Prog. Theor. Exp. Phys. 2014, 103B01 (21 pages)
DOI: 10.1093/ptep/ptu126
Relativistic remnants of non-relativistic electrons
Taro Kashiwa∗ and Taisuke Yamaguchi∗
Department of Physics, Graduate School of Science and Engineering, Ehime University, Matsuyama
790-8577, Japan
∗
E-mail: [email protected], [email protected]
Received May 17, 2014; Revised July 24, 2014; Accepted August 4, 2014; Published October 8 , 2014
...............................................................................
Electrons obeying the Dirac equation are investigated under the non-relativistic c → ∞ limit.
General solutions are given by derivatives of relativistic invariant functions that possess discontinuity at the light-cone, yielding the delta function of (ct)2 − x2 . This light-cone singularity does
survive in this limit to show that the charge and the current densities of electrons travel at the
speed of light in spite of their massiveness.
...............................................................................
Subject Index
1.
A64, B30
Introduction
It is well known that the Dirac electron has a piece vibrating at light velocity, called “Zitterbewegung”,
whose origin is supposed as a mixing of the positive and negative components. Discussions have
been given on the Heisenberg equation of motion for the Dirac Hamiltonian [1] or on the momentum representation for the Dirac field (x); see, e.g., Ref. [2]. An electron moving with a zigzag
motion at light speed also appears on the stage of the pilot-wave approach to quantum field theory
[3,4] (Feynman had already discussed the zigzag motion at light velocity in the context of the path
integral [5].)
We shall, in this paper, focus on the electron field (x) itself, to show that electrons obeying the
Dirac equation inevitably bear portions traveling at light speed in the non-relativistic limit. As a
preliminary, let us recall the Dirac Hamiltonian
HD = c (α · p + βmc) ;
with
μ ν
γ , γ = 2ημν ,
α ≡ γ 0γ ; β ≡ γ 0,
diag(ημν ) = (1, −1, −1, −1); (μ, ν = 0, 1, 2, 3),
being the 4 × 4 gamma matrices represented as
I 0
0
0
γ =
,γ =
0 −I
−σ
σ
;
0
σ : Pauli matrices.
(1)
(2)
(3)
In order to investigate the non-relativistic limit, it is useful to perform a unitary (called the
Foldy–Wouthuysen) transformation [6],
γ · p
|p|
γ ·p
|p|
θ (p) = cos
θ (p) +
sin
θ (p) ,
(4)
U ≡ exp
mc
mc
|p|
mc
with
|p|
|p|
,
tan 2 θ (p) =
mc
mc
© The Author(s) 2014. Published by Oxford University Press on behalf of the Physical Society of Japan.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/),
which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.
PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
such that
U HDU = β H ;
H ≡ c p2 + m 2 c2 .
†
(5)
Here the lower component corresponds to the negative energy state, which should be discarded in
the non-relativistic world. Therefore we shall pick up the positive energy part and study its wavemechanical structure in the next section. In Sect. 3, we shall treat covariant solutions of the free Dirac
equation, which will be extended to interacting cases in Sect. 4. The final section is devoted to the
discussion. Some of the detailed calculations in Sect. 4 are relegated to the appendices.
Wave mechanics of H = c p2 + m2 c2
2.
Consider a single-component wave mechanics governed by the Hamiltonian (5), i.e., a wave function
(t, x) obeying the Schrödinger equation
∂
i (t, x) = c −(∇)2 + m 2 c2 (t, x).
(6)
∂t
The solution reads
(t, x) =
d 3 x K (x; t) ψ(0, x ),
where ψ(0, x ) is an arbitrary function,
ct 2
2 2
P̂ + m c |x K (x; t) ≡ x| exp −i
i
d 3p
2
2
2
p · x − ct p + m c
;
exp
=
(2π )3
(7)
x ≡ x − x ,
(8)
is the kernel1 with P̂ designating the momentum operator, and the integration range from −∞ to ∞
is omitted here and hereafter unless otherwise specified.
In order to calculate the kernel (8), introduce k ≡ p/ and write
mc
,
(9)
μ≡
to obtain
1 ∂ I (r, t)
d 3k
2
2
;
(10)
exp ik · x − ict k + μ = − 2
K (x; t) =
3
(2π )
4π r ∂r
∞
I (r, t) ≡
dk exp i kr − ct k 2 + μ2 ;
r ≡ |x|,
(11)
−∞
where use has been made of the polar coordinates in the final expression. Put k = μ sinh to find
∞
I (r, t) = μ
d cosh exp [−iμ (ct cosh − r sinh )] ,
(12)
−∞
whose exponent reads, with the aid of an addition theorem,
⎧
r
⎪
⎨ xμ2 cosh( − α) : tanh α = ; ct > r
ct
ct cosh − r sinh =
ct
⎪
2
⎩− −xμ sinh( − β) : tanh β = ; r > ct
r
1
,
(13)
In the path integral formalism, this is a typical example in that the Hamiltonian prescription is more general
[7,8] than Feynman’s [9], whose Euclidean case, i.e., t → −it in (8), has been discussed in Ref. [10].
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T. Kashiwa and T. Yamaguchi
with
xμ2 ≡ (ct)2 − r 2 .
(14)
Make shifts → + α, β and again utilize the addition theorem to obtain
I (r, t) = μθ (ct − r ) cosh α
× cosh β
∞
−∞
∞
−∞
d cosh e−iχ+ cosh + μθ (r − ct)
d cosh e
with
iχ− sinh + sinh β
∞
−∞
d sinh e
iχ− sinh ,
χ± ≡ μ ±xμ2 .
(Here we have discarded the odd function part:
∞
−∞
d cosh e
iχ− sinh =
∞
−∞
(15)
(16)
∞
−∞ d sinh e
−iχ+ cosh = 0.) By noting that
d(sinh )eiχ− sinh = 2π δ(χ− ) =
2π 2 δ
−xμ ,
μ
and cosh β = r/ −xμ2 in view of (13), the second term of (15) reads
μθ (r − ct) cosh β
⎛
∞
−∞
d cosh eiχ− sinh ⎞
r ⎠ 2π 2 ⎝
δ
= θ (r − ct) μ −xμ
μ
−xμ2
= 4πr θ (r − ct)δ(xμ2 ) = θ (r − ct)
2πr
δ(ct − r ) = π δ(ct − r ),
ct
where we have used the relations
δ
−xμ2 = 2 −xμ2 δ xμ2 ,
(17)
1
θ (0) = .
