This question already has an answer here:
- Why are Only Real Things Measurable? 7 answers
Can the spectrum of a quantum mechanical operator contain both real and complex numbers?
-1
$\begingroup$
$\endgroup$
marked as duplicate by AccidentalFourierTransform, Jon Custer, John Rennie, Cosmas Zachos, Qmechanic♦ Jul 28 '18 at 16:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
-
$\begingroup$ 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
2
$\begingroup$
$\endgroup$
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 $\{1, i, -i\}$.
1
$\begingroup$
$\endgroup$
I would think the boson annihilation operator can be called a "quantum mechanical operator", and it has both real and complex eigenvalues.
-
$\begingroup$ 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
