Post Undeleted by José Figueroa-O'Farrill
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Let me try to answer this question. I apologise if my notation is slightly different, since I will work in some more generality, since the equivariance properties of the creation and annihilation operators are actually more transparent, I believe, relative to the the general linear group instead of the orthogonal group. Also the fact that this is the harmonic oscillator is a red herring. In the case of the harmonic oscillator, we have to introduce more structure, reducing the group of symmetries.It is also in this case, where the grading to be discussed below coincides (up to a choice of scale) with the energy of the system.

The automorphism group of $\mathfrak{h}$ is the group $\operatorname{O}(E\oplus \operatorname{Sp}(E\oplus E^*)$ of linear transformations of $E\oplus E^*$ which preserve the split-signature symplectic inner product defined by the dual pairing:$$\langle \omega\left( (x,\alpha), (y,\beta) \rangle right) = \alpha(y) -\alpha(y) + \beta(x).$$

The subgroup of $\operatorname{O}(E\oplus \operatorname{Sp}(E\oplus E^*)$ which acts on $V_k$ is the general linear group $\operatorname{GL}(E)$ and hence $V_k$ becomes a $\operatorname{GL}(E)$-module. In fact, $V_k$ is graded (by the grading in the symmetric algebra of $E^*$ or equivalently the degree of the polynomial):

Both of these maps are $\operatorname{GL}(E)$-equivariant, and this is perhaps the most invariant statement I can think of concerning the creation and annihilation operators.

Now, of particular interest are the unitary representations of $\mathfrak{h}$ and this is in any case the original setting of this question. I will not do this in detail, but it is not hard. Just introduce a positive-definite inner product on $E$, hence on $\operatorname{Sym}^pE^*$ for every $p$, whence ultimately on $V_k$ after a choice of inner product on $W_k$. It is usual to make $W_k$, and hence $V_k$, into a complex representation, in which case the inner product extends to a hermitian inner product on $V_k$. Finally, it is then usual to consider the Hilbert space completion of $V_k$.Then it is the orthogonal group $\operatorname{O}(E)$ preserving the inner product on $E$ which acts on $V_k$ preserving the inner product. The inner product on $E$ also defines musical isomorphisms between $E$ and $E^*$, whence we can attach to every $x \in E$ not just an annihilation operator, but also a creation operator.The equivariance properties I mentioned above now restrict to equivariance under $\operatorname{O}(E)$.

Post Deleted by José Figueroa-O'Farrill
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