Article 1st

Relativistic Di¤ration in Time
Salvador Godoy and Karen Villay
Universidad Nacional Autónoma de México,
Facultad de Ciencias, Depto de Física,
México D.F. 04310.
May 13, 2015
Abstract
We discuss the exact solution for the Klein-Gordon shutter’s prob-
lem. We …nd that the wave function does not resemble the optical
expression that appears in optical theory of di¤raction. However, the
exact relativistic charge density, when plotted versus time, shows tran-
sient oscillations which, apart from some relativistic features, clearly
resemble a di¤raction pattern. We claim that di¤raction in time does
exist in the relativistic realm.
1 Introduction
Similarities between optics and quantum mechanics have long been recog-
nized. [1] One example of this symmetry was obtained by M. Moshinsky
[2]. Moshinsky addressed the following non-relativistic quantum 1D shut-
ter problem: consider a monoenergetic beam of free particles, ! ~k2
=2m,
moving parallel to the x-axis. For negative times, the beam is interrupted
at x = 0 by a perfectly absorbing (no re‡ected wave) shutter perpendicular
E-mail: sgs@ciencias.unam.mx
y
This research is in partial ful…llment of the Bachelor Degree in Physics at the Univer-
sidad Nacional Autónoma de México.
1

to the beam. Suddenly, at time t = 0, the shutter is opened, allowing for
t > 0 the free time-evolution of the beam of particles. What is the transient
density observed at a distance x from the shutter? The shutter problem im-
plies solving, as a boundary-value problem, the time-dependent Schrödinger
equation with an initial condition given by
(x; 0) = eikx
( x); (1)
where (x) denotes the Heaviside step function de…ned as: (x) = (1 if x > 0)
or (0 if x < 0).
For t > 0 , Moshinsky proved that the free propagation of the beam has
an exact solution given by:
(x;t) = exp[i(kx !t)]
1
p
2i
f
1 + i
2
+ F[ (x; t)]g; (2)
where F( ) denotes the complex Fresnel integral: F( )
R
0
exp(i u2
=2) du
C( ) + iS( ) and is given by
(x; t)
r
m
~t
(
~k
m
t x): (3)
For the beam, the probability density is then
(x;t) =
1
2
C( ) +
1
2
2
+
1
2
S( ) +
1
2
2
: (4)
The right-hand side in (4) is identical to the optical expression for the light
intensity in the Fresnel di¤raction by a straight edge [3]. For a …xed position
x, the plot of the probability density (x;t) as a function of time is shown
in Fig. 1. This quantum transient behavior has been given the name of
di¤raction in time. A good measure of the ‘width’in time of this di¤raction
e¤ect, can be obtained from the di¤erence T t2 t1 between the …rst two
times at which takes the classical (mean) value. We obtain T ( mx=~k2
).
2

Figure 1: Probability density for non-relativistic di¤raction in time.
As an example, for thermal neutrons at a distance of x = 1 m, the di¤raction
width would be T 10 9
sec. The di¢ cult experimental evidence of this
quantum prediction has been con…rmed until very recently by Szriftigiser,
Guéry-Odelin, Arndt, and Dalibard [4].
Notice that the transient density (4) increases monotonically from the
very moment in which we open the shutter, and therefore, an observer at a
distance x from the shutter could detect particles before a time (x=c), where
c is the velocity of light. This would imply that some of the particles travel
with velocities larger than c. So, for very short times all Schrödinger’s results
have to be disregarded. The non-relativistic character of the Schrödinger’s
equation, is the cause of this erroneous prediction.
The analogy between the transient densities (4) and the optical di¤rac-
tion, raises the question of whether the transient densities for other types
3

