# Spectrum of the complex harmonic oscilllator

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

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

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

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.