The wave function of the electron 2016

Abstract

In this paper, by solving the equations of the relativistic M2 [2] it will show that elementary particles, such as electrons, can contain the wave process with very specific properties. The wave model of a stationary electron is represented in the form of a spherical wave process.

Full text

The wave function of the electron.
Arayik Danghyan
EXHLAB Inc.
Glendale, United States.
arayik@exhlab.com
Abstract
In this paper, by solving relativistic equation M2, the wave nature of the electron is
revealed. The validity of the initial hypothesis of Louis de Broglie about the existence
of matter waves is proved. The connection between the wave characteristics of an
electron and the Compton wavelength is revealed. The problem of dispersion and
spreading of the wave packet is solved. It is demonstrated that theoretical and
experimental results correlate.
1. Introduction.
The basic equation from which Louis de Broglie's hypothesis about the wave
properties of matter [1] followed was a formula written by him based on Planck's
hypothesis of the smallest energy quantum and Einstein's equation relating mass
and energy
2
00
mc

=
, accordingly, each chunk of energy with a mass of
0
m
corresponds to a periodic process with a frequency equal to
2
0
mc
.
Applying the de Broglie hypothesis to elementary particles, mainly the electron, will
mean that the electron contains an internal wave process. However, this wave
process was not identified, and de Broglie waves were only related to the mechanical
motion of the particles.
Modern physics interprets De Broglie waves as probability waves with no material
manifestation.
However recently, there were experimental evidence to support the initial
hypothesis of the existence of de Broglie matter waves [2];[3].
These experimental results have stimulated numerous attempts at a theoretical
explanation. [4]; [5].
In this work, through the solution of the relativistic equation M2 [6], it will be
demonstrated that elementary particles, in particular the electron, may include a

wave process with very specific properties. The wave model of a stationary electron
is represented as a spherical wave process.
2. Time-dependent equation M2.
Let us transform the stationary relativistic equation M2 (1.1) obtained in a study [6]
into a time-dependent form.
46 22
22
10
( ( ))
mc mc
E U r

 − −  =

−

(1.1)
To do this, let us remove the potential energy from the equation
()Ur
and substitute
the energy square operator
22
2
2
Et

=−
into the equation.
As a result, we get:
4 6 2 22
22
2
0
mc mc
t
+  + =
 

(1.2)
Now we need to find the dispersion relation for the resulting equation (1.2)
To do this, we substitute a spherical wave
( )
1i t kr
e
r

−−
=
(1.3) into the equation.
We define the second derivative with respect to time:
( )
22
2
i t kr
e
tr


−−
=−

(1.4)
The Laplacian operator in a spherical coordinate system is as follows:
2,
,, 22
2
rr r r r




 = + +

(1.5)
Let us define the result of the action of the Laplacian operator on a spherical wave:
( )
2i t kr
ke
r

−−
 = −
(1.6) Substitute the spherical wave (1.3) and the calculated
values (1.4) and (1.6) into the initial time-dependent equation (1.2).
As a result, we obtain the following form for the dispersion relation:
46 2 2 2 2
22 0
mc k m c

− − + =
(1.7).

3. Wave without dispersion and non-spreading wave packet.
For the resulting dispersion relation, we determine the phase and group velocities
of wave propagation.
As we know, the phase velocity is determined through the circular frequency

and
wavenumber
k
by the following formula:
f
Vk

=
(2.1)
From equation (1.7) lets determine the circular frequency

.
( )
46
2 2 2 2 2
mc
c m k

=−
(2.2)
Now let us substitute the obtained value (2.2) into the formula (2.1)
( )
46
2 2 2 2 2 4
f
mc
Vk c m k
=−
(2.3)
Group velocity is determined by the following formula:
gr
d
Vdk

=
(2.4). Let us
define the derivative using formula (2.2).
( )
46
2
2 2 2 2 2
2 2 2 2
gr
mc
kc m k
Vc m k
−
=−
(2.5).
Let's plot graphs showing the phase and group velocity's dependence on the
wavenumber
k
. In what follows, the Hartree atomic system of units will be utilized.
00
1, 1, 1, 1, 137.03599971,4 1a m e c

= = = = = =

Figure 1. A plot of the phase velocity
f
V
from the wave number
k
.
Figure 2. A plot of the phase velocity
gr
V
from the wave number
k
.

