# Eigenvalues of position operator in higher dimensions is vector, not scalar?

If we accept that eigenvalues of operators show correspond with possible values measurements by that operator can take, then it makes sense to define the position operator $$X$$ on a "deterministic"/eigenvector ket "state" $$|x_0 \rangle$$ to be $$x_0$$. This is exactly what the Wikipedia page says https://en.wikipedia.org/wiki/Position_operator. However, this seems to only make sense when the position is a real number. Going to higher dimensions, the Wiki page https://en.wikipedia.org/wiki/Position_operator#Three_dimensions says to take $$\hat{\mathbf x} \psi := \mathbf x \psi$$. But it looks like the eigenvalue should now be the vector quantity $$\mathbf x$$!

Comments in this post How to define the position operator in higher dimensions? seem to indicate that we should think of the "eigenvalues" in the context of a tensor product. But to be honest, I have never heard eigenvalues taking values as vectors in some vector space; eigenvalues have always just been elements of the underlying scalar field.

If you want to model a measurement as a hermitian operator, you must assign distinct real numbers to the possible outcomes, whether they naturally correspond to real numbers or not. This is a limitation of the operator formalism and not a fundamental property of measurement in quantum mechanics.

If your position measurement amounts to noting which grain of a photographic plate darkened, you could number the grains 1, 2, ... in an arbitrary order, and that would be a valid representation of the measurement, though perhaps not a very useful one.

If you want the outcome of the measurement to be represented by a point in $$\mathbb R^d$$, you have to split the measurement into $$d$$ measurements of a single coordinate and imagine that you are performing all of them in sequence (which creates no problems since they commute). This is just to make the formalism happy, and has no meaning beyond that.

• Ok thanks for making clear what is and is not just formalism. I think this is often the most confusing and least explained part.
– D.R
May 1, 2023 at 23:11

Don't overthink it - the simplest way to think about the vector of position operators $$\hat{\vec x}$$ is just as notation: When someone writes that $$\hat{\vec x} \lvert \vec x_0\rangle = \vec x_0\lvert \vec x_0\rangle$$, all that means is that the equation holds in the components, i.e. for each of the position operators $$\hat{x}_i$$, we have $$\hat{x}_i\lvert \vec x_0\rangle = x_{0,i}\lvert \vec x_0\rangle$$ where $$x_{0,i}$$ is the $$i$$-th component of $$\vec x_0$$. Since all the $$\hat{x}_i$$ commute, such simultaneous eigenstates exist.

You can think of the position operator as a set of N operators that transform as a vector, furthermore since all these operators conmute among each other is it possible to have fixed eigenvalues of each of them (these eigenvalues also transform as a vector). Note the difference with the angular momentum vector, it also transform as a vector but since the underlying algebra is not trivial you can not have well defined eigenvalues of each one of them simultaneously.

• The fact that the set of eigenvalues of individual position operator components, ${\mathbf x} \equiv (x_1, \ldots, x_n)$ transforms as a vector is actually an important point May 2, 2023 at 10:10