2
Finally the Bessel function formulas [11]
∞
(2)
d cosh e−iχ+ cosh = −π H1 (χ+ ) ,
−∞
∞
−∞
d sinh eiχ− sinh = 2i K 1 (χ− )
lead us to
I (r, t) = π δ(ct − r ) − 2ctμ2
(2)
π H1 (χ+ )
θ (ct − r )
2
χ+
K 1 (χ− )
.
− θ (r − ct) i
χ−
(18)
Here note that I consists of different functions in the regions ct > r and ct < r , which causes the
delta function singularity when a differentiation is made. (The μ-independent delta function emerges
from the huge momentum domain p mc.)
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T. Kashiwa and T. Yamaguchi
The kernel is obtained, from (10), by differentiating (18) with respect to r , as
iμ2
1 ∂
δ (ct − r ) −
δ (ct − r )
4πr ∂r
4π 2
⎤
⎡
(2)
2
2
H
x
−x
K
μ
μ
2
2
μ
μ
ctμ ⎣
π 2
⎦,
− iθ (r − ct)
+
θ (ct − r )
2
2
2
2π
2
xμ
xμ
K (x; t) = −
where use has been made of
∂
∂
θ (ct − r ) = −δ(ct − r );
θ (r − ct) = δ(ct − r ),
∂r
∂r
and then (see Appendix A)
(2)
ctμ2
π H1 (χ+ )
K 1 (χ− )
iμ2
−
δ
−
r
+
i
δ (ct − r ) ,
=
−
(ct
)
2π 2r
2
χ+
χ−
4π 2
as well as (see Ref. [11], p. 909, 8.444-2, where they use Yn for Nn , and 8.446),
d
Z 1 (z)
Z 2 (z)
= − 2 , Z n : Jn , Nn , (also Hn(2) ) and K n ,
zdz
z
z
(19)
(20)
(21)
by noting
1 ∂ (16)
1 ∂
= ∓μ2
.
r ∂r
χ± ∂χ±
By taking the non-relativistic limit c → ∞, i.e., μ → ∞ (9), the kernel (19) reads
(22)
3/2 iμ2
,
(23)
δ
−
|x|)
+
O
μ
(ct
4π 2
where we have employed the asymptotic expansion of the Bessel function (see Ref. [11], p. 910,
8.451-1, 2, 4, and 6)
−z z→∞
z→∞ e
Z n (z) = O z −1/2 , Z n : Jn , Nn , (also Hn(2) ); K n (z) = √ 1 + O z −2 . (24)
z
μ→∞
K (x; t) = −
Now put
ψ(0, x ) =
1
aπ
3/4
x 2
exp −
2a
,
then substitute (23) into (7) while changing the variables as x → x to find
iμ2 1 3/4
(x + x)2
3
(t, x) = −
d (x)δ (ct − |x|) exp −
+ O μ3/2 ,
4π aπ
2a
which, by introducing the polar coordinates, yields
(ct − |x|)2
(ct + |x|)2
iμ2 a 1/4 ct
exp −
− exp −
.
(t, x) = − 3/4
π
|x|
2a
2a
√
When ct, |x| a, the second term fades away and around the peak,
Equation (26) becomes
(25)
(26)
ct − |x| ≈ 0,
(27)
iμ2 a 1/4
(ct − |x|)2
(t, x) ≈ − 3/4 exp −
,
π
2a
(28)
which apparently travels at the speed of light in spite of its massiveness.
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T. Kashiwa and T. Yamaguchi
The origin lies in the delta function δ(ct − |x|) (19) emerging from the discontinuity of I (18)
between the time-, ct > |x|, and the space-like, ct < |x|, regions. We shall call this a light-cone
singularity. (It should be emphasized that the light-cone singularity cannot become visible under the
momentum representation [12].)
As a necessary consequence, any wave written as
∀
ψ(y),
(x) = d 4 y D(x, y)ψ(y);
must have the light-cone singularity, if D contains derivatives to some function with a discontinuity
on the light-cone. The solutions of the Dirac equation meet with this requirement, so, in the following
sections, we shall study those.
3.
The charge and the current density of free electrons
First let us summarize the relativistic invariant functions that participate in solving a relativistic
equation. The D-dimensional scalar Klein–Gordon field is given as
+ μ2 φ(x) = J (x),
≡ ∂ μ ∂μ =
D−1
" ∂2
∂2
−
,
2
∂ x02
∂
x
j
j=1
(29)
where μ is defined by (9) and J (x) is a source, a complicated function of φ describing interactions.
Here and hereafter the repeated indices always imply the summation. When J = 0, the solution is
φ0 (x) = d D y(x − y)ϕ(y),
(30)
where (x) is an invariant function defined by
+ μ2 (x) = 0,
2
i
D
2 −ikx
d
e
k(k
)δ
k
−
μ
(x) ≡ −
0
(2π ) D−1
D−1 k −i(ωk t−k·x)
d
ic
i(ωk t−k·x)
=−
−e
e
, ωk ≡ c k2 + μ2 ,
(2π ) D−1
2ωk
(31)
and ϕ(y) is an arbitrary function. The notations
k 2 ≡ (k0 )2 − k2 ;
kx ≡ k0 x0 − k · x,
(x0 ≡ ct),
with k and x being a (D − 1)-dimensional vector, should be understood. When J = 0, the solution is
φ(x) = φ0 (x) − d D yF (x − y)J (y),
(32)
where φ0 (x) is (30) and F (x) is the Feynman propagator,
+ μ2 F (x) = −δ D (x) (≡ −δ(x0 )δ(x1 ) · · · δ(x D−1 )) ,
e−ikx
d Dk
.
F (x) ≡
(2π ) D k 2 − μ2 + i
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PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
These are shown as [13]
−ν
μ2ν
D−2
2
2
,
θ
(x
)(x
)
μ
x
J
μ x2 ; ν ≡
0
−ν
ν
2(2π )
2
−ν #
$
μ2ν
2
2
2 − iN
θ
(x
F (x) = −
)
μ
x
x
J
μ
μ x2
−ν
−ν
4(2π )ν
−ν
2
2
2
2
,
θ (−x ) μ −x
K −ν μ −x
+i
π
(x) = −
(34)
(35)
with x 2 ≡ (x0 )2 − x2 . By noting J−n (z) = (−)n Jn (z), N−n (z) = (−)n Nn (z), and K −n (z) = K n (z),
they become in D = 4(ν = 1) [14–16]
√ ⎤
⎡
2
μ x2
J
1
μ
(x0 ) ⎣ 2
2
⎦,
θ (x )
(x) = −
(36)
δ(x ) −
√
2π
2
μ x2
√ ⎤
√
⎡ √ 2
2
2
2
2
2
2
N
x
x
K
−x
J
μ
μ
μ
1
1
1
1
μ θ (x ) ⎣
⎦ − i μ θ (−x )
−i
.
