I'm new to quantum mechanics, but not quite as new to linear algebra and operator theory, and trying to understand the nature of observations from a mathematical perspective.

Consider a two-dimensional Hilbert space with an orthonormal basis $|0\rangle$, $|1\rangle$. Define the Hermitian operator $P = |0\rangle\langle 0|$. $P\psi$ projects the state vector $\psi$ onto the line spanned by the ket $|0\rangle$. It has eigenvalue 1 corresponding to eigenstate $|0\rangle$, and eigenvalue 0 corresponding to eigenstate $|1\rangle$.

Suppose the state vector begins as $\psi = |1\rangle$. When we measure $\psi$ with the observable corresponding to $P$, by the Born rule we'll observe the eigenvalue 0 with probability 1.

But what eigenstate does $\psi$ take on after this measurement? It has to be projected onto $\text{span}\{|0\rangle\}$, but it's orthogonal to that subspace and such a projection produces the zero vector. That doesn't make any sense, because quantum states are rays (or equivalently unit vectors) in the Hilbert space.

There's an error in this logic somewhere, but I can't figure it out:

  • Is my definition of an observable too broad? I understand that not all observables have physical meaning, but I don't see any sources specifically tell me that an observable can result in an unnormalizable state. Requiring observables' Hermitian operators to be full-rank would repair this but I've never seen this stated explicitly.
  • Have I misconceived what an observable does to the quantum state upon measurement? I thought it projects $\psi$ onto the eigenbasis of the corresponding Hermitian operator.
  • Is this setup just completely artificial with no physical consequences, requiring astronomical luck to have $\psi = e^{i\theta}|1\rangle$?
  • $\begingroup$ What do you mean with your last point/question? $\endgroup$ Commented Jun 10, 2023 at 6:02
  • $\begingroup$ Why do you say the post-measurement state has to be projected onto $|0\rangle$? $\endgroup$ Commented Jun 10, 2023 at 6:23
  • $\begingroup$ @TobiasFünke Ah, good point. For almost all initial $\psi$, the measurement does actually project onto $|0\rangle$, right? And if initially $\langle 0 | \psi \rangle = 0$, does the measurement just not change $\psi$? $\endgroup$
    – wwww
    Commented Jun 10, 2023 at 6:50
  • $\begingroup$ No. The post-measurement state is (up to nornalization) the state projected on the eigensubspace of the measurement result. $\endgroup$ Commented Jun 10, 2023 at 6:52
  • $\begingroup$ You’re confusing the Born rule. If your initial state is $|\phi>$, then the chance of observing $i$ is $|<i|\phi>|^2$. In this case, since you start in $|1>$, you are guaranteed to stay in $|1>$ and you’ll observe $<1|P|1>=0$. Also, you’re choosing a strange convention here - we often label our states based on their eigenvalues so that we don’t have to specify them separately. $\endgroup$
    – Eric
    Commented Jun 10, 2023 at 7:26

1 Answer 1


I had a cascade of misunderstandings here:

  1. A linear operator generally does not have a single eigenspace. There is actually a different eigenspace for every eigenvalue. This was the core mistake on my part.
  2. The operator $P$ has two eigenspaces, one spanned by $|0\rangle$ with eigenvalue 1 and one spanned by $|1\rangle$ with eigenvalue 0.
  3. I got confused between the projection operator $P$ itself and the projection operators onto the eigenspaces, which are different things. The projection operator for the eigenvalue 1 is $|0\rangle\langle 0|$, and the projection operator for the eigenvalue 0 is $|1\rangle\langle 1|$. It is incidental that $P$ is equal to one of the projection operators, and this isn't true for general observables, which may not be projections at all.
  4. I misremembered the Born rule, which in this case actually results in the final state being $|1\rangle$ with probability 1.

I'm a bit embarrassed now to have made such silly mistakes, but hopefully this is useful to someone else.


Your Answer

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

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