Modified wave equation for spinless particles and its solutions in an external magnetic field

0.12 MB·English

by  S. I. Kruglov

Modified wave equation for spinless particles and its solutions in
an external magnetic field
S. I. Kruglov
Department of Chemical and Physical Sciences,
2 University of Toronto Mississauga,
1
0 3359 Mississauga Rd. North, Mississauga, Ontario, Canada L5L 1C6
2

l
u
Abstract
J

5 The wave equation for spinless particles in the framework of de-
2
formed special relativity is considered. We formulate the third-order

] in derivatives wave equation leading to the modified dispersion rela-
h
tion. The first-order formalism is considered and the density matrix
t
-
p is obtained. The Schro¨dinger form of equations is presented and the
e quantum-mechanical Hamiltonian is found. Exact solutions of the
h
wave equation are obtained for particles in the constant and uniform
[
external magnetic field. The change of the synchrotron radiation ra-
1
dius due to quantum gravity corrections is calculated.
v
3
7
5 1 Introduction
6
.
7
The deformed dispersion relations, which result in the Lorentz violation, can
0
2 be caused by quantum gravity [1], [2], [3], [4], [5], [6], [7], [8], [9]. According
1
to [5], [6], we consider the special case of the modified dispersion relation
:
v
(the speed of light in vacuum c equals unit in our notations):
i
X
r p2 = p2 +m2 Lp p2, (1)
a 0 − 0
where p isanenergy andpisa momentum ofa particleandLis aparameter
0
with the dimension of “length”. The positive value of L (L > 0) corresponds
to the subluminal propagation of particles. We imply that the last term in
Eq.(1), violating the Lorentz symmetry, is due to quantum gravity correc-
tions, and L is of the order of the Plank length L = M−1 (M = 1.22 1019
P P P ×
GeV is the Plank mass). The modified dispersion relation (1) appears in
space-time foam Liouville-string models [10], [11]. Constrains on quantum
gravity corrections were estimated from the Crab Nebula synchrotron radi-
ation [12], [13], [14]. The goal of this paper is to describe spinless fields
1
realizing the deformed dispersion relation (1) in third-order and first-order
formalisms. Also we obtain equation solutions of free particles and particles
in the external magnetic field.
The paper is organized as follows. In Sec.2, we formulate the wave equa-
tion with modified dispersion relation in the third-order and first-order for-
malisms and obtain the density matrix. The Schro¨dinger form of equations
is presented and the quantum-mechanical Hamiltonian is found in Sec.3. In
Sec.4 we obtain exact solutions of the wave equation for particles in the con-
stant and uniform external magnetic fields. The synchrotron radius with
quantum gravity corrections is estimated. A conclusion is made in Sec.5. In
Appendix, Sec.6, useful products of the equation matrices are obtained.
The Euclidean metric is explored and the system of units h¯ = c = 1 is
used. Greek letters run 1,2,3,4 and Latin letters run 1,2,3.
2 Field equation for spinless particle
2.1 Wave equation in the first-order formalism
Let us consider the wave equation for spinless particles:
∂2 m2 iL∂2∂ Φ(x) = 0, (2)
µ − − i t
(cid:16) (cid:17)
where ∂ = ∂/∂x = (∂/∂x ,∂/(i∂t)), x = t is a time. The plane-wave
µ µ i 0
solutionforpositive energy Φ(x) = Φ exp[i(px p x )] toEq.(2) leads tothe
0 0 0
−
modified dispersion relation (1). The last term in (2) violates the invariance
under theLorentz transformations. To present thehigher derivative equation
(2)inthefirst-orderformalism, wefollowthemethodof[15]. Letusintroduce
the system of first order equations which are equivalent to Eq.(2)
∂ Ψ +mΦ+∂ Φ = 0,
µ µ 4
∂ Φ+mΨ = 0, (3)
µ µ e
L∂ Ψ Φ = 0.
