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Let's define the harmonic oscillator $H = -\Delta+x^2$ in a domain $\Omega$ of $\mathbb R^d$. Thus, we consider the Dirichlet eigenvalue problem $$ (H - \lambda^2)u (x) = 0, \ x \in \Omega ; \ \text{ and } \ u(x) = 0 \ x \in \partial \Omega.$$
Denote the spectral projection to the eigenvalue $ k = \lambda^2 $ by $P_k$.

My question is: Are there any $L^p$ bounds results for eigenfunctions of the Hermite operator $H$ that have the form
$$ \| P_k u \|_{L^p(\Omega)} \leq k^{\rho(p)} \| u\|_{L^2(\Omega)},$$ where $p$ would be in $[2, 2d/(d-2)]$ and the exponent $\rho$ as function of $1/p$.

Note that in the case of $\Omega = \mathbb R^d$ there are many results, Karadzhov 1994, Thangavelu 1998, Koch and Tataru 2004.

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    $\begingroup$ The nonnegative integers are the eigenvalues on the whole space but not in $\Omega$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 10, 2023 at 12:43
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    $\begingroup$ You denoted by $P_k$ the spectral projection relative to the eigenvalue $k=\lambda^2$...I am referring to that. $\endgroup$
    – Giorgio Metafune
    Commented Aug 10, 2023 at 14:32
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    $\begingroup$ Please correct, then. Do you mean that $P_k$ is the spectral projection relative to the eigenvalue $\lambda_k$? $\endgroup$
    – Giorgio Metafune
    Commented Aug 10, 2023 at 18:29
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    $\begingroup$ Where you wrote $−∆+x^2$, the spacing surrounding the $\text{“}{+}\text{”}$ sign was not there, and I edited it to say $-\Delta+x^2.$ Sometimes MathJax is not as good as LaTeX when you use code like ∆ instead of \Delta. $\endgroup$
    – Michael Hardy
    Commented Aug 10, 2023 at 21:27
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    $\begingroup$ On bounded domains, this question is entirely related to the asymptotics of $\lambda_k$ as a function of $k$: you have that $\int |\nabla P_k u|^2 \leq \lambda_k \int u^2$ by definition, and then you apply Sobolev embeddings (unless you are interested in the sharp constant term by term). As I understand it, the heuristic is usually that if your potential is nice, the asymptotics of $\lambda_k$ are the same as without the potential. The problem on $\mathbb{R}^n$ (or unbounded domains), on the other hand, seems quite different and more challenging. $\endgroup$
    – user378654
    Commented Aug 11, 2023 at 1:36

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