How do we obtain the π molecular orbitals for allene via Hückel theory? For a conjugated system, for instance 1,3-butadiene, we can try to obtain the π molecular orbital energies and wavefunctions via Hückel theory. In our given molecule, there are 4 interacting Pz orbitals. By creating a 4x4 Hückel matrix and solving for the energy eigenvalues, we can arrive at 4 possible linear combinations (of p-orbitals) to get 4 molecular orbitals. 

However, how can we use Hückel theory to deal with a molecule like 1,2-propadiene (allene)? 

My attempt: I thought that since this molecule has essentially 2 separate π-systems:
 

we may be able to apply Hückel theory to each one individually. Then, we can take all the possible combinations of the individual energies to arrive at the possible total energies. But I am not sure if this is correct. What would be the correct way of approaching this problem?
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.