Let us analyze once again what happens at a potential step.
The standard plane wave solution for the Schrödinger equation is
exp(-i (E t - p x)).
Let us match the solutions for a potential step of height V. The incoming wave is 1, the reflected wave B, and the transmitted wave is 1 + C.
p q
1 --> 1 + C -->
-p <-- B __________
________| V
x = 0
From the continuity we get:
1 + B = 1 + C => B = C.
From the continuous derivative d/dx:
p - B p = q + B q
=>
B = (p - q) / (p + q).
The "current" is proportional to p times the square of the amplitude. We should have
(1 - B^2) p = (1 + B)^2 q
<=>
(1 - B) p = (1 + B) q
which is true.
If V is high, then q is imaginary and
B = exp(-i * 2 φ),
where φ is the phase of the complex number (p, q). The plane wave solutions of the Schrödinger equation work very well over a potential step.
Potential step of the Schrödinger equation
The standard plane wave solution for the Schrödinger equation is
exp(-i (E t - p x)).
Let us match the solutions for a potential step of height V. The incoming wave is 1, the reflected wave B, and the transmitted wave is 1 + C.
p q
1 --> 1 + C -->
-p <-- B __________
________| V
x = 0
From the continuity we get:
1 + B = 1 + C => B = C.
From the continuous derivative d/dx:
p - B p = q + B q
=>
B = (p - q) / (p + q).
The "current" is proportional to p times the square of the amplitude. We should have
(1 - B^2) p = (1 + B)^2 q
<=>
(1 - B) p = (1 + B) q
which is true.
If V is high, then q is imaginary and
B = exp(-i * 2 φ),
where φ is the phase of the complex number (p, q). The plane wave solutions of the Schrödinger equation work very well over a potential step.
Potential step of the new Dirac equation
The standard plane wave solution for our new 1+1-dimensional Dirac equations is
(1, p / (E + m + V)) exp(-i (E t - p x)),
and its conserved current (divided by 2) is
J = p / (E + m + V).
Let us solve the potential step in the diagram of the previous section. From the continuity of component 1 we get
1 + B = 1 + C.
Continuity of component 2 gives:
(1 - B) p / (E + m) = (1 + B) q / (E + m + V).
<=> (def.)
(1 - B) J_p = (1 + B) J_q
<=>
B = (J_p - J_q) / (J_p + J_q).
The conservation of currents requires
(1 - B^2) J_p = (1 + B)^2 J_q
<=>
(1 - B) J_p = (1 + B) J_q,
which is true. Thus, our new Dirac equation behaves very well.
Let us then study what happens when V is large. The energy-momentum relation is
E^2 = q^2 + (m + V)^2.
If (m + V)^2 > E^2, then q is imaginary and also J_q is imaginary.
B = (J_p - J_q) / (J_p + J_q)
= exp(-i * 2 φ),
where φ is the phase of the complex number (J_p, J_q).
We have shown that our new Dirac equation is very well-behaved. There is no Klein paradox in our equation.
Could it be possible to solve the hydrogen atom with the new equation?
We have already discussed the case m + V < 0, in which we think an imaginary i |m + V| is the correct interpretation for the inertial mass, and the electron is a tachyon very close to the proton. Another possibility is that the electron moves at exactly the speed of light but fails to see the force of the proton.
Yet another possibility is that the electron changes into a positron close to the proton and sees a repulsive force from the proton.
If the electron is superluminal, some observers see it right-handed, some observers left-handed (= positron). We may think that it is in an intermediate stage of turning into a positron. In pair annihilation, the electron would turn back in time and be a positron.
Does the new Dirac equation cast light on this? What happens if m + V < 0 or when m + V is imaginary?
t
^
| 511 keV 511 keV photons
| ^ ^
| \ /
| \ ___/
| / \
| / \
| e- e+
--------------------------------> x
When an electron approaches a positron to annihilate, then at some point, m + V = 0 and p = E. The plane wave in that case is
(1, 1) exp(-i (E t - p x)).
That would be a good point to glue it to the positron plane wave, since the vector is (1, 1) for both particles?
But the electron has to emit photons. If the electron emits a 511 keV photon at that point, the plane wave after that might be something like
(1, 2p / (i 2p)) exp(-i (0 t - 2p x)).
The mass of the electron at that stage is i 2p and it moves at an infinite speed because E = 0 and the wave fronts in a spacetime diagram "move" horizontally.
The electron emits a 511 keV photon again and the wave is
(1, 1) exp(-i (-E t - p x)).
That is, the electron has become a positron. In the annihilation, the electron became a tachyon in the intermediate stage. If you need to turn back in time, it is not that strange to be a tachyon during one line segment.
If we flip the roles of t and x, the diagram above shows an electron in an elastic collision with a 511 keV photon.
Annihilation allows superluminal communication. That may be ok if we cannot measure the speed accurately enough to prove that superluminal communication took place. If the electron and the positron collide with a high energy, then the superluminal phase has to be shorter.
We may need a superluminal phase to make the wave function smooth in the spacetime diagram.
A basic principle of special relativity is that time and space are similar entities. Allowing superluminal travel makes them even more similar, because the path of a particle is not confined inside the light cone. Allowing traveling back in time still improves the similarity.
The diagram above suggests that at short distances, time and space become interchangeable, they have an "euclidian metric". But why at macroscopic scales we are bound inside the light cone and time and space most definitely cannot change places?
