, s13]

E2 − m2, (5.5)

P1 = (1 + s33μ)(s�13s13 − s�

14s14)

|s14|2 − s213

− sσ11 − s33, B1 = (1 + s11μ)W [s13, s14]

|s14|2 − s213

. (5.6)

Q1 = |s14|2 − s213 + s11s33 − s�

11 +(

μ + s11(1 + s33μ)

|s14|2 − s213

)(s�

14s14 − s�13s13). (5.7)

Since Eqs. (5.2) may be obtained from (5.1) by transformation m → −m, s11 ↔ s33, ψ1 ↔ψ3, ψ2 ↔ ψ4 from (5.5) we get

ψ��3 (t) + P2ψ

�3 (t) + Q2ψ3(t) = B2(t)(ψ�

4 (t) − s33ψ4(t))+

[s13(ψ�2 (t) − s11ψ2) + s14(ψ�

1 (t) − s11ψ1))]μW [t, s14, s13]

E2 − m2, (5.8)

where

P2(t) = (1 + s11μ)(s�13s13 − s�

14s14)

|s14|2 − s213

− s11 − sσ33, B2(t) = (1 + s33μ)W [s13, s14]

|s14|2 − s213

, (5.9)

Q2(t) = |s14|2 − s213 + s11s33 − s�

33 +(

μ + s33(1 + s11μ)

|s14|2 − s213

)(s�

14s14 − s�13s13). (5.10)

Assuming (2.5) from (5.5), (5.8) we get Eqs. (2.6), in view of

W [s13, s14](t) = B1(t) = B2(t) ≡ 0. (5.11)

Further from (2.5) assuming (2.10) we get (2.11), (2.12) since

p�1 (t) − eA�

1 (t)

p1(t) − eA1(t)= p�

2 (t) − eA�2 (t)

p2(t) − eA2(t)= p�

3 (t) − eA�3 (t)

p3(t) − eA3(t). (5.12)

The solutions ψ1(t), ψ3(t) of Eqs. (2.6j) we seek in the Euler form

ψ1(t) = C1e1(t) + C2e2(t), j = 1, ψ3(t) = C3e3(t) + C4e4(t), j = 2. (5.13)

By substitution (5.13) in (5.4) we get

ψ2(t) = C3(θ3 − s33)e3(t) + C4(θ4 − s33)e4(t) − s13C1e1(t) − s13C2e2(t)

s14,

ψ4(t) = C1(θ1 − s11)e1(t) + C2(θ2 − s11)e2(t) − s13C3e3(t) − s13C4e4(t)

s14,


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


or (2.17), (2.18). Solving (2.17) for Cs we get

Cs = [�−1ψ]s = (−1)s−1 (s11 − θ3−s)ψ1 + s13ψ3 + s14ψ4

(θ1 − θ2)es, s = 1, 2.

In view of (5.1), (5.2) these formulas are simplified

Cs = ψ�s (t) − θ4 j−1−s(t)ψs(t)

(θs(t) − θ4 j−1−s(t))es(t), s = 1, 2, 3, 4. (5.14)

Defining non-negative energy functionals by formulas

Js(t, ψ) = CsCs, J5(t, ψ) = |C1C2|, J6(t, ψ) = |C3C4| s = 1, 2, 3, 4, (5.15)

we get (2.21), (2.22). Denoting

ξs(t) = θs(t) − Pj (t), s = 1, 2, 3, 4, (5.16)

formula for characteristic functional (2.15) is simplified,

C L(θs) = ξσs (t)ξs(t) + ξ�

s (t)(1 − μPj ) + I j (t), I j (t) = Q j (t) − P2j (t) − P�

j (t). (5.17)

VI. PROOFS

Lemma 6.1: (Ref. 4) Assume there exist a matrix function �(t) ∈ Crd such that �σ is invertibleand (3.2) is satisfied, where

M0(s) = ‖M(s)‖, M(s) ≡ (�(σ (s))−1[��(s) − S(s)�(s)

]. (6.1)

Then every solution of (2.1) can be represented in form (2.19), where

‖δ(t)‖ ≤ ‖C‖(

−1 + exp∫ ∞

t‖M(s)‖�s

), (6.2)

and ‖ · ‖ is the Euclidean vector (or matrix) norm: ‖δ(t)‖ =√

δ21(t) + . . . + δ2

4(t).

