The problem is a fundamental one and deserves a blog post of its own: how to add a potential term to the Dirac equation?
http://rspa.royalsocietypublishing.org/content/117/778/610
Dirac himself, in his famous 1928 article, added the potential energy of the electron as a simple term V Ψ to his equation. His choice is, at least in some cases, equivalent to subtracting the potential energy from the total energy of the electron. But that is a strange choice because the potential energy "moves" along with the electron, and should be added to the rest mass of the electron.
Intuitively, an electron under a constant potential should act like a free electron.
How is the matter in the Schrödinger equation? A particle which moves to a lower potential zone gains kinetic energy and acts like a free particle with more kinetic energy. Since the Schrödinger equation is nonrelativistic, we do not need to add the (small) potential energy to the rest mass of the particle.
In the Dirac equation, we may have something like
dE / dx = -dV(x) / dx,
d/dx sqrt(p(x)^2 + m^2) = -dV(x) / dx,
2 p(x) dp(x) / dx
* 1/2 * 1/sqrt(...) = -dV(x) / dx.
We get a complicated formula for d^2 Ψ / dx^2, that is, dp(x) / dx.
dp(x) / dx = -dV(x) / dx
* sqrt(p(x)^2 + m^2)
/ p(x).
If m = 0, we get simply dp(x) / dx = -dV(x) / dx, which implies p(x) = -V(x) + C.
The massless case m = 0
The Dirac equation with m = 0 is
α * -i dΨ/dx = i dΨ/dt,
or in the component form:
-i dΨ_2/dx = i dΨ_1/dt
-i dΨ_1/dx = i dΨ_2/dt.
Let us try a solution where Ψ_1 = exp(-i (E t - p x))
dΨ_2/dx = i E exp(-i (E t - p x))
-i p exp(-i (E t - p x)) = dΨ_2/dt.
We get
Ψ_2 = i E / (i p) exp(-i (E t - p x))
Ψ_2 = i p / (i E) exp(-i (E t - p x)).
This implies E^2 = p^2 => E = +-p.
Thus, the solution is:
Ψ_1 = Ψ_2 = exp(-i E (t - x))
or
Ψ_1 = exp(-i E (t + x))
Ψ_2 = -exp(-i E (t + x)).
For a light-speed particle, E = |p|, assuming c = 1.
https://en.wikipedia.org/wiki/Klein_paradox
The Wikipedia article uses these two solutions to calculate the reflection and the transmission at a potential step. In the massless case, the equation
α p Ψ + VΨ = i dΨ/dt
does yield a somewhat reasonable solution, but it produces the Klein paradox.
Should we rotate the phase of the wave function?
Suppose that a single electron approaches a step potential at x = x_0. If we represent the electron as a wave of a type
exp(-i (E t - p x)),
there is no way we can make a continuous wave function by gluing together that wave with
exp(-i (E' t - p' x))
at x_0. Should we allow a discontinuous wave function? Or is there a total reflection of the wave?
exp(-i p x),
which we can easily glue together with a function
exp(-i p' x)
at x_0, and we obtain a continuous wave function. But does this make sense? Let us consider a more realistic setup.
A path integral approach
e- ----->
\ |
\ |
proton O x
/ |
e- ------>
screen
The 1+1-dimensional case is too restricted. Let us look at a 1+2-dimensional case. Suppose that there is a static 1 / r potential of a proton. A flux of electrons hits it and forms an interference pattern on a screen. We should calculate the interference pattern.
If the flux consists of electrons of a fixed momentum p, then there are just 2 paths through which an electron can end up at a point x on the screen. We can calculate the phase of each of these 2 paths and their interference.
From this case we see that we are not allowed to rotate the phase of the incoming flux arbitrarily at different spacetime points. It would spoil the interference pattern.
Does the above diagram give us a clue how to add a potential to the Dirac equation?
The two paths to the point x are classical in the sense that we know the initial momentum p of the electron exactly and we can measure its position at x with an arbitrary precision. It is like a classical particle with a precisely defined rotating phase hitting the screen. Is there any sense in trying to derive a differential equation which describes the phase at x? The phase is a function of two classical paths. It does not matter what happens outside those paths, as long as no new paths to x appear.
Wave equations describe a smooth process where what happens at x is determined in a smooth way from what happens in its surroundings.
We may imagine that the flux which hits the proton is just a single electron, whose momentum we know exactly, but whose position has a huge uncertainty. Then it is obvious that the different paths of the electron are independent. The paths do not interact in any way. This suggests that there is really no sense in describing the process as a wave equation. Why does the Dirac equation work then?
A potential step
Let us consider in 1+1 dimensions the case where a flux of electrons of momentum p hits a low potential step at x =0, of height V.
^ t X
| | /
| | /
|/
/|
/ |
/ |
/ |
e-
0 ----> x
step
The flux of free electrons, or a single free electron whose position we do not know, is described by the 2-vector
(1, p / (E + m)) * exp(-i (E t - p x)),
where E^2 = p^2 + m^2. How does the electron behave once it has crossed the potential step? Its kinetic energy decreases by V. A crucial question is what happens to E?
Only one path leads to X. If we translate the path upward by Δt, the phase of the start of the path advances by E Δt. The end of the path is a spacetime point X'. How much does the phase advance in X' relative to X?
If we assume that the phase of the electron rotates at a fixed speed per distance after crossing 0, then the phase in X' relative to X has advanced the same E Δt as at the start of the path.