F (x) = − δ(x 2 ) +
√
√
√
2
2
4π
8π
4π 2
μ −x 2
μ x
μ x
(37)
Note that they have a discontinuity on the light-cone. (Any relativistic invariant function does.) In
view of (30) and (32), however, there are no derivatives so that we cannot have light-cone singularities
for scalar fields.2 (We do not care about the O(1) delta function in (36) and (37) under the nonrelativistic limit μ → ∞.)
The free electron field 0 (x) obeys the Dirac equation
(i ∂/ − μ)0 (x) = 0,
∂/ ≡ γ μ ∂μ ,
(38)
where μ is (9) and 0 (x) is given as
ϕ0 (x)
0 (x) ≡
,
χ0 (x)
with ϕ0 and χ0 being a two-component spinor.
The solution of (38) reads
0 (x) = d 4 y S(x − y)ψ(y),
(39)
(40)
where S(x) is the invariant function for the Dirac field
(i ∂/ − μ)S(x) = 0,
S(x) ≡ (i ∂/ + μ)(x),
(41)
with (x) (36) and ψ(y) being an arbitrary four-component spinor. From (40) and (41), 0 (x) must
own the light-cone singularity.
As for complex scalars, the current density Jμ (x) ≡ ie φ ∗ (x)∂μ φ(x) − (∂μ φ ∗ (x))φ(x) does have a lightcone singularity.
2
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T. Kashiwa and T. Yamaguchi
For the sake of simplicity, the initial electron configuration is assumed to be
ξ0
ψ(y) = δ(y0 ) f (y)
,
0
where ξ0 is a constant two-component spinor. Since from (31) (with D = 4)
%
%
%
%
%
(x)%
= 0; ∂0 (x)%%
= −δ 3 (x),
x0 =0
(42)
(43)
x0 =0
(40) with (42) implies the initial condition
ξ0
0 (x0 = 0, x) = −i f (y)
.
0
In the following we consider three cases:
⎧
2
1
y
⎪
(3)
⎪
⎪
⎪ f (y) ≡ a 3/4 exp − 2a ,
⎪
⎪
⎪
⎪
⎪
⎨ (2)
y22
1
2
2
2
f (y) =⇒ f (y) ≡ a 1/4 δ(y3 ) exp − 2a ; y2 ≡ y1 + y2 ,
⎪
⎪
⎪
⎪
⎪
2
⎪
y
⎪
(1)
1/4
1
⎪
⎪
⎩ f (y) ≡ a δ(y2 )δ(y3 ) exp − 2a ,
which are called 3, 2, and 1D packets respectively.3
In view of (41), the solution (40) becomes
(k)
x +μ
(x)
ϕ
i∂
(k)
0
0
0 ≡
= d 3y
(x0 , x − y) f (k) (y)ξ0 ; (k = 1, 2, 3).
(k)
−iσ · ∇ x
χ0 (x)
In the following, we shall discuss the charge and the current densities defined by4
e (k) † (k) e
(k) †
(k)
(k) †
(k)
ϕ0 (x)ϕ0 (x) + χ0 (x)χ0 (x) ,
ρ (k) ≡ 0 0 =
c
c
(k) †
(k)
(k) †
(k)
(k) †
(k)
J (k) ≡ e0 γ 0 γ 0 = e ϕ0 (x)σ χ0 (x) + χ0 (x)σ ϕ0 (x) .
Now take the non-relativistic limit c → ∞ (μ → ∞) to find that (x) (36) reduces to
√ J1 μ x 2
μ→∞ μ2 (x 0 )
(x) ≈
θ (x 2 )
,
√
4π
μ x2
whose derivatives are calculated as follows. First note
∂0 (x0 )θ (x 2 ) = 2|x0 |δ(x 2 ),
(44)
(45)
(46)
(47)
(48)
(49)
by use of
∂0 (x0 ) = 2δ(x0 );
δ(x0 )θ (x 2 ) = 0;
x0 (x0 ) = |x0 |,
In terms of the length scale L, the dimension of a spinor field (x) is L −3/2 . (x) is L −2 then S(x) is L −3
from (41). Hence ψ(x) (40) is L −5/2 so that f (x) (42) is L −3/2 .
4
In relativity, a current 4-vector reads J μ ≡ (cρ, J ) in MKS (meter-kilogram-second) units. The Dirac
particle induces a current J μ ≡ eγ μ with ≡ † γ 0 .
3
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T. Kashiwa and T. Yamaguchi
as well as ∂0 = 2x0 ∂/∂ x 2 and
∂
θ (±x 2 ) = ±δ(x 2 ).
∂x2
Thus
(50)
√ ⎛ √ ⎞
2
2
J1 μ x 2
∂ ⎝ J1 μ x ⎠
μ
2
2
∂0 (x) =
2|x0 |δ(x )
(x0 )θ (x )2x0 2
+
,
√
√
4π
4π
∂x
μ x2
μ x2
μ2
whose second term reads
⎛ √ ⎞
√
J1 μ x 2
2
3/2 ∂
μ2 2 ⎠ z≡μ= x μ2 ∂z d J1 (z) (21)
,
μ2 2 ⎝
=
−
J
x
μ
=
O
μ
√
2
∂x
∂ x 2 dz
z
2x 2
μ x2
√
z→∞
where Jn (z) = O(1/ z) (24) has been considered in the final expression. Meanwhile the first
%
term becomes, with the aid of J1 (z)/z %z=0 = 1/2 (A5),
∂0 (x) =
μ2
|x0 |δ(x 2 ) + O μ3/2 .
4π
(51)
Similarly, by ∂k = −2xk ∂/∂ x 2 and (50),
∂k (x) = −
μ2
xk (x0 )δ(x 2 ) + O μ3/2 .
4π
(52)
From these, we can convince ourselves that the leading terms in the non-relativistic approximation
are nothing but the light-cone singularities. Therefore, by noting that
δ(x 2 ) =
1 δ(x0 − |x|) + δ(x0 + |x|) ,
2|x|
(53)
as well as μ(x) = O μ3/2 , (46) reads
(k)
0
iμ2
=
8π
⎞
1
d 3 yδ(x0 − |y|) ⎝ −y · σ ⎠ f (k) (y + x)ξ0 + O μ3/2 ,
|y|
⎛
where we have noticed x0 > 0 and made a shift y → y + x.
