1
$\begingroup$

I am familiar with the traditional definition of the position and of the momentum operator in several dimensions. I do not like it, as it uses from the very beginning a concrete Cartesian basis. What I am looking for is a definition in the following form:

Let $V$ be a real vector space. Let $\mu$ be a measure on $V$. Let $L^2(V,{\mathbb C})$ be the Hilbert space of square integrable $L^2$ functions with respect to the measure $\mu$. Denote the Hilbert space scalar product by $\langle \phi | \psi \rangle = \int_V \bar{\phi} \cdot \psi d \mu$. Then the position operator $Q$ is the function $Q\colon L^2 (V, {\mathbb C}) \to L^2 (V, {\mathbb C})$ defined by ... and the momentum operator ...

Question: How can I fill in the dots in above definition? I am interested in an abstract construction from the very beginning.

This means: Using ${\mathbb R}^3$ or a specific basis in $V$ and then abstracting away from the specifics by using morphisms is considered cheating in the context of this question. Using some fundamentally covariant structure on $V$ is fine. Using categorial constructions which later allow for natural isomorphisms between them in the sense of category theory is fine. Using a different Hilbert space is fine.

Goal: I want to understand multidimensionality in QM along the same lines as I understand it in SRT and GRT. I do not want to start with some concrete coordinates, then add Lorentz-transformation or general coordinate transformations - I want to start from the abstract clean mathematical concept of a Minkowski space with a (3,1) index quadratic form or from an abstract differentiable manifold, where the Lorentz-invariance or the general covariance is already encoded into the mathematical object as such.

$\endgroup$
8
  • 1
    $\begingroup$ Surely it's more like $Q:L^2(V,\mathbb C)\to V\otimes L^2(V,\mathbb C)$? (I did some Googling and apparently you might need a bit more than typical $\otimes,$ so I'm not answering.) Also, I don't think taking standard QM and replacing $\mathbb R^3$ with Minkowski space is physically relevant. If you have a particle in spacetime, you want it to exist at all times, not at just one event. You need QFT to really do relativistic quantum physics, in which the position operator is demoted to a parameter for a family of operators. $\endgroup$
    – HTNW
    Apr 27, 2022 at 1:23
  • $\begingroup$ How would I do spectral analysis when I have an operator where domain and codomain are different? Some hints to your Google results would be helpful since I did not succeed with my Googling attