Usually, if we have a state $|\psi\rangle$, and have to measure an observable $A$, then all we do is expand $|\psi\rangle$ in terms of the eigenvectors of observable A, and then the probability of measuring an eigenvalue is basically $|\langle u_i|\psi\rangle|^2$ where $|u_i\rangle$ is the eigenstate of $A$. Basically, it is possible to write down the eigenvalue equation $A|\psi\rangle = a|\psi\rangle$.

What if we are given a state $|\psi\rangle$ explicitly in a problem (say, we had $|\psi\rangle = (6, 3i, 4+5i)^T$) such that it can never be written as an eigenvalue equation for an observable $A$? How would we measure the observable for that given state, then? Most of the time, if a state is given explicitly, it's usually that it will be a superposition of eigenvectors of the given observable to measure. What if it isn't?

In short, we can only use $|\langle u_i|\psi\rangle|^2$ if $|\psi\rangle$ can be written as a superposition of eigenkets of $A$. What if $|\psi\rangle$ is such that we cannot write it as a superposition of the eigenkets of the observable we wanna measure?

  • 2
    $\begingroup$ Do you have a single example for such a case? (See e.g. physics.stackexchange.com/q/54154/50583 or any of the many other questions discussing the spectral theorem for why this can't happen) $\endgroup$
    – ACuriousMind
    Dec 31, 2022 at 15:15
  • 3
    $\begingroup$ Well, ignoring some mathematical subtleties, it is always possible to write $|\psi\rangle$ in terms of the eigenvectors of a self-adjoint operator/ of an observable. Why do you think otherwise? $\endgroup$ Dec 31, 2022 at 15:16
  • $\begingroup$ @ACuriousMind I don't have any particular example. This was just a doubt I had thought about while speculating on various questions that may come for the exam. Here, I was worried about a possibility of a question appearing wherein it asks us to measure an observable for a state and that state is 'explicitly' given to us such that it cannot be written as a superposition of eigenstates of the observable. $\endgroup$ Dec 31, 2022 at 15:21
  • $\begingroup$ @TobiasFünke Say we have already been given a state (as mentioned in the question) explicitly (say [3, 4i, 5+6i]) then the question is to measure an observable A in the state [3, 4i, 5+6i], and it turns out that while solving you find that you cannot find a way to express the state as a superposition of eigenkets of that observable. So, my question was is such a case physically possible or are projective measurements aren't enough to solve such problems? $\endgroup$ Dec 31, 2022 at 15:26
  • 3
    $\begingroup$ ignoring mathematical subtleties, observables always have a complete set of eigenstates, so it is always possible to expand any state in terms of eigenstates. $\endgroup$ Dec 31, 2022 at 15:27

1 Answer 1


As $A$ is an observable, it can always be diagonalized by the spectral theorem. Furthermore, the eigenvectors of $A$ will always form an orthonormal basis from the space you’re studying ($\ell^2$ in this case). As the set of eigenvectors of $A$ forms an orthonormal basis, any vector from the space can be written as a linear combination of this set of vectors.

As long as you define $|\psi \rangle$ and $A$ correctly, you can always do it.

  • $\begingroup$ I understood that, but my question was if |ψ⟩ was already numerically given, then if a question was asked to measure an observable in that state, then is the question wrong if that state cannot be broken down as superposition of eigenkets of the observable? $\endgroup$ Dec 31, 2022 at 15:32
  • 1
    $\begingroup$ I think such a question cannot exist. $A$ must be an hermitian operator. $\endgroup$
    – ErikLAndre
    Dec 31, 2022 at 15:34
  • $\begingroup$ I read somewhere about non-hermitian observables (in the field of quantum information theory). Could it be possible that this is the case here? $\endgroup$ Dec 31, 2022 at 15:47
  • 2
    $\begingroup$ @ErikLAndre Your second sentence should be more precise: In the case of a degenerate eigenvalue, the corresponding (normalized) eigenvectors will not automatically form an orthonormal basis. (Take the unit operator as a simple example.) An exact formulation would be that "one can always find an orthonormal basis of eigenvectors of a hermitean operator" (disregarding the additional subtleties in the case of a continuous spectrum). Apart from that, your answer is, of course, correct. $\endgroup$
    – Hyperon
    Dec 31, 2022 at 15:55
  • 2
    $\begingroup$ @PrathamHullamballi unless you provide a specific example, the discussion is moot. There are PT-symmetric theories (see here ) to evade hermiticity but you can still always expand everything in terms of eigenstates. $\endgroup$ Dec 31, 2022 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.