Now proceed to individual cases: from (45) and (54) the 3D solution is
⎞
⎛
2
2
1
iμ
x
(3)
⎠ I (3) (x0 , x)ξ0 ,
⎝a
exp −
0 =
σ ·∇
8πa 3/4
2a
x0
where
I
(3)
(x0 , x) ≡
y2
x·y
−
,
d yδ(x0 − |y|) exp −
a
2a
(54)
(55)
3
which becomes with the aid of the polar coordinates
x02
|x|x0
2πax0
|x|x0
(3)
I (x0 , x) =
exp −
− exp −
.
exp
|x|
2a
a
a
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(56)
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PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
Hence
⎡⎛
(3)
0
When x0 , |x| ⎞
(x0 − |x|)2
⎠
a
=
exp −
1−
4|x|
2a
x0 |x|
⎞
⎤
⎛
2
1
+
|x|)
(x
ξ
0
0
⎦
a ⎠ exp −
,
∓⎝
1+
n · σ ξ0
2a
x0 |x|
iμ2 a 1/4 x
0
⎣⎝
1
n≡
x
.
|x|
(58)
√
a, we have
(3)
0
(x0 − |x|)2
iμ2 a 1/4 x0
ξ0
exp −
≈
,
n · σ ξ0
4|x|
2a
which further turns out to be
(3)
0
(x0 − |x|)2
iμ2 a 1/4
ξ0
exp −
≈
n · σ ξ0
4
2a
(59)
around the peak
x0 − |x| ≈ 0,
(60)
which apparently travels at the speed of light.
Insert (59) into (47) to find
2
2e μ2 a 1/4
(x0 − |x|)2
†
(3)
ξ0 ξ0 , J (3) = cnρ (3) ,
exp −
(61)
ρ =
c
4
a
where use has been made of the anti-commutation relations σ j , σk = 2δ jk in J (3) . Those travel at
the speed of light.
It would be easier to prepare packets restricted in the 2 or 1D regions. The former reads, by inserting
(45) into (54),
⎞
⎛
1
2
2
iμ
x
2
(2)
⎝a
∂ ⎠ I (2) (x0 , x2 )ξ0 ,
exp −
(62)
0 =
σ2 ·
8πa 1/4
2a
x0
∂x2
where σ 2 ≡ (σ1 , σ2 ), x2 ≡ (x1 , x2 ) and
x 2 · y2 y2 2
(2)
2
−
.
(63)
I (x0 , x2 ) ≡ d y2 δ(x0 − |y2 |) exp −
a
2a
Here we have put x3 = 0, since the observation should also be made in the x3 = 0 plane.
Equation (63) becomes, under the polar coordinates,
∞
2π
y2
|x2 |y
(2)
cos θ −
I (x0 , x2 ) =
dyy
dθ δ(x0 − y) exp −
a
2a
0
0
2
2π
x
|x2 |x0
cos θ .
= x0 exp − 0
dθ exp −
2a
a
0
√
When |x2 |, x0 a, the saddle point θ = π gives an asymptotic value such that
&
x02
2πax0
|x2 |x0
(2)
I (x0 , x2 ) ≈
exp − +
.
(64)
|x2 |
2a
a
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Thus (62) reads
(2)
0
iμ2 a 1/4
(x0 − |x2 |)2
ξ0
≈ √
.
exp −
n2 · σ 2 ξ0
2a
4 2π
(65)
Here the final expressions have been obtained by putting x0 /|x2 | → 1 in the coefficient, since the
peak is now given as
x0 − |x2 | ≈ 0,
(66)
whose velocity is again light speed. The charge and the current densities (47) read
2
2e μ2 a 1/4
(x0 − |x2 |)2
†
(2)
ξ0 ξ0 , J (2) = cn2 ρ (2) , 0 .
exp −
ρ =
√
c 4 2π
a
Finally we discuss the 1D case written, after putting x2 = 0, x3 = 0, as
⎞
⎛
2
2 a 1/4
1
x
iμ
(1)
⎠ I (1) (x0 , x1 )ξ0 ,
exp − 1 ⎝ a
0 =
σ1 ∂1
8π
2a
x0
where
I
(1)
∞
y2
x1 y1
− 1
(x0 , x1 ) ≡
dy1 δ(x0 − |y1 |) exp −
a
2a
−∞
which turns out, by recalling x0 > 0, to be
0
(1)
I (x0 , x1 ) =
dy1 δ(x0 + y1 ) +
(67)
(68)
,
y12
x1 y1
−
dy1 δ(x0 − y1 ) exp −
a
2a
−∞
0
x x x x x2
0 1
0 1
= exp
+ exp −
exp − 0 .
a
a
2a
(69)
∞
(70)
Thus
2 a 1/4 2
2 −
x
)
+
x
)
(x
(x
iμ
ξ
0
1
0
1
0
(1)
exp −
± exp −
.
0 =
σ1 ξ0
8π
2a
2a
√
When x0 , x1 a, it becomes
(x0 − x1 )2
iμ2 a 1/4
ξ0
(1)
exp −
,
0 ≈
σ1 ξ0
8π
2a
(71)
(72)
whose peak is
x0 − x1 ≈ 0,
around which the charge and the current densities (47) read
2
2e μ2 a 1/4
(x0 − x1 )2
†
(1)
ξ0 ξ0 ,
exp −
ρ =
c
8π
a
(73)
J (1) = cρ (1) , 0, 0 .
(74)
In view of (61), (67), and (74), the free electrons travel at the speed of light in the non-relativistic
world.5
Note that the relation between the current and the charge density J = cnρ in (61), (67), and (74) is merely
a kinematical consequence of taking the upper component ξ0 only.
5
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4.
T. Kashiwa and T. Yamaguchi
Electrons in a laboratory
In a realistic situation, electrons interact with the electromagnetic field Aμ (x) such that
(i ∂/ − μ)(x) =
whose solution is
(x) = 0 (x) +
e μ
γ Aμ (x)(x),
c
d 4 y SF (x − y)eγ μ Aμ (y)(y),
(75)
(76)
where 0 (x) is the free field discussed in the previous section and SF (x) is the Feynman propagator
for the Dirac field,
(i ∂/ − μ)SF (x) = δ 4 (x),
SF (x) ≡ (i ∂/ + μ)F (x),
with F (x) given in (37).
