...
Our previous blog post showed that treating potential energy as rest mass, or rest mass as potential energy, makes sense. The Schrödinger equation becomes more logical when rest mass and the potential energy are identified. The hamiltonian operator H then includes all energy of the particle, also the energy in the rest mass.
This link contains an open-access .pdf file of Dirac's original 1928 paper:
http://rspa.royalsocietypublishing.org/content/117/778/610
The Dirac equation has been called the "square root" of the Klein-Gordon equation. The 1+1-dimensional Dirac equation which we treated in our blog post about zitterbewegung uses the second component of the wave function to store the current value of p / (E + m).
We know that if |p| is small, then the Schrödinger equation is the correct one.
Make Ψ_2 = dΨ_1/dx
Another approach to remove a second derivative is to introduce a new variable which contains the first spatial derivative of the wave function. We can make two differential equations which are formally first order, though, in reality, second order.
The two equations for the Schrödinger equation would be:
-i * dΨ_2/dx + (m + V) Ψ_1 = i dΨ_1/dt
-i * dΨ_1/dx - 2(m + V) Ψ_2 = 0.
The Dirac two equations:
-i * dΨ_2/dx + (m + V) Ψ_1 = i dΨ_1/dt
-i * dΨ_1/dx - (m + V) Ψ_2 = i dΨ_2/dt.
Let us try to write Schrödinger's two equations in a form which is Lorentz covariant:
(S1) -i * dΨ_2/dx + (m + V) Ψ_1
= i dΨ_1/dt
(S2) -i * dΨ_1/dx + (E - 2(m + V)) Ψ_2
= i dΨ_2/dt.
These are the same equations as Dirac's, but there is an extra E_kin * Ψ_2 in the lower equation. We denote E_kin = E - (m + V). E is the total energy of the particle.
If V gives a preferred coordinate system, like in the hydrogen atom, can we define that E is the energy in that coordinate system? Then E stays constant in the Lorentz transformation. If E is not constant, it may spoil Lorentz covariance.
If the two equations are symmetric in dx and dt, they maybe are Lorentz covariant, because we know that the similar Dirac equations are.
Lorentz transformation of the Dirac equations
Convention 1. Throughout this article we assume that v is small and that we can ignore the gamma factor of a Lorentz transformation, and also all terms of the form v^2.
Suppose that we have two wave functions Ψ_1 and Ψ_2 and there is no potential. An inertial observer A measures the function values at every spacetime point.
> tilt
^ t
|
|
------------> x ^ tilt
-------------- string 1
-------------- string 2
The wave functions may, for example, be wave heights in two strings which are stacked on top of each other. That is a very concrete example.
Let us have an observer B who is moving at a constant speed v relative to A.
If v is small, the coordinate system of B has the coordinate axis x' tilted upward the angle v relative to x and the axis t' is tilted right the angle v relative to t.
The Lorentz transformation is:
γ = 1 / sqrt(1 - v^2)
x' = γ (x - vt)
t' = γ (t - vx).
If v is small, we may assume γ = 1.
Let us solve the above for t' = 0, x' = 1. We get:
1 = x - vt
0 = t - vx
=>
x = 1
t = v,
when v is small. That is, the axes of B are tilted by the angle v but they are not contracted.
If v is small, the coordinate system of B has the coordinate axis x' tilted upward the angle v relative to x and the axis t' is tilted right the angle v relative to t.
The Lorentz transformation is:
γ = 1 / sqrt(1 - v^2)
x' = γ (x - vt)
t' = γ (t - vx).
If v is small, we may assume γ = 1.
Let us solve the above for t' = 0, x' = 1. We get:
1 = x - vt
0 = t - vx
=>
x = 1
t = v,
when v is small. That is, the axes of B are tilted by the angle v but they are not contracted.
If A has an equation on the directional derivatives of the wave functions along the axes of A, we may ask if the same equations hold for the directional derivatives along the axes of B. If they do, then the equations are Lorentz covariant.
Let us try to prove the Lorentz covariance of the Dirac equations in the case V = 0. Dirac equations according to observer A are schematically:
-x_2 + P1 = t_1
-x_1 - P2 = t_2.
B measures the directional derivatives in a different way if his coordinate axes are tilted by the small angle v:
-x_2 - v t_2 + P1 = t_1 + v x_1
-x_1 - v t_1 - P2 = t_2 + v x_2.
We should show that the two latter equations hold. We get:
-x_2 + P1 + v P2 = t_1
-x_1 - P2 - v P1 = t_2.
