Thursday, April 18, 2019

The Dirac equation follows from "curvature" of space under a magnetic field?

Let us start from the energy-momentum relation of special relativity:

       E^2 = p^2 + m^2.

The usual interpretations of the corresponding operators are from the Schrödinger equation:

       E = i d/dt
       p = -i ∇
       p_x = -i d/dx
       p_y = -i d/dy
       p_z = -i d/dz,

where ∇ is the gradient, that is, the vector (d/dx, d/dy, d/dz).


The wave operator in the Klein-Gordon equation


We obtain the Klein-Gordon equation by interpreting the symbols in the energy-momentum relation as operators:

       (-d^2/dt^2 + ∇^2) ψ = m^2 ψ

Let us call -d^2/dt^2 + ∇^2 the wave operator of the Klein-Gordon equation. The solutions of the Klein-Gordon equation are then eigenfunctions of the wave operator and the eigenvalues are squares of the rest mass m of the particle.

The wave operator determines the "behavior" of the wave ψ in spacetime. It ties the partial derivatives to x, y, z directions to the partial derivative in the t direction.

In the Schrödinger equation, the wave operator can be split to the kinetic energy term p^2/(2m) and the time evolution term i d/dt.

Orthogonal commuting momentum operators


For a free particle, the momentum operators E, p_x, p_y, p_z commute, and are orthogonal in the sense that they share a common eigenbase, and each operator has an eigenfunction whose eigenvalue is non-zero, but whose eigenvalues for the other operators are zero.

The common eigenbase of these operators is the plane waves

       exp(-i (Et - p r)).

Postulate 1. If we are working in empty space with no external fields, we postulate that the correct way to describe a quantum mechanical particle is to make the wave operator a sum of (+, -) squares of momentum operators, such that E^2 has a different sign from p_x^2 and other spatial operators.


The Klein-Gordon equation follows from this recipe. The intuitive reason for forming such a wave operator is that empty space is a flat Minkowski space, and the operators p_x, p_y, p_z are symmetric and orthogonal. It is a good guess that the wave operator should look much like the Pythagorean theorem.

A free electron or a positron is described by the Klein-Gordon equation. In the plane wave, a positive energy

       E = sqrt(p^2 + m^2)

describes an electron, and a negative energy

       E = -sqrt(p^2 + m^2)

describes a positron.


Non-commuting, non-orthogonal momentum operators


Let us add an external magnetic field with a vector potential A to our spacetime. The minimal coupling replaces p with p - qA, where q is the charge of the particle.

But

       p_x - qA_x, p_y - qA_y, p_z - qA_z

do not commute if A is non-zero. Since they do not commute, they are not orthogonal either. What should the wave operator look like then?

There should probably be some "cross terms" of partial derivatives.

In the case of a free particle, the wave operator was a sum of (+-) squares of momentum operators.

Postulate 2. We postulate that the wave operator should be a square of another operator, and if the magnetic field is zero, the wave operator should be "reduced", in some sense, to the wave operator of the free particle.


https://en.wikipedia.org/wiki/Dirac_operator

Let us denote by KG the Klein-Gordon wave operator. The corresponding Dirac operator D is such that

       KG I = D^2,

where I is the identity matrix for some dimension n. We have to let n be at least 4, to find a solution for D.

https://en.wikipedia.org/wiki/Dirac_equation

The solutions for D are obtained through the well-known trick of Paul Dirac, where we add "degrees of freedom" to the equation by making the wave function 4-place, and take a "square root" of the Klein-Gordon operator by using the gamma matrices as coefficients.

The standard Dirac equation is

       D ψ = m ψ,

where ψ is a 4-place complex-valued wave function. If D satisfies KG I = D^2, then also -D satisfies it. We get the second Dirac equation

       D ψ = -m ψ.

For now, we do not add an electric potential to the equation. The 4-place vector is called a Dirac spinor.

http://physics.gu.se/~tfkhj/TOPO/DiracEquation.pdf

Let the particle have a momentum along the z axis. The eigenfunctions of the standard Dirac equation are schematically of the form

     / electron spin_z up    \
     | electron spin_z down |
     | positron spin_z down |  exp(-i (Et - pr)).
     \ positron spin_z up     /

The eigenvalues are 4-component spinors. For the positron solutions, E is negative.

If p is zero, then the solution for the spin up electron is

       /1 \
      | 0  |
      | 0  |   exp(-i Et).                                            (1)
      \ 0 /

For the second Dirac equation, we may define that the positron with the spin down is the same formula (1) as above, and E is negative.

Question 3. The second Dirac equation above produces a second set of solutions for an electron spin up, and so on. Why is there such a degeneracy of the solutions? Is it certain that the second solutions represent the same particles as the first solutions?


If we think of the positron as an electron traveling "backward in time", and the electron may scatter so that it turns back in time, then it is natural to represent both the electron and the positron with the formula (1) above, so that we do not need to switch the spinor

       (1, 0, 0, 0)

suddenly to the spinor

       (0, 0, 0, 1).

The positron is then a solution of the second Dirac equation.

If the magnetic field is zero (A = 0), then the Dirac equation does, in a sense, reduce to the Klein-Gordon equation. The exponent part of a Dirac eigenfunction is a plane wave and is an eigenfunction of the Klein-Gordon equation. We can discard the spinor part because for a free particle there is no way to interact with the spin of the particle. With no interaction, the spinor stays the same at all times and we can ignore that degree of freedom.


The electron is fundamentally a wave phenomenon and cannot be explained as a classical point particle or a billiard ball


We wrote in our previous blog post that our four months of hard thinking did not produce a model where the electron would be described, in some sense, as a classical point particle or a classical billiard ball, and the wave associated with the electron would be derived from a path integral on some "lagrangian density".

The electron does certainly exhibit some features of a classical billiard ball or a point particle. We can measure its location with a great precision. The electron carries angular momentum in its spin, like a rotating ball.

We hypothesize that these classical features emerge from the electron wave, and the wave is the fundamental "structure" of the electron. Saying that the electron is a "point particle" is wrong, because that conveys the impression that it is a classical particle. It is better to say that the electron is a wave.

Question 4. Why does the electron wave exhibit classical features, like collapsing to a point when its location is measured, or carrying angular momentum?


Question 5. The electron cannot be described through a path integral. Does this have any bearing on Feynman diagrams, where we do calculate path integrals?

No comments:

Post a Comment