The interaction is assumed to take place for a finite interval, giving
e μ
ξ1
(k)
γ Aμ (y)(y) ≡ h (y)
, (k = 1, 2, 3),
0
c
(77)
(78)
where the 3, 2, and 1D packets are
2
2
y
y
h (3) (y0 , y) ≡
exp − 0 −
,
(τ + b)1/4 τ 1/4 b3/4
2τ
2b
2
2
y
y
1
δ(y3 ) exp − 0 − 2 ; y22 ≡ y12 + y22 ,
h (2) (y0 , y) ≡
(τ + b)1/4 τ 1/4 b1/4
2τ
2b
y02
y12
b1/4
(1)
h (y0 , y) ≡
δ(y2 )δ(y3 ) exp − −
,
(τ + b)1/4 τ 1/4
2τ
2b
1
(79)
with ξ1 being a constant two-component spinor. (Note that the dimension of h(y) is L −5/2 . See
footnote 3.) If we write
(k) ϕh (x)
i∂0x + μ
(k)
4
(k)
(k = 1, 2, 3),
(80)
≡ d y
h ≡
x F (x − y)h (y)ξ1 ;
(k)
−iσ
·
∇
χh (x)
the general solution (76) reads
(k)
(k)
(k) = 0 + h ,
(81)
(k)
with 0 given by (46).
In the non-relativistic limit μ → ∞, F (x) (37) reduces to
√ ⎤
√
⎡ √ 2
2
2
2
2
2
2
N1 μ x
K 1 μ −x
J1 μ x
μ→∞ μ θ (x )
⎦ − i μ θ (−x )
⎣
F (x) ≈
−i
√
√
√
8π
4π 2
μ −x 2
μ x2
μ x2
√ √
(2)
2
2
2
2
2
2
x
K
−x
μ
μ
H
1
μ θ −x
μ θ (x ) 1
=
−i
,
√
√
2
8π
4π
μ −x 2
μ x2
where we have utilized (A2) in the final expression whose form reminds us of (18) in Sect. 2.
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PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
The derivative reads
√ √ √
⎞⎤
⎛
2
2
K
x
−x
N
μ
μ
J1 μ x 2
1
2i π 1
⎠⎦ + O μ3/2 ,
− ⎝
−
∂μ F (x) = 2xμ δ(x 2 ) ⎣
√
√
√
8π
π 2
μ −x 2
μ x2
μ x2
⎡
μ2
where use has been made of (50) and the asymptotic behavior (24). Noting (A5) and (A8) in
Appendix A, we have the light-cone singularity,
iμ2
(83)
δ(x 2 ) + O μ3/2 .
2
4π
In contrast to the previous situation, we now need both terms in (53), since y0 cannot always be
positive, so that
∂μ F (x) = xμ
(
3/2 −μ2 '
δ(x
,
(84)
−
y
−
|x
−
y|)
−
δ(x
−
y
+
|x
−
y|)
+
O
μ
0
0
0
0
8π 2
(
−μ2 σ · (x − y) '
−iσ · ∇ x F (x − y) =
δ(x0 − y0 − |x − y|) + δ(x0 − y0 + |x − y|) + O μ3/2 .
2
8π |x − y|
(85)
(i∂0x + μ)F (x − y) =
Then (80) turns out to be
⎛
⎞
1
2
−μ
(k)
h =
d 3 y h (k) (x0 − |x − y|, y) ∓ h (k) (x0 + |x − y|, y) ⎝ −y · σ ⎠ ξ1 ,
8π 2
|y|
where a shift y → y + x has been made.
Let us examine individual cases: with the 3D packet, (86) reads
1/4
x02
x2
1
−μ2
τ ∂0
(3)
(3)
h =
exp − −
I (x0 , x)ξ1 ,
bσ · ∇ h
8π 2 τ b3 (τ + b)
2τ
2b
where
(86)
(87)
)
*
x0 |y|
x0 |y|
(τ + b)y2 x · y
d 3y
exp
+ exp −
exp −
−
≡
|y|
τ
τ
2τ b
b
3
2
x0
(τ + b)y
x·y
d y
exp −
cosh
|y| exp −
=2
,
(88)
|y|
2τ b
τ
b
which, with the aid of the polar coordinates and the error function (see Ref. [11], p. 880, 8.250-1
(instead of (x) we write erf(x))),
x
2
dt exp(−t 2 )
(89)
erf(x) ≡ √
π 0
yields
&
(bx0 + τ |x|)2
(2π b)3 τ 1
bx0 + τ |x|
(3)
exp
erf √
Ih (x0 , x) =
τ + b |x|
2τ b(τ + b)
2τ b(τ + b)
2
(bx0 − τ |x|)
bx0 − τ |x|
exp
.
(90)
− erf √
2τ b(τ + b)
2τ b(τ + b)
√ √
Applying the differentiations in (87) and assuming x0 , |x| τ , b, we obtain
1/4
2 τ b3
2
−
|x|)
−μ
(x
ξ
0
1
(3)
exp −
(91)
h ≈ √
n · σ ξ1
2(τ + b)
2 2π (τ + b)3
(3)
Ih (x0 , x)
around the peak (60). (The details are relegated to Appendix B).
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T. Kashiwa and T. Yamaguchi
(3)
(3)
Therefore, in view of 0 (59) and h (91), the total (81) is obtained as
2 a 1/4
τ
+
a
+
b
iμ
ξ
exp −
(x0 − |x|)2
(3) =
,
n · σξ
4
4a(τ + b)
where we have introduced a two-component spinor
τ −a+b
2
(x0 − |x|) ξ0
ξ ≡ exp −
4a(τ + b)
+ 1/4
2
τ b3
τ −a+b
2
(x0 − |x|) ξ1 .
+i
exp
π a(τ + b)3
4a(τ + b)
(92)
(93)
The charge and the current densities are
2
e (3) † (3) 2e μ2 a 1/4
τ +a+b
(3)
2
ρtot ≡ =
(x0 − |x|) ξ † ξ
exp −
c
c
4
2a(τ + b)
⎡
&
2
τ b3
2e μ2 a 1/4 ⎣
2
(x0 − |x|)2
(x0 − |x|)2
†
†
=
ξ0 ξ0 +
ξ 1 ξ1
exp −
exp −
c
4
a
π a(τ + b)3
2(τ + b)
+ 1/4
2
τ b3
τ +a+b
†
†
(x0 − |x|)2 (ξ0 ξ1 − ξ1 ξ0 ) ,
+i
exp −
π a(τ + b)3
2a(τ + b)
(3)
†
(3)
J tot ≡ e (3) γ 0 γ (3) = cnρtot .
(94)
Since each term has a peak at the light-cone, the electron signal traveling at light speed would be
observed.
Next we consider the 2D case: again we restrict x to x3 = 0 so that (86) with (79) are found as
⎞
⎛
1/4
2
2
τ ∂0
2 x
x
1
−μ
(2)
∂ ⎠ Ih(2) (x0 , x2 )ξ1 ,
h =
exp − 0 − 2 ⎝
(95)
bσ 2 ·
8π 2 τ b(τ + b)
2τ
2b
∂x2
where
2
(τ + b) 2 x2 · y2
x0 |y|
d y2
(2)
exp −
y −
cosh
,
(96)
Ih (x0 , x2 ) ≡ 2
|y|
2τ b 2
b
τ
which yields
&
(2π )2 τ b2
(τ |x2 | + bx0 )2
(2)
exp
Ih (x0 , x2 ) ≈
|x2 |(τ |x2 | + bx0 )
2τ b(τ + b)
&
(2π )2 τ b2
(τ |x2 | − bx0 )2
+
exp
.
