The Dirac equation has been called the "square root" of the Klein-Gordon equation. If we take this seriously, then the electron and the positron are "square roots" of a Klein-Gordon particle.
e+ <-------> e-
The electron and the positron are always born together. Maybe we should always treat them together, as a combination of two particles. The wave function of the combined particle might be the product of individual wave functions.
We had the classical model where the electron at the speed of light makes a circle whose radius is 2 * 10^-13 m. That would give the right spin angular momentum.
The strange thing was that the plane wave representation of the electron only completes half a cycle in one round. Why there is no destructive interference which would spoil the wave? The reason may be that the complete system includes also the positron. The product of the electron and the positron waves does complete a full cycle when the electron makes a single circle. There is a constructive interference when we look at the whole system electron & positron.
In quantum mechanics, there is no interference of the wave of a single particle if enough information of the wave is copied somewhere else. The cardinal example is the double slit experiment.
When the electron rotates around its axis, we may imagine that information of the rotation is present in the positron, too.
When an electron and a positron are born, we may think that their spin rotation is entangled in some sense. We cannot treat them as separate particles when it comes to the rotation.
The positron may annihilate with another electron B, but then the information about the rotation might be preserved in the created photons, or alternatively, in some way in the positron which is the pair of electron B.
The photon has some similarity to the Klein-Gordon equation. Maybe the electron and the positron are square roots of the photon in some sense?
Does this new idea cast light on the question in which way the positron is an electron traveling back in time?
Taking the square root of the Klein-Gordon equation
Let us try to figure out in which sense the solutions of the Dirac equation are square roots of solutions for the Klein-Gordon equation. We work in 1+1 dimensions. Paul Dirac in his book Principles of quantum mechanics (1930) writes a "relativistic hamiltonian":
(p_0^2 - p_1^2 - m^2) Ψ = 0,
where p_n = i d / dx_n is the momentum operator. It is the Klein-Gordon equation.
Dirac realized that he can obtain an equation with just first derivatives by taking a "square root" of the operator:
(p_0 - α_1 p_1 - β m) Ψ = 0,
where we can choose
0 1
α_1 =
1 0
1 0
β =
0 -1
By choosing α_1 and β appropriately, the square of the operator above is equal to the operator in the upper equation. Suppose that we find a solution Ψ for the lower equation. It is trivially a solution of the upper equation, too. What about Ψ^2? Under what condition it is a solution of the upper equation?
Let us denote t = x_0 and x = x_1. The standard plane wave solution of the Dirac electron is
Ψ_1 = (1, p / (E + m)) * exp(-i (E t - p x)),
where E^2 = p^2 + m^2. Let a positron solution be
Ψ_2 = (-p / (E + m), 1) * exp(-i (-E t - p x)).
The squares Ψ_1^2, Ψ_2^2, as well as
(Ψ_1 + Ψ_2)^2 = Ψ_1^2 + Ψ_2^2
are solutions for the Klein-Gordon equation for mass 2m. The electron and the positron are orthogonal solutions of the Dirac equation.
The gyromagnetic ratio 2
The Dirac equation apparently implies that the gyromagnetic ratio g of the spinning electron is 2. That is, if we imagine the electron as a classical particle making a loop at a slow speed, its magnetic moment relative to its angular momentum is 2X of what we would expect from a classical particle.
But if the electron moves at a fast speed, then the Lorentz transformation of its electric field has the fields multiplied by γ = 1 / sqrt(1 - v^2 / c^2). If the electron would move at the speed sqrt(3) / 2 * c, then the magnetic field would be double of what the nonrelativistic formula gives.
The electron magnetic moment from the Dirac equation
The link above contains a derivation which compares the Dirac equation under an electromagnetic field to a hypothetical Klein-Gordon equation under an electromagnetic field.
The electromagnetic field potential is incorporated into a free field wave equation by converting the partial derivatives to a "gauge covariant form":
D_μ = ꝺ_μ - i e A_μ.
For the free field equation, the "square" of the operator of the Dirac equation
(i D-slash - m) * (-i D-slash - m)
is the Klein-Gordon equation operator
D^2 + m^2.
But if the electromagnetic field is non-zero, then off-diagonal elements appear in the square of the Dirac operator:
e F_μν S^μν.
The Klein-Gordon equation is a generalization of a 1+1-dimensional wave equation. The discussion above raises the question if a better generalization of the wave equation to 1+3 dimension should contain all second order derivatives and not just the diagonal derivatives d^2/dx^2. Then the new Klein-Gordon equation might be equivalent to the Dirac equation also in the case where an electromagnetic field is present.
