If $[\hat A,\hat B] = 0$, where $\hat A$ and $\hat B$ are operators, then the operators commute. This also means that, when applied to a wavefunction, that one can measure observables $A$ and $B$ in any order and in any sequence that one would like without getting a different answer. However, if one computes:

$$[\hat x, \hat p]\psi = i\hbar \ \psi$$

This implies that momentum and position do not commute, and this then implies that they cannot be measured together. However, this doesn't seem obvious to me. All it seems to me is it just multiplies $i \hbar$ to the eigenstate. It multiplies a constant to an eigenstate, just an observable like $\hat p$ does. It makes sense to me that, for order of measurement not to be matter, that they should commute, but it doesn't seem obvious to me that this particular answer implies that they are incompatible, if that makes sense.

I've been trying to understand the answer to my confusion in terms of linear algebra, since I feel relatively comfortable approaching learning this discipline in as much linear algebra recourse possible, but I haven't worked it out yet.

I suppose the exact question(s) I have is (are):

$1)$ Why does $[\hat x, \hat p]\psi = i\hbar \ \psi$ directly imply that $\hat x$ and $\hat p$ are incompatible? It just seems like a linear transformation to me with eigenvalue $i\hbar$ to me, just like $\hat p$ is a linear transformation with eigenvalue $p$.

$2)$ What kind of linear mappings are quantum mechanics operators, anyway? I understand that they map vectors to the same Hilbert space they all live in, but are observables operators change of bases for the Hilbert space? I am told that some eigenstates can be eigenfunctions of more than one observable -- does this mean they are members of a subspace of the Hilbert space that describe eigenstates that are eigenfunctions of those two operators? I'm used to understanding linear transformations as a matrix detailing the changes the bases take, and so trying to imagine observables operators in this respect seem obscure to me. What exactly does applying an operator to a wavefunction do to it in terms of a transformation other than mapping the vector to the same space?

  • $\begingroup$ A general operator $A$ acting on a state $| \alpha \rangle$ gives $| \beta \rangle$, where $| \beta \rangle$ is a new state, in general. Only when $| \alpha \rangle$ is an eigenvector of $A$, does $A |\alpha \rangle$ end up being equal to its eigenvalue $\times | \alpha \rangle$, and then $A$ is called an observable. $\endgroup$ – Avantgarde Oct 22 '18 at 15:41

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