Now let's combine the two graphs into a single figure.
Figure 3. Combined graph of phase and group velocities.
As we can see, at a particular value of the wave number, the phase velocity is equal
to the group velocity.
Equating the corresponding formulas
f gr
VV=
and solving the resulting equation,
determine the value of the stationary-state wave number
02
cm
k=
(2.6). In case of this value following speed will be obtained
. Corresponding circular frequency value is (2.7). Thus,
we have obtained a spherical wave without dispersion (standing wave). Since under
the condition of equality of the phase and group velocities dispersion disappears.
The resulting speed value is twice the speed of light. However, this is not the speed
of the electron. Since we are considering an electron in a stationary state, this is the
speed of propagation in the space of a spherical de Broglie wave (for the initial
version of the hypothesis), that is, a wave of matter. Furthermore, since the obtained
matter wave has an unknown nature so far, we will not limit its characteristics,
particularly the propagation speed.
Before continuing, let us note the following. We define the electron rest energy in
accordance with the formula
0
E

=
. Then we get
2
2E mc=
. As we can see, the

resulting value of the energy in
2
times more than the expected value, the electron
rest energy
2
E mc=
.
To eliminate the resulting inconsistency, we introduce the concept of the bare mass
of an electron, which, by entering internal wave processes, forms rest energy
2
E mc=
. From these considerations, we obtain the value of the bare mass
02
m
m=
. In further calculations, we will utilize the bare mass instead of the
electron mass. Then we finally obtain the value of the wave number for the steady
state
02
cm
k=
(2.8) and the value of the circular frequency
2
0
mc

=
(2.9).
Integrating spherical waves in a small neighborhood


of the steady state (region
1, Figure 3). We get a spherical wave packet, without the spreading:
( )
( )
00
0
ki t kr
k
e
A k dk
r


+−−
−
= 
(2.10)
Now, after obtaining the parameters of the stationary state of the electron, it makes
sense to go over to the stationary equation and finally obtain the wave function of a
bare stationary electron.
To do this, we substitute the obtained value of the bare mass
02
m
m=
and the rest
energy value
2
E mc=
into the original stationary equation (1.1) without potential
energy. Then we get:
22
2
10
4
mc
 +  =
(2.11)
The resulting equation, in the theory of differential equations, is known as the
Helmholtz equation [7].
4. The solution of the Helmholtz equation.
To solve equation (2.11), we apply the standard method of separation of variables in
a spherical coordinate system.
Let us represent the wave function as a product of the radial and angular parts:
( , , ) ( ) ( , )r R r Y
   
=
(3.1)

The Laplacian operator in spherical coordinates looks like this:
2,
,, 22
2
rr r r r




 = + +

(3.2) Let us substitute the product (3.1) into the
initial equation (2.11).
2 2 2
,
2 2 2
21
0
4
Y
R R m c
Y Y R RY
r r r r



+ + + =

. Let us
multiply the resulting equation by a fraction
2
r
RY
.
2 2 2 2 2 ,
2
22
21
4
Y
r R r R m c r
R r R r r Y



+ + = −

(3.3). As you can see, the left side of
equation (3.3) depends only on the variable
r
, and the right of the variables

and

. Therefore both parts are equal to some constant number

. This allows us
to separate the radial part of the equation from the angular part.
2 2 2
2 2 2
21 0
4
R R m c RR
r r r r


+ + − =

(3.4)
,YY
 
 = −
(3.5) Next, we
represent the function
( )
,Y

as a product
( ) ( ) ( )
,Y
   
=  
(3.6). The
angular part of the Laplacian operator has the following form:
22
,2 2 2
1
sin
ctg
 
   
  
 = + +
  
(3.7). By substituting product (3.6) in the
angular equation (3.5) we get:
22
2 2 2
1
sin
ctg

   
    
 +  +  = − 
  
(3.8).
Multiplying equation (3.8) by the fraction
2
sin


we will get:
2 2 2 2
2
22
sin sin 1
sinctg

  
  
    
+ + = −
     
(3.9). The left side of equation (3.9)
depends only on the variable

while the right side only on the variable

.
Therefore, both parts are equal to some constant number, which we denote
2
m
(not
to be confused with the mass of an electron). As a result, we obtain the two
equations:
22
22
0
sin
m
ctg

  
  
+ +  −  =

(3.10) and
22
2
10m

+=

(3.11)
The solution of

equation (3.11) is well known
( ) ( )
expA im

 = 
(3.12). Since
for identical values of the

(
0
and
2

) the function must have the same value,
which is
( ) ( )
exp 0 exp 2A im A im A

 =  =
and
( )
exp 2 1im

=
. Using Euler's formula
for complex numbers:
( ) ( )
cos 2 sin 2 1m i m

=
, we obtain
0, 1, 2, 3,......m=   
In this way
m
can only take integer values. The constant is determined from the
normalization condition for the function

. However, since the standard
normalization and probabilistic interpretation of the wave function are no longer
valid, this issue will not be discussed.

