Take the Hamiltonian you write at end of your question as the example(where your statement about the eigenvector is not right), the procedure to obtain the eigenvalue and eigenvectors of other Hamiltonian is nearly the same.

$$H=\sum_{ij}A_{ij}a_i^\dagger a_j$$

Writing in matrix form you have(suppose it is a $n\times n$ matrix):

$$H=\alpha^\dagger A \alpha$$

with:

$$ \alpha^\dagger=(a_1^\dagger, a_2^\dagger,\dotsb,a_n^\dagger )$$

Now you are ready to diagonalize Matrix analytically or numerically:

$$A=X^\dagger D X$$

Where $D=[E_1,E_2,\dotsb,E_n]$ is the eigenvalue you want, and $X$ is just a **Unitary** matrix that diagonalize $A$. Indeed, it is the **normalized** mathematical eigenvectors of $A$ write in **column** way(which is same in Fortran if you do it numerically).

Now substitute this to the original Hamiltonian:

$$H=\alpha^\dagger A \alpha=\alpha^\dagger X^\dagger DX\alpha=\beta^\dagger D \beta$$

where $\beta$ is $\beta =X\alpha=(\beta_1,\beta_2,\dotsb,\beta_n)^T$. Because this is a unitary transformation, you can easily check that the commute or anti-commute relation still holds for $\beta_i,\beta_2,\dotsb$.

Now you can consider the system is consisting of non-interacting qusi-particles $\beta_i,\beta_2,\dotsb$, then the eigenstate of the system is easily obtained:
$$|\psi \rangle=(\beta_1^\dagger)^{n_1}(\beta_2^\dagger)^{n_2}\dotsb(\beta_n^\dagger)^{n_n}|0\rangle$$

$n_i=0,1$ for fermions and $n_i=0,1,2,3\dotsb$ for bosons.

Also, see this thread with the very similar question.

This post imported from StackExchange Physics at 2014-04-05 11:04 (UCT), posted by SE-user luming