Even though p^2 + m^2 is reduced after the potential step, the potential energy V seems to make for it. Thus, we may write:
E = sqrt(p^2 + m^2) + V.
What is the phase of the electron with respect to x, if x > 0? In the Schrödinger equation, it is determined by p x, where p is the momentum. We may guess that the same holds for the Dirac case.
The phase in the area x > 0 would then be:
exp(-i (E t - p x)),
where
p^2 = (E - V)^2 - m^2
= E^2 - m^2 - 2VE + V^2.
p^2 = (E - V)^2 - m^2
= E^2 - m^2 - 2VE + V^2.
Let us try to add the potential term to the Dirac equation in the same way as in the Schrödinger equation:
α * -i dΨ/dx + β m Ψ + VΨ = i dΨ/dt.
Let us try a solution exp(-i (E t - p x)). From our zitterbewegung blog post, we get the formula
p^2 / (E + m) + m + V = E,
p^2 + m^2 + VE + Vm = E^2,
p^2 = E^2 - m^2 - VE + Vm.
The formula does not look right. The relation between E, m, p, V is complicated.
Has anyone found a natural way to add V to the Dirac equation? Maybe we should look at the paper by Oskar Klein from 1929, which introduces the Klein paradox.
/|
/\
=======================
- - - - - - - - - - - - - -
x y
Suppose that we have a fixed planar negative charge. An inhabitant living on the plane pulls an electron close to the plane. The electron acquires a potential energy V. Does it then behave like a particle whose rest mass is m + V, where m is the rest mass of the electron in empty space?
In the pulling, the inhabitant can use an energy store of size V at location x. The inhabitant carries the electron to location y and fills another energy store there with energy V by letting the electron move away.
In the process, energy m + V moved from x to y.
The conservation of the center of mass requires that the electron has to behave just like a particle with a rest mass of m + V. Otherwise the center of mass of the system the plane & the electron would change.
This suggests that we should define
m(x) = m + V(x).
The Dirac equation would then be
α * -i dΨ/dx + β (m + V(x)) Ψ = i dΨ/dt.
That would make a lot of sense, because if we move the electron under a potential V > 0, we move an inertial mass which would imply a higher rest mass.
We dropped the term V(x) Ψ because the purpose of that term was to subtract the potential energy from the total energy. We add the potential energy to the rest mass, and the total energy remains the same.
The rest mass of the electron itself can be considered as potential energy which it gains when it in pair production is separated from the positron.
But what if the electron descends into a negative potential whose absolute value exceeds its rest mass? If the electron would be able to give up its kinetic energy in that case, there would be a breach of energy conservation.
In the plane example, the inhabitant can let a positron approach the plane and fill an energy store V at location x. We assume V > m. Then he carries the positron to y where he uses an energy store V to push the positron away. Energy V moved y -> x and energy m moved x -> y. The positron should have a negative rest mass in this process, which would be an utterly strange experience for the inhabitant.
The energy-momentum relation suggests that the mass becomes imaginary if the kinetic energy exceeds the rest mass of a particle. The particle is then a tachyon.
α * -i dΨ/dx + β m Ψ + VΨ = i dΨ/dt.
Let us try a solution exp(-i (E t - p x)). From our zitterbewegung blog post, we get the formula
p^2 / (E + m) + m + V = E,
p^2 + m^2 + VE + Vm = E^2,
p^2 = E^2 - m^2 - VE + Vm.
The formula does not look right. The relation between E, m, p, V is complicated.
Has anyone found a natural way to add V to the Dirac equation? Maybe we should look at the paper by Oskar Klein from 1929, which introduces the Klein paradox.
Add the potential energy to the rest mass of the electron?
e-
O//|
/\
=======================
- - - - - - - - - - - - - -
x y
Suppose that we have a fixed planar negative charge. An inhabitant living on the plane pulls an electron close to the plane. The electron acquires a potential energy V. Does it then behave like a particle whose rest mass is m + V, where m is the rest mass of the electron in empty space?
In the pulling, the inhabitant can use an energy store of size V at location x. The inhabitant carries the electron to location y and fills another energy store there with energy V by letting the electron move away.
In the process, energy m + V moved from x to y.
The conservation of the center of mass requires that the electron has to behave just like a particle with a rest mass of m + V. Otherwise the center of mass of the system the plane & the electron would change.
This suggests that we should define
m(x) = m + V(x).
The Dirac equation would then be
α * -i dΨ/dx + β (m + V(x)) Ψ = i dΨ/dt.
That would make a lot of sense, because if we move the electron under a potential V > 0, we move an inertial mass which would imply a higher rest mass.
We dropped the term V(x) Ψ because the purpose of that term was to subtract the potential energy from the total energy. We add the potential energy to the rest mass, and the total energy remains the same.
The rest mass of the electron itself can be considered as potential energy which it gains when it in pair production is separated from the positron.
But what if the electron descends into a negative potential whose absolute value exceeds its rest mass? If the electron would be able to give up its kinetic energy in that case, there would be a breach of energy conservation.
In the plane example, the inhabitant can let a positron approach the plane and fill an energy store V at location x. We assume V > m. Then he carries the positron to y where he uses an energy store V to push the positron away. Energy V moved y -> x and energy m moved x -> y. The positron should have a negative rest mass in this process, which would be an utterly strange experience for the inhabitant.
The energy-momentum relation suggests that the mass becomes imaginary if the kinetic energy exceeds the rest mass of a particle. The particle is then a tachyon.
No comments:
Post a Comment