Vector Operators in Quantum Mechanics: Adapting to Different Coordinate Systems

[David Olivier]

Thanks. That's more than I expected!

In the statement that describes a PVM, what is ##\mathcal B(\mathbb R)##? The set of measureable subsets of ##\mathbb R##? It's not clear to me why we can't have ##P^A(\Delta)## for ##\Delta \in \mathcal B(\mathcal E)##.

“and work purely within the image space, which is a structured set of projection operators”. I suppose this is ##\mathcal P(\mathcal H)##. How is this defined? Is it the set of linear ##p: \mathcal H \to \mathcal H## such that ##p \circ p = p##, or that are self-adjoint with eigenvalues 0 and 1, or something else?

I've found a Hilbert lattice defined in substance as a set of closed linear subspaces of ##\mathcal H##, that form a lattice for inclusion. This is a lot for me! Do you have any references that explain some of this?

Is the resultant picture independent also of the origin? And - better still - of the Galilean frame of reference?

 

[Physics Footnotes]

@David Olivier The reason I left out the details is that you had indicated that you didn't want to deal with the subtleties in this thread. So here's a bit more:

• Yes, ##\mathcal B (\mathbb R)## is the Borel Algebra of measurable subsets of ##\mathbb R##. You can set up a PVM over any measurable space you like (e.g. ##\mathcal B (\mathcal E)##, as long as you define what that means!). I was just pointing out that you get self-adjoint operators when you make that algebra ##\mathcal B (\mathbb R^n)##. I suppose leaving ##\mathcal E## abstract, there might be a theorem that gives a representation of a coordinate-free vector operator of sorts, but that is a problem I've never explored. This might be the right way for you to think about this issue though.
• The set ##\mathcal P (\mathcal H)## denotes the orthogonal projection operators on ##\mathcal H##. That is, ## P^2=P=P^{\dagger}##; the first equality defining projections, while the second guarantees orthogonality. For convenience, I call an orthogonal projection a projector. And yes, the spectrum of any projector is ##\{0,1\}##
• Note that the image of a projector is a closed subspace of ##\mathcal H##, and it is common practice to work with the lattice of closed subspaces of a Hilbert Space, rather than with ##\mathcal P (\mathcal H)##.
• There is no concept of a spatial origin until you impose one. So yes, the group of automorphisms includes translations.

This way of studying QM is very old now. I learned it from the now classic books by Mackey (Mathematical Foundations of QM), Piron (Foundations of Quantum Physics), Jauch (Foundations of Quantum Mechanics), Varadarajan (Geometry of Quantum Mechanics; 2 vols), and others I can't remember off hand. There would surely be some simpler more modern introductory tomes by now, but I'm afraid I'm not familiar with them.

 

[vanhees71]

It is a rather strange and exceptional fact that quantum mechanics are usually expressed with reference to a coordinate system. I was asking if there are known ways to do so. Despite having no answer to that effect, I believe that there probably are, because it is quite normal to want to do so. If you have no interest in the issue, no problem, though.

I try once more to understand what you want. Of course, you can formulate classical mechanics in terms of abstract (Euclidean) vectors, which are simply invariant under rotations.

In QT, observables are represented by self-adjoint operators, which of course you can group into vector components and also operator valued invariant vectors. This is so, because by construction the QT must admit the fundamental space-time symmetry group (at least the part continuously connected to the unit operator).

This is the Galilei group for non-relativistic, i.e., the semidirect product of the temporal and spatial translations, rotations, and Galiei boosts. The symmetries are represented, and this is very important, by socalled unitary ray representations, which can be lifted to a unitary representation of a central extension of the covering group. This results in an 11-dimensional Lie group, where the mass operator is introduced as a central charge of the Lie algebra of the Galilei group and where the rotation group SO(3) is substituted by its coverning group SU(2). The former results in the mass superselection rule in non-relativistic QT the latter in the possibility of half-integer spin.

By definition the operatorvalued vectors are invariant under rotations (i.e., the components transform as given in my previous posting). It's, however important to remember that these vectors are in general not determined, except if all 3 components commute (like for position or momentum). E.g., the vector ##\vec{J}## of angular momentum is almost always never determined but only one component and ##\vec{J}^2##.