(97)
|x2 |(τ |x2 | − bx0 )
2τ b(τ + b)
√ √
Therefore, when x0 , |x2 | τ , b,
1/4 &
τ |x2 | + bx0
τ b3
(x0 − |x2 |)2
−μ2
ξ1
(2)
exp −
h ≈
.
(98)
n2 · σ 2 ξ1
4π
(τ + b)5
|x2 |
2(τ + b)
(The details are relegated to Appendix B.) This further reduces to
1/4
2 τ b3
2
−
|x
|)
−μ
(x
ξ
0
2
1
(2)
h ≈
exp −
n2 · σ 2 ξ1
4π
(τ + b)3
2(τ + b)
around the peak (66).
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PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
From (65) and (99), the total (2) (81) is given by
(2)
ξ (2)
(τ + a + b)
iμ2 a 1/4
ξ (2)
2
exp −
(x0 − |x2 |)
= √
;
n2 · σ 2 ξ (2)
4a(τ + b)
4 2π
(τ − a + b)
2
(x0 − |x2 |) ξ0
≡ exp −
4a(τ + b)
+ 1/4
2
τ b3
(τ − a + b)
2
(x0 − |x2 |) ξ1 .
+i
exp
π a(τ + b)3
4a(τ + b)
(100)
(101)
The charge and the current densities are thus found as
2
e
2e μ2 a 1/4
(τ + a + b)
†
†
(2)
(x0 − |x2 |)2 ξ (2) ξ (2)
exp −
ρtot ≡ (2) (2) =
√
c
c 4 2π
2a(τ + b)
2 1/2
τ b3
2e μ2 a 1/4
2
(x0 − |x2 |)2
†
ξ0 ξ0 +
=
exp −
√
c 4 2π
2a
π a(τ + b)3
+ 1/4
2
τ b3
(x0 − |x2 |)2
†
ξ 1 ξ1 + i
× exp −
2(τ + b)
π a(τ + b)3
(τ + a + b)
†
†
2
(x0 − |x2 |)
× exp −
ξ0 ξ1 − ξ 1 ξ0 ,
2a(τ + b)
†
(2)
(2)
J tot ≡ e (2) γ 0 γ (2) = cn2 ρtot , 0 .
(102)
Finally we consider the 1D case: by restricting x to x2 = x3 = 0 (79) brings (86) to
1/4
x02
b
exp − −
τ (τ + b)
2τ
(τ + b) 2
dy1
(1)
exp −
y −
Ih (x0 , x1 ) ≡
|y1 |
2τ b 1
(1)
h
−μ2
=
8π 2
τ ∂0
(1)
I (x0 , x1 )ξ1 ,
bσ1 ∂1 h
x0 |y1 |
x1 y1
2 cosh
.
b
τ
x12
2b
After a little calculation (see Appendix B) it reads, when x0 , x1 (1)
h
−μ2
≈ √
4 2π
τ b3
(τ + b)3
1/4
(103)
(104)
√ √
τ , b,
(x0 − x1 )2
exp −
2(τ + b)
ξ1
.
σ1 ξ1
(105)
From (72) and (105) the total (1) (81) is obtained as
(1)
ξ (1)
(1) (τ + a + b)
iμ2 a 1/4
ξ
2
exp −
(x0 − x1 )
=
;
σ1 ξ (1)
8π
4a(τ + b)
+ 1/4
2
τ b3
(τ − a + b)
2
(x0 − x1 ) ξ0 + i
≡ exp −
4a(τ + b)
π a(τ + b)3
(τ − a + b)
(x0 − x1 )2 ξ1 .
× exp
4a(τ + b)
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PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
The charge and the current densities are6
2
e (1) † (1) 2e μ2 a 1/4
(τ + a + b)
†
(1)
2
(x0 − x1 ) ξ (1) ξ (1)
ρtot ≡ =
exp −
c
c
8π
2a(τ + b)
1/2
2 1/4 2 τ b3
2e μ a
π
(x0 − x1 )2
(x0 − x1 )2
†
†
ξ0 ξ0 +
ξ 1 ξ1
=
exp −
exp −
3
c
8π
2a
2 a(τ + b)
2(τ + b)
+ 1/4
2
τ b3
(τ + a + b)
†
†
2
(x
+i
exp
−
−
x
)
ξ
−
ξ
ξ
,
ξ
0
1
1
0
0
1
π a(τ + b)3
2a(τ + b)
†
(1)
(1)
J tot ≡ e (1) γ 0 γ (1) = cρtot , 0, 0 .
(108)
5.
Discussion
In this paper, first we discuss the wave mechanics of H = c p2 + m 2 c2 , which tells us that solutions of the Schrödinger equation inevitably possess a light-speed portion in the non-relativistic limit
c → ∞. The reason for this is that the kernel contains a derivative acting on a function that owns the
discontinuity on the light-cone. The solutions of the Dirac equation are also expressed by differentiations to the invariant functions (x) and F (x), which consist of different functions in the timeand space-like regions, thus yielding the light-cone singularity, which was the content of Sects. 3
and 4. In relativistic field theories, c appears as μ = mc/ so that the non-relativistic limit implies
μ → ∞, which, however, also interprets the semiclassical → 0 or an infinite mass limit m → ∞.
According to the last case we can convince ourselves of the survival of the light-cone singularity
for massive particles. We should emphasize that our conclusion has been derived exclusively in the
x-representation of (x), not in the momentum representation.
The situation is unchanged if the source (78) would have a velocity v: consider, e.g.,
y02
v
(3)
3
; β≡ .
(109)
h (y) ∼ δ (y − β y0 ) exp −
2τ
c
Then from (86),
y2
h ∼ d 4 yδ(x0 − y0 ∓ |x − y|)δ 3 (y − β y0 ) exp − 0
2τ
y2
= dy0 δ(x0 − y0 ∓ |x − β y0 |) exp − 0 .
2τ
(The spinor part is irrelevant.) Since the zeros in the delta function are given by
,
(x0 + |x|) (1 − β · n) + O(β 2 )
x0 − y0 = ±|x − β y0 | =⇒ y0 =
(x0 − |x|) (1 + β · n) + O(β 2 )
√
it reads when x0 , x τ ,
(x0 − |x|)2 (1 + β · n)2
h ∼ exp −
,
2τ
(110)
,
(111)
which again shows that the maximum signal travels at the speed of light.