Lemma 6.2: If (2.4), (2.5) are satisfied then

‖M(t)‖ ≤√

8 maxs=1,2,3,4

p=s,4 j−s−1

∣∣∣∣∣ es(t)C L(θs(t)

eσp (t)(θs − θ4 j−s−1)σ (t)

∣∣∣∣∣ . (6.3)

Proof of Lemma 6.2: To calculate delta derivative �� we will use the formula(ab

c

)�

=[

a�

a

bσ

b+ b�

b− c�

c

]ab

cσ.

In view of (e1s13

−s14

)�

=(

sσ13θ1

s13+ s�

13

s13− s�

14

s14

)e1s13

−sσ14

= − sσ13θ1e1

sσ14

= − s13θ1e1

s14,

(e1(θ1 − s11)

s14

)�

=(

(θ1 − s11)σ θ1

θ1 − s11+ (θ1 − s11)�

(θ1 − s11)− s�

14

s14

)(θ1 − s11)e1

sσ14

,


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


we get from (2.18)

�� =

⎛⎜⎜⎜⎜⎝

θ1e1 θ2e2 0 0

− s13s14

θ1e1 − s13s14

θ2e2 N3e3 N4e4

0 0 θ3e3 θ4e4

N1e1 N2e2 − s13s14

θ3e3 − s13s14

θ4e4

⎞⎟⎟⎟⎟⎠ , (6.4)

where

N j (t) = (θ j − s11)

sσ14

(θ j + (1 + μθ j )

(θ j − s11)�

(θ j − s11)− s�

14

s14

), j = 1, 2,

N j (t) = (θ j − s33)

sσ14

(θ j + (1 + μθ j )

(θ j − s33)�

(θ j − s33)− s�

14

s14

), j = 3, 4. (6.5)

Further from (6.1) we get (6.3) since by direct calculations

M(t) =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

− e1(t)C L(θ1)eσ

1 (θ1−θ2)σ (t) − e2(t)C L(θ2)eσ

1 (θ1−θ2)σ (t) 0 0

e1(t)C L(θ1)eσ

2 (θ1−θ2)σ (t)e2(t)C L(θ2)

eσ2 (θ1−θ2)σ (t) 0 0

0 0 − e3(t)C L(θ3)eσ

3 (θ3−θ4)σ (t) − e4(t)C L(θ4)eσ

3 (θ3−θ4)σ (t)

0 0 e3(t)C L(θ3)eσ

4 (θ3−θ4)σ (t)e4(t)C L(θ4)

eσ4 (θ3−θ4)σ (t)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

. (6.6)

�Proof of Theorem 2.1: From (2.18) in view of (2.4) we get

det(�) = −(s14(t))2

e1e2e3e4(t)(θ1 − θ2)(θ3 − θ4)= [p1 − eA1 − i(p2 − eA2)]2

e1e2e3e4(θ1 − θ2)(θ3 − θ4)�= 0. (6.7)

So �σ (t) is invertible and Lemma 6.1 is applicable. From (6.3), in view of (2.24)∫ ∞

t1

‖M(τ )‖�t =∫ ∞

t1ε

1

ε‖M(τ )‖�τ ≤ C0ε

m, τ = tε ∈ Tε, (6.8)

and from (6.2), (2.25) we have

‖δ(t)‖ ≤ ‖C‖ (eC0εm − 1

) ≤ ‖C‖eC0εm ≤ C3ε

m . (6.9)

From (2.19) and (6.9) we get (2.26) since for s = 1, 2, 3, 4

Cs + δs(t) = (�−1(t)ψ(t))s, Js(tk, ψ) = |(�−1ψ)s(tk)|2 = |Cs + δs(tk)|2, (6.10)

Js(t1) − Js(t2) = |C j + δs(t1)|2 − |C j + δs(t2)|2 = O(εm), ε → 0. (6.11)

�Proof of Theorem 2.2: To prove Theorem 2.2 it is enough to check that condition (2.24) of

Theorem 2.1 is followed from (2.31). Solutions of characteristic Eqs. (2.15) we look in form (2.27).By substitution (2.27) in (2.15) we get for s = 1, 2, 3, 4

C L(θs) = εθ�s (τ )(1 + μθs) + θ2

s + Pjθs + Q j =∞∑

k=0

bksεk,


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


or

C L(θs) =m∑

k=0

εk+1ζ�ks + μ

m∑p=0

m∑q=0

ε p+q+1ζpsζ�qs(τ ) +

m∑p=0

m∑q=0

ε p+qζpsζqs+

Pj

m∑k=0

εkζks + Q j =∞∑

k=0

bksεk, (6.12)

where

b0s = ζ 20s + Pjζ0s + Q j , s = 1, 2, 3, 4, (6.13)

bks = ζ�k−1,s(τ )(1 + μζ0s) + (2ζ0s + Pj )ζks +

k−1∑p=1

ζps(ζk−p,s + μζ�k−1−p,s), k = 1, 2, . . . .