of wave equations show this analogy. Professor Moshinsky was the …rst to
address this question and answer it [5]. Working with the ordinary-wave
equation and the Klein-Gordon equation, Moshinsky arrived to the conclu-
sion that only for the Schrödinger equation (non-relativistic case) the wave
function does resemble the expression that appear in Sommerfeld’s theory
of di¤raction. In his conclusions [5], Moshinsky emphatically denies the ex-
istence of di¤raction in time in the relativistic case. However, taking into
account that his reported relativistic solution was an approximation, and
that no attempt was made to plot the charge density, we distrusts his …nal
conclusions.
The purpose of this article is a revision of the Klein-Gordon shutter’s
problem. First, we get the exact solution for the Klein-Gordon shutter’s
problem using similar discontinuous initial conditions as in the Schrödinger
shutter’s problem. Our solution looks quite di¤erent from the one reported
Moshinsky . In agreement with Moshinsky’s conclusions, we …nd that the
Klein-Gordon wave function does not resemble the mathematical expression
that appears in Sommerfeld’s theory of di¤raction. However, and this is our
main contribution, when the exact relativistic charge density is plotted versus
time, the plot shows transient oscillations which, apart from some relativistic
features, clearly resemble a di¤raction pattern. For this reason in this article
we claim that di¤raction in time does exist in the relativistic realm. We show
the time intervals in the transient region. In a typical relativistic case of let
say (v=c = 0:7) a -meson has oscillation periods of the order of 10 23
sec.
This relativistic periods are by far smaller than the corresponding periods in
the Schrödinger solution.
4

2 The relativistic shutter’s problem
For a beam of particles at relativistic energies, our main interest concerns
the discussion of the transient e¤ects of the discontinuous initial conditions
of the di¤raction in time problem. For a beam of electrons, we would have, of
course, to replace the Schrödinger equation by the Dirac’s equation. However,
since the features of spin and statistics and hole theory are irrelevant to our
analysis, for simplicity, instead of Dirac’s equation we shall use the Klein-
Gordon equation.
The wave function (x; t) satis…es now the equation:
@2
@x2
1
c2
@2
@t2
= 2
(5)
where mc=~ = 1= C and C is the Compton wavelength.
For a bounded (x; t), the following discontinuous initial conditions truly
de…ne the di¤raction in time problem. In the shutter problem, for t < 0, when
the shutter was closed, we assume as in the previous section, an incident plane
wave given by
(x; t) =
exp[i(kx !t)] if x < 0
0 if x > 0
= ei(kx !t)
( x); (6)
where the relativistic ! is given by: ! = c(k2
+ 2
)1=2
. This implies that at
t = 0, when the shutter is suddenly opened, we have the initial conditions:
(x; 0) = eikx
( x);
@ (x; 0)
@t
= i! eikx
( x): (7)
Before solving the di¤raction in time problem, notice, for the sake of
simplicity, that the Klein-Gordon equation (5) may be rewritten as
@2
@( x)2
@2
@( ct)2
= : (8)
5

Therefore, if we use the Compton length, C , and the Compton time, C=c,
to de…ne dimensionless variables for ‘position’ and ‘time’ given by
x
C
= x;
t
C=c
= ct (9)
the Klein-Gordon equation may be recast as
@2
@ 2
@2
@ 2
= : (10)
The initial conditions for di¤raction in time may also be rewritten in terms
of these dimensionless variables:
(x; 0) = exp(i
k
x) ( x) ei
( ); (11)
@ (x; 0)
@( ct)
= i
!
c
exp(i
k
x) ( x) i ei
( ): (12)
where we have de…ned the dimensionless ‘wavenumber’: k= and the
dimensionless ‘angular frequency’: != c =
p
1 + 2. For simplicity, all
results that follow will be written in terms of dimensionless variables.
The exact solution of this relativistic problem is calculated in appendix
A. At a distance x, on the right side of the shutter, x > 0, where the particle
detector is located, we have the Klein-Gordon wave function:
( ; ; ) =
1
2
( )[ ei ( )
Z
du ei (u )
f
J1
p
u2 2
p
u2 2
+ i J0
p
u2 2 g] (13)
which holds only for x 0.
Notice in Eq.(13) the outstanding presence of ( ). It means, as
expected, that the wave function vanishes for t < x=c, where x is the distance
from shutter to the particle detector. The presence of ( ) is the hallmark
of a relativistic process; this property is missing in the Schrödinger solution.
6

