One can derive the Schrodinger equation as follows:
$U(\delta t)|\psi(t)\rangle = (I - i \delta tH)|\psi(t)\rangle = |\psi(t+\delta t)\rangle \rightarrow i\frac{|\psi(t+\delta t) - |\psi(t)\rangle}{\delta t} = i\frac{\partial |\psi(t)\rangle}{\partial t} = H|\psi(t)\rangle$
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I am now interested in deriving the Dirac equation using the same methodology.
My idea is instead of $U=e^{-itH/\hbar}$, I would start with this:
$$ G=\exp -i \frac{mc}{4\hbar}(t\gamma_0+x\gamma_1+y\gamma_2+z\gamma_3) $$
Now, I apply the same method as the Schrodinger equation (one for each variable):
$$ G(\delta t, x,y ,z) | \psi(t,x,y,z)\rangle=(I-i\frac{mc}{4\hbar}\gamma_0 \delta t)|\psi(t,x,y,z)\rangle =|\psi(t+\delta t,x,y,z)\rangle \\ G(t, \delta x,y ,z) | \psi(t,x,y,z)\rangle=(I-i\frac{mc}{4\hbar}\gamma_1 \delta x)|\psi(t,x,y,z)\rangle = |\psi(t,x+\delta x ,y,z)\rangle\\ G(t, x, \delta y ,z) | \psi(t,x,y,z)\rangle=(I-i\frac{mc}{4\hbar}\gamma_2 \delta y)|\psi(t,x,y,z)\rangle = |\psi(t,x,y+\delta y,z)\rangle\\ G(t, x, y , \delta z) | \psi(t,x,y,z)\rangle=(I-i\frac{mc}{4\hbar}\gamma_3 \delta z)|\psi(t,x,y,z)\rangle = |\psi(t,x, y,z + \delta z)\rangle $$
Each one of them produces a Schrodinger-like equation for its respective component:
$$ \gamma_0 i \frac{\partial |\psi(t,x,y,z)\rangle }{\partial t} = \frac{mc}{4\hbar}|\psi(t,x,y,z)\rangle\\ \gamma_1 i\frac{\partial |\psi(t,x,y,z)\rangle }{\partial x} = \frac{mc}{4\hbar}|\psi(t,x,y,z)\rangle\\ \gamma_2 i\frac{\partial |\psi(t,x,y,z)\rangle }{\partial y} = \frac{mc}{4\hbar}|\psi(t,x,y,z)\rangle\\ \gamma_3 i\frac{\partial |\psi(t,x,y,z)\rangle }{\partial z} = \frac{mc}{4\hbar}|\psi(t,x,y,z)\rangle $$
Finally, I add them together and I get the Dirac equation
$$ i \hbar \gamma^\mu \partial_\mu \psi - mc \psi =0 $$
Does any of this works? Can we understand the Dirac equation as a 4 parameter evolution of $G$, which would be an extension of the one-parameter evolution of U for the Schrodinger equation?