The Klein-Gordon equation is a generalization of a 1+1-dimensional wave equation. The discussion above raises the question if a better generalization of the wave equation to 1+3 dimension should contain all second order derivatives and not just the diagonal derivatives d^2/dx^2. Then the new Klein-Gordon equation might be equivalent to the Dirac equation also in the case where an electromagnetic field is present.
Pauli matrices and squeezing n + 1 dimensions into n
The Pauli equation describes the electron spin interaction in the nonrelativistic case. The wave function is a two-component one. Though, the full freedom of having two independent wave functions apparently does not come into reality. All examples seem to contain a two-component complex-valued vector multiplying a single wave function.
The Pauli matrices operate on the two-component wave function. They are the spin operators in the x, y, z directions. The eigenvalues for each of them are -1 and 1. An eigenfunction seems to be any product of an eigenvector and any function.
The spin axis should, by default, require a 3-component vector to specify its direction. Our uncertainty of the direction of the spin axis seems to be so profound that we can squeeze the mathematical structure of the spin axis into a 2-component vector. This is probably the mathematical origin of the strange 720 degree rotation symmetry of spinors.
The wave functions in the Pauli machinery do not contain any spinning or rotating motion. The electron seems to be a pointlike magnetic dipole or a magnetic needle.
We suggested in an earlier post that the electron spin comes from our uncertainty about the rotation of the electron's electric field. It is not clear if the Pauli machinery somehow incorporates this idea.
If we transform the Klein-Gordon equation into a group of equations that only contain first derivatives, the standard way is to introduce a new component for each derivative on t, x, y, z. Then we would have a 5-component wave function. The Dirac equation manages the same feat with a 4-component wave function. Somehow are 5 dimensions squeezed into 4. This reminds us of the Pauli machinery which squeezes 3 dimensions into 2. We need to find out what is going on with Pauli and Dirac.
The Klein-Gordon equation in 1+1 dimensions is
d^2 Ψ/dx^2 - m^2 Ψ - d^2 Ψ/dt^2 = 0.
It is Lorentz covariant. The standard way to remove the second derivatives with new functions is (if m is not zero):
dΨ_2/dx - mΨ + dΨ_1/dt = 0
mΨ_1 + dΨ/dt = 0
-dΨ/dx + mΨ_2 = 0.
The Dirac equations in 1+1 dimensions are
-i dΨ_2/dx + mΨ_1 - i dΨ_1/dt = 0
-i dΨ_1/dx - mΨ_2 - i dΨ_2/dt = 0.
Any solution of the Dirac equations is also a solution of the Klein-Gordon equation, but the converse is probably not true. Dirac adds some extra constraints which remove solutions of the Klein-Gordon equation. As if the Klein-Gordon equation would allow too much freedom for solutions if m is not zero.
Using the operator symbols p and E, the Klein-Gordon equation is
(p^2 + (m + V)^2 - E^2) Ψ = 0.
We have added a scalar potential V to the rest mass, like we did in our earlier blog posts. If |p^4 / m^2| is very small and |V| is small, we can factor it like this:
(p^2 / (2m) + m + V + E)
* (p^2 / (2m) + m + V - E) Ψ = 0.
The lower factor is the Schrödinger equation. The upper factor of the operator would have Ψ rotating in time to the opposite direction. Is it the Schrödinger equation for the antiparticle?
Suppose that V is a linear potential which decreases to the right. Let us initialize the system with a wave function
exp(-i (Et - px))
where p = 0. The particles start to slide to the right, which means that after some time, there will be components like
exp(-i (Et - px)),
where p > 0, if we solve the Schrödinger equation.
If we solve the upper factor equation (we remove the Schrödinger operator from the equation above), it will have components like
exp(-i (-Et - px)).
If we increase t by dt, then to restore the phase in the exp() above, we need to decrease x by some dx. We see that the component describes a particle sliding to the left. The potential V works to those particles in the opposite way than in the Schrödinger equation. We can call them antiparticles.
Let us factorize the energy-momentum relation
E^2 - p^2 - m^2 = 0
with the matrices α_1, β above:
1 0 0 1 1 0
( E - p - m)
0 1 1 0 0 -1
*
1 0 0 1 1 0 0 0
( E + p + m) =
0 1 1 0 0 -1 0 0.
In the Schrödinger factorization, we had
E^2 = p^2 + m^2
<=> (p^2 small)
E = +- (p^2 / (2m) + m).
Using the identity matrix I, we can write the above as
I^2 E^2 = I^2 p^2 + I^2 m^2
<=> (p small)
I E = +- (I p^2 / (2m) + I m).
In the Dirac factorization, we have
I^2 E^2 = α_1^2 p^2 + β^2 m^2
= (α_1 p + β m)^2
<=
I E = +- (α_1 p + β m).