3 Transient charge density
Given the Klein-Gordon wave function (x; t), we have a charge density given
by (charge q = 1)
=
i~
2mc2
(
@
@t
@
@t
) =
~
mc2
Im[
@
@t
] (14)
Since we know the exact Klein-Gordon wave function for the shutter prob-
lem, we will now show that the plot of the charge density looks similar to
the density plot predicted in the Schrödinger theory. In fact, independently
of some relativistic details, the charge density shows, in time, damped os-
cillations very similar to the ones we …nd in the Schrödinger solution, but in
a smaller time scale. For this reason we still call the oscillations, ‘relativistic
di¤raction in time’.
Let us rewrite the charge density in Eq. (14) as: = Im[ @ =@( ct)],
therefore, in dimensionless variables we have:
( ; ; ) = Im[ ( ; ; )
@ ( ; ; )
@
] (15)
In Eq. (13) we have the exact wave function for the relativistic shutter
problem; so it is an straightaway procedure to plot, as a function of time,
the exact charge density for the right side of the shutter. In Figure 2, for
…xed values of x > 0 and , we show a model of a Klein-Gordon di¤raction
in time plot.
The main features of this plot are:
1) The charge density vanishes for times 0 t x=c; this is an evident
relativistic property which is nonexistent in the Schrödinger solution.
2) The down oscillation, immediately following, t = x=c, is also a relativis-
tic property. In the Schrödinger solution for all times before the oscillations
the probability density is a monotonous increasing function.
7

Figure 2: Example of a typical Klein-Gordon di¤raction in time charge den-
sity.
3) The asymptotic behavior 0 is not 1, as it occurs in the Schrödinger
solution. In the particular case of = 1 (v=c = 0:7) shown in Fig. 2, the
stationary density is 1:4, which is the correct prediction for the shutter’s
initial conditions (7). In fact
( ; 0) = Im[ ( ; 0)
@ ( ; 0)
@
] =
p
1 + 2 ( ) (16)
Therefore, 0( ) =
p
1 + 2, and for = 1 the predicted stationary density
is: 0 = 1:4.
4) The theoretical value for 0 implies that, in the non-relativistic limit,
1, the stationary limit is 0 ! 1, which is the predicted value in the
Schrödinger solution.
8

5) The relativistic periods of oscillation are by far smaller than the cor-
responding periods in the Schrödinger solution. From Fig. 2 we see that, for
our particular value of = 1 (v=c = 0:7), the half period is about, 2 1 9,
which implies a time interval of: T t2 t1 9~=mc2
. For a -meson we
have T 10 23
sec. And if we use the Klein-Gordon as an approximation
to describe a relativistic proton we have T 10 24
sec. which is impressively
small. We will show next that in the non-relativistic limit, the time interval
T will dramatically increase as the value of decreases.
4 Di¤erent values of
So far we have plotted a charge density with a single value of . How does
the plot look like for di¤erent values of ? In Fig. 3 we plot the charge
density for three di¤erent values of = 0:9, 0:3 and 0:07.
The main features of this plot are:
1) As expected, the density’s stationary value is a function of : 0 =
p
1 + 2.
2) The smaller the the bigger the oscillation periods are; For = 0:3
the period of these oscillations, for a -meson, is about T 10 22
sec.
3) In the non-relativistic limit, the Schrödinger oscillations with 1,
are expected to have oscillation periods which are orders of magnitude bigger
than the Klein-Gordon ones. This is proved next.
5 The non-relativistic limit
By de…nition of
k
=
~k
mc
=
v=c
(1 v2=c2)1=2
: (17)
9