Maybe at short distances, the uncertainty of quantum mechanics allows particles to "tunnel" back in time. This might be the explanation for zitterbewegung. If we paint a fuzzy line of width one Compton wavelength to the the spacetime diagram above, it looks quite natural to allow the electron path zigzag not just in space but also in time. During the trips back in time it is a positron.
Thus, the electron is allowed to make superluminal trips as well as trips back in time as long as it stays inside the fuzzy path whose width is one Compton wavelength 2 * 10^-12 m or the corresponding time, 7 * 10^-21 s.
The annihilation case does not tell us what the electron does in the vicinity of a proton. The process might be an uneventful superluminal trip.
Note that in a head-on collision with a superluminal particle, the proton cannot exchange momentum with the particle. This is because the proton does not have time to react before the collision is already over. The potential of the proton will appear static to the colliding particle, that is, it stays in the same place as if the proton were infinitely heavy. If the superluminal particle passes from the side, then the proton will exchange momentum with it.
1 + B = 1 + C.
Continuity of component 2 gives:
(1 - B) p / (E + m) = (1 + B) q / (E + m + V).
<=> (def.)
(1 - B) J_p = (1 + B) J_q
<=>
B = (J_p - J_q) / (J_p + J_q).
The conservation of currents requires
(1 - B^2) J_p = (1 + B)^2 J_q
<=>
(1 - B) J_p = (1 + B) J_q,
which is true. Thus, our new Dirac equation behaves very well.
Let us then study what happens when V is large. The energy-momentum relation is
E^2 = q^2 + (m + V)^2.
If (m + V)^2 > E^2, then q is imaginary and also J_q is imaginary.
B = (J_p - J_q) / (J_p + J_q)
= exp(-i * 2 φ),
where φ is the phase of the complex number (J_p, J_q).
We have shown that our new Dirac equation is very well-behaved. There is no Klein paradox in our equation.
Could it be possible to solve the hydrogen atom with the new equation?
Is the electron very close to the proton a tachyon, a positron, or a neutral light-speed particle?
UPDATE Nov 15, 2018: In our Nov 15, 2018 post we try to analyze the impact of a negative potential energy more carefully. The inertial mass may stay positive, after all.
…
Yet another possibility is that the electron changes into a positron close to the proton and sees a repulsive force from the proton.
If the electron is superluminal, some observers see it right-handed, some observers left-handed (= positron). We may think that it is in an intermediate stage of turning into a positron. In pair annihilation, the electron would turn back in time and be a positron.
Does the new Dirac equation cast light on this? What happens if m + V < 0 or when m + V is imaginary?
t
^
| 511 keV 511 keV photons
| ^ ^
| \ /
| \ ___/
| / \
| / \
| e- e+
--------------------------------> x
When an electron approaches a positron to annihilate, then at some point, m + V = 0 and p = E. The plane wave in that case is
(1, 1) exp(-i (E t - p x)).
That would be a good point to glue it to the positron plane wave, since the vector is (1, 1) for both particles?
But the electron has to emit photons. If the electron emits a 511 keV photon at that point, the plane wave after that might be something like
(1, 2p / (i 2p)) exp(-i (0 t - 2p x)).
The mass of the electron at that stage is i 2p and it moves at an infinite speed because E = 0 and the wave fronts in a spacetime diagram "move" horizontally.
The electron emits a 511 keV photon again and the wave is
(1, 1) exp(-i (-E t - p x)).
That is, the electron has become a positron. In the annihilation, the electron became a tachyon in the intermediate stage. If you need to turn back in time, it is not that strange to be a tachyon during one line segment.
If we flip the roles of t and x, the diagram above shows an electron in an elastic collision with a 511 keV photon.
Annihilation allows superluminal communication. That may be ok if we cannot measure the speed accurately enough to prove that superluminal communication took place. If the electron and the positron collide with a high energy, then the superluminal phase has to be shorter.
We may need a superluminal phase to make the wave function smooth in the spacetime diagram.
A basic principle of special relativity is that time and space are similar entities. Allowing superluminal travel makes them even more similar, because the path of a particle is not confined inside the light cone. Allowing traveling back in time still improves the similarity.
The diagram above suggests that at short distances, time and space become interchangeable, they have an "euclidian metric". But why at macroscopic scales we are bound inside the light cone and time and space most definitely cannot change places?
Maybe at short distances, the uncertainty of quantum mechanics allows particles to "tunnel" back in time. This might be the explanation for zitterbewegung. If we paint a fuzzy line of width one Compton wavelength to the the spacetime diagram above, it looks quite natural to allow the electron path zigzag not just in space but also in time. During the trips back in time it is a positron.
Thus, the electron is allowed to make superluminal trips as well as trips back in time as long as it stays inside the fuzzy path whose width is one Compton wavelength 2 * 10^-12 m or the corresponding time, 7 * 10^-21 s.
The annihilation case does not tell us what the electron does in the vicinity of a proton. The process might be an uneventful superluminal trip.
Note that in a head-on collision with a superluminal particle, the proton cannot exchange momentum with the particle. This is because the proton does not have time to react before the collision is already over. The potential of the proton will appear static to the colliding particle, that is, it stays in the same place as if the proton were infinitely heavy. If the superluminal particle passes from the side, then the proton will exchange momentum with it.
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