The factorizations are completely different but both yield sensible results. Why?
The Schrödinger factorization produces the (almost) right result because of cross terms in the product. The Dirac factorization has cross terms zero. It has only implication into one direction.
The Klein-Gordon equation in 1+1 dimensions is
d^2 Ψ/dx^2 - m^2 Ψ - d^2 Ψ/dt^2 = 0.
It is Lorentz covariant. The standard way to remove the second derivatives with new functions is (if m is not zero):
dΨ_2/dx - mΨ + dΨ_1/dt = 0
mΨ_1 + dΨ/dt = 0
-dΨ/dx + mΨ_2 = 0.
The Dirac equations in 1+1 dimensions are
-i dΨ_2/dx + mΨ_1 - i dΨ_1/dt = 0
-i dΨ_1/dx - mΨ_2 - i dΨ_2/dt = 0.
Any solution of the Dirac equations is also a solution of the Klein-Gordon equation, but the converse is probably not true. Dirac adds some extra constraints which remove solutions of the Klein-Gordon equation. As if the Klein-Gordon equation would allow too much freedom for solutions if m is not zero.
Deriving the Schrödinger equation from the Klein-Gordon equation
Using the operator symbols p and E, the Klein-Gordon equation is
(p^2 + (m + V)^2 - E^2) Ψ = 0.
We have added a scalar potential V to the rest mass, like we did in our earlier blog posts. If |p^4 / m^2| is very small and |V| is small, we can factor it like this:
(p^2 / (2m) + m + V + E)
* (p^2 / (2m) + m + V - E) Ψ = 0.
The lower factor is the Schrödinger equation. The upper factor of the operator would have Ψ rotating in time to the opposite direction. Is it the Schrödinger equation for the antiparticle?
Suppose that V is a linear potential which decreases to the right. Let us initialize the system with a wave function
exp(-i (Et - px))
where p = 0. The particles start to slide to the right, which means that after some time, there will be components like
exp(-i (Et - px)),
where p > 0, if we solve the Schrödinger equation.
If we solve the upper factor equation (we remove the Schrödinger operator from the equation above), it will have components like
exp(-i (-Et - px)).
If we increase t by dt, then to restore the phase in the exp() above, we need to decrease x by some dx. We see that the component describes a particle sliding to the left. The potential V works to those particles in the opposite way than in the Schrödinger equation. We can call them antiparticles.
Deriving the Dirac and Schrödinger equations from the energy-momentum relation
Let us factorize the energy-momentum relation
E^2 - p^2 - m^2 = 0
with the matrices α_1, β above:
1 0 0 1 1 0
( E - p - m)
0 1 1 0 0 -1
*
1 0 0 1 1 0 0 0
( E + p + m) =
0 1 1 0 0 -1 0 0.
In the Schrödinger factorization, we had
E^2 = p^2 + m^2
<=> (p^2 small)
E = +- (p^2 / (2m) + m).
Using the identity matrix I, we can write the above as
I^2 E^2 = I^2 p^2 + I^2 m^2
<=> (p small)
I E = +- (I p^2 / (2m) + I m).
In the Dirac factorization, we have
I^2 E^2 = α_1^2 p^2 + β^2 m^2
= (α_1 p + β m)^2
<=
I E = +- (α_1 p + β m).
The factorizations are completely different but both yield sensible results. Why?
The Schrödinger factorization produces the (almost) right result because of cross terms in the product. The Dirac factorization has cross terms zero. It has only implication into one direction.
Does every solution of the Schrödinger equation also solve the Dirac equation?
Let us use a two-component wave function Ψ and use the identity matrix I in our Klein-Gordon equation:
I (p^2 + (m + V)^2) Ψ = I E^2 Ψ.
Above p and E are operators and V is a scalar potential. Both components of Ψ are individually solutions of the same one-component equation.
The corresponding Schrödinger equations are
+- I (p^2 / (2m) + m + V) Ψ = I E Ψ.
For very small |p^4 / m^2| and small |V|, any solution of our Schrödinger equation is a solution of our Klein-Gordon equation. The converse is probably not true.
Our Dirac equations are
+- (α_1 p + β (m + V)) Ψ = I E Ψ.
If we have a solution of our Dirac equation, then it is a solution of our Klein-Gordon equation.
But are solutions of our Schrödinger equation solutions of our Dirac equation, and vice versa? Yes if the solutions only depend on time by a factor exp(-i E t), where E is constant, that is, they are energy eigenfunctions.
But are solutions of our Schrödinger equation solutions of our Dirac equation, and vice versa? Yes if the solutions only depend on time by a factor exp(-i E t), where E is constant, that is, they are energy eigenfunctions.
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