The equations are different from the first ones. The wave functions in the frame of B have to differ from the ones in the frame of A. This is ok if all observable physical quantities are the same.
The waves of the electron need to be Lorentz transformed along with t and x. This is like the electromagnetic field (E, B) which needs to be transformed when we move to new coordinates. If there were an electric potential in our equations, the potential should be transformed.
In our string example, it would not be ok to transform the waves because the displacement of a string is an observable quantity.
Let us study the Dirac covariance in more detail.
Hsin-Chia Cheng from UC Davis:
http://cheng.physics.ucdavis.edu
has a field theory course online. The link:
Let us try to prove the Lorentz covariance of the Dirac equations in the case V = 0. Dirac equations according to observer A are schematically:
-x_2 + P1 = t_1
-x_1 - P2 = t_2.
B measures the directional derivatives in a different way if his coordinate axes are tilted by the small angle v:
-x_2 - v t_2 + P1 = t_1 + v x_1
-x_1 - v t_1 - P2 = t_2 + v x_2.
We should show that the two latter equations hold. We get:
-x_2 + P1 + v P2 = t_1
-x_1 - P2 - v P1 = t_2.
The equations are different from the first ones. The wave functions in the frame of B have to differ from the ones in the frame of A. This is ok if all observable physical quantities are the same.
The waves of the electron need to be Lorentz transformed along with t and x. This is like the electromagnetic field (E, B) which needs to be transformed when we move to new coordinates. If there were an electric potential in our equations, the potential should be transformed.
In our string example, it would not be ok to transform the waves because the displacement of a string is an observable quantity.
Let us study the Dirac covariance in more detail.
Hsin-Chia Cheng from UC Davis:
http://cheng.physics.ucdavis.edu
has a field theory course online. The link:
http://cheng.physics.ucdavis.edu/teaching/230A-s07/rqm4_rev.pdf
contains a derivation of the transformation S for the wave function. For a boost in the x direction in the 1+3-dimensional case:
S = I cosh(η / 2) - α_1 sinh(η / 2),
where η = -v for small v. The Lorentz transformation matrix is
cosh(v) sinh(v)
sinh(v) cosh(v).
We can try our own α matrix for the α_1 in the transformation S:
0 1
1 0
Let us try a transformation:
P1' = P1 - v/2 P2
P2' = P2 - v/2 P1.
We need to calculate the directional derivatives again. We can ignore terms with v^2 in them. Schematically, we get:
-x_2 - v/2 t_2 + P1 - v/2 P2 = t_1 + v/2 x_1
-x_1 - v/2 t_1 - P2 - v/2 P1 = t_2 + v/2 x_2.
An easy manipulation shows that the above is equivalent to the untransformed equations. We showed that, assuming v^2 is very small and γ = 1, the Dirac equation in 1+1 dimensions has a natural-looking Lorentz transformation for the wave function.
We should still show that the transformed function is physically sensible. Let us check that the momentum of the electron in the transformed wave function makes sense.
The standard plane wave of an electron in the laboratory frame of A is
(1, p / (E + m)) * exp(-i (E t - p x)).
The frame of B is moving with a speed v. We denote the speed of the electron by v_e in the frame of A. Let us assume that p and v are positive and v is small:
v_e = p / E
v_e' = (v_e - v) / (1 - v v_e)
= (v_e - v) (1 + v v_e)
= v_e - v (1 - v_e^2)
v_e'^2 = v_e^2 - v * 2 (v_e - v_e^3)
E'^2 = v_e'^2 E'^2 + m^2
E'^2 = m^2 / (1 - v_e'^2)
= m^2 / (1 - v_e^2
+ v * 2(v_e - v_e^3))
= E^2 (1 - v * 2(v_e - v_e^3)
/ (1 - v_e^2))
= E^2 (1 - 2 v v_e)
E' = E - vp
p' = (E - vp) (p/E - v + v p^2/E^2)
= p - vE
We showed that energy and momentum transform like time and space.
Let us calculate first what is the standard plane wave of the electron in the frame of B:
(1, (p - vE) / (E - vp + m))
* exp(-i (Et' - vpt' - px'+ vEx')).
From
t' = t - vx
x' = x - vt
we get the exponent (we ignore v^2):
Et - vEx - vpt - px + vpt + vEx
= Et - px.
The exponent part is just like the wave of A.
The vector part is
(1, (p - vE) / (E - vp + m)).
t
^
|
|
--------------> x
e- e-8
Suppose that A sees a density of ρ of electrons moving at speed v_e. A sees an electric current of J = ρ v_e.
