# On the spectrum of a quantum mechanical system [duplicate]

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

## marked as duplicate by AccidentalFourierTransform, Jon Custer, John Rennie, Cosmas Zachos, Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 28 '18 at 16:06

• Can you clarify what part of your question isn't answered here: en.wikipedia.org/wiki/… – Rob Jul 21 '18 at 14:38
• 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
• @J. Murray, yes i mean a quantum mechanical operator. Any operator. Can its spectrum contain both complex and real values ? – OneTwoOne Jul 21 '18 at 14:43
• 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

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

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

• The OP might not know about coherent states yet. – Cosmas Zachos Jul 28 '18 at 16:32
• @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
• Of course. But readers are lazy... A direct link might bring facts to them... – Cosmas Zachos Jul 28 '18 at 18:18
• @CosmasZachos: Writers can be even lazier:-) – akhmeteli Jul 28 '18 at 18:29