https://en.wikipedia.org/wiki/Fine_structure#Darwin_term
The Darwin term in the spectrum of hydrogen was derived by Charles Galton Darwin (1887 - 1962), who was a grandson of the famous Charles Darwin. The Darwin term is explained by the "smearing" which zitterbewegung causes in the location of the electron. In an s-orbital, the electron sometimes overlaps the proton in the center. If the electron behaves like a charge cloud of the size of the Compton wavelength, it cannot fall as low in the potential of the proton, as it could if it were a point particle. The energy level of the s orbital is therefore higher than one would expect from a point particle model. The energy difference for state 2s is 9 * 10^-5 eV.
https://en.wikipedia.org/wiki/Foldy–Wouthuysen_transformation
The Darwin term can be derived more rigorously (?) with a technique of Foldy and Wouthuysen.
A free spin-up electron is described in the Dirac equation as a 4-vector
1
0
c p_z / (E_p + mc^2)
c(p_x + ip_y) / (E_p + mc^2)
times
exp(-i (E t - p • x) / h-bar).
If we have a free electron, then we can build a wave packet from pure positive frequencies, and the packet will not exhibit zitterbewegung. But there is a strange thing: to build a packet of a "typical" waveform, we need to use also negative frequencies. As if the packet would contain some mixture of a positron.
The two upper components in the 4-vector are called the "major" components for an electron and the two lower "minor". But for a relativistic electron, the minor components are not minor at all! Maybe we should accept the fact that the thing which differentiates an electron from a positron is the sign of the frequency in time, not the components of the 4-vector.
Is there zitterbewegung in the hydrogen atom?
Zitterbewegung is a result of a mixture of positive and negative frequencies in a Dirac wave packet.
But, actually, are there any negative frequency components in an electron orbital? If we use a reduced mass for the proton, then the electron moves in a static potential. A static potential does not make a positive frequency wave a chirp after the electron has left the collision area. That is, no negative frequencies are produced if the electron escapes.
If there are negative frequencies, they represent a positron. Why the positron does not escape from the atom? The proton repels it.
The wave is spatially distorted. It looks like a chirp spatially. But that does not make it negative frequency. A chirp in time contains negative frequencies, but not a chirp in space. Though what about different coordinate systems? A spatial chirp does appear as a time chirp to an observer moving at a constant speed. But there is a preferred frame in the hydrogen atom, the center of mass.
If there are negative frequencies, they represent a positron. Why the positron does not escape from the atom? The proton repels it.
The wave is spatially distorted. It looks like a chirp spatially. But that does not make it negative frequency. A chirp in time contains negative frequencies, but not a chirp in space. Though what about different coordinate systems? A spatial chirp does appear as a time chirp to an observer moving at a constant speed. But there is a preferred frame in the hydrogen atom, the center of mass.
Since there are no apparent negative frequencies in the hydrogen atom, zitterbewegung may be a wrong explanation for the Darwin term.
The Foldy-Wouthuysen transformation
The idea in the transformation is to write the Dirac hamiltonian in terms of a new wave function "variable", so that the hamiltonian is diagonalized, that is, the 4 different components in the 4-vector, which are electron spin up/down, positron spin up/down, no longer interact in the hamiltonian.
Hsin-Chia Cheng has a field theory course online, with notes about the Foldy-Wouthuysen transformation:
http://cheng.physics.ucdavis.edu
Hsin-Chia Cheng has a field theory course online, with notes about the Foldy-Wouthuysen transformation:
http://cheng.physics.ucdavis.edu
The new wave function variable is related to the old through a unitary transformation exp(i S):
ψ' = exp(i S) ψ,
where S is hermitian. The transformation is not exact. Only a couple of leading terms are used.
where S is hermitian. The transformation is not exact. Only a couple of leading terms are used.
Once there is no interaction in the hamiltonian between the components of the 4-vector, we can set the major component 1 and all other components zero. The transformed hamiltonian H' turns out to be the Schrödinger hamiltonian plus some smallish terms. If we have a solution for the Schrödinger equation, we can perturb its wave function so that it is an approximate eigenfunction of H'. In that way we get the perturbed energy levels.
Cheng's manuscript says that the Darwin term is actually of the form
-e / (8 m^2) * ∇•E.
That is, it depends on the charge density ∇^2 Φ. It is not associated with the singular -1 / r potential of a pointlike charge.
It is not due to the electron charge being "smeared" over an area of size one Compton wavelength. But why does the wrong derivation from zitterbewegung yield the right answer?
What if we take more terms from the Foldy-Wouthuysen formula? Does the series still give the same Darwin term?
The operator p in the Dirac hamiltonian causes the time evolution of the wave function to mix the major and minor components in the 4-vector of the wave function. The transformation has to be chosen such that H' = exp(i S) H exp(-i S) does not cause such mixing in the time evolution of the modified wave function.
