Proving that Eigenvectors Span Hilbert Space I have a specific problem I am trying to solve, but I would like to learn general principles, so I will start my question pretty general and add specifics later. Please answer the most general form of the question possible and if you are really nice, answer the specifics in this context.
General
Given a Hilbert space operator for which I have found some eigenvectors, how do I prove/disprove that these vectors span my Hilbert space?
Specific Problem
Again, please teach me the principles because I will be working on harder problems later.
I have a Hamiltonian operator of the form,
$$
\hat{H} \equiv \frac{-\hbar^2}{2m} \left(\partial_x^2 + \partial_y^2\right) + V_0 \mathbb{1}_{(x, y) \in \mathbb{R}^2_{>0}}
$$
It represents a potential step in the first quadrant.
Solving for outside of the first quadrant, I can make a guess at the eigenvector for the hamiltonian as,
$$
\psi(x, y) = \psi(x)\psi(y) = (Ae^{\alpha x} + Be^{-\alpha x})(Ce^{\beta y} + De^{-\beta y})
$$
which gives eigenvalues for $ (x, y) \notin \mathbb{R}^2_{>0} $ of $ \frac{-\hbar^2}{2m} (\alpha^2 + \beta^2) $. I can do similarly for the points inside the potential step and fit boundary conditions to find vectors for the full space.
Question How do I prove/disprove that the eigenvectors I guessed span my Hilbert space? What if I were working with more complicated hamiltonians?
I am worried that I might be missing vectors. I want to write simulations, so I need all of them (I think).
 A: I think I found my answer, please correct it if I am wrong.
In many Physics books, they mention the "completeness relation",
$$
 \int \partial E \ \ | E \rangle \langle E | = \mathbb{1}
$$
where $ \mathbb{1} $ is the identity operator.
So I think I need to prove this in order to demonstrate completeness. I have worked with wavelets, so below lead to my intuition.
Relation to Wavelets
I remember reading in Daubechies "Ten Lectures on Wavelets" that the vector group is a discrete frame if,
$$
A\mathbb{1} \leq \sum\limits_\alpha | \psi_\alpha \rangle \langle \psi_\alpha | \leq B\mathbb{1}
$$
for $ 0 < A \leq B < \infty $. Intuitively, this means that no vector will  be projected to null in this frame nor will the projection diverge.
Daubechies would prove this by taking the projection of arbitrary functions. For a continuous wavelet transform, she would prove that the group integral (my vocab might be off) resolved to a scaled identity for normalized vectors.
In math notation,
$$
\int \partial \phi(\alpha)\ \|\langle \psi_\alpha | f \rangle \|^2 = A \|f\|^2
$$
for $ \| \psi_\alpha \|^2 = 1 $
Where $ \phi(\alpha) $ is the left invariant group measure and for the unitary representation $ | \psi_\alpha \rangle = \hat{U}(\alpha) | \psi \rangle $. I am not a group theorist, so I might be a bit confused, but I think I am right.
Proving Completeness Convergence in the Norm
One must prove the similar relation,
$$
\int \partial \phi(\alpha)\ \|\langle \psi_\alpha | f \rangle \|^2 =  \|f\|^2
$$
for $ \| \psi_\alpha \|^2 = 1 $
And prove that the vectors are orthogonal, ie. $ \langle \psi_\alpha | \psi_\beta \rangle = 0 $ iff $ \alpha \ne \beta $.
