Dirac’s Electron

On page 620 of Penrose’s “Road to Reality” there is a version of Dirac’s equation as follows:
∂'ψ = −iMψ

I write ‘∂'’ where the book uses ‘∂’ with a slash ‘/’ thru it. In this equation:
∂' = γ^a∂/∂x^a = γ^a∂_a
ψ is the electron field over Minkowski space
i = the square root of −1
M = μ/ℏ
γ^a = g^abγ_b
g^ab is the contravariant metric tensor.

The coordinates x^a are real. γ_b are four basis elements of the vector space of a Clifford algebra with signature (− + + +). That vector space is a tangent space to the manifold (here supposed flat) for which the x^a are coordinates. g^ab = diag(−1, 1, 1, 1) for now. Penrose does not say whether the vector space of the Clifford algebra is real or complex. The signature of a Clifford algebra over a complex vector space serves only to indicate which basis vectors we choose. γ₀γ₀ = 1 but had iγ₀ been chosen in place of γ₀ then we would have had γ_jγ_j = −1 for all j and then the signature would have been (+ + + +). See too this connection between Dirac and Clifford.

Schrödinger’s original equation requires complex numbers to keep it simple. Penrose adds Clifford algebra to simplify Dirac’s equation. I want to see if Clifford algebra can supplant the complex numbers and thus remove i from the Dirac equation, thus further simplifying it. We must also decide which sort of Clifford algebra is needed.

I have not yet said what sorts of values ψ takes on. Penrose is silent. M is real; −iM is complex; it would seem that ψ is forced to take on values that are complex Clifford values—that is values in a Clifford algebra over a 4D vector space over the complex numbers. This is a 32 dimensional space. I think this is like Dirac’s original 1920 equation but don’t quote me.

Alternatively the “i” in “∂'ψ = −iMψ” can be explained away by viewing the complex Clifford ψ field as a pair of real Clifford fields over a real vector space (ψ1, ψ2). This gives:
∂'ψ1 = Mψ2
∂'ψ2 = −Mψ1.
In this form the field is still 32D and this formulation is isomorphic with the above version.

In either of these formulations it is unclear if the 32D space is entirely populated. I suspect that only 16D are necessary—perhaps a 16D subspace. It is likewise unclear to me what equivalence classes there are as in “For real θ the field ψ and e^iθψ have the same physical meaning.” which certainly holds here. If 32D are really necessary then I cannot achieve the simplification I want.

Simple Solutions to Dirac’s equation

I will start with a particle with somewhat known velocity and location:
ψ =

First solve the non relativistic 1D Schrödinger equation
ψ = e^{i(x−vt)−αx²} = e^i(x−vt)e^−αx²
∂ψ/∂x = (i − 2αx)e^{i(x−vt)−αx²}
This is not what I remember. I will refer to this cheat sheet (or this) and try for a Gaussian distribution of both location and velocity—i.e. an electron whose expected position and velocity are both 0. Then we can do a relativistic transform and see if it still solves Dirac’s equation.

I see (under “The Gaussian Wavepacket”) ψ(x,t=0) = Ae^ik₀xe^−x²/2Δ²
where the local Schrödinger equation is
∂ψ/∂t = (iℏ/2m)∇²ψ.
To normalize we get
ψ(x,t=0) = π^−1/4Δ^−1/2 e^ik₀xe^−x²/2Δ² = π^−1/4Δ^−1/2 e^{ik₀x−x²/2Δ²}

I see the problem with my earlier stab. If I know the location within Δ when t=0, I will know it less well later. It would seem that Δ grows with t, just as in the classic diffusion situation. There is a difference, however, in classic diffusion Δ² grows with t; here Δ grows with t. Later the velocity uncertainty will decrease, at least when a new position is known. I am not ready to guess a solution to the time dependent ψ.