Again note that J = cnρ in (94), (102), and (108) does not imply that the current’s velocity is c. See
footnote 5.
6
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T. Kashiwa and T. Yamaguchi
In order to widen the possibility we finally consider the case in which the initial configuration is
given with a definite momenta p, i.e., instead of the packets (42) take
1
i
y2
ξ̃0
†
ψp (y) ≡
δ(y0 ) exp
p·y−
(112)
,
ξ̃0 ξ̃0 = 1,
0
(aπ )3/4
2a
in the solution
p (x) =
d 4 y S(x − y)ψp (y),
since, at x0 = 0, as in (44), (112) implies an initial configuration,
−i
i
x2
ξ̃0
p (x0 = 0, x) =
exp
p·x−
,
0
(aπ )3/4
2a
(113)
(114)
whose momentum reads
d 3 x p† (x) (−i ∇) p (x) = p.
Following a similar procedure from (46) to (56), we find
⎞
⎛
2 + x2
2
1
x
iμ
i
⎝a
⎠ I ξ̃ ,
p (x) =
exp
p·x− 0
σ ·∇ p 0
8π(aπ )3/4
2a
x0
with
Ip ≡
i
x·y
,
d yδ(x0 − |y|) exp
p·y−
a
(115)
(116)
3
which is further rewritten as
x · y
ia
Ip = exp − p · ∇
d 3 yδ(x0 − |y|) exp −
a
|x|x0
|x|x0
ia
1
= 2πax0 exp − p · ∇
exp
− exp −
,
|x|
a
a
(117)
(118)
with the help of (57). By noting
∇=n
∂
;
∂|x|
n=
x
,
|x|
∂
and that exp − ia
p
·
n
∂|x| is a shift operator, it reads
|x|x0
p·n
p·n
|x|x0
−i
+i
exp
x0 − exp −
x0
a
a
Ip = 2πax0
.
|x| − iap · n/
(119)
(120)
Therefore
p (x) =
x0
iμ2 a 1/4
3/4
4π
|x| − iap · n/
⎡⎛
⎞
2
1
⎠ exp − i(p · nx0 − p · x) − (x0 − |x|)
a
× ⎣⎝
1−
2a
x0 (|x| − iap · n/)
⎞
⎤
⎛
2
−1
−
p
·
x)
+
|x|)
(x
i(p
·
nx
ξ̃
0
0
0
⎠ exp
⎦
a
−
+⎝
.
1+
σ · nξ̃0
2a
x0 (|x| − iap · n/)
(121)
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When x0 , |x| T. Kashiwa and T. Yamaguchi
√
a; |x| − iap · n/ ≈ |x|, it yields a plane wave,
iμ2 a 1/4
i(p · nx0 − p · x)
ξ̃0
p (x) ≈
exp −
,
σ · nξ̃0
4π 3/4
(122)
around the peak x0 − |x| ≈ 0, which implies that the energy–momentum relation is given by
E = c|p|.
(123)
Therefore an alternative way to observe the relativistic remnant is to measure the energy and the
momentum of electrons in the vacuum.
Acknowledgements
The authors are grateful to Hiroto So for discussions. T.K. also thanks Tadashi Toyoda for useful comments.
Appendix A. The derivation of (20)
Owing to the delta function δ(ct − r ),
χ± = μ ± (ct)2 − r 2 → 0,
so that
%
(2)
π H1 (χ+ ) %%
%
%
2
χ+
χ+ =0
%
K 1 (χ− ) %%
+i
%
χ
−
%
π J1 (χ+ ) %%
=
−i
2
χ+ %χ+ =0
,
χ− =0
%
%
π N1 (χ+ ) %%
K 1 (χ− ) %%
−
,
2
χ+ %χ+ =0
χ− %χ− =0
(A1)
where we have noticed
Hn(2) (z) = Jn (z) − i Nn (z),
(n = 0, 1, 2, . . .).
(A2)
From the definition of Jn and In (see Ref. [11], p. 908, 8.440 and p. 909, 8.445),
Jn (z) =
In (z) =
we have
While
∞
z n "
2
z n
2
k=0
∞
"
k=0
%
J1 (χ+ ) %%
1
= ;
%
χ+ χ+ =0 2
(−)k z 2k
,
k!(n + k)! 2
(A3)
z 2k
1
,
k!(n + k)! 2
(A4)
%
I1 (χ− ) %%
1
= .
%
χ− χ− =0 2
%
%
%
π N1 (χ+ ) %%
1 %%
J1 (χ+ ) 1
χ+ %%
=
− −
,
γ + ln
2 χ+ %χ+ =0
χ+
2 %χ+ =0 4 χ+2 %χ+ =0
%
%
%
K 1 (χ− ) %%
1 %%
I1 (χ− ) 1
χ− %%
=
+ +
,
γ + ln
χ− %χ− =0
χ−
2 %χ− =0 4 χ−2 %χ− =0
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(A5)
PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
with Euler’s constant γ , in view of (see Ref. [11], p. 909, 8.444-2 and 8.446, where they use C for
Euler’s constant, and also, for a more complete expression, Ref. [17])
k
∞
n+k
"
1
π
z 1 z n " (−)k z 2k " 1
Nn (z) = Jn (z) γ + ln
+
−
2
2
2 2
k!(n + k)! 2
m
m
m=1
k=0
m=1
n−1
1 2 n " (n − k − 1)! z 2k
;
−
2 z
k!
2
(A6)
k=0
k
∞
n+k
(−)n z n "
z 2k "
k
"
(−)
1
1
z
+
K n (z) = (−)n+1 In (z) γ + ln
−
2
2
2
k!(n + k)! 2
m
m
k=0
+
1
2
m=1
m=1
n "
n−1
(n − k − 1)! z 2k
2
,
z
k!
2
(A7)
k=0
then
%
%
π N1 (χ+ ) %%
K 1 (χ− ) %%
−
2 χ+ %χ+ =0
χ− %χ− =0
%
%%
%
1
1 2
1
1
%
=
− −
−
ln xμ − ln −xμ2 %%
2x2
2x2 %
2
2
μ
μ
2
μ
μ
xμ =0
xμ2 =0
%
(%
iπ
1
1'
1
ln(ct − r ) + ln(ct + r ) − ln(r − ct) − ln(r + ct) %%
− ,
=
− =−
4
2
4
2
ct=r
(A8)
where we have used ln(r − ct) = iπ + ln(ct − r ) in the final expression.
Therefore (A1) reads
%
%
(2)
π H1 (χ+ ) %%
π
i
i
K 1 (χ− ) %%
π
+i
= − + = .