(6.14)We define {ζks}m

k=0 as solutions of bks = 0, k = 0, 1, . . . m. Solving equations b0s = ζ 20s + Pjζ0s

+ Q j = 0, for ζ0s we get (2.28) in view of 2ζos(t) = 2iνs − Pj . Solving bks = 0, k = 1, . . . , mfor ζks , we obtain (2.29). From

bks(t) = ζps(t) = 0, k ≤ m, p ≥ m + 1, (6.15)

it follows

bks ≡ 0, k ≤ m, k ≥ 2m + 2. (6.16)

bks = ζ�k−1,s(τ )(1 + μζ0s)δk,m+1+

k−1∑p=1

ζps(ζk−p,s + μζ�k−1−p,s(τ )), k = m + 1, m + 2, . . . , 2m + 1. (6.17)

Using notation Zs =∑2m+1k=m+1 εk−m−1bks s = 1, 2, 3, 4, we have

Zs(t) = ζ�ms(τ )(1 + μζ0s) +

2m+1∑k=m+1

εk−m−1

⎛⎝k−1∑

p=1

ζps(ζk−p,s + μζ�k−p−1,s(τ ))

⎞⎠ ,

or, in view of (6.16), we get (2.30). From (6.12) in view of (6.16) we have

C L(θs) =2m+1∑

k=m+1

bksεk = εm+1

2m+1∑k=m+1

bksεk−m−1εm+1 Zs(t), s = 1, . . . , 4. (6.18)

Theorem 2.2 is followed from Theorem 2.1 in view of (6.18). �If m = 1 condition (2.24) could be simplified. Indeed in this case

θ1,2 = ±iν1 − P1

2+ ε

(∓ i P�

1

4ν1− ν�

1 (τ )

2ν1

)(1 ± iμν1 − μP1

2

),

Zs = ζ�1s (τ )(1 + μζ0s) + ζ0sμζ�

1s (τ )) + ζ1s(ζ1s + μζ�0s (τ )) + εζ1sμζ�

1s (τ ),

Zs = ζ�1s (τ )(1 + 2μζ0s + εμζ1s) + ζ1s(ζ1s + μζ�

0s (τ )). (6.19)

Further for m = 1, k = 2 we have ζ2s ≡ 0, and condition (2.24) turns to

∫ ∞

tεmax

s=1,...,4,

p=s,4 j−s−1

∣∣∣∣∣∣es(t)[ζ�

1s (τ )(1 + 2μζ0s + εμζ1s) + ζ1s(ζ1s + μζ�0s (τ ))]

2eσp (t)ν1

(1 − εP�

1 (τ )4ν2

1− εμν�

1 (τ )2ν1

+ εμP�1 (τ )P1

8ν21

)∣∣∣∣∣∣�τ ≤ C. (6.20)


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


Under assumptions (2.32) condition (6.20) turns to (2.33) and adiabatic invariant

Js(t) = |ψ�s − θ4 j−s−1ψs |2

4|ν2s e2

s (t)|∣∣∣1 − εP�

j (τ )

4ν2s

− εμν�s (τ )

2νs+ εμP�

j (τ )Pj

8ν2s

∣∣∣2 , s = 1, 2, 3, 4

in estimate (2.26) may be simplified to (2.34).Assuming (2.35) for continuous time scale we have θ1 + θ1 = − [P1] − ν ′

1(t)ν1(t) , |e1(t)|

= ν1(t0)(p3−eA3)(t0)||ν1(t)(p3−eA3)(t)| = C

|ν1(t)(p3−eA3)(t)| , and we get adiabatic invariant (2.36).

