The issue on existence of inverse operations of $a$ and $a^{\dagger}$ I have asked a question at math.stackexchange  that have a physical meaning.

My assumption: Suppose $a$ and $a^\dagger$ is Hermitian adjoint operators and $[a,a^\dagger]=1$. I want to prove that there are no inverse operators for $a$ and $a^\dagger$.

I thought that this assumption purely mathematical, but I have no answers there. Maybe I am missing something?
I will clarify that $a$ and $a^\dagger$ are just Creation and annihilation operators for quantum harmonic oscillator.
 A: The ground state of the harmonic oscillator $|0\rangle$ obeys
$$a|0\rangle = 0$$
which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you will never get $0$ back so it implies that there's no operator $a^{-1}$ that would satisfy
$$ a^{-1}a = {\bf 1}.$$
On the other hand, there is an inverse from the opposite side that obeys
$$aa^{-1} = {\bf 1}.$$
The action of this $a^{-1}$ on $|n\rangle$ is simply defined as $|n+1\rangle/\sqrt{n+1}$ or whatever coefficient is needed for it to be inverse. I can write this one-sided inverse operator as $a^\dagger (aa^\dagger)^{-1}$ which is well-defined because $aa^\dagger$ only has nonzero eigenvalues.
For $a^\dagger$, the claims are reverted, of course. There exists an inverse that obeys
$$ (a^{\dagger})^{-1} a^\dagger = {\bf 1}$$
but not the other one.
