# If states cannot be written as superposition of eigenkets of observable, then how do we measure an observable for that state?

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?

• 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) Dec 31, 2022 at 15:15
• 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? Dec 31, 2022 at 15:16
• @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. Dec 31, 2022 at 15:21
• @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? Dec 31, 2022 at 15:26
• ignoring mathematical subtleties, observables always have a complete set of eigenstates, so it is always possible to expand any state in terms of eigenstates. Dec 31, 2022 at 15:27

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.
• I think such a question cannot exist. $A$ must be an hermitian operator. Dec 31, 2022 at 15:34