%
%
2
χ+
χ−
4
4
2
2
χ+ =0
χ− =0
(A9)
Appendix B. Derivations of h(3) , h(2) , and h(1)
Equation (90): With the aid of the polar coordinates, (88) becomes
(3)
Ih (x0 , x)
x |x|
(τ + b)y 2
0
cosh
y sinh
y
dy exp −
2τ b
τ
b
0
bx0 + τ |x|
bx0 − τ |x|
4π b ∞
(τ + b)y 2
sinh
y − sinh
y ,
=
dy exp −
|x| 0
2τ b
τb
τb
(B1)
8π b
=
|x|
∞
where the addition theorem for the hyperbolic function has been used. In view of the error function
formula (89) we have
∞
dye
0
−Ay 2
1
sinh By =
2
+
2
B
π
B
erf √
,
exp
A
4A
2 A
so that (B1) becomes (90).
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(B2)
PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
(3)
Equation (91) h : In view of (90) and (87), note that
x02
x2
(bx0 ± τ |x|)2
(x0 ∓ |x|)2
+
= exp −
,
exp − −
2τ
2b
2τ b(τ + b)
2(τ + b)
then apply ∂0 and ∇ to obtain
1/4 (x0 − |x|)2
τ b3
bx0 + τ |x|
bx0 + τ |x|
exp −
erf √
(τ + b)7
|x|
2(τ + b)
2τ b(τ + b)
(x0 + |x|)2
bx0 − τ |x|
bx0 − τ |x|
exp −
erf √
,
−
|x|
2(τ + b)
2τ b(τ + b)
1/4
τ b3
−μ2
(3)
χh (x) = √
2 2π (τ + b)7
(x0 − |x|)2
bx0 + τ |x|
bx0 + τ |x| b(τ + b)
exp
−
−
erf
×
√
|x|
x2
2(τ + b)
2τ b(τ + b)
(x0 + |x|)2
bx0 − τ |x|
bx0 − τ |x| b(τ + b)
exp −
+
erf √
+
|x|
x2
2(τ + b)
2τ b(τ + b)
+
x2
2τ b(τ + b) 2
x2
exp − 0 −
+
(n · σ )ξ1 .
π
|x|
2τ
2b
(3)
ϕh (x)
−μ2
= √
2 2π
When x0 , |x| √ √
τ , b, the terms vanish, except the first one on the right-hand side, to give
(3)
h
1/4
bx0 + τ |x|
τ b3
−μ2
≈ √
7
|x|
2 2π (τ + b)
(x0 − |x|)2
bx0 + τ |x|
ξ1
exp −
× erf √
,
n · σ ξ1
2(τ + b)
2τ b(τ + b)
which further, by noting
lim erf(X ) = 1;
X →∞
becomes
(3)
h
−μ2
≈ √
2 2π
τ b3
(τ + b)7
1/4
bx0 + τ |x|
X≡√
,
2τ b(τ + b)
bx0 + τ |x|
(x0 − |x|)2
ξ1
exp −
,
n · σ ξ1
|x|
2(τ + b)
yielding (91) around the peak (60).
Equation (97): Eq. (96) reads, by use of the polar coordinate,
∞
x y 2π
(τ + b) 2
|x2 |y
0
(2)
y cosh
cos θ .
dy exp −
dθ exp −
Ih (x0 , x2 ) = 2
2τ b
τ
b
0
0
√
When |x2 | b, the saddle point method around θ = π brings the angular part to
&
2π
2π b
|x2 |y
1
|x2 |y
cos θ =
exp
1+O
.
dθ exp −
b
|x2 |y
b
|x2 |
0
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(B3)
PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
Then
&
(2)
Ih (x0 , x2 )
2π b ∞ dy
(τ + b) 2 (τ |x2 | + bx0 )
y +
y
≈
√ exp −
|x2 | 0
y
2τ b
τb
(τ + b) 2 (τ |x2 | − bx0 )
y +
y ,
+ exp −
2τ b
τb
which is further rewritten, in terms of a dimensionless quantity Y ,
y=
as
τ |x2 | ± bx0
Y,
√
τb
&
2π b (+)
+ F (−) ,
(B4)
F
|x2 |
&
-
,
√
∞
2 Y2
(τ
+
b)(τ
|x
τ
|x
|
±
bx
dY
|
±
bx
)
τ
b
2
0
2
0
−
Y
.
F (±) ≡
√
√ exp −
(τ b)2
2
τ +b
τb
Y
0
√
√ √
Since |x2 |, x0 τ , b, the saddle point method around Y = τ b/(τ + b) gives
&
2π τ b
1
(τ |x2 | ± bx0 )2
(±)
1+O
.
F
=
exp
τ |x2 | ± bx0
2τ b(τ + b)
(τ |x2 | ± bx0 )2
(2)
Ih (x0 , x2 ) =
Inserting this into (B4), we have (97).
(2)
Equation (98) h : Apply differentiations in (95) and note the relation (B3) by putting x → x2 to
obtain
1/4 &
2 τ b3
τ |x2 | + bx0
(x0 − |x2 |)2
−μ
(2)
h ≈
exp −
4π
(τ + b)5
|x2 |
2(τ + b)
&
τ |x2 | − bx0
(x0 + |x2 |)2
ξ1
∓
exp −
,
(B5)
n2 · σ 2 ξ1
|x2 |
2(τ + b)
where we have omitted terms of
b(τ + b)
τ b(τ + b)
.
,
O
O
(τ |x2 | ± bx0 )2
|x2 |(τ |x2 | ± bx0 )
√ √
Under x0 , |x2 | τ , b, (B5) becomes (98).
(1)
Equation (105) h : Eq. (104) becomes (we have put y1 → y)
∞
dy
bx0 + τ x1
(τ + b) 2
bx0 − τ x1
(1)
Ih (x0 , x1 ) = 2
cosh
exp −
y
y + cosh
y .
y
2τ b
τb
τb
0
(B6)
Apply the differentiation in (103), with the aid of the error function formula (B2), to find
1/4 τ b3
bx0 + τ x1
(x0 − x1 )2
−μ2
(1)
h (x0 , x1 ) = √
erf
exp −
2τ b(τ + b)
2(τ + b)
4 2π (τ + b)3
(x0 + x1 )2
bx0 − τ x1
ξ1
exp −
± erf
,
σ1 ξ1
2τ b(τ + b)
2(τ + b)
√ √
which yields, when x0 , x1 τ , b, (105).
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(B7)
PTEP 2014, 103B01
T. Kashiwa and T. Yamaguchi
References
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[3]
[4]
[5]
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[8]
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[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
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