Let $$ H_\lambda=-\frac{d^2}{dx^2}+\lambda^2 x^2,\quad\lambda>0. $$ It is known that the spectrum of $H_\lambda$ is the set $\{(2n-1)\lambda,n\in \Bbb N^*\}$. Now put $$ (U_\mu \phi)(x)= e^{\mu\over 2}\phi (e^{\mu}x)\mu \in \Bbb R. $$ It is easy to check that $\{U_\mu,\mu\in\Bbb R\}$ forms a one-parameter unitary group and that $$ U_\mu H_1 U^{-1}_\mu = e^{-2\mu}\bigl( -\frac{d^2}{dx^2}+ e^{4\mu}x^2 \bigr) ,\quad\mu\in\Bbb R $$ and can be analytically continued into regions of complex $\mu$. Hence for $\lambda,\mu\in\Bbb C$, we have $$U_\mu (H_1-\lambda) U^{-1}_\mu = e^{-2\mu}\biggl( -\frac{d^2}{dx^2}+ e^{4\mu}x^2-\lambda e^{2\mu}\biggr).$$ This seems to imply that the spectrum of the operator $-\frac{d^2}{dx^2}+ e^{4\mu}x^2$, $\mu\in \Bbb C$ is the set $\{(2n-1)e^{2\mu},n\in \Bbb N^*\}$: is this result true? And if the answer is affirmative, how can we rigorously prove it?

@AlexandreEremenko. Here is the definition of the spectrum Let $T$ be a closed linear operator from a complex Banach space $X$ into $X$ with dense domain $D(T)$. Then the resolvent set $\rho(T)$ of $T$ is defined to be the set of all complex numbers $\lambda$ for which $T-\lambda I: D(T)\to X$ is bijective and $(T-\lambda I)^{-1}:X\to D(T)$ is a bounded operator, where $I$ is the identity operator on $X$. The spectrum $\sigma (T)$ is simply the complement of $\rho(T)$ in $\Bbb C$.

the point spectrum $\sigma_p (T)$ of $T$ is the set of all complex numbers $\lambda$ such that $T-\lambda I$ is not injective. The continuous spectrum $\sigma_c (T)$ of $T$ is the set of all complex numbers $\lambda$ such that the range $R(T-\lambda I) $ of $(T-\lambda I)$ is dense in $X, (T-\lambda I)^{-1}$ exists, but is unbounded. The residual spectrum $\sigma_r (T)$ of $T$ is the set of all complex numbers $\lambda$ such that $(T-\lambda I)^{-1}$ is bounded, but the range $ R(T-\lambda I)$ is not dense in $X$. It is easy to see that $\sigma_p (T), \sigma_c (T)$ and $\sigma_r (T)$ are mutually disjoints and

$$ \sigma(T)=\sigma_p (T)\sqcup \sigma_c (T)\sqcup \sigma_r (T) .$$

  • $\begingroup$ The answer depends on what do you exactly mean by "the spectrum". For complex $\mu$ your operator may not have any eigenvalues in $L^2(R)$. $\endgroup$ Sep 25, 2022 at 13:39
  • $\begingroup$ @AlexandreEremenko. See above the definition $\endgroup$ Sep 25, 2022 at 19:17
  • $\begingroup$ You have to define what $D(T)$ is, and on which space does your operator act. $\endgroup$ Sep 25, 2022 at 20:26
  • $\begingroup$ It's domain is given by $D_\lambda=\{u \in L^2(\Bbb R): H_\lambda u \in L^2(\Bbb R)\}$ $\endgroup$ Sep 25, 2022 at 21:20
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    $\begingroup$ MR1204365 Bender, Carl M., Turbiner, Alexander Analytic continuation of eigenvalue problems. Phys. Lett. A 173 (1993), no. 6, 442–446. $\endgroup$ Sep 26, 2022 at 19:28

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Indeed, this is the result of Davies - Pseudo-Spectra, the Harmonic Oscillator and Complex Resonances (1982): The resolvent operator $(H-zI)^{-1}$ of $$H=-d^2/dx^2+cx^2,\;\;\operatorname{Re}c>0,\;\; \operatorname{Im}c>0,$$ is compact for all $z$ not in the spectrum consisting of the set $\{(2n-1)\sqrt c,\;\;n=1,2,3,\dotsc\}$. The spectrum is referred to as a "pseudo-spectrum", because the associated eigenfunctions do not form a basis of the Hilbert space.

  • $\begingroup$ Thank you a lot @Carlo Beenakker. My question is my proof is correct? $\endgroup$ Sep 25, 2022 at 19:19
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    $\begingroup$ I think so, with your definition of "spectrum". $\endgroup$ Sep 25, 2022 at 19:37

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