Firstly, let's dispense with the sub-thread started by
@andrewkirk concerning the (im)possibility of joint measurements of separate coordinates of a particle in QM. As a bonus, this will segue into the primary thread anyway.
For any (strictly speaking,
measurable) region ##\Delta \subset \mathbb R^3## one can construct a projection operator ##P_{\Delta}## with the property that for any pure state ##\psi##, ##\langle \psi | P \psi \rangle## represents the probability that the particle is found in the region ##\Delta##. In practice, ##P_{\Delta}## represents a particle (location) detector and there are no theoretical limits whatsoever on the shape or size of the region ##\Delta##. Any eigenvector of this projector will correspond to a particle localized within ##\Delta##, and this makes the use of a 3-dimensional Dirac delta function to describe the limiting case no more controversial than in one dimension.
Note that none of this requires talk of simultaneous measurements from different directions, or anything like that. We just have a single detector which does or doesn't fire. And this is exactly how things work in the real experimental situation too. No coordinates necessary! Of course if you
do want to construct this projector with coordinates you can do that too. You can represent a small cuboidal detector, for example, with the projector ##P_\Delta := P_{\Delta X}P_{\Delta Y}P_{\Delta Z}##.
And this talk of coordinates provides the segue I referred to above ;-)
In particular, if you want to do QM in a coordinate-free way (and you are absolutely right about the importance of doing this, as well as in your analogy of Minkowski Space), you have to choose the right mathematical language. It turns out that the vector space structure of Euclidean Space ##\mathcal E## is just the wrong way to go about it (even though it raises an interesting mathematical question, which I'm not sure of the answer to). For one thing, ##\mathcal E## just won't capture the non-commutative structure of QM, which is absolutely fundamental. (You still have to accommodate the structure of ##\mathcal E##, of course, because that's where the experiment lives; but you don't do that by making the fundamental theory Euclidean.)
So How To Remove Coordinates From The Picture?
The instant you commit to using self-adjoint operators to model observables, you are committing to coordinates, for the following reason. A self-adjoint operator is, at its base, a special kind of mapping called a Projection Valued Measure (PVM) $$P^A:\mathcal B(\mathbb R) \ni \Delta \mapsto P^A(\Delta) \in \mathcal P (\mathcal H)$$It is the projection operators on the right which represent the coordinate-free objects, while the mapping calibrates these questions against a copy of the real number line, based on your choice of coordinate system. The secret to a coordinate-free representation is to strip this arbitrary calibration away, and work purely within the image space, which is a structured set of projection operators.
Specifically, a real observable in QM is a calibration of a coordinate-free Boolean Algebra of projection operators. All the Boolean Algebras of all possible experimental arrangements mesh together in a non-commutative way to form a special type of structure I'll call a Hilbert Lattice, which is more often described as Quantum Logic (although you should not think of the word 'Logic' here as anything more profound that labeling a particular type of algebraic structure.)
Where Does The Euclidean Structure Fit In?
As I mentioned earlier, all experiments live in Euclidean Space (ignoring relativity for now!) and so its structure has to enter the picture somehow. But this is not done directly, by trying to glue quantum objects together to form vectors, but rather by representation theory. That is, the structure of ##\mathcal E## is imposed on the Hilbert Lattice by a representation of its automorphisms (which happen to manifest as unitary operators).
That is why
@vanhees71 comment, below, is important (even though it doesn't do what you were trying to do):
vanhees71 said:
As any vector operator the three operators representing the Cartesian components of the position operator transform as a vector (that's why it's called a vector operator in the first place), i.e., for ##R \in \mathrm{SO}(3)## you have a unitary representation ##\hat{U}(R):\mathcal{H} \rightarrow \mathcal{H}## (such a unitary representation must exist since a Galileian invariant QT must admit rotations as symmetry operators). The operators for the Carstesian components of the position operator thus transform under rotations as
$$\hat{U}(R) \hat{\vec{x}} \hat{U}^{\dagger}(R)=R \hat{\vec{x}}.$$
(Although I would have phrased it in terms of coordinate-free projection operators rather than coordinatized self-adjoint operators.)
So if 'coordinate-free' is your real ambition, I would forget about vector operators (except as an interesting mathematical puzzle) and focus on Boolean Algrebras, Lattices, and their Automorphisms (which happen to be induced by unitary operators).
If, on the other hand, you really want to know how to represent operator vectors abstractly, I'm afraid I may have wasted your time with all this.;-)