Proof of Theorem 3.1: Equation (2.6) we can rewrite as a system

v� = S1(t)v(t), v(t) =(

u(t)

u�(t)

), S1(t) =

(0 1

−Q(t) −P(t)

). (6.21)

To apply Lemma 6.1 to this system we calculate the matrix M(t)

M(t) = (�σ (t))−1[��(t) − S1(t)�(t)

], � =

(u1(t) u2(t)

u�1 (t) u�

2 (t)

), (6.22)

M(t) = 1

W σ [u1, u2]

(uσ

2 L[u1] uσ2 L[u2]

−uσ1 L[u1] −uσ

1 L[u2]

). (6.23)

Now Theorem 3.1 is followed from Lemma 6.1 since (6.1) turns to (3.3), and asymptotic represen-tation (2.19), (6.2) is equivalent to (3.4), (3.5). �

Proof of Theorem 3.2: Theorem 3.2 is followed from Theorem 3.1 by choosing the phasefunctions

θs(t) = (−1)s im − Pj (t), ξs(t) = (−1)s im, s = 1, 2, 3, 4,

eθ2 (t)

eθ1 (t)= eq1 (t),

eθ4 (t)

eθ3 (t)= eq2 (t), q j (t) = −2im

1 + imμ − μPj, j = 1, 2.

Since − [Pj ], j = 1, 2 are regressive θ j , q j are regressive as well,

|1 + μθ1,2|2 = (1 − μ [Pj ])2 + (�[Pj ] ± m)2μ2 > 0, |1 + q jμ| =

∣∣∣∣∣1 − imμ − μPj

1 + imμ − μPj

∣∣∣∣∣ > 0.

By direct calculations from (5.17)

es(t)C L(θs)

eσp (t)(θσ

s − θσ4 j−s−1)

= e±1q (t)(I j − m2)

m(1 + μθp),

es(t) + e4 j−s−1(t)

e−Pj

= e im1−μP j

+ e −im1−μP j

= 2 cosρ(t),es(t) − e4 j−s−1(t)

e−Pj

= 2i sinρ(t),

and we get asymptotic representation (3.4), (3.6) with error estimate (3.5). �Proof of Theorem 3.3: By choosing

ξs(t) = A(t)√

I j (t), G(t) =(

I j (t)

I σj (t)

)1/4

, s = 1, 2, 3, 4, (6.24)


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


we have

ξ�s = A

√I j

� + A�√

I σj = AG2 I �

j√I j (1 + G2)

+ A�√

I j

G2. (6.25)

Further from (5.17) we get

C L(θs) = I j

G2

(A2 + 2p j A + G2 + A�√

I j(1 − Pjμ + μA

√I j )

), (6.26)

where

p j (t) = −1 − Pjμ

2(I −1/2

j )� = −1 − Pjμ

2(w2)�(t) = (Pjμ − 1)w�

j (w j + wσj )

2. (6.27)

Choosing JWKB approximation: A = (−1)s−1i − p j or

ξs = [(−1)s−1i − p j (t)]√

I j , θs = i(−1)s−1√

I j − p j

√I j , (6.28)

we deduce Theorem 3.3 from Theorem 3.1. Indeed from

2iC L(θs)

(ξ1 − ξ2)σ= −1

w2j

(p2

j + μ(w2j )

�

w2j

+ p�j

[w2

j (1 − Pjμ) + μ(−1)s−1i − μp j])

, (6.29)

we get (3.15), where M j (t) = − 2iC L(θs )(ξ1−ξ2)σ (t) , Formula (3.16) is followed from eq1 (t) = e2(t)

e1(t) , q1(t)

= θ2−θ11+μθ1

. �Proof of transition probability formula (4.4): From initial conditions (4.1) and

(2.17), (2.18) by taking t0 = 0, e j (0, 0) = 1 we get C1 + C2 = 1, C3 + C4 = 0,(C1+C2)s13(0)+C3(s33−θ3)(0)+C4(s33−θ4)(0)

s14(0) = 0, C1(s11−θ1)(0)+C2(s11−θ2)(0)+s13(0)(C3+C4)s14

= 0. Solving thissystem we get

C1 = s11(0) − θ2(0)

θ1(0) − θ2(0), C2 = 1 − C1, C3 = s13(0)

θ3(0) − θ4(0)C4 = −C3.

From (2.17), (2.18)

ψ1 = C1e1(t) + (1 − C1)e2(t) = C1(e1(t) − e2(t)) + e2(t), ψ3 = C3(e3(t) − e4(t)).