0
A sees successive electrons at coordinates (0, 0), (0, 1 / ρ). B will see them at coordinates (0, 0) and
(-v / ρ, 1 / ρ).
B sees the first electron still move for a distance v v_e / ρ before B thinks its time coordinate is zero. B sees after that the first electron at coordinates
(0, 1 / ρ + v v_e / ρ)
B sees an electron density
ρ' = ρ / (1 + v v_e) = ρ - v ρ v_e = ρ - v J.
A sees successive electrons at (0, 0) and (1 / (ρ v_e), 0). B sees them at (0, 0) and
(1 / (ρ v_e) - v / (ρ v_e)).
It takes the electron time v / (ρ v_e^2) to reach x' = 0. The current which B sees is
1 / (1 / (ρ v_e) + v / (ρ v_e^2))
= ρ v_e / (1 + v / v_e)
= J - v ρ
The conserved current (ρ, J) of a Dirac wave is
(Ψ† Ψ, Ψ† α_1 Ψ).
Note that in our notation, α_0 is the identity matrix. Our matrices have some signs flipped relative to the standard Dirac gamma matrices.
The current for the standard plane wave of observer A is
(1 + p^2 / (E + m)^2,
2 p / (E + m))
= (1 + b^2, 2b),
where we denote b = p / (E + m).
The current seen by B in his own standard plane wave is:
(1 + (p - vE)^2 / (E + m - vp)^2,
2 (p - vE) / (E + m - vp)).
The Lorentz transformation of the standard plane wave of A is
(1 - v / 2 b, b - v / 2) * exp(...).
Its current is
(1 - 2 v b + b^2,
2 b - v - v b^2).
The above is also the Lorentz transformation of the current of the standard plane wave of A. The values match, which means that the Dirac wave function transforms in a sensible way.
(S1) -i * dΨ_2/dx + (m + V) Ψ_1
= i dΨ_1/dt
(S2) -i * dΨ_1/dx + (E_kin - m - V) Ψ_2
= i dΨ_2/dt.
We will study if we can make a Lorentz transformation work with these equations.
Also, we need to study how to Lorentz transform the electric potential V. We conjecture that the rest mass m is electric potential energy, too. How to transform m?
Our two Schrödinger equations have the obvious error that for the standard plane wave solution they claim:
E = p^2 / (2m) + m
<=>
2 mE - m^2 = p^2 + m^2.
That is not the correct energy-momentum relation for large |p|:
E^2 = p^2 + m^2.
The electromagnetic four-potential (ϕ, A) is the correct Lorentz covariant way to treat a potential:
https://en.m.wikipedia.org/wiki/Electromagnetic_four-potential
If we identify the rest mass with the scalar potential, we get an energy-momentum relation
E^2 = (p + A)^2 + ϕ^2,
where ϕ includes the rest mass m. The first term in the right side is the energy of the magnetic field of the electron and the second term is the energy of the electric field of the electron.
The Lorentz transformation for the potentials is:
ϕ = ϕ - v A
A = A - v ϕ.
In 1+1 dimensions we cannot really treat the rest mass as potential energy because the electric force in one spatial dimension is constant. We need to move to 1+3 dimensions.
Actually, if rest mass is electric potential energy, then it is doubtful that we can describe a single electron in a Lorentz covariant way. Suppose that we have a universe with just one electron and one positron. The rest mass of the electron is potential energy in the electric field of the positron. The positron sets a preferred frame of reference. Lorentz covariance bans preferred frames.
The system the electron & the positron is Lorentz covariant, but individually, the particles are not.
A correct Lorentz transformation must act on all particles in the system.
A big question is that since the electron can be annihilated by any positron, what particles should we consider when we describe an electron? If we have two positrons flying to opposite directions, what is the preferred frame where we should describe the electron?
The energy-momentum relation for the static particle is
E^2 = ϕ^2 = m^2,
where m denotes what we traditionally regard as the rest mass.
Let the electron move to the positive x direction at a slow speed v_e. The magnetic fields cause an additional attraction between the particles.
contains a derivation of the transformation S for the wave function. For a boost in the x direction in the 1+3-dimensional case:
S = I cosh(η / 2) - α_1 sinh(η / 2),
where η = -v for small v. The Lorentz transformation matrix is
cosh(v) sinh(v)
sinh(v) cosh(v).
We can try our own α matrix for the α_1 in the transformation S:
0 1
1 0
Let us try a transformation:
P1' = P1 - v/2 P2
P2' = P2 - v/2 P1.
