General Solution of Klein Gordon Equation

Let us discuss something about the general solution of the Klein Gordon equation which will be later useful when we adopt for the field formulation of our theory.  
Assuming our solution as a general waveform solution in the form,$$ \phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \tilde{\phi}(k) \,e^{-(ikx = ik_{\mu}x^{\mu})}$$ where 4 vector notation is generally implied unless it is specified. 
We substitute this general form in our equation to obtain the conditions to be satisfied from the KG equation. It gives, $$ \frac{1}{(2\pi)^{2}}\int \,d^4k \,e^{-ikx}\left(-k^2+m^2\right)\tilde{\phi}(k) $$ [Normalization factor $\sqrt{2\pi}$ for four integrations]. 

From the above equation, the solution should be of the form such that, if $k^2=m^2$, $\tilde{\phi}(k)$ can take any arbitrary value and if $k^2\neq {m^2}$ then $\tilde{\phi}(k)$ should be zero, 
therefore it should be in the form, $$\tilde{\phi}(k) = \delta(k^2-m^2)\,\tilde{f}(k)$$ 
Thus, our general wave solution cannot take any arbitrary form but should always satisfy the energy momentum relation - which is imposed by the dirac delta function. It is equivalent to saying that the energy can take only positive and negative values of $\omega_k$ given by $\delta(k^2-m^2) = \delta(k_0^2-|\vec{k}|^2-m^2) = \delta(k_0^2-\omega_k^2)$ where $\omega_k^2=|\vec{k}|^2+m^2$ 
Using the identity, $$\delta(f(x)) = \sum_{k}\frac{\delta(x-x_k)}{f'(x_k)}$$, we can rewrite our solution as,
$$\phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \left[e^{-ikx}\delta(k_0-\omega_k)\tilde{f}(k) + e^{-ikx}\delta(k_0+\omega_k)\tilde{f}(k)\right]$$ Now, integrating the zeroth component of momentum vector and using the fact that the integration is symmetric under $+\vec{k}$ and $-\vec{k}$ we can arrive at the form, $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx}\frac{\tilde{f}(\omega_k,\vec{k})}{\sqrt{4\pi\omega_k}}+ e^{ikx}\frac{\tilde{f}(-\omega_k,-\vec{k})}{\sqrt{4\pi\omega_k}}\right]$$
Now we identify the terms other than exponential as $ a(k), b^*(k)$ just for a general convenience. $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx} a(k) + e^{ikx} b^*(k)\right]\vert_{k_0=\omega_k} = \phi_+ +\phi_-$$ 
The complex conjugate is, $$\phi^*(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{ikx} a^*(k) + e^{-ikx} b(k)\right]\vert_{k_0=\omega_k} $$ 

For real fields, one can show that, $a(k) = b(k)$ and $a^*(k) = b^*(k)$

Similarly, the momentum space terms can be written as, $$a(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) +i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p} $$ and $$b^*(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) - i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p}$$ 
Hint: One can prove from RHS to LHS by integrating with respect to $d^3x$, making use of the definition of Dirac delta function, integrating again with respect to $d^3k$ and finally using the fact that $\omega_k=\omega_{-k}=\omega_p=\omega_{-p}$ - the energy is independent of the direction of the momentum vector. 
It is always a good practice to use different symbols to discuss it in a general form of the derivation. 
Finally, one can obtain $a^*(p), b(p)$ by taking the complex conjugate. 

Dirac Equation

The essential fact about the Klein Gordon equation is that it is a non-linear equation. But, equations are consistent with Lorentz transformation when they are linear in space and time. Dirac came forward with a new way of resolving the problem of this non-linearity by factorizing the relativistic energy momentum relation into two linear equations. The equation is given by the format, 
$$ i\frac{\partial\psi}{\partial{t}} = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right)\psi $$
But, still it should satisfy the energy momentum relation $$ E^2 \psi = \left(m^2+|\vec{p}|^2\right)\psi $$
that is,
$$ E (E\psi) = i \frac{\partial}{\partial{t}}\left(\frac{i\partial\psi}{\partial{t}}\right) \\ = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \psi $$ [the differentials commute]
In summation convention,
$$\frac{\partial^2\psi}{\partial{t^2}} = \left[\alpha_i\alpha_j\partial_i\partial_j + i(\alpha_i\beta+\beta\alpha_i)m\partial_i - \beta^2m^2\right]\psi $$
[It is assumed $\beta$ commutes with differentials]

The above can be evaluated separately from Klein Gordon equation as, $$\frac{\partial^2\psi}{\partial{t^2}} = \left[\vec{\nabla}^2 - m^2\right]\psi $$

Comparing both of them, we can see that $\alpha_i,\beta$ shouldn't commute and so they cannot be numbers. Dirac proposed the idea of matrices instead of numbers. This leads to the following properties that should be satisfied by $\alpha_i$ and $\beta$ as, 

$$\{\alpha_i,\alpha_j\} = 2 \delta_{ij} \\ \{\alpha_i,\beta\} = 0 \\ \alpha_i^2=\beta^2 =1$$
$$Tr[\alpha_i^2{X}\, (or)\, \beta^2{X}=\lambda^2{X}=IX]$$ gives us the eigenvalues to be $\pm 1$ 

$$Tr[\alpha_i] \\ = Tr[\alpha_i{\beta^2}] = Tr[\beta\alpha_i{\beta}] (using \,the \,property\, Tr(AB)=Tr(BA)) \\ = Tr[-\beta\alpha_i\beta] (using \,anti-commutation \,relation)$$ 

Doesn't matter whether we start with $\alpha_i$ or $\beta$, this proves the trace of the matrices $\alpha_i,\beta$ should be zero. 
From the above two properties, dimension of the matrix should be even. 

Starting with the dimension N=2, there are no sufficient number of independent matrices to describe our four matrices (since pauli matrices span the space completely).

Thus, we are left with the next possibility of N=4 by constructing our matrices as, 

$$ \alpha_i = \left(\begin{matrix}0 & {\sigma_i}\\ {\sigma_i}&0\end{matrix}\right) \\ \beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0 & \mathbb{-I}\end{matrix}\right) $$

To make it simple, we introduce the symbol $\gamma$ and defined everything in terms of new Dirac $\gamma$ matrices. And, the Dirac equation is finally given in the simplest form as, 

$$\gamma^0 =\beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0& \mathbb{-I}\end{matrix}\right) \\ \gamma^i = \beta\alpha_i = \left(\begin{matrix}0 &{\sigma_i}\\ {-\sigma_i}&0\end{matrix}\right)$$

with the properties,

$$\gamma^0 = (\gamma^0)^\dagger \\ (\gamma^0)^2 = \mathbb{I} \\ \gamma^i = -(\gamma^i)^\dagger \\ (\gamma^i)^2=-\mathbb{I} \\ \{\gamma^\mu,\gamma^\nu\} = 2g_{\mu\nu} $$

Using this new representation, the Dirac equation is written as, $$\left(i\gamma^\mu\partial_\mu - m \right)\psi = 0 $$

In slashed notation, any Lorentz vector is $\slashed{x} = \gamma^\mu{x}_\mu$ and so, Dirac equation is, $$ \left(i\slashed{\partial}-m\right)\psi = 0$$