Does every representation of the harmonic oscillator Lie algebra necessarily admit a basis of eigenfunctions? It is well-known in quantum mechanics that the harmonic oscillator Hamiltonian given by $\mathcal{H} = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}$ admits a basis of eigenfunctions on $L^2(\Bbb R,dx)$. There are many proofs of this ranging from hard analysis (brute force proving density) to Bargmann transform techniques to showing that the resolvent is compact. An important part of the analysis is that the Gaussian $e^{-\frac{x^2}{2}}$ is an eigenfunction with eigenvalue $0$.
Associated to this system is a very nice Lie algebra. Define the operator $a$ by
$$a = \frac{1}{\sqrt{2}}\left(\frac{d}{dx}+x\right)$$
and its (formal - I'm not going through the awful nitty gritty details of domains of definition, etc) adjoint
$$a^* = \frac{1}{\sqrt{2}}\left(-\frac{d}{dx}+x\right).$$
$\mathcal{H}$ has the following representation:
$$\mathcal{H} = a^*a.$$
Moreover, this triple of operators satisfies (on sufficiently nice functions):
$$[\mathcal{H},a^*] = a^*, \qquad [\mathcal{H},a] = -a, \qquad [a,a^*] = -1.$$
The quadruplet $\mathcal{H},a^*,a,1$ generates a Lie algebra. Because of these relations, applying $a^*$ repeatedly to the Gaussian leads to successive eigenfunctions (that they are in $L^2$ needs to be checked, but it is straightforward by induction) and indeed these eigenfunctions form a basis.

My question is this: does a representation of the Lie algebra
$$[a^*a, a^*] = a^*, \qquad [a^*a, a] = -a, \qquad [a,a^*] = -1$$
on an abstract separable Hilbert space $\mathfrak{H}$ (not necessarily $L^2(\Bbb R,dx)$) necessarily admit a basis of eigenfunctions for $a^*a$ assuming that $\ker a\neq \{0\}$? That is, if $\{\psi_i:i\in I\}$ denotes a basis for the kernel of $a$, does $\{(a^*)^m\psi_i:m\in\Bbb N_0,i\in I\}$ form a basis for $\mathfrak{H}$? Or is this somehow a happy accident of working on $\Bbb R$ (or, as we know, more generally $\Bbb R^n$)?
 A: Assume a unit $e$, and assume there is an underlying Hilbert space $H$. Then
$$
                        (a^*a)a^*-a^*(a^*a)=a^* \\
                    (a^*a-e)a^* = a^*(a^*a) \\
                    (a^*a-e-\lambda e)a^* =a^*(a^*a-\lambda e) \\
                      a^*(a^*a-\lambda e)^{-1}=(a^*a-(\lambda+1)e)^{-1}a^*
$$
If $E$ is the spectral measure of $a^*a$, Stone's formula for real $r < s$ is
$$
     \frac{1}{2}(E(r,s)+E[r,s]) \\
  =s\mbox{-}\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{r}^{s}(a^*a-(u+i\epsilon)e)^{-1}-(a^*a-(u-i\epsilon)e)^{-1}du,
$$
which leads to
$$
               a^*E(r,s) = E(r+1,s+1)a^*.
$$
Taking the adjoint yields
$$
               a E(r,s) = E(r-1,s-1)a.
$$
$E(r,s)H$ is in the domain of $a$ and $a^*$ for all finite $r,s$. The intervals can be open, closed, or half-open. $E$ is the spectral measure of the positive operator $a^*a$; hence $E(-\infty,0)=0$. Therefore
$$
     aE[0,1)=E[-1,0)a=0, \\
     a^{n}E[n-1,n)=0,\;\; n=1,2,3,\cdots.
$$
Then $a^*aE[0,1)=0$ forces $E[0,1)=E\{0\}$, which is the classical case where the spectrum of $a^*a$ is $\{0,1,2,3,\cdots\}$ and $a^*,a$ act as ladder operators. In the classical case, one also has $\mbox{dim}E\{0\}=1$; in your case, there may be multiple identical and orthogonal copies of the classical one-dimensional space.
