PX262 - N2 - dirac equation

the klein-gordon equation contains a second order time derivative, as opposed to the schrödinger has just a first order time derivative
special relativity implies that the wave equation must treat space and time on an equal footing
ie. should be linear in both space and time derivatives
dirac proposed:

\hat{H} = α_{x} c {\hat{p}}_{x} + α_{y} c {\hat{p}}_{y} + α_{z} c {\hat{p}}_{z} + β m c^{2}

$\hat{H} ψ = i ℏ \frac{\partial ψ}{\partial t}$ has the same form as the schrödinger equation
choosing $α_{x}, α_{y}, α_{z}, β$ such that $\hat{H} ψ = - ℏ^{2} \partial_{t}^{2} ψ$ becomes the klein-gordon equation
this requires

\begin{matrix} α_{x}^{2} = α_{y}^{2} = α_{z}^{2} = β^{2} = 1 \\ α_{x} α_{y} = - α_{y} α_{x} \\ α_{x} α_{z} = - α_{z} α_{x} \\ α_{y} α_{z} = - α_{z} α_{y} \\ α_{i} β = - β α_{i} i \in {x, y, z} \end{matrix}

ie: $α_{x} α_{z} + α_{z} α_{x} = 0$ and $α_{x} β + β α_{x} = 1$ , etc
dirac proposed that the coefficients be taken as $4 \times 4$ matrices and find the set of matrices to meet these conditions
the results from the pauli spin matrices look promising:

\begin{matrix} σ_{x} σ_{z} + σ_{z} σ_{x} = 0 \\ σ_{x}^{2} = I \end{matrix}

these meet the conditions:

\begin{matrix} α_{x} = (\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}) = (\begin{matrix} 0 & σ_{x} \\ σ_{x} & 0 \end{matrix}) \\ α_{y} = (\begin{matrix} 0 & σ_{y} \\ σ_{y} & 0 \end{matrix}) \\ α_{z} = (\begin{matrix} 0 & σ_{z} \\ σ_{z} & 0 \end{matrix}) \\ β = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{matrix}) = (\begin{matrix} I & 0 \\ 0 & - I \end{matrix}) \end{matrix}

the dirac equation is:

[- i ℏ c (α_{x} \frac{\partial}{\partial x} + α_{y} \frac{\partial}{\partial y} + α_{z} \frac{\partial}{\partial z}) + β m c^{2}] ψ = i ℏ \frac{\partial ψ}{\partial t}

$\vec{α} = (α_{x}, α_{y}, α_{z})$ and $β$ are matrices
$ψ (\vec{r}, t)$ must have four components, called thedirac spinors:

ψ = (\begin{matrix} ψ_{1} \\ ψ_{2} \\ ψ_{3} \\ ψ_{4} \end{matrix}) = (\begin{matrix} u_{1} (p) \\ u_{2} (p) \\ u_{3} (p) \\ u_{4} (p) \end{matrix}) \exp (i \frac{(\vec{p} \cdot \vec{r} - \vec{k} t)}{ℏ})

in the non-relativistic limit ( $1 / c^{2} \to 0$ ), the dirac equation reduces to the schrödinger equation, but now, for a particle with a spin, $ℏ / 2$

ψ (x, y, z, t) = (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) \exp (i \frac{(\vec{p} \cdot \vec{r} - E t)}{ℏ})

where, $u_{3}, u_{4} \to 0$ as $c^{2} \to \infty$ , and $u_{1}$ and $u_{2}$ become independent of the momentum

thus, spin is explained
by looking at the leading relativistic corrections, spin-orbit coupling is also described
by comparing with the continuity equation:

\vec{\nabla} \cdot \vec{j} = - \frac{\partial ρ}{\partial t}

the probability density:

ρ = | ψ_{1}^{2} | + | ψ_{2} |^{2} + | ψ_{3} |^{2} + | ψ |^{4}