Can the spectrum of a quantum mechanical operator contain both real and complex numbers?
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Can you clarify what part of your question isn't answered here: https://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics#Measurement_probabilities_and_wave_function_collapse – Rob Jul 21 '18 at 14:38
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A spectrum is a property of an operator, not of a system. Do you mean the spectrum of the Hamiltonian operator? – J. Murray Jul 21 '18 at 14:39
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@J. Murray, yes i mean a quantum mechanical operator. Any operator. Can its spectrum contain both complex and real values ? – Q_p Jul 21 '18 at 14:43
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If you want a good answer, you should edit your question to reflect what you're asking. – 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 $\{1, i, -i\}$.
Javier
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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.
akhmeteli
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@CosmasZachos : The OP can always look up "annihilation operator". Furthermore, I often have to repeat that answers are not just for the OP. – akhmeteli Jul 28 '18 at 18:10
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Of course. But readers are lazy... A direct link might bring facts to them... – Cosmas Zachos Jul 28 '18 at 18:18
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