Undergrad Vector Operators in Quantum Mechanics: Adapting to Different Coordinate Systems

[andrewkirk]

It's not relevant, because I don't even assume that the “operator” has to have eigenvectors. An ordinary, scalar operator $\mathcal H \to \mathcal H$ will have eigenvectors, but for instance it is well known that the angular momentum “operator” $\vec J$ doesn't have eigenvectors.

This paragraph doesn't seem to relate to what I wrote. I didn't mention eigenvectors in that latest post, and they play no role in the argument it presents. The argument is based on rotational symmetries of the post-measurement wavefunction in the location basis.

What I am asking is something else. What should be the expression of such vector (or, for $\vec J$, pseudovector) operators; what kind of objects are they, and what conditions do they satisfy (how do you write that they are self-adjoint, for instance)?

The trouble is that thequestion is not well-defined. You ask what 'should' the expression of the vector $(X,Y,Z)$ be. 'Should' according to what principle? We already have an expression of it. It is $(X,Y,Z)$.

In your OP you said you would like to be able to refer to that triple in a way that does not refer to any spatial coordinate system. Implicitly you asked whether that is possible and if so, how it could be done.

My answer is that I am reasonably* confident that it is impossible, and my post above indicates why. You may not agree, but I don't think you can say it is not an answer or not relevant.

I'm very happy to have logical flaws in the above argument pointed out. That will be a learning opportunity for me, and I always relish those. But in the absence of such flaws being identified, it looks like a tough barrier to a coordinate-free representation of the location triple.

*Let's make that 'moderately'. it doesn't do to be too confident of anything in QM.

 

[vanhees71]

Sorry, vanhees71, but it's a mess mainly because almost no one has seriously attempted to answer the original question.

Your answer essentially adds nothing to what I had stated in the original post. I am not asking how to express the three Cartesian components of a vector operator in a particular basis of the 3D space, nor how they transform under $SO(3)$; I am asking how to express them without any choice of a basis in the first place.

It should really not be that difficult to read the question before answering it.

I have no clue what you want then. The components are always defined with respect to a basis. If you refer to Cartesian vector and tensor components then for a physicists these are a operator valued set $\hat{T}_{jk\ldots}$ that transform under rotations $R \in \mathrm{SO}(3)$ via a given unitary representation of this group (or its covering group $\mathrm{SU}(2)$) as
$$\hat{U}(R) \hat{T}_{jkl\ldots} \hat{U}^{\dagger}(R)=R_{ja} R_{kb} R_{lc} \cdots \hat{T}_{abc\ldots}.$$
It's not my fault, if you don't use the usually used terminology.

 

[David Olivier]

This paragraph doesn't seem to relate to what I wrote. I didn't mention eigenvectors in that latest post, and they play no role in the argument it presents.

I think it's basic QM that you can in principle perform a collection of measurements if there are eigenvectors corresponding to the resultant state. This in turn is possible if the observables commute. Since $X$, $Y$ and $Z$ do commute, I think this settles the issue.

You don't see the point in obtaining an expression of quantities such as position, momentum or angular momentum independently from a coordinate system; I do. See below.

I have no clue what you want then. The components are always defined with respect to a basis.

Thanks for informing me that components are defined with respect to a basis.

My question seems quite clear, and I've restated it enough. There is nothing strange or nonstandard in using mathematical objects other than numbers and collections of numbers. There are other things out there, such as vector spaces, tensors, topological spaces and so on. There is nothing unusual in expressing theorems, definitions and physical laws independently from collections of numbers. A vector space, in particular, is not a collection of numbers.

In all domains of physics, there is a natural strive to express laws and concepts independently from any choice of spatial basis. This means expressing them not with coordinates, but with more complex, albeit more abstract, objects.

In special relativity, for instance, the Minkowski vector space can be described as a four-dimensional real vector space $V$ equipped with a bilinear mapping $g: V \times V \to \mathbb R$ such that 1) there exists at least one vector $\vec v$ with $g(\vec t, \vec t) < 0$ and 2) for any two vectors $\vec t, \vec v$ with $g(\vec t, \vec t) < 0$, $\vec v \neq \vec 0$ and $g(\vec t, \vec v) = 0$, we have $g(\vec v, \vec v) > 0$. This is a full description of the metric properties of a Minkowski vector space without the slightest reference to a basis. The whole of special relativity can be expressed in such a way, and there are good reasons to do so.

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.

 

[andrewkirk]

I think it's basic QM that you can in principle perform a collection of measurements if there are eigenvectors corresponding to the resultant state.

Again, this does not address the argument I presented in #29. If anything it adds support to the argument by acknowledging that a collection of measurements is made rather than a single one.

You don't see the point in obtaining an expression of quantities such as position, momentum or angular momentum independently from a coordinate system

I never said that, and it is not true. I greatly prefer coordinate-free presentations, always using them when they are available. That's why I became interested in this thread. But they are not always available, and I'm not going to assume they must be available in every case just because I'd like the world to be that way. If you can disprove my argument above and go on to demonstrate that the post-measurement state of an object will be identical regardless of the directions from which its location is measured, I will be delighted.

 

[Physics Footnotes]

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):

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.;-)

 

[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 $\vec{J}^2$.