Introducing (4.6) and using properties of exponential functions on time scales ep

eq= e p−q

1+μq, epeq

= ep+q+μpq , we have

es(t) = eθs (t) = e2ηs

(t), s = 1, 2, 3, 4 1 + μ(t)ηs(t) =√

1 + μ(t)θs(t), (6.30)

e1(t) − e2(t) = eη1 eη2

(eη1

eη2

− eη2

eη1

)= eη1+η2+μη1η2

(eiq1 − 1

eiq1

). (6.31)

Since {θs}2s=1 are regressive the functions {ηs}2

s=1 are regressive as well. Definingtrigonometric functions sintp(t), costp(t) by (4.2) we get e1 − e2 = 2ieη1 eη2 sintq1 (t),e3 − e4

= 2ieη3 eη4 sintq2 (t), e2 − e3 = 2ieη2 eη3 sintq3 (t). Further from (2.18)

ψ1 = C1e1(t) + (1 − C1)e2(t) = C1(e1(t) − e2(t)) + e2(t) = 2iC1eη1 eη2 sinq1 (t) + e2(t)

ψ1 = C5eη1 (t)eη2 (t) sinq1 (t) + e2(t), C5 = 2iC1 = 2m(0) + 2eϕ(0) − 2iθ2(0)

θ1(0) − θ2(0),

ψ2 = (C1e1 + C2e2)s13 + C3(s33 − θ3)e3 + C4(s33 − θ4)e4

−s14= R1(t)eη1 (t)eη2 (t)sintq1 (t)+


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25


−R2(t)eη3 (t)eη4 (t)sintq2 (t) + R3(t)eη2 (t)eη3 (t)sintq3 (t) + R4(t)e3(t),

ψ3 = C3(e3(t) − e4(t)) = 2iC3eη3 eη4 sinq2 (t) = 2(p3(0) − eA3(0))

θ3(0) − θ4(0)eη3 eη4 sinq2 (t),

and we get (4.4), (4.5). If p3 − eA3 ≡ 0 then we get (4.8).In the case 2m

E � 1,2|p3−eA3|

E � 1, from (4.15) we have P Rep = 12 . �

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4 Hovhannisyan, G.,“Asymptotic stability for dynamic equations on time scales,” Adv. Differ. Equ. 2006, 17 (2006).5 Hovhannisyan, G. R. and Liu, W., “On non-autonomous Dirac equation,” J. Math. Phys. 50, 123507 (2009).6 Hovhannisyan, G.,“Levinson theorem for 2 x 2 system and applications to the asymptotic stability and Schrodinger

equation,” Int. J. Evol. Equ. 3:2, 181–203 (2007) ISSN:1549-2907.7 Hovhannisyan, G.,“WKB estimates for 2 × 2 linear dynamic systems on time scales,” Adv. Differ. Equ. 2008, 12 (2008).8 Kevorkian, J. and Cole, J. D., Multiple Scale and Singular Perturbation Methods (Springer, New York, 1996), ISBN

0-387-94202-5.9 Landim, R. R. and Guedes, I.,“Wave functions for a Dirac particle in a time-dependent potential,” Phys. Rev. A 61, 054101

(2000).10 Littlewood, J. E., “Lorentz’s pendulum problem,” Ann. Phys. (N.Y.) 21, 233–242 (1963).11 Spigler, R. and Vianello, M., A Survey on the LiouvilleGreen (WKB) Approximation for Linear Difference Equations of

the Second Order, Advances in Difference Equations: Proceedings of the Second International Conference on DifferenceEquations, VeszprA C©m, Hungary, 7–11 August 1995, edited by S. Elaydi, I. GyÅri, and G. E. Ladas (CRC Press, London,1998), pp. 567–577. ISBN 90-5699-521-9.

12 Thaller, B., The Dirac Equation (Springer-Verlag, Berlin, 1992).13 Zhang, Z.-G., “Complete solutions for a class of time-dependent Dirac equations,” Phys. Scr. 76(4), 349–353 (2007).


128.114.34.22 On: Tue, 25 Nov 2014 06:43:25

http://dx.doi.org/10.1155/ADE/2006/18157

http://dx.doi.org/10.1063/1.3265922

http://dx.doi.org/10.1155/2008/712913

http://dx.doi.org/10.1103/PhysRevA.61.054101

http://dx.doi.org/10.1016/0003-4916(63)90107-6

http://dx.doi.org/10.1088/0031-8949/76/4/012

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