Figure 3: Charge density for = 0:9, 0:3 and 0:07.
In the nonrelativistic limit (v c), we have v=c 1, and this implies
that: = (1 + 2
)1=2
1. In this limit, the Klein-Gordon wave function
becomes factorized as
( ; ) = e i
'( ; ) e i
'( ; ): (18)
We know that in the non-relativistic limit, the function '( ; ) must satisfy
the Schrödinger equation. For our Klein-Gordon di¤raction in time function
(18) let us …nd this limit.
In order to get the Schrödinger wave function, we will use a mathematical
technique that, as far as we know, it has never been used before.
We begin taking the Laplace transform of the Klein-Gordon wave func-
10

tion, Eq.(18).
L[ ( ; )](s) L[e i
'( ; )](s) = L['( ; )](s + i): (19)
This relation suggests a method to …nd the Schrödinger wave function. If we
rewrite the Laplace transform L[ ( ; )](s) as a function of s + i , and get
let say, F(s + i), then in the nonrelativistic limit of small velocities, 1,
and long times, jsj 1, the function F(s) must be the Laplace transform of
the Schrödinger wave function, '( ; ), that is
L 1
[F(s)] = '( ; ) for 1 and s 1 (20)
Beginning from the Klein-Gordon di¤raction in time wave function, we will
use this method to …nd the corresponding Schrödinger wave function.
1) In the appendix A, in Eq.(42), we already have the Laplace transform
of the Klein-Gordon wave function
L[ ( ; )](s) >( ; s)
=
1
2
1
s + i
(1
iK
p
s2 + 1
) exp(
p
s2 + 1) (21)
2) Next, we rewrite >( ; s) as a function F(s + i), we have
s + i = (s + i) + i( 1)
s2
+ 1 = (s + i i)2
+ 1 = (s + i)2
2i(s + i)
3) We get F(s) as
F(s) =
1
2
1
s + i( 1)
(1
iK
p
s2 2is
) exp(
p
s2 2is) (22)
4) In this function F(s) we get now the non-relativistic limit of small
velocities, 1, and long times, jsj 1, we have
s2
2is ! 2is; and 1 ! 2
=2 (23)
11

therefore
F(s) !
1
2
1
s + i 2=2
(1
iK
p
2is
) exp(
p
2is) '( ; s) (24)
We claim that this function '( ; s) is the exact Laplace transform of the
Schrödinger wave function. Indeed, if we take the inverse Laplace transform
we get:
L 1
['( ; s)]( ) =
1
2
ei( 2 =2)
[1 erf(
r
1
2i
r
2i
)] (25)
which in terms of natural variables (x; t) is given by
(x; t) =
1
2
ei(kx ~k2t=2m)
[1 erf(
r
m
2i~t
(x
~k
m
t))] (26)
To get the more familiar expression for (x; t), the error function can be
expressed in terms of Fresnel Integrals [6]:
C(z) + iS(z) =
r
i
2
erf(
r
2i
z) (27)
and we get
(x; t) = ei(kx ~k2t=2m) 1
p
2i
[
r
i
2
+ C( ) + iS( )] (28)
where (x; t)
p
m= ~t(~kt=m x). This is the exact Moshinsky’s solution
for the non-relativistic shutter problem. This proves that our exact Klein-
Gordon wave function (42) has the correct non-relativistic limit.
-
A The Exact Klein-Gordon Solution
Let us consider the Klein-Gordon equation for ( ; ):
@2
@ 2
@2
@ 2
= ; (29)
12

de…ned in the in…nite domain 1 < < 1. The boundary conditions are:
lim
jxj! 1
j ( ; )j < 1, and the initial conditions, corresponding to di¤raction
in time, are:
( ; 0) = exp(i ) ( ) (30)
@ ( ; 0)
@
= i exp(i ) ( ) (31)
where
p
1 + 2 . We begin by taking the Laplace transform of Eq.(29)
and …nd the di¤erential equation
d2
( ; s)
d 2
(s2
+ 1) ( ; s) = (s i )ei
( ) (32)
which holds in the domain 1 < < 1. In Eq. (32) the second derivative
generates a step function ( ), therefore, the origin = 0 is a singular
point where the …rst derivative has di¤erent slopes at right and left of that
point. Nevertheless, both the …rst derivative and the function itself must
be continuous at the origin. This facts suggest to break the in…nite domain
into two domains, ( 0) and ( 0), having the corresponding di¤erential
equation on each domain. For the left side of the shutter, 0, we de…ne
<( ; s) as the solution of the di¤erential equation:
d2
<
d 2
(s2
+ 1) < = (s i )ei
; (33)
and for the right side, 0, we de…ne >( ; s) as the solution of
d2
>
d 2
(s2
+ 1) > = 0: (34)
Here both functions < and > must be bounded: ( < at 1) and ( > at
+1). Other important boundary condition is that the two functions and
their derivatives must be continuous at the interface, = 0.
13

