quantum mechanics - How do we know that we have captured the entire spectrum of the Harmonic Oscillator by using ladder operators?

Consider standard quantum harmonic oscillator, $H = \frac{1}{2m}P^2 + \frac{1}{2}m\omega^2Q^2$.

We can solve this problem by defining the ladder operators $a$ and $a^{\dagger}$. One can show that there is a unique "ground state" eigenvector $\psi_0$ with $H\psi_0 = \frac{1}{2}\hbar\omega\psi_0$ and furthermore that given any eigenvector $\psi$ of $H$ with eigenvalue $E$, the vector $a^{\dagger}\psi$ is also an eigenvector of $H$ with eigenvalue $E + \hbar\omega$.

However, it is usually stated that we now have all eigenvectors of $H$ by considering all vectors of the form $(a^{\dagger})^n\psi_0$.

How do we know that we have not missed any eigenvectors by this process? e.g. how do we know that eigenvalues are only of the form $E_n = (n+\frac{1}{2})\hbar\omega$?

Also a slightly more technical question, how do we know that the continuous spectrum of $H$ is empty?

The technical details I am operating with are that $\mathcal{H} = L^2(\mathbb{R})$ and all operators ($H, P, Q$) are defined on Schwartz space, so that they are essentially self-adjoint with their unique self-adjoint extensions corresponding to the actual observables.

Answer

It is sufficient to prove that the vectors $|n\rangle$ form a Hilbert basis of $L^2(\mathbb R)$. This fact cannot be completely established by using the ladder operators. To prove that the span of the afore-mentioned vectors is dense in the Hilbert space, one should write down the explicit expression of the wavefunctions of the said vectors recognizing that they are the well-known Hilbert basis of Hermite functions. Since the vectors $|n\rangle$ are a Hilbert basis, from standard results of spectral theory, the operator

$$\sum_n \hbar \omega(n +1/2 ) |n\rangle \langle n | \tag{1}$$ (using the strong operator topology which defines the domain of this operator implicitly) is self-adjoint and its spectrum is a pure point spectrum made of the numbers $\hbar \omega(n +1/2 )$ with $n$ natural. This fact proves that the initial symmetric Hamiltonian operator you described in your post and defined on the Schwartz space admits at least one self-adjoint extension with the said spectrum (in particular no continuous spectrum takes place). To prove that it is the unique self-adjoint extesion, i.e., that the initial symmetric operator is essentially self-adjoint, the shortest way is to observe that the vectors $|n\rangle$ are necessarily analytic vectors of the initial Hamiltonian (notice that all the afore-mentioned vectors stay in the Schwartz space which is the initial domain) because they are eigenvectors. Since they are a Hilbert basis, their span is dense. Under these hypotheses, a celebrated theorem by Nelson implies that the initial symmetric Hamiltonian operator is essentially self-adjoint and thus (1) is the only self-adjoint extension of the initial symmetric Hamiltonian operator. As a final comment, it is interesting to remark that (1) is not a differential operator differently from the naive initial Hamiltonian which is a differential operator but it is not self-adjoint.