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Can the spectrum of a quantum mechanical operator contain both real and complex numbers?

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  • $\begingroup$ Can you clarify what part of your question isn't answered here: en.wikipedia.org/wiki/… $\endgroup$ – Rob Jul 21 '18 at 14:38
  • $\begingroup$ A spectrum is a property of an operator, not of a system. Do you mean the spectrum of the Hamiltonian operator? $\endgroup$ – J. Murray Jul 21 '18 at 14:39
  • $\begingroup$ @J. Murray, yes i mean a quantum mechanical operator. Any operator. Can its spectrum contain both complex and real values ? $\endgroup$ – OneTwoOne Jul 21 '18 at 14:43
  • $\begingroup$ If you want a good answer, you should edit your question to reflect what you're asking. $\endgroup$ – J. Murray Jul 21 '18 at 14:53
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The spectrum of an operator can be complex but not the spectrum of an observable, pretty much by definition. Consider a two state system with the operator

$$A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$

This operator has eigenvalues $\pm i$, which obviously are complex. But it is not an observable, because an observable by definition has to be self-adjoint, and $A$ isn't; we demand that an observable be self-adjoint precisely because it guarantees that all eigenvalues will be real.

Edit: I just realized you might be asking whether a single operator can have both real and non-real eigenvalues. In that case,

$$B = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

has spectrum $\{1, i, -i\}$.

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I would think the boson annihilation operator can be called a "quantum mechanical operator", and it has both real and complex eigenvalues.

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  • $\begingroup$ The OP might not know about coherent states yet. $\endgroup$ – Cosmas Zachos Jul 28 '18 at 16:32
  • $\begingroup$ @CosmasZachos : The OP can always look up "annihilation operator". Furthermore, I often have to repeat that answers are not just for the OP. $\endgroup$ – akhmeteli Jul 28 '18 at 18:10
  • $\begingroup$ Of course. But readers are lazy... A direct link might bring facts to them... $\endgroup$ – Cosmas Zachos Jul 28 '18 at 18:18
  • $\begingroup$ @CosmasZachos: Writers can be even lazier:-) $\endgroup$ – akhmeteli Jul 28 '18 at 18:29

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