We need to calculate the directional derivatives again. We can ignore terms with v^2 in them. Schematically, we get:
-x_2 - v/2 t_2 + P1 - v/2 P2 = t_1 + v/2 x_1
-x_1 - v/2 t_1 - P2 - v/2 P1 = t_2 + v/2 x_2.
An easy manipulation shows that the above is equivalent to the untransformed equations. We showed that, assuming v^2 is very small and γ = 1, the Dirac equation in 1+1 dimensions has a natural-looking Lorentz transformation for the wave function.
We should still show that the transformed function is physically sensible. Let us check that the momentum of the electron in the transformed wave function makes sense.
The standard plane wave of an electron in the laboratory frame of A is
(1, p / (E + m)) * exp(-i (E t - p x)).
The frame of B is moving with a speed v. We denote the speed of the electron by v_e in the frame of A. Let us assume that p and v are positive and v is small:
v_e = p / E
v_e' = (v_e - v) / (1 - v v_e)
= (v_e - v) (1 + v v_e)
= v_e - v (1 - v_e^2)
v_e'^2 = v_e^2 - v * 2 (v_e - v_e^3)
E'^2 = v_e'^2 E'^2 + m^2
E'^2 = m^2 / (1 - v_e'^2)
= m^2 / (1 - v_e^2
+ v * 2(v_e - v_e^3))
= E^2 (1 - v * 2(v_e - v_e^3)
/ (1 - v_e^2))
= E^2 (1 - 2 v v_e)
E' = E - vp
p' = (E - vp) (p/E - v + v p^2/E^2)
= p - vE
We showed that energy and momentum transform like time and space.
Let us calculate first what is the standard plane wave of the electron in the frame of B:
(1, (p - vE) / (E - vp + m))
* exp(-i (Et' - vpt' - px'+ vEx')).
From
t' = t - vx
x' = x - vt
we get the exponent (we ignore v^2):
Et - vEx - vpt - px + vpt + vEx
= Et - px.
The exponent part is just like the wave of A.
The vector part is
(1, (p - vE) / (E - vp + m)).
The conserved current
t
^
|
|
--------------> x
e- e-8
Suppose that A sees a density of ρ of electrons moving at speed v_e. A sees an electric current of J = ρ v_e.
0
A sees successive electrons at coordinates (0, 0), (0, 1 / ρ). B will see them at coordinates (0, 0) and
(-v / ρ, 1 / ρ).
B sees the first electron still move for a distance v v_e / ρ before B thinks its time coordinate is zero. B sees after that the first electron at coordinates
(0, 1 / ρ + v v_e / ρ)
B sees an electron density
ρ' = ρ / (1 + v v_e) = ρ - v ρ v_e = ρ - v J.
A sees successive electrons at (0, 0) and (1 / (ρ v_e), 0). B sees them at (0, 0) and
(1 / (ρ v_e) - v / (ρ v_e)).
It takes the electron time v / (ρ v_e^2) to reach x' = 0. The current which B sees is
1 / (1 / (ρ v_e) + v / (ρ v_e^2))
= ρ v_e / (1 + v / v_e)
= J - v ρ
The conserved current (ρ, J) of a Dirac wave is
(Ψ† Ψ, Ψ† α_1 Ψ).
Note that in our notation, α_0 is the identity matrix. Our matrices have some signs flipped relative to the standard Dirac gamma matrices.
The current for the standard plane wave of observer A is
(1 + p^2 / (E + m)^2,
2 p / (E + m))
= (1 + b^2, 2b),
where we denote b = p / (E + m).
The current seen by B in his own standard plane wave is:
(1 + (p - vE)^2 / (E + m - vp)^2,
2 (p - vE) / (E + m - vp)).
The Lorentz transformation of the standard plane wave of A is
(1 - v / 2 b, b - v / 2) * exp(...).
Its current is
(1 - 2 v b + b^2,
2 b - v - v b^2).
The above is also the Lorentz transformation of the current of the standard plane wave of A. The values match, which means that the Dirac wave function transforms in a sensible way.
Can we prove the Lorentz covariance of the Schrödinger equations?
We introduced the two Schrödinger equations which only differ from the Dirac equations by the term E_kin in the lower equation.
(S1) -i * dΨ_2/dx + (m + V) Ψ_1
= i dΨ_1/dt
(S2) -i * dΨ_1/dx + (E_kin - m - V) Ψ_2
= i dΨ_2/dt.
We will study if we can make a Lorentz transformation work with these equations.
