Graduate Can imaginary position operators explain real eigenvalues in quantum mechanics?

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Imaginary position operators may yield real eigenvalues in quantum mechanics when applied to certain non-hermitian operators. Specifically, using the imaginary position operator, represented as ix, in integrals can result in real (though negative) eigenvalues, contrasting with the expected complex eigenvalues from the normal position operator, x. The discussion highlights a lack of existing literature on this unconventional approach, raising questions about the validity and origins of using imaginary position eigenvalues. The concept appears to stem from non-hermiticity in operator theory, suggesting that further exploration into complex operators may be warranted. Overall, the use of imaginary position operators presents a novel perspective in quantum mechanics that merits additional investigation.
SeM
Hello, some operators seem to "add up" and give real eigenvalues only if they are applied on the imaginary position, ix, rather than the normal position operator, x, in the integral :

\begin{equation}
\langle Bx, x\rangle
\end{equation}

when replaced by:\begin{equation}
\langle Bix, ix\rangle
\end{equation}

or\begin{equation}
\langle Bix, x\rangle
\end{equation}

So using (2) or even (3) I get real (but negative) eigenvalues, instead of complex eigenvalues. I have not found any literature on such an absurd thing, as the imaginary position, however, being the only answer to these operators, I am wondering if anyone can point to some further literature on complex operators and complex eigenvalues in QM and whether (2) and (3) make any sense at all

Thanks
 
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What's ##B##? Without clear definitions, I cannot make any sense of the expressions above. Where does this idea of complex/imaginary position eigenvalues come from?
 
It's an idea that comes from a calculation that sums up only if I use an imaginary position operator. Apparently, it is so non-heard of that I will leave it as it is. It has to do with non-hermiticity, where B (not hermitian) only gives a real value on that integral if the position is imaginary.

Thanks!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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