Spectrum of the complex harmonic oscilllator

Question

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

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

Answer 1

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.