Also, we need to study how to Lorentz transform the electric potential V. We conjecture that the rest mass m is electric potential energy, too. How to transform m?
Our two Schrödinger equations have the obvious error that for the standard plane wave solution they claim:
E = p^2 / (2m) + m
<=>
2 mE - m^2 = p^2 + m^2.
That is not the correct energy-momentum relation for large |p|:
E^2 = p^2 + m^2.
The electromagnetic four-potential (ϕ, A) is the correct Lorentz covariant way to treat a potential:
https://en.m.wikipedia.org/wiki/Electromagnetic_four-potential
If we identify the rest mass with the scalar potential, we get an energy-momentum relation
E^2 = (p + A)^2 + ϕ^2,
where ϕ includes the rest mass m. The first term in the right side is the energy of the magnetic field of the electron and the second term is the energy of the electric field of the electron.
The Lorentz transformation for the potentials is:
ϕ = ϕ - v A
A = A - v ϕ.
In 1+1 dimensions we cannot really treat the rest mass as potential energy because the electric force in one spatial dimension is constant. We need to move to 1+3 dimensions.
Actually, if rest mass is electric potential energy, then it is doubtful that we can describe a single electron in a Lorentz covariant way. Suppose that we have a universe with just one electron and one positron. The rest mass of the electron is potential energy in the electric field of the positron. The positron sets a preferred frame of reference. Lorentz covariance bans preferred frames.
The system the electron & the positron is Lorentz covariant, but individually, the particles are not.
A correct Lorentz transformation must act on all particles in the system.
A big question is that since the electron can be annihilated by any positron, what particles should we consider when we describe an electron? If we have two positrons flying to opposite directions, what is the preferred frame where we should describe the electron?
Energy-momentum relation for an electron-positron pair
We need to understand the Lorentz transformation properly before we can do quantum mechanics. Let us consider an electron-positron pair which is static in the laboratory frame symmetrically on the y axis.
^ x
|
|
|
----------------------------> y
e+ e+
When the distance between the particles was just 10^-15 m, the total energy of the system was zero. The whole rest mass of the particles can be defined to be potential energy in their mutual electric attraction.
The energy-momentum relation for the static particle is
E^2 = ϕ^2 = m^2,
where m denotes what we traditionally regard as the rest mass.
Let the electron move to the positive x direction at a slow speed v_e. The magnetic fields cause an additional attraction between the particles.
The energy-momentum relation, obtained through a Lorentz transformation with v_e, for the electron is
E^2 = (v_e E + v_e ϕ)^2 + ϕ^2
= 4 (m v_e)^2 + ϕ^2
= 4 p^2 + ϕ^2.
That is, the inertial mass seems to be twice the rest mass of the electron. That is strange. Could this have something to do with the fact that the magnetic field generated by an electric current tends to keep the current going? It seems to store momentum which it will give back to the current if the current decreases.
Let us think. The potential energy of the system the electron & the positron is 1.022 MeV. We showed in an earlier blog post that the division of the inertial mass on the parts of the system depends on the configuration. In our example, all inertial mass was on the electron.
If the positron is at a potential which makes its total energy zero, then the whole inertial mass of 1.022 MeV is on the electron.
The Lorentz transformation might not be the right tool.
= 4 (m v_e)^2 + ϕ^2
= 4 p^2 + ϕ^2.
That is, the inertial mass seems to be twice the rest mass of the electron. That is strange. Could this have something to do with the fact that the magnetic field generated by an electric current tends to keep the current going? It seems to store momentum which it will give back to the current if the current decreases.
Let us think. The potential energy of the system the electron & the positron is 1.022 MeV. We showed in an earlier blog post that the division of the inertial mass on the parts of the system depends on the configuration. In our example, all inertial mass was on the electron.
If the positron is at a potential which makes its total energy zero, then the whole inertial mass of 1.022 MeV is on the electron.
The Lorentz transformation might not be the right tool.
The new positron equations
Dirac's astonishing find "explained" why the electron has a spin, and why the positron exists.
The Dirac equations in our 1+1 case do not have a spin, but they still predict the existence of the positron as the negative energy solution.
Do our new Schrödinger equations have negative energy solutions? The basic positive energy solution of a free particle is
Ψ = (1, p / (2m)) * exp(-i (E t - p x)).
From (S1) and (S2) we get:
p^2/(2m) + m = E
p - 2m * p / (2m) = 0.
If we assume that m is negative, then E is negative. That is probably the positron equation.
