$$H \; = \; \left[\begin{array}{cc} \alpha & \beta \\ \beta & \alpha \\ \end{array}\right] \; \; . $$
But, how do we explicitly find what $\alpha$ and $\beta$ are?
First, let's find alpha. We will start by looking at the operation
$$H\vert 1 \rangle \; = \; \left[\begin{array}{cc} \alpha & \beta \\ \beta & \alpha \\ \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ \end{array}\right] \; = \; \left[\begin{array}{c} \alpha \\ \beta \\ \end{array}\right]. $$
Changing this operation into function notation will give:
$$H \phi(x-a) \; = \; \alpha \phi(x-a) + \beta \phi (x-2a) \; \; .$$
Multiplying both sides by $\int \phi^{*}(x-a)$ will then give:
$$\int \phi^{*}(x-a)H \phi(x-a) \; = \; \int \phi^{*}(x-a)\alpha \phi(x-a) + \int \phi^{*}(x-a)\beta \phi (x-2a) \; \; .$$
We can then pull out $\alpha$ and $\beta$ from the integrals because they are constants.
$$\int \phi^{*}(x-a)H \phi(x-a) \; = \; \alpha\int \phi^{*}(x-a) \phi(x-a) + \beta\int \phi^{*}(x-a) \phi (x-2a) \; \; .$$
The basis states are orthogonal to each other, so evaluating the integral next to $\beta$ will give zero while the integral next to $\alpha$ will equal 1. Doing so shows that
$$\alpha \; = \; \int \phi^{*}(x-a)H \phi(x-a) \; \; . $$
Note that in bra-ket notation, the expression looks like
$$\alpha \; = \; \langle 1 \vert H \vert 1 \rangle \; \; . $$
$$\beta \; = \; \int \phi^{*g}(x-a) \, H \, \phi^{g}(x-2a)dx \; \; . $$
Notice that $\beta$ is an overlap function. As such, if the potential wells are very far apart, we can expect that the value for $\beta$ will be very small. In either case, it is expected that $\beta < \alpha$.
$$H=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{atom}\left(x-a\right) + V_{atom}\left(x-2a\right) \; \; ,$$
and evaluate the integral over all space.