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?


1 Answer 1


Chapter 9 of my book https://www-users.cse.umn.edu/~garrett/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. :)

  • $\begingroup$ Thank you, this looks perfect. Much appreciated. $\endgroup$ Sep 28 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.