I saw in this MO post: Is there a machinery describing all the irreducible representations ? that classifying irreducible representations of the Weyl algebra is essentially intractable. My question is if it becomes more reasonable if one restricts to unitary irreps.
To be completely precise: By the Weyl algebra I mean the complex algebra generated by $1$, $a_{+}$, and $a_{-}$ subject to the relations $[a_{+},a_{-}] = 1$. A unitary representation is a complex vector space $V$ with a Hermitian form, on which the Weyl algebra acts in such a way that $a_{+}$ and $a_{-}$ are adjoint pairs (and $1$ acts as $1$). There is no topological component: $V$ need not be complete, the operators must be everywhere defined, and irreducible means no subrepresentations (not just 'no closed subrepresentations').
I know that the Fock space (generated by a vector $v$ with $a_{-}v = 0$) is one such irrep, and one can conjugate by automorphisms to the algebra to obtain other irreps: $a_{+} \mapsto \alpha a_{+} + \beta a_{-}$; $a_{-} \mapsto \overline{\beta}a_{+} + \overline{\alpha} a_{-}$; where $|\alpha|^2 - |\beta|^2 = 1$. The automorphisms where $\beta = 0$ do not give distinct irreps, so one gets a two-dimensional space of irreps (unless there are other redundancies here).
Any futher explanations or references would be great, thanks!
EDIT: Since I think I was unclear, my question is specifically about how unitary irreps of a Lie algebra are different from unitary irreps of the corresponding Lie group. The corresponding Lie group here (exponentiating the anti-self-adjoint elements) is the Heisenberg group, which has a unique unitary irrep (fixing a nonzero central action). But it seems that the Weyl algebra has multiple nonequivalent unitary irreps.