How do we obtain the π molecular orbitals for allene via Hückel theory?
For p-orbitals $i$ and $j$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero:
$$\langle i | H | j \rangle = 0$$
If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form:
$$\mathbf{Hc} = E\mathbf{c},$$
or in explicit form, using the fact that the matrix elements of $\mathbf{H}$ are $\mathbf{H}_{ij} = \langle i | H | j \rangle$,
$$\begin{pmatrix}
\alpha &amp; \beta &amp; 0 &amp; 0 \\
\beta &amp; \alpha &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; \alpha &amp; \beta \\
0 &amp; 0 &amp; \beta &amp; \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}
= 
E
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}$$
This matrix has block-diagonal form, so solving for the eigenvalues of the entire matrix simply reduces to solving for the eigenvalues of each individual $2\times 2$ block (lots of discussion can be found on Google). For a physical interpretation, this means that both π systems have no interaction with each other, so you can indeed solve them individually.
Each block gives rise to one MO with energy $\alpha + \beta$ (&quot;bonding&quot;) and one with energy $\alpha - \beta$ (&quot;antibonding&quot;), so all in all you will have two degenerate MOs with energy $\alpha + \beta$ and two degenerate MOs with energy $\alpha - \beta$.

* If you don't ignore the overlap matrix $\mathbf{S}$, whose matrix elements are defined by $\mathbf{S}_{ij} = \langle i | j \rangle$, this becomes a &quot;generalised eigenvalue equation&quot;
$$\mathbf{Hc} = E\mathbf{Sc}$$
However, because the overlap matrix also has the same block-diagonal form, ignoring the overlap matrix doesn't change the fact that the problem is separable into two $2\times 2$ blocks. It merely changes the exact eigenvalues and eigenstates by a small amount.