It is no wonder that the final H' contains a term ∇^2 Φ. Namely, the operator p applied twice to Φ ψ will produce such a term, among others. Cheng's manuscript contains the detailed calculation.
Let us calculate the electric potential at the surface of the proton.
V = k e / r
= 10^10 * 1.6 * 10^-19 / 10^-15
= 1.6 * 10^6 V.
The term Φe is there 3 * 10^-13 J.
Let us then calculate the Darwin term inside the proton.
h-bar^2 e / (8m^2 c^2 ε_0)
* ∇• E
= 10^-68 * 1.6 * 10^-19
/ (10^-59 * 10^17 * 10^-11)
* 1.6 * 10^-19 / (4/3 * 3.14 * 10^-45)
= 10^-8 J.
That is, the "perturbation" in the Darwin term is 30,000 times larger than the major term of the electric potential!
We cannot treat such a huge term as a perturbation. If we could "smear" the proton charge over a radius of 10^-13 m, then it would make sense as a perturbation. But there is no known mechanism to smear the charge of a proton.
https://www.amazon.com/Quantum-Field-Theory-Dover-Physics/dp/0486445682
The treatment of the Darwin term in the textbook of Itzykson and Zuber is essentially identical to the manuscript of Cheng. It contains no comment that the Darwin term cannot really be treated as a perturbation.
We know that treating the electron as a charge cloud of size one Compton wavelength reproduces the experimentally measured Darwin term. But neither the zitterbewegung derivation nor the Foldy-Wouthuysen derivation is mathematically sound. Maybe we need to develop a "regularization" scheme for the central potential well of the proton, to create a robust way of deriving the Darwin term?
A general observation about solutions of wave equations in a potential
We are phenomenally lucky to have an exact solution for the Schrödinger hydrogen atom. Generally, wave equations are too complex to have exact solutions at all. The Schrödinger helium atom does not have an exact solution. The waveform may be, in some sense, chaotic.
Some sources claim that Walter Gordon in 1928 found an exact solution for the Dirac hydrogen atom. But on the Internet there is no mention of Gordon deriving the Darwin term.
https://www.uni-due.de/~sfjooert/downloads/msc_thesis.pdf
The Master's thesis of Johannes Oertel (Freiberg 2014) lists only a few very simple cases where an exact solution for the Dirac equation has been found. Oertel finds several more exact solutions by solving the potential for a given wave function.
The charge singularities inside the proton
The proton contains two up quarks with the charge +2/3 and one down quark with the charge -1/3. There are three electric potential singularities in the proton. In the case of a singularity, it does not make sense to talk about charge density at all.
In the classical Schrödinger solution of the hydrogen atom, the electron moves faster than light if its distance is less than 6 * 10^-15 m from the proton. The classical solution obviously needs fixing.
The solution of the 1s orbital is
2 * exp(r / a_0) * exp(i E t).
A very simple formula. The potential singularity at r = 0 affects the solution very little.
The classical solution does not bother much about the infinitely deep potential well. But why would the relativistic Dirac equation essentially erase the potential well at distances less than 10^-12 m from the proton?
Our discussion about the vacuum atom and singularities in other blog posts suggests a solution to the Darwin term problem. A path integral solution to the hydrogen atom might have a characteristic energy of 511 keV. The paths have to be relatively straight. They cannot contain abrupt curves. This means that the paths do not see the attractive force of the proton at all when the electron is close to the proton.
The wavelength of the electron Dirac wave in the time direction in spacetime corresponds roughly to 511 keV. Suppose that the electron gains kinetic energy in a deep potential well. Then the wavelength in the space direction might correspond to much more than 511 keV. Then the wave would propagate superluminally because its wavefronts would be close to vertical in a spacetime diagram where time is vertical and space is horizontal.
t
^
|
|
------------------> x
Maybe we should use the rest mass plus the kinetic energy to calculate the wavelength in the time direction? But this is against the Schrödinger solution where the wave function is a stationary product of a function in space and exp(i E t).
If an electron moves to a lower potential which is in a wide spatial area, and gains kinetic energy, then we probably should use the rest mass plus the kinetic energy to determine the wavelength in the time direction. The electron Dirac wave would propagate within the light cone then.
The kinetic energy gained close to the proton is not "free energy" because the electron cannot give up that energy.
Can we find a reasonable principle which would entirely remove the potential well that the proton has withing the radius 1/2 * the Compton wavelength of an electron? If the electron is in the lowest energy state, then it has no free kinetic energy at all. Why would the area within 1/2 Compton wavelength be special?
Definition 1. The free kinetic energy of an electron is energy which it can give up without breaking the uncertainty principle and other rules of quantum mechanics. The electron is in an "excited state" and can move to a lower energy state.
