How Should I Think About the Dirac Equation? In Weinberg's QFT Vol. 1 he says the Dirac equation is not a true generalization of Schrodinger's equation, that it does not stand up to inspection when viewed in this light. He says it should be viewed as an approximation to a true relativistic quantum field theory of photons and electrons.


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*I do not understand what this means, would someone mind filling me
in? (Assuming he's saying something more general & subtle than merely saying Dirac is not a one-particle equation)


One of Dirac's motivations for his derivation was that Klein-Gordon was not first order in time like the non-relativistic Schrodinger equation is.


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*Since the 2'nd order in time Klein-Gordon equation does actually
describe something physical, does this mean Dirac's point about not
being first-order in time is actually flawed, and that Dirac's equation worked for some other reason?


From browsing Cartan's Spinor's book it seems the Dirac equation holds for any spinor, it apparently relates left & right representations of a spinor or something, thus it holds in GR etc... There is also this great quote from Atiyah that a spinor is a square root of a geometry.


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*What is the Dirac equation & how does this explain why Dirac's
derivation worked, why it relates representations of a spinor &
explains this square root of a geometry business?

 A: 'Square root of geometry'
A Dirac spinor field $\psi^\alpha(x)$ under Lorentz transformations behaves as,
$$\psi^\alpha(x)\to S[\Lambda]^\alpha_\beta \psi^\beta(\Lambda^{-1}x)$$
where $\Lambda = \exp(\frac{1}{2} \Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$ and $S[\Lambda] = \exp(\frac{1}{2}\Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$, where $\mathcal{M}$ are the generators of the Lorentz transformations; they are anti-symmetric and obey the Lorenz Lie algebra. The matrices $S$ satisfy the spinor representation of the Lorenz group (exercise!), and are defined by,
$$S^{\mu\nu} = \frac{1}{4}[\gamma^\mu,\gamma^\nu]$$
A general rotation matrix acting on a spinor may be written as,
$$\left( 
\begin{array}{cc}
e^{i\varphi \cdot \sigma /2} && 0\\
0 && e^{i\varphi \cdot \sigma /2}
\end{array}
\right)$$
A rotation by $2\pi$, i.e. an entire turn, around the $x^3$ axis is given by the choice $\varphi = (0,0,2\pi)$, and the rotation matrix becomes,
$$\left( 
\begin{array}{cc}
e^{i\pi  \sigma^3} && 0\\
0 && e^{i\pi  \sigma^3}
\end{array}
\right) = -1$$
Hence, under a full turn, a spinor transforms as, $\psi^\alpha(x) \to -\psi^\alpha(x)$, in other words it does not return to its original state. That's why we call them the 'square root of geometry.' You in fact have to rotate them fully twice rather than once.