Equations (33) and (34) are ordinary second order di¤erential equations
with constant coe¢ cients and their solution is readily obtained. Taking into
account the boundary conditions at 1 we have:
<( ; s) = A exp(+
p
s2 + 1) +
1
s + i
exp(i ) (35)
which holds for ( 0) and
>( ; s) = B exp(
p
s2 + 1) (36)
which holds for ( 0).
The constants A and B are …xed from the boundary conditions at the
interface: < and >, and their derivatives, d <=d and d >=d , must be
continuous at = 0. We have then a set of coupled algebraic equations:
B = A +
1
s + i
(37)
B
p
s2 + 1 = A
p
s2 + 1 +
iK
s + i
(38)
which have the solution:
A =
1
2
1
s + i
(1 +
iK
p
s2 + 1
) (39)
B =
1
2
1
s + i
(1
iK
p
s2 + 1
) (40)
Substituting Eq.(39) into Eq.(35) and Eq.(40) into Eq.(36) we get the ex-
act solution for relativistic di¤raction in time in the dimensionless variables
( ; s).
For ( 0) we have the solution:
<( ; s) =
1
2
1
s + i
(1 +
iK
p
s2 + 1
) exp(+
p
s2 + 1)
+
1
s + i
exp(i ); (41)
14

and for ( 0) we have another solution:
>( ; s) =
1
2
1
s + i
(1
iK
p
s2 + 1
) exp(
p
s2 + 1): (42)
Obviously to get the solution in the time variable, , we must invert the
Laplace transforms. We …nd in Laplace Transforms Tables [6] the following
results.
L 1
[
1
s + i
] = e i
(43)
L 1
[e j j
p
s2+1
] = ( j j) j j
J1
p
2 2
p
2 2
( j j) (44)
L 1
[
e j j
p
s2+1
p
s2 + 1
] = J0
p
2 2 ( j j) (45)
Therefore, by the convolution theorem, and after some simpli…cations we
have:
L 1
[
e j j
p
s2+1
(s + i )
p
s2 + 1
] = ( j j)
Z
j j
du e i ( u)
J0
p
u2 2 (46)
and
L 1
[
e j j
p
s2+1
s + i
] = ( j j)[ e i ( j j)
j j
Z
j j
du e i ( u)
J1
p
u2 2
p
u2 2
] (47)
With the help of Eqs. (46) and (47) we can get the inverse Laplace transform
of Eqs. (41) and (42), we have the …nal solutions:
For 0 we get the incident and re‡ected wave:
<( ; ) = ei( ) 1
2
( + )[ e i ( + )
+
Z
du ei (u )
J1
p
u2 2
p
u2 2
+ i
Z
du ei (u )
J0
p
u2 2 ]; (48)
15

and for 0 we get the transmitted wave:
>( ; ) =
1
2
( )[ ei ( )
Z
du ei (u )
J1
p
u2 2
p
u2 2
i
Z
du ei (u )
J0
p
u2 2 ]; (49)
We claim that Eqs.(48) and (49) are the exact Klein-Gordon wave functions
for the di¤raction in time problem. The presence of the Bessel functions
J0
p
u2 2 and J1
p
u2 2 =
p
u2 2 is not a surprise. They are
just the Green’s function and its derivative for the Klein-Gordon equation
[7].
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16

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17

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