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In quantum mechanics, an operator $\hat{O}$ is related to its eigenkets $|o_i\rangle$ via the relation $$ \hat{O}|o_i\rangle = o_i |o_i\rangle$$ The eigenvalues $o_i$ gives the result of measuring the quantity represented by $\hat{O}$ on the state represented by $|o_i\rangle$.

At least in elementary presentations, the eigenvalues are taken to just be real numbers. But there are plenty of cases where it's somewhat more natural to take the possible outcomes of measurements to be represented by something other than some real numbers: for example, if we were measuring the position of a particle, then a coordinate-free representation of the possible outcomes would be a three-dimensional Euclidean space. So is there any way of doing the formalism so that we use (say) the points of a three-dimensional Euclidean space as the set of eigenvalues?

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  • $\begingroup$ Sure, you just make a vector of operators $(\hat{x},\hat{y},\hat{z})$ with eigenvalues $\mathbf{r} = (x,y,z)$. One can get more fancy and look at how such a vector of operators transforms under rotations or more general transformations, which would lead you to the study of tensor operators. But you might not want such mathematical detail. $\endgroup$
    – Mark Mitchison
    Commented Sep 9, 2015 at 19:15
  • $\begingroup$ So, this is close to what I want. But even if we have (say) vectors as the eigenvalues of our operator, that's still only appropriate if we make a choice of origin (in physical space), so that positions are representable as vectors. Is there some way of merely having eigenvalues which take values in a three-dimensional affine space, rather than a vector space? $\endgroup$
    – anygivenpoint
    Commented Sep 14, 2015 at 17:56

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Yes.

If you want to make a vector out of some operator do make sure the operator has for its components operators that mutually commute. And don't look at an operator with a name like $\hat L^2$ and mistakenly think it is the square of an operator $\vec{\hat L}$ and a vector of operators isn't really an operator even if it is an observable.

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  • $\begingroup$ Note that the components of the (pseudo)vector angular momentum operator do not commute with each other. $\endgroup$
    – Mark Mitchison
    Commented Sep 9, 2015 at 23:36
  • $\begingroup$ @MarkMitchison That is one of two reasons you don't want to think $L^2$ is the square of an operator, firstly a function from a Hilbert space to a tensor product of the Hilbert space shouldn't be composed with itself because the domain and codomain don't match and secondly the three "components" of angular momentum don't commute so it is not an observable vector of operators since it doesn't have eigenvectors in the Hilbert Space. $\endgroup$
    – Timaeus
    Commented Sep 9, 2015 at 23:50

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