I wouldn't say that there is a derivation of the Dirac equation starting with the Pauli equation. But the trick that makes the Pauli equation work can be used as a heuristic for developing a relativistic version that is equivalent to the Dirac equation.
Start with the nonrelativistic formula relating energy and momentum for a particle:
\frac{p^2}{2m} = E
This is for a free particle, but to get a particle in an electromagnetic field, you can do the substitutions:
\vec{p} \Rightarrow \vec{p} - e \vec{A}
E \Rightarrow E - e \Phi
where \vec{A} is the electromagnetic vector potential, and \Phi is the scalar potential, and e is the particle's charge.
To get the Schrodinger equation, you do the following:
- Replace \vec{p} by the operator -i \vec{\nabla} (I'm using units where \hbar = c = 1)
- Replace E by the operator i \frac{d}{dt}
- Interpret the operator equation as applying to a scalar wave function \psi(\vec{r}, t).
To get the Pauli equation, you do a twist on the above:
- Instead of 1 above, you replace the \vec{p} by the operator \vec{p} \cdot \vec{\sigma}, where \vec{\sigma} is the three Pauli spin matrices.
- Instead of 3 above, you interpret the operator equation as applying to a two component spinor U(\vec{r}, t).
The use of the spin matrices makes no difference for free particles, since (\vec{p} \cdot \vec{\sigma})^2 = (\vec{p})^2. But once you introduce the electromagnetic field, this makes a difference, because you get an extra term, -e \vec{B} \cdot \vec{\sigma} where \vec{B} = \vec{\nabla} \times \vec{A}
Now, we can do the same thing to almost get the Dirac equation for the relativistic case.
Instead of starting with the equation:
\frac{(\vec{p})^2}{2m} = E
Start with a relativistically correct equation:
E^2 - \vec{p}^2 = m^2
Now introduce the substitution: \vec{p} \Rightarrow \vec{p} \cdot \vec{\sigma} to get:
E^2 - (\vec{p} \cdot \vec{\sigma})^2 = m^2
Here comes an unmotivated step: Write the above in a "factored" form.
(E - \vec{p} \cdot \vec{\sigma})(E + \vec{p} \cdot \vec{\sigma}) = m^2
Factoring it this way makes no difference for free particles, but will make a difference when you introduce the electromagnetic field. Of course, this is interpreted as applied to a two-component spinor, U. But here's the huge benefit of factoring:
Let's define a new two-component spinor, V via the equation:
V = \frac{1}{m} (E + \vec{p} \cdot \vec{\sigma}) U
Then the second-order factored equation can be written as a pair of coupled first-order equations:
(E - \vec{p} \cdot \vec{\sigma}) V = m U
(E + \vec{p} \cdot \vec{\sigma}) U = m V
These two equations can be written as a 4-component matrix equation:
(\gamma^0 E - \vec{\gamma} \cdot \vec{p}) \Psi = m \Psi
where \Psi = \left( \begin{array}\\ U \\ V \end{array} \right) and where \gamma^0 = \left( \begin{array}\\ 0 & 1 \\ 1 & 0 \end{array} \right) and \gamma^j = \left( \begin{array}\\ 0 & \sigma_j \\ -\sigma_j & 0 \end{array} \right)
(All the "1"s and "0"s really mean the 2x2 unit matrix and 2x2 zero matrix, respectively)
That is not the usual choice for \gamma^0, but the result is equivalent to the usual covariant form of the Dirac equation.
This was definitely not the way that Dirac came up with his equation. He didn't start with the second-order Pauli equation, but just tried to guess a first-order equation that produced the right energy-momentum relationship.