m m
−
Indeed, replacing Ψ and Φ from Eqs.(3) into the first equation of (3), one
µ e
obtains Eq.(2). The fields Ψ , Φ, Φ possess the same dimension. It is conve-
µ
e
nient to introduce the wave function
e
Φ(x)
Ψ(x) = Ψ (x) = Ψ (x) , (4)
A  µ 
{ }
Φ(x)
 
 
e
2
and index runs A = (0,µ,0), Ψ = Φ, Ψ = Φ. With the help of the elements
0 0
of the entire matrix algebra εA,B, with properties [16]
e e
εM,N = δ δ , e εM,AεB,N = δ εM,N, (5)
MA NB AB
AB
(cid:16) (cid:17)
where A,B,M,N = (0,µ,0), the system of equations (3) may be presented
in the matrix form as follows:
e
∂ εµ,0 +ε0,µ +δ ε0,0 mLδ ε0,m
µ µ4 µm
−
(cid:20) (cid:18) (cid:19) (6)
+m ε0,0 +εµ,µ +ε0,0 e Ψ (x) =e 0.
B
(cid:18) (cid:19)(cid:21)AB
withthesummationoverallrepeatedinedeices. Weintroducethe6 6matrices
×
β = εµ,0 +ε0,µ +δ ε0,0 mLδ ε0,m, I = ε0,0 +εµ,µ +ε0,0. (7)
µ µ4 µm 6
−
Taking into account these eqeuations Eq.e(6) becomes the first-oredeer wave
equation
(β ∂ +m)Ψ(x) = 0, (8)
µ µ
where we have used the unit 6 6-matrix I . We note that the 5-dimensional
6
×
matrices
β(0) = εµ,0 +ε0,µ (9)
µ
enter the Lorentz covariant wave equation for scalar particles
β(0)∂ +m Ψ(0)(x) = 0, (10)
µ µ
(cid:16) (cid:17)
where the wave function reads
Φ(x)
Ψ(0)(x) = . (11)
Ψ (x)
µ !
Matrices (9) obey the Duffin Kemmer Petiau algebra [16]
− −
β(0)β(0)β(0) +β(0)β(0)β(0) = δ β(0) +δ β(0). (12)
µ ν α α ν µ µν α αν µ
The Lorentz group generators in the 5-dimension representation space are
given by
J = β(0)β(0) β(0)β(0) = εµ,ν εν,µ, (13)
µν µ ν − ν µ −
and obey the commutation relations
[J ,J ] = δ J +δ J δ J δ J ,
ρσ µν σµ ρν ρν σµ ρµ σν σν ρµ
− −
(14)
β(0),J = δ β(0) δ β(0).
λ µν λµ ν − λν µ
The form-invariance ofhEq.(8) uinder the Lorentz transformations is broken
due to terms containing quantum gravity parameter L.
3
2.2 The density matrix
In the momentum space, for the positive energies Ψ(x) exp[i(p,x p x )]
0 0
∼ −
and Eq.(8) becomes
(ip+m)Ψ(p) = 0, (15)
where p = β p . The matrix p obeys the equation as follows (see Appendix):
µ µ b
b p4 bp2p2 +mLp4p2p = 0, (16)
−
where p2 = p2−p20, p4 =bip0. Wbith the help obf (16), one may prove that the
matrix
Λ = ip+m (17)
obeys the matrix equation
b
Λ4 4mΛ3 +Λ2 p2 +6m2 Λ 2mp2 +4m3 +imLp p2 = 0. (18)
4
− −
(cid:16) (cid:17) (cid:16) (cid:17)
It follows from Eq.(18) that solutions to Eq.(15) in the form of the projection
matrix are given by
Π = N Λ3 4mΛ2 +Λ p2 +6m2 2mp2 4m3 imLp p2 . (19)
4
− − − −
h (cid:16) (cid:17) i
so that (ip+m)Π = 0, and N is the normalization constant. The require-
ment that Π is the projection matrix [17] gives
b
Π2 = Π. (20)
FromEq.(20), with thehelp ofEq.(18), we obtainthe normalizationconstant
1
N = . (21)
−m(2m2 +Lp p2)
0
Equation (19) for the matrix Π can be simplified using Eq.(16),(21), and the
result is
ip(p2 +imp Lp p2)
0
Π = − . (22)
m(2m2 +Lp p2)
0
b b b
The density matrix (22) can be used for calculating some processes with
scalar particles obeying Eq.(8) in the perturbation theory. Every column
of the matrix Π is the solution to Eq.(15). The wave function Ψ(p) in
Eq.(15) also can be written as Ψ(p) = ΠΨ , where Ψ is arbitrary non-zero
0 0
6-component vector.