To solve the

equation (3.10), we use the mathematical online
resource WolframAlpha http://www.wolframalpha.com/.
After solving we obtain:
( )
( )
( )
( )
( )
1 1 2 1
4 1 1 4 1 1
22
cos cos
mm
c P c Q

  
+ − + −
 = +
(3.13)
That is, the solution is the sum of the associated Legendre functions of the first and
second kind. Let us make a designation
( )
14 1 1
2
s

= + −
(3.14). The variable
s
will
be associated with the spin quantum number. Let us determine the range of possible
values of the quantum number
s
. From formula (3.14) it follows that
4 1 0

+
so
1
2
s−
(3.15). Rewriting dependence (3.14) with respect to
s
we obtain
( )
1ss

=+
(3.16) as expected. Thus, the angular part of the wave function will
have the form:
( ) ( ) ( ) ( )
12
, cos cos exp
mm
ss
Y c P c Q im
    

= + 

(3.17)
Since our electron is in a stationary state, the usual conditions for quantization of
the orbital momentum do not hold. Therefore, we cannot yet say that the quantum
number
s
can only take integer or half-integer values. Range of possible values
s
will be limited only by the ratio (3.15) and the condition of continuity of the angular
part of the wave function (3.17). Specific values of the quantum numbers
m
and
s
and integration constants
1
c
and
2
c
will depend on external factors.
That is, the boundary conditions of a specific task. In the works of the author [8] is
an approach that the quantum numbers
m
and
s
can take as whole and half-integer
values. We believe this approach is reasonable.
Here, for clarity, some graphic examples of the angular part of the wave function at
different values of the quantum numbers
m
and
s
.

We now turn to the solution of the radial equation (3.4). For this we use a
mathematical online resource WolframAlpha http://www.wolframalpha.com/.
We rewrite the equation (3.4) with the notation (3.16).
2 2 2
2 2 2
2 1 ( 1) 0
4
R R m c s s
RR
r r r r
  +
+ + − =

(3.18)
0
m
=
2
s
=
1
m
=
2
s
=

The solution has this form:
( )
12
[ , ] [ , ]
22
cmr cmr
R r k SphericalBesselJ s k SphericalBesselY s=+
(3.19)
That is, it is a combination of spherical Bessel functions of the first and second kind.
As is known, the spherical Bessel function of the second kind is not bounded for
0r→
. Therefore, we accept
20k=
. And finally, the radial part of the wave
function is of the form:
( )
1
[ , ]
2
cmr
R r k SphericalBesselJ s=
(3.20)
Let us plot of the radial wave function in Hartree atomic units, for the first few values
of the quantum number
13
0, ,1, , 2
22
s=
. Since we have not yet discussed the issue
of normalization, for the reasons given, we will accept
11k=
.
In Hartree units, the electron mass is
1m=
, Planck's constant
1=
and speed of
light
137.03599971c=
.
Figure 4. Graph of the radial part of the wave function at values
13
0, ,1, ,2
22
s=
.
Therefore, the total wave function of an electron will be the product of the radial,
angular, and time-dependent parts.

( ) ( ) ( ) ( ) ( )
, , ,r t R r t
   
 =   
( ) ( ) ( ) ( )
2
12
, , , [ , ]* cos cos exp exp( )
2
mm
ss
cmr mc
r t SphericalBesselJ s c P c Q im i t
    

 = +  −

(3.21)
Results and discussion:
As has been shown, the electron has an internal wave process with a frequency
2
0
mc

=
. The wave process extends outside and forms matter waves.
And the wave number is equal to
02
cm
k=
, hence the wavelength is
0
24
k cm


==
, and the speed of propagation in space is twice the speed of light.
Consequently, matter waves are not electromagnetic in nature.
As is known, an electron is characterized by an experimentally found value:
Compton wavelength, which is
2
Comp cm


=
. Therefore, the resulting wavelength is
twice the Compton wavelength
2Comp

=
. From here, it can be assumed that the
Compton wavelength is related to the wave properties of the electron.
By that time, E. Schrödinger disagreed with the purely corpuscular explanation of
the Compton effect. Moreover, he wrote an article where he expressed his views. [9].
"Schrödinger argued that X-rays can be diffracted by a standing "charge density
wave" created by an incident and reflected electron, just as light can be diffracted by
a standing wave of ultrasound. (Born and Wolf 1959).”
Further, it was found that not the entire mass of the electron participates in the
oscillatory process.
As a result, the concept of bare mass was introduced
02
m
m=
. In other words, the
mass of the electron is formed as a result of the entry of the bare mass into the
internal wave process.
By analogy, it can be assumed that the seed mass is also formed because of the wave
process of the second stage. Moreover, consequently, the matter has a multi-stage,
nested wave structure.

It has been established that the spin of an electron does not have to have a value of
1
2
. The range of possible values of electron quantum numbers is still to be studied.
The results obtained open new horizons for theoretical and experimental research.
They allow taking a fresh look at many experimental results in which the wave
properties of an electron appear.
References
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[5] David Hestenesa. Hunting for Snarks in Quantum Mechanics Physics
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