Conjecture 2. In a path integral, the energy scale is the rest mass of the particle plus its free kinetic energy. The energy scale determines a cutoff of how sharply paths can bend in spacetime. The particle totally ignores potential wells which are too small to "see" at the energy scale. In the case of the hydrogen atom, the electron ignores the potential well whose size is less than the Compton wavelength of the electron. In the hydrogen atom, the electron does not have free kinetic energy at all if it is in the ground state.
In a hydrogen atom, the electron does not see features which are smaller than 2 * 10^-12 m, that is, the Compton wavelength. In a particle accelerator, we can give an electron a free kinetic energy of, say, 500 GeV. Such an electron will see features of size 2 * 10^-18 m.
In the Schrödinger equation, the electron does see even small features of the potential. Conjecture 2 claims that the electron does not see such small features at all. It will move past the miniature features on an essentially straight path, which is equivalent to not being affected by the potential at all.
The annihilation of an electron by a positron may break Conjecture 2. The electron must get within 10^-15 m of the positron, so that they can free 1.022 MeV of electromagnetic energy. Though, since the electron and the positron are actually the same particle, this cannot be directly compared to the hydrogen atom.
Conjecture 2 implies that we can think of the electron as a charge cloud of size one Compton wavelength. Its origin is in the uncertainty principle. Conjecture 2 says that we cannot circumvent the uncertainty principle by using non-free kinetic energy which the electron gains from the attraction of the proton.
Why is Conjecture 2 a consequence of applying special relativity to quantum mechanics? Abrupt curves in the path of a particle may involve short wavelengths in the spatial direction. If the wavelength in the time direction is determined by the rest mass plus the free kinetic energy, then those short spatial waves would move at a superluminal speed. That may be the fundamental reason why a relativistic particle must totally ignore potential wells which are smaller than its wavelength.
e- ---> small kinetic energy
_____ _____
| |
| |
|_| deep potential well
Let us think more about Conjecture 2. Suppose that we have a very deep step potential well which is also narrow. If the well is deep enough, the electron can drop into it and release energy. This requires that the wavelength of the electron inside the well is short enough.
A Dirac electron wave is
exp(-i (E t - p x)).
If we assume that a crest of a wave can only move within the light cone, then p cannot be larger than E. The question is what is E in various potentials? How does a relativistic particle behave in a box potential?
If the electron moves permanently from the potential 0 to a lower potential -V, then the kinetic energy of the electron grows by that amount V.
If we want to exhibit small features with a Dirac wave, the momentum p has to be large. But what if the potential well is not deep enough to give enough energy for a large p?
The big difference between the Schrödinger equation and the Dirac equation is that the former contains a second derivative with respect to space, d^2/dx^2, while the latter has the first derivative d/dx. If we try to fit a sine waves to the potential in the diagram above, then in the first approximation the Dirac wave will contain a discontinuity in the first spatial derivative, while the Schrödinger wave will have the discontinuity in the second derivative.
If we do a Fourier decomposition, it will probably contain more disturbance in the Dirac case. Does that imply that in the final fit, the Dirac wave should largely ignore a narrow potential well?
Schrödinger and Dirac waves meet a potential well
Let us try to calculate how each wave behaves.
i dΨ / dt = (-d^2/dx^2 + V) Ψ.
The Schrödinger case is well known. Part of the wave will reflect back at the discontinuities of the potential. The main wave has a shorter wavelength in the potential well. Its phase changes relative to a wave which did not fall into the potential well.
i dΨ / dt = (-d^2/dx^2 + V) Ψ.
The Schrödinger case is well known. Part of the wave will reflect back at the discontinuities of the potential. The main wave has a shorter wavelength in the potential well. Its phase changes relative to a wave which did not fall into the potential well.
How do we adjust a Dirac wave for a potential well?
i dΨ / dt = (α d/dx + β m + V) Ψ.
Suppose that we have a solution:
E^2 = (pc)^2 + (m_0 c^2)^2,
Suppose that we have a solution:
E^2 = (pc)^2 + (m_0 c^2)^2,
exp(-i (E t - p x)).
How do we adjust it if the wave meets a potential well? In the Schrödinger case, E is constant. If we find a solution for t = 0, then the time dependence is simply the product with exp(-i E t).
The Dirac case is harder. If we simply increase p inside the well, then the
exp(-i E t)
term will make the wave discontinuous at the temporal border of the potential well. If we glue together wave functions so that the wave is continuous in x at time t = 0, the wave will not be continuous at t = 1.
Let us consider the wave equation with two spatial dimensions and a time dimension for a while.
Something similar happens when water waves hit a shallow area in an otherwise open sea. The waves in the area move slower. The end result is a complex pattern of bent waves moving to various directions and interfering. If |V| is small, then maybe we should consider the term
VΨ
in the Dirac equation as a source term which generates new waves?
What if |V| is large, but only in a small area relative to the wavelength of Ψ? Then it is like scattering from a small object.
In scattering in classical mechanics, if |V| is relatively large in a small area, it will scatter particles coming to that area to an arbitrary angle. It does not matter much if we make |V| a lot larger still. Thus, the depth of the proton potential well does not matter much as long as it is deep.
The scattering argument is the same for the Schrödinger wave as well as for the Dirac wave. Why would the Dirac wave be blind to small detail in V?
In the Schrödinger case, we get a good approximate solution just by increasing p inside the well. The solution is then the original solution, except that the phase of the waves has advanced in those waves that went through the shallow area. Equivalently, we can convert part of the main wave to a wave which has advanced 90 degrees relative to the main wave. The source in the wave equation is VΨ inside the well, and the produced wave is the difference of the new advanced wave and the old wave.
The source term solution of a Schrödinger wave
Let us return back to 1 spatial dimension and look at the problem of a narrow well from the point of a source term. An incoming wave exp(-i (E t - p x)) hits the potential well.
In the Schrödinger case, we have the equation
d^2 Ψ/dx^2 + i dΨ/dt = V δ(0) exp(-i E t),
where on the right, |V| is small and it multiplies a standard plane wave solution of the equation. V is negative close to the origin; we have the Dirac delta there. E is the kinetic energy p^2 / (2m).
The term on the right is the source term. What kind of a wave does it produce? Everyday experience teaches us that if we exert a periodic force on a point in a string, then waves will start to propagate to the left and the right.
The solution should be something like
C exp(-i (E t - p x - π/2))
for x > 0, and with + p for x < 0. Let us denote the solution by Ψ_V.
The function is not differentiable at x = 0. We may imagine that at x = 0 we have a Dirac delta function as the second derivative.
Close to the origin, the second derivative dominates the term dΨ/dt. The first derivative is
C * i p * exp(-i (E t - p x)) * i,
where we have used the fact that exp(-i π/2) = i. The second derivative at x = 0 is
-2C p δ(0) exp(-i (E t - p x)).
____ ____ V
\/
/\ Ψ
We see that
C = -V / (2p)
gives us the solution. C has to be much smaller than 1 for the solution to make sense. Produced new waves cannot contain more energy than the incoming wave.
Let us analyze the sum Ψ + Ψ_V for x > 0. Looking at the behavior in time, se wee that Ψ_V rotates to the same phase as Ψ at a later time. But looking at x, Ψ_V attains the phase of Ψ at a smaller value of x.
If we look at the real component of the wave function, Ψ + Ψ_V attains the maximum value at a smaller x than Ψ. This is expected since the wavelength is smaller inside the potential well.
A Dirac wave in 1 + 1 dimensions
Suppose that we have a flux of electrons
Ψ = (1, p / (E + m)) * exp(-i (E t - p x)).
We denote a 2-component vector wave function by (Ψ_1, Ψ_2). We denote W = exp(-i (E t - p x)).
α * -i * dΨ/dx + β mΨ = i dΨ/dtis the 1-dimensional Dirac wave equation. The matrices
0 1
α =
1 0
1 0
β =
0 -1
have α^2 and β^2 equal to the identity matrix, and αβ + βα is the zero matrix. Let us write the two equations explicitly:
-i dΨ_2/dx + mΨ_1 = i dΨ_1/dt
-i dΨ_1/dx - mΨ_2 = i dΨ_2/dt.
Let us check that the above electron plane wave function satisfies the Dirac equation.
The first product in the Dirac equation with the given Ψ is
(p^2 / (E + m), p) * W.
The second is
(m, -m p / (E + m)) * W.
The right hand side of the equation is
(E, E p / (E + m)) * W.
From the first component we get the equation:
E = m + p^2 / (E + m)
<=>
E^2 - m^2 = p^2.
From the second we get:
p - mp / (E + m) = E p / (E + m)
<=>
E p + mp - mp = E p <=> 0 = 0.
The solution is E = +- sqrt(p^2 + m^2).
How to add a potential term to the Dirac equation?
How should we add a potential to the Dirac equation? E is the energy of a relativistic particle when we consider the potential V(x) zero.
In the Foldy-Wouthuysen calculation, a term V(x) Ψ is simply added to the Dirac equation to describe the electric potential.
If an electron permanently moves into a lower potential V, then we should obviously describe it as a free electron. In that case, the potential V should be ignored in the Dirac equation.
https://en.wikipedia.org/wiki/Klein_paradox
In the Klein paradox, the Wikipedia authors make a time-independent version of the massless (m = 0) Dirac equation by assuming that Ψ(t, x) = exp(-i E_0 t) * Ψ(x), where E_0 is constant.
Wikipedia then adds a term V(x) Ψ(x) to describe a potential barrier. The Klein paradox follows.
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