According to Feynman, a positron is a "negative energy electron going backward in time". But in real life, a positron certainly is a positive energy particle going forward in time. The idea that a positron is actually an electron scattered back in time is beautiful. In annihilation, the wave function of the electron might continue (after a short superluminal trip) as the wave function of the positron. The wave of the electron rotates clockwise with time. Then the wave of the positron should rotate anticlockwise with time.
The relativistic positron equations might be (S1) and (S2) where we have inverted the sign of (m + V).
t _ _ superluminal section
^ / \
| / \
| e- e+
-----------------> x
In the diagram, the electron is a wave which rotates clockwise with time and anticlockwise with space:
Ψ_e- = (1, p / (2m)) * exp(-i (E t - p x))
When viewed visually, the spatial wave moves right with increasing time.
It annihilates with the positron whose wave function is
Ψ_e+ = (1, p / (2m)) * exp(-i (-E t - p x)).
The wave rotates in the positron anticlockwise with time, and anticlockwise with space. When viewed visually, the spatial wave moves left with increasing time.
We have conjectured in our earlier blog posts that there is a short superluminal section in the path when the potential energy between the particles is < -1.022 MeV. The energy-momentum relation
E^2 = p^2 + (m + V)^2
would have the sum m + V imaginary if p^2 > E^2. An imaginary mass corresponds to a tachyon, and a tachyon has to move superluminally.
Ψ = (1, p / (2m)) * exp(-i (E t - p x)).
From (S1) and (S2) we get:
p^2/(2m) + m = E
p - 2m * p / (2m) = 0.
If we assume that m is negative, then E is negative. That is probably the positron equation.
According to Feynman, a positron is a "negative energy electron going backward in time". But in real life, a positron certainly is a positive energy particle going forward in time. The idea that a positron is actually an electron scattered back in time is beautiful. In annihilation, the wave function of the electron might continue (after a short superluminal trip) as the wave function of the positron. The wave of the electron rotates clockwise with time. Then the wave of the positron should rotate anticlockwise with time.
The relativistic positron equations might be (S1) and (S2) where we have inverted the sign of (m + V).
t _ _ superluminal section
^ / \
| / \
| e- e+
-----------------> x
In the diagram, the electron is a wave which rotates clockwise with time and anticlockwise with space:
Ψ_e- = (1, p / (2m)) * exp(-i (E t - p x))
When viewed visually, the spatial wave moves right with increasing time.
It annihilates with the positron whose wave function is
Ψ_e+ = (1, p / (2m)) * exp(-i (-E t - p x)).
The wave rotates in the positron anticlockwise with time, and anticlockwise with space. When viewed visually, the spatial wave moves left with increasing time.
We have conjectured in our earlier blog posts that there is a short superluminal section in the path when the potential energy between the particles is < -1.022 MeV. The energy-momentum relation
E^2 = p^2 + (m + V)^2
would have the sum m + V imaginary if p^2 > E^2. An imaginary mass corresponds to a tachyon, and a tachyon has to move superluminally.
Relativistic corrections in the hydrogen atom in the non-relativistic Schrödinger equation
The energy-momentum relation without a potential is
E^2 = p^2 + m^2.
What is the kinetic energy if p is smallish?
E_kin = sqrt(p^2 + m^2) - m
= m sqrt(p^2/m^2 + 1) - m
= 1/2 * p^2/m - 1/8 * p^4/m^3.
The term -1/8... is the leading relativistic correction. It reduces 1s by 9 * 10^-4 eV.
Correcting the mass for m + V should raise 1s. Is this the real reason for the Darwin term?
A very brief ballpark calculation gives that the correction raises the hydrogen energy levels by 1 meV. How precisely do we know the absolute energy -13.6 eV of the 1s-orbital? The Rydberg constant has a value with 11 decimals, but what actual measurements have shown?
Since s-orbitals have their probability amplitude largest at the proton, their energy is raised by some 10^-4 eV. This is the same order of magnitude as the Darwin term.
The term -1/8... is the leading relativistic correction. It reduces 1s by 9 * 10^-4 eV.
Correcting the mass for m + V should raise 1s. Is this the real reason for the Darwin term?
A very brief ballpark calculation gives that the correction raises the hydrogen energy levels by 1 meV. How precisely do we know the absolute energy -13.6 eV of the 1s-orbital? The Rydberg constant has a value with 11 decimals, but what actual measurements have shown?
Since s-orbitals have their probability amplitude largest at the proton, their energy is raised by some 10^-4 eV. This is the same order of magnitude as the Darwin term.
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