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Theorem 1. The correct energy-momentum relation under a potential is
E^2 = p^2 + (m + V)^2.
That is, the potential energy is added to the rest mass.
Proof. See our example at the end of our previous blog post, of a man living on a plane of negative charge. If the electron in his hand would not behave like an object with a rest mass m + V, then the conservation of the center of mass would be broken. If we assume that very basic conservation law of newtonian mechanics, the potential energy has to be added to the rest mass.
Also, the rest mass of the electron is actually potential energy that it acquired in pair creation. QED.
The Dirac equation which we introduced in the previous blog post should give reasonable results when a flux of electrons hits a step potential. For now, we omit the term V Ψ. We calculate the momentum after a potential step with the energy-momentum relation and do not use the wave equation in that.
Let us use the 1+1 dimensional equation. Let there be a step potential which is 0 if x < 0 and V if x > 0. Let a flux of electrons hit the step from the left. The rest mass of the electron is m.
A solution to the Dirac equation is
Ψ(p, m) = (1, p / (E + m)) exp(-i (E t - p x)).
Let us then solve the step potential exactly. We assume that the wave function at x < 0 is a sum
A Ψ(p, m) + B Ψ(-p, m),
where A and B are complex numbers. The incoming flux is A and the reflected flux is B.
The transmitted flux is
C Ψ(q, m + V),
where C is complex and q may be real or imaginary.
The energy-momentum relation for the incoming flux is
E^2 = p^2 + m^2,
and for the transmitted flux
E^2 = q^2 + (m + V)^2.
We may assume A = 1. The continuity of component 1 of the wave function at 0 requires:
1 + B = C,
and component 2:
p / (E + m) - B p / (E + m)
= C q / (E + m + V)
=>
1 - B = C q (E + m) / (p (E + m + V)).
Let us denote C = 1 + C'. We have:
1 + B = 1 + C'
1 - B = (1 + C') (q / p) (E + m) / (E + m + V).
Let us denote
1 - 2 R = (q / p) (E + m) / (E + m + V).
R comes from "Reflection". If V is small and positive, then R is small and positive.
We have B = C' and
1 - C' = (1 + C') (1 - 2 R)
=>
2 = (1 + C') (2 - 2 R)
=>
1 + C' = 1 / (1 - R)
=>
C' = R / (1 - R).
To summarize, the reflected flux is
B = R / (1 - R),
and the transmitted flux is
C =1 + R / (1 - R).
What if we require dΨ/dx to be continuous?
If the potential V(x) is continuous, then it might make sense to require our wave function to have a continuous derivative in space. Let us see what equations we get for A, B, C, where A = 1.
Component 1 gives:
p (1 - B) = C q.
Component 2 gives:
p^2 (1 + B) / (E + m)
= C q^2 / (E + m + dV),
where we assume that dV is close to zero. Let us again denote C = 1 + C'.
p (1 - B) = q (1 + C'),
(1 + B) p^2 / (E + m)
= (1 + C') q^2 / (E + m + dV).
Continuity of the wave function earlier produced equations:
1 + B = 1 + C',
1 - B = (1 + C')
* (q / p) (E + m) / (E + m + dV).
Thus, B = C'. Let us see if the equations yield anything sensible.p (1 - B) = C q.
Component 2 gives:
p^2 (1 + B) / (E + m)
= C q^2 / (E + m + dV),
where we assume that dV is close to zero. Let us again denote C = 1 + C'.
p (1 - B) = q (1 + C'),
(1 + B) p^2 / (E + m)
= (1 + C') q^2 / (E + m + dV).
Continuity of the wave function earlier produced equations:
1 + B = 1 + C',
1 - B = (1 + C')
* (q / p) (E + m) / (E + m + dV).
q = p (1 - B) / (1 + B),
q^2 = p^2 (1 - dV / (E + m)),
q = p (1 - B) / (1 + B).
In the two last equations we used the fact that dV is close to zero. The energy-momentum relation gives:
q^2 = p^2 + m^2 - (m + dV)^2
= p^2 (1 - 2m dV)
The energy-momentum relation, of course, gives a sensible result. On the other hand, the requirement that the derivative of the second component of the wave function be continuous produces a nonsensical result. If we used the old Dirac equation, then this continuity requirement would imply q = p, which does not make sense, either.
Destructive interference in wave equations
Suppose that the potential step is rectangular:
____
_____| |____
If the width of the step is suitable, then the reflections from each end of the rectangle will have a destructive interference. Then the reflected flux is zero, and transmission is 100%. This phenomenon is used in antireflective coatings of eyeglasses.
But how do we describe this in a linear wave equation? What mechanism causes the transmitted flux to increase if there is a destructive interference at the other end of the barrier? The next blog post concerns this fundamental problem.
The Klein paradox
We also need to study what happens when V goes to infinity. The energy-momentum relation gives then
q = i V.
A wide rectangular barrier will let through an infinitesimal flux because q is imaginary. That is a consequence of our new energy-momentum relation where we add the potential energy to the rest mass.
The physical setup of a discontinuous potential is not realistic. We should calculate the result with a steep continuous potential.
https://www.hep.phy.cam.ac.uk/theory/webber/GFT/gft_handout2_06.pdf
The Klein paradox for the Klein-Gordon equation matches the wave function value and the spatial derivative at the barrier borders.
Our new energy-momentum relation does remove the Klein paradox there, because the momentum inside barrier is imaginary, while it is real for the traditional energy-momentum relation.
Old "solutions" to the Klein paradox claimed that a pair is created at the other end of the wall and that a positron mysteriously finds the incoming electron and annihilates it. The positron is bound inside the strong field of the barrier. Why and how would it find an incoming electron?
Our solution says that the transmission is, indeed, very small. The Dirac equation and the Klein-Gordon equation do not describe pair creation. There should be no tunneling by pair creation if we solve the problem using those equations.
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