Why don't non-Hermitian operators with all real-eigenvalues correspond to observables? [duplicate]

Suppose you could construct an operator that was non-Hermitian but had all real eigenvalues or could at least be restricted in a way to create only real eigenvalues, why would this operator not correspond to an observable quantity?

marked as duplicate by user36790, Bill N, CR Drost, Qmechanic♦Sep 15 '15 at 5:24

Post measurement you want the eigenspaces to be orthogonal (and to have the projection onto the eigenspace to be entangled with a state of the measurement device).

So you want the different eigenspaces to be orthogonal. And you want to be able to evolve to a post measurement state that has the right kinds of states. So really it is about the kinds of end states you can evolve to. So that's what you want, let's talk about why your proposal doesn't get you that.

Just because the eigenvectors are real does not mean that eigenvectors with different eigenvalues are orthogonal.

The orthogonal decomposition is the first essential thing you want. The requirement that the eigenvalues be real isn't as big a deal, you might even want to model decay (poorly) with a complex energy. But you still want the different eigenspaces to be orthogonal.

And you do want more than that. You really want the eigenvectors to contain a maximal orthonormal set, and for a maximal set of commuting operators you get a unique such set, the set of mutual eigenvectors.

And it is after you have this that you can, for instance, finally bring up probability theory (and make a sample space and all that).

Now there is a whole branch of physics (with thousands of papers) based on the operators you mention. But when it comes time to do probability and compute expectation values, they ... change the geometry .. thus making the eigenspaces orthogonal in the new geometry, thus making the operators exactly the normal operators (pun not intended).

So considering non-Hermitian operators gains you nothing in the end. To get the probability part, you pick a geometry where the observables are Hermitian anyway so you could have just worked with that geometry and the Hermitian operators on it. Now, it doesn't necessarily make it wrong if you get the right answers and if doing it in a particular way happens to be computationally easier or otherwise nice then good for you. But non-orthogonality of the eigenvectors is a problem.

And since I mentioned Hermitian and non-Hermitian operators I do think it is a good time to point out then when you look for a geometry to make the eigenvectors orthogonal you could be doing this because you are aiming for a self-adjoint operator rather than just aiming for a Hermitian operator. Mostly on that front I just wanted you to know that both are out there. And having real eigenvalues won't give you either if the eigenvectors aren't orthogonal.

I haven't seen a paper that says that any self adjoint operator has an experimental realization. So it could be easier to catalog how things don't work rather than to think every nice enough operator can be measured.

• There are self-adjoint operators with no eigenvectors at all, or with eigenvectors that do not form an orthonormal basis. And not in some obscure mathematical setting, it is sufficient to take a particle in a non-confining potential (e.g. a step potential, a $\sin$ potential,no potential,...). Self-adjointness is a necessary but not sufficient condition to have an orthonormal basis of eigenvectors. On the other hand, self-adjointness of the Hamiltonian is sufficient for the existence and uniqueness of the unitary dynamics. – yuggib Sep 15 '15 at 8:37
• Or more in general, self-adjointness of the generator is sufficient to have a unitary representation of the symmetry group. Therefore your answer, implying that self-adjointness is sufficient to get orthonormal bases of eigenvectors, is highly inaccurate. Sadly, general QM is set in infinite dimensional spaces and there linear operators are not matrices. In addition, the so-called continuous spectrum (the one where there are no eigenvectors) is related to the existence of unbound states, while the discrete spectrum is associated to bound states. And that is quite important in physics. – yuggib Sep 15 '15 at 8:43
• As I often experienced, you are always ready to criticize answers given by others. But you should care about the quality of your own answers first. – yuggib Sep 15 '15 at 8:46
• @yuggib I apologize if my post was unclear. I tried to say you might be going for orthogonal eigenvector because you are aiming for self adjoint rather han aiming for Hermitian, I wasn't saying that was sufficient. I've edited. And I never said this was a good answer. The first paragraph is the answer and it is pretty poor, the rest was just criticism and trying to point out the literature and prior work. I suspected the question was already asked earlier and fleshing out the first paragraph would be the best way to answer. Being terse there meant someone else could expand in their own answer. – Timaeus Sep 15 '15 at 15:50
1) If all the eigenvalues of an operator are real, then it is Hermitian. You can see this by writing the operator (call it A) in the eigenvector basis. Then A has all real eigenvalues along its diagonal and zeros everywhere else. Therefore, $A^\dagger = A$ which means it is Hermitian.
2) Many of the operators that we call "observables" are the generators of transformations. For example: J (angular momentum) is the generator of rotations $e^{i\theta J}$, P is the generator of translations $e^{ixP}$, and H is the generator of time translation $e^{itH}$. If these transformations are to be unitary to conserve probability, then the generator must be Hermitian. For example unitary means (where $\theta,x,t$ are real): $$(e^{ixP})^\dagger e^{ixP}=I$$ $$e^{-ixP^\dagger} e^{ixP}=I$$ $$-P^\dagger + P =0$$
• To make at Javier's comment clearer: for the reasons in Timaeus's answer, you only change basis through orthonormal transformations. But not every operator with real eigenvalues can be diagonalized by such a transformation. Therefore your first statement is wrong; real eigenvalues are a necessary but not sufficient condition of self-adjointness. Use the matrix $\left(\begin{array}{cc}2&3\\4&5\end{array}\right)$ as a simple example: it has real eigenvalues, but is not Hermitian. – WetSavannaAnimal Sep 15 '15 at 2:38
• In fact, in part 2 of my answer, I again assumed that the operator P was diagonalizable so in the second equation $[P^\dagger,P]=0$ and I could slide them into the same exponent. – Gary Godfrey Sep 15 '15 at 3:54