Where do the particle and antiparticle wavefunctions originate from in the Klein Gordon equation? In my textbook (Sakurai) it is given that 
$$\left(D_\mu D^\mu+m^2\right)\Psi(\mathbf{x},t)=0$$
where $D_\mu=\partial_\mu+ieA_\mu$ is the covariant derivative.
It states that since it is a second order differntial equation we must specify the wavefunction at its initial time as well as its first derivative. Alternatively we can reduce the second-order KG equation to two first order equations and intepret the result in terms of the sign of the electric charge.
We can use that $D_\mu D^\mu=D_t^2-\vec{D}^2$ to get the two new functions
$$\phi(\mathbf{x},t)=\frac{1}{2}\left(\Psi(\mathbf{x,t})+\frac{i}{m}D_t\Psi(\mathbf{x,t})\right)$$
$$\chi(\mathbf{x},t)=\frac{1}{2}\left(\Psi(\mathbf{x,t})-\frac{i}{m}D_t\Psi(\mathbf{x,t})\right)$$
but where do these two functions arise from? How do you get them from this information?
 A: The two equations you wrote at the bottom of are just definitions of $\phi$ and $\chi$. They have no physical interpretation at this point in Sakurai's argument. The physical interpretation comes later when he rewrites the Klein-Gordon 'probability' density in terms of them (see 8.1.20)
There is a standard way to turn a higher order differential equation into a system of first order equations. For instance if we give $D_t \Psi(x,t)$ a new name, $\Pi(x,t)$,
$$D_t \Psi \equiv -i m\, \Pi$$
then the Klein Gordon equation becomes a first order system in two functions $\Psi,\Pi$
$$D_t \Pi = \frac{i}{m}(\vec{D}^2 - m^2)\Psi$$
Substitute the definition of $\Pi$ to get the original Klein Gordon equation.
Now instead of the degrees of freedom $\Psi,\Pi$ we are free to define two independent linear combinations instead:
$$\phi\equiv\frac{1}{2}(\Psi+\Pi)\quad\chi\equiv\frac{1}{2}(\Psi-\Pi)$$
Then adding and subtracting the first two equations we get first order equations for $\phi$ and $\chi$ instead (see 8.1.16, or try it yourself).
