I have seen an equivalence claimed in a few places, but I do not know of a reference that actually proves it with details and it has been a while since I took graduate courses on all this. Apologies if this is something standard. A reference would be appreciated, particularly if it illuminates the general picture.


The simplest concrete version of my question is the following. Let


$$\mathcal{H} = -\Delta + |x|^2$$


be the quantum harmonic oscillator Hamiltonian in $\mathbb{R}^d$ and for $k\in\mathbb{N}$ define a Hilbert space by taking the closure of the Schwarz class functions $\mathcal{S}(\mathbb{R}^d)$ in the norm defined by
$$
\|f\|\_k^2 = \int\_{\mathbb{R}^d}|\mathcal{H}^k f(x)|^2dx
$$
I have seen it claimed in a few places (for example Proposition 2.3 in <https://aip.scitation.org/doi/abs/10.1063/1.5048726?journalCode=jmp>) that this norm is equivalent to the norm given by a combination of a weighted $L^2$ norm and an ordinary Bessel-Sobolev norm
$$
|||f|||\_k^2 = \|\mathcal{F}^{-1}(1+|\xi|^2)^{k}\mathcal{F}f\|^2\_{L^2(\mathbb{R}^d)} + \| |x|^{2k}f \|^2\_{L^2(\mathbb{R}^d)}
$$
I think I see that $\|\cdot\|\_k$ is controlled by $|||\cdot|||\_k$, but I do not know how to prove the reverse inequality.


Based on how I have seen this talked about (for example, the proof of that proposition in that reference) my understanding is that there is a simple proof of this that begins by viewing $\mathcal{H}$ as the operator with Fourier symbol given by $a(x,\xi) = 4\pi^2|\xi|^2+|x|^2$.


Does anyone here know how such a proof would go? Or of any proof of this claim?



Equivalence of Hilbert space norm associated to the harmonic oscillator and a sum of Sobolev and weighted $L^2$ norms
Chapter 9 of my book [https://www-users.cse.umn.edu/~garrett/m/v/current\_version.pdf](https://www-users.cse.umn.edu/%7Egarrett/m/v/current_version.pdf)
(that is mostly aimed at applications to automorphic forms) considers this. See section 9.8, in particular.


I really don't know of other references for this sort of computation/comparison, but, while it's a bit annoying, it's not toooo subtle. :)