4
3 The Schro¨dinger form of the equation
Introducing interactions of scalar particles under consideration with external
electromagnetic fields by replacing ∂ D = ∂ ieA (A is the vector-
µ µ µ µ µ
→ −
potential of the electromagnetic fields, e is the charge of the particle), we
rewrite Eq.(8) as follows:
iβ ∂ Ψ(x) = β D +m+eA β Ψ(x). (23)
4 t m m 0 4
(cid:18) (cid:19)
The matrix β obeys the matrix equation (see Appendix)
4
β4 = β2. (24)
4 4
Thus, the matrix Σ = β2 is the projection operator, Σ2 = Σ. The matrix Σ,
4
acting on the wave function Ψ(x), retains only the dynamical components
φ(x) = ΣΨ(x) of the wave function Ψ(x). To separate the dynamical com-
ponents of the wave function Ψ(x) from Eq.(23), we consider the projection
operator
Ω = I Λ = ε0,0 +εm,m ε4,0. (25)
6
− −
OnecanverifythatΩ2 = Ω. Non-dynaemeicalcomponeentsofthewavefunction
Ψ(x) are defined by χ = ΩΨ(x). After multiplying Eq.(23) by the matrix β ,
4
we obtain the equation
i∂ φ(x) = β β D Ψ+mβ Ψ+eA φ. (26)
t 4 m m 4 0
One can use the relation Ψ(x) = φ(x)+χ(x) as Σ+Ω = I . With the help
6
of equations (see Appendix) β Ω = 0, β β Σ = 0, we find from Eq.(26)
4 4 m
i∂ φ(x) = β β D χ+(mβ +eA )φ. (27)
t 4 m m 4 0
Non-dynamical components χ(x) can be eliminated from Eq.(27). Indeed,
multiplying Eq.(23)bythematrixΩ, andtakingintoconsiderationtheequal-
ity Ωβ = 0, one finds the equation as follows:
4
Ωβ D Ψ+mχ = 0. (28)
n n
Eliminating χ from Eq.(28) and replacing it into Eq.(27), with the help of
the relation β β Ωβ Ω = 0, we obtain the Schro¨dinger form of the equation
4 m n
1
i∂ φ(x) = β β D Ωβ D +mβ +eA φ(x). (29)
t 4 m m n n 4 0
−m
(cid:18) (cid:19)
5
It is easy to verify that the wave function φ possesses only two non-zero
components:
Φ(x)
φ(x) = . (30)
Ψ (x)+Φ(x)
4 !
The wave function (30) corresponds to two states with positive and negative
e
energies and does not contain auxiliary components. Then Eq.(29) takes the
form
i∂ φ(x) = φ(x), (31)
t
H
where the Hamiltonian is given by (see Appendix)
1
= m ε4,0 +ε0,4 +eA ε0,0 +ε4,4 ε4,0 mLε0,0 D2 . (32)
H 0 − m − m
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Using (5), the quantum-mechanical Hamiltonian becomes
eA +LD2 m
= 0 m . (33)
H m−(1/m)Dm2 eA0 !
One can rewrite Eq.(31), with the help of Eqs.(30),(33), in the component
form
i∂ Φ(x) = eA Φ(x)+m Ψ (x)+Φ(x) +LD2 Φ(x),
t 0 4 m
(cid:16) (cid:17) (34)
e 1
i∂ Ψ (x)+Φ(x) = mΦ(x)+eA Ψ (x)+Φ(x) D2 Φ(x).
t 4 0 4 − m m
(cid:16) (cid:17) (cid:16) (cid:17)
One can check thateEqs.(34) can be obtained from Eeqs.(3), after the replace-
ment ∂ D and the exclusion of non-dynamical components Ψ (x) =
µ µ m
→
(1/m)D Φ(x). Eqs.(34) and (31) contain only components with the time
m
−
derivatives. The Schro¨dinger equation (31) with the Hamiltonian (33) can be
used for solving different quantum mechanical problems. The matrix Hamil-
tonian (33) for free space (A = 0) in the momentum space ∂ ip obeys
µ µ µ
→
the equation
2 +Lp2 p2 +m2 I = 0, (35)
2
H H−
(cid:16) (cid:17)
where I is the unit 2 2-matrix. From Eq.(35), one finds that the eigenvalue
2
×
of the Hamiltonian (33) p satisfies the dispersion equation (1).
0
6
4 Particle in an external magnetic field
Introducing the electromagnetic interaction of particles in the standard way
by substitution ∂ D = ∂ ieA , Eq.(2) becomes 1
µ µ µ µ
→ −
(∂ ieA )2 m2 iL(∂ ieA )2(∂ +ieA ) Φ(x) = 0, (36)
µ µ i i t 0
− − − −
h i
and t = x is the time. It is obvious that the gauge invariance of Eq.(36) is
0
preserved under the transformations:
A (x) A (x)+∂ Λ(x), Φ(x) Φ(x)exp(ieΛ(x)).
µ µ µ
→ →
For a uniform and static magnetic field, we can take the 4-potential in the
form
1 1
A = x H, x H,0,0 , (37)
µ 2 1
−2 2
(cid:18) (cid:19)
and the magnetic field becomes along the x axis, H = (0,0,H). For
3
the potential (37) the Lorentz condition ∂ A = 0 and Coulomb condition
µ µ
∂ A = 0 are satisfied, and A = 0. Let us consider the motion of negative
m m 0
particles, e = e , e > 0. Then Eq.(36) for a spinless particle with the
0 0
−
potential (37) reads
e2H2
∂2 e HL 0 x2 +x2 (1 iL∂ ) ∂2 m2 Φ(x,t) = 0,
(" m − 0 3 − 4 1 2 # − t − t − )
(cid:16) (cid:17)
(38)
where the projection operator of the angular momentum on the x axis is
3
given by L = i(x ∂ x ∂ ). To solve Eq.(38), we follow very closely to
3 2 1 1 2
−
[18], [19]. The solution to Eq.(38) can be obtained by the substitution [21]
1
Φ(x,t) = Φ(x ,x )exp[i(p x p t)], (39)
√λ 1 2 3 3 − 0
where p = 2πn /λ, n is a vertical quantum number, λ is a cut-off of the
3 3 3
integration over the x . Replacing Eq.(39) into Eq.(38), one finds
3
e2H2η
η ∂2 +∂2 e HηL 0 x2 +x2
1 2 − 0 3 − 4 1 2
(cid:20) (cid:16) (cid:17) (cid:16) (cid:17) (40)
1
It should be noted that authors of the paper [14] chose another coupling of a particle
with the potential.
7
ηp2 +p2 m2 Φ(x ,x ) = 0,
− 3 0 − 1 2
(cid:21)
where η = 1 Lp . It is convenient to introduce cylindrical coordinates
0
−
x = rcosϕ, x = rsinϕ, and then L = i∂/∂ϕ. The solution to Eq.(40)
1 2 3
−
in cylindrical coordinates exists in the form
exp(ilϕ)
Φ(x ,x ) = Ψ(r), (41)
1 2 √2π
with l being an orbital quantum number, l = ..., 2, 1,0,1,2,.... Introduc-
− −
ing new variable ρ = e Hr2/2, Eq.(40) in cylindrical coordinates, and with
0
taking into account (41), becomes
d2 d l2 ρ
ρ + +P Ψ(ρ) = 0, (42)
dρ2 dρ − 4ρ − 4 !
where P = (p2 ηp2 m2 e Hlη)/(2e Hη). The finite solution (at ρ = 0
0 − 3 − − 0 0
and ρ ) to Eq.(42) is given by [19]
→ ∞
N ρ
Ψ(ρ) = 0 exp ρl/2Ql(ρ), (43)
√n!s! −2 s
(cid:18) (cid:19)
where N is the normalization constant, s = 0,1,2,... is the radial quantum
0
number, Ql(ρ) is the Laguerre polynomial [20].
s
ds ρs+le−ρ
Ql(ρ) = eρρ−l . (44)
s (cid:16) dρs (cid:17)
The energy p is quantized and is given by
0
p2 = ηp2 +m2 +e Hη(2n+1), (45)
0 3 0
where n = l + s = 0,1,2,... is a principal quantum number. The orbital
quantum number runs the values < l < n. For negative orbital quantum
−∞
number l, one can use the relation
( 1)lρ−lQ−l (ρ) = Ql(ρ). (46)
− s+l s
Eq.(45) at η = 1 (L = 0) is converted into the known expression correspond-
ing to the Klein Gordon equation [21]. Eq.(45) is consistent with Eq.(1)
−
because as for the Klein Gordon equation for scalar particles in external
−
8
magnetic fields one has to make the replacement p2 +p2 e H(2n+1) in
1 2 → 0
the dispersion equation 2. From (39),(41),(43), one obtains
N eilϕ exp[i(p x p t)]
Φ(x,t) = 0 3 3 − 0 e−ρ/2ρl/2Ql(ρ). (47)
√n!s!√2π √λ s
The coefficient N can be obtained from the normalization [21]:
0
p ∞ λ/2 2π
0 ∗
rdr dx Φ (x)Φ(x)dϕ = 1. (48)
3
m Z0 Z−λ/2 Z0
Calculating integrals in Eq.(48) with the help of the relation
∞
e−ρρl Ql(ρ) 2dρ = s!Γ(l +s+1),
s
Z0 h i
where Γ(x) is the Gamma function, and using the wave function (47), we
find the normalization constant:
e Hm
0
N = . (49)
0
s p0
Eqs.(45)(47),(49) allow us to investigate the synchrotron radiationof spinless
particles with modified dispersion relation (1). Charged particles moving in
an external magnetic field (in helical orbits) emit the synchrotron radiation
with frequency depending on the radius of the orbit [21]. The orbit radius
can be estimated by the classical relation [21]:
βp
0
R = , (50)
e H
0
where β = v is a particle velocity (in our notations c = 1). Let us consider
the case p = 0 when particles rotate in cycles with ultra-relativistic energies
3
(p m, v 1). Then Eq.(45) becomes
0
≫ ≈
p2 e H (2n+1)(1 Lp ). (51)
0 ≈ 0 − 0
From quadratic equation (51), we obtain the approximate solution for posi-
tive energy implying that Lp 1, L e H(2n+1) 1:
0 0
≪ ≪
q
1
p e H(2n+1) Le H n+ . (52)
0 0 0
≈ − 2
q (cid:18) (cid:19)
2
Authors of [14] obtained different from (45) expressionbecause of their non-standard
coupling with the electromagnetic fields.
9
With the help of Eq.(52), we obtain from Eq.(50) the radius of the orbit
R R Ln, (53)
0
≈ −
where
2n
R (54)
0
≈ se0H
is the radius of the orbit within the Klein Gordon equation, and we took
−
into consideration that for ultra-relativistic energies n 1 (n + 1/2 n).
≫ ≈
Eq.(53) indicates that the Lorentz-violating term reduces the radius of the
orbit (L > 0). Thus, for high energies the second term in Eq.(53) should
be taken into account. As a result, the angular orbital frequency ω = v/R
is greater than in Lorentz-invariant theory, ω > ω (ω = v/R ). To clear
0 0 0
up the physical meaning of the radial quantum number s, we calculate the
quantum average square radius
p
r2 = 0 Ψ∗(x)r2Ψ(x)d3x. (55)
quant
m
Z
Evaluating integral in Eq.(55) using the equality
∞
e−ρρl+1 Ql(ρ) 2dρ = n!s!(n+s+1),
s
Z0 h i
and n = l+s, one obtains [21]
2
r2 = (n+s+1). (56)
quant
e H
0
comparing the macroscopic classical average square radius [21] r2 = R2+a2,
cl
where a is the distance between the center of the trajectory and the origin,
with the quantum average square radius (56), we find [21]
2s
a . (57)
≈ se H
0
From Eqs.(54),(57), one obtains l e H(R2 a2)/2 so that at R > a the
≈ 0 0 − 0
orbital quantum number is positive, l > 0, and at R < a, we have l < 0.
0
One can construct the coherent states of a spinless particle at non-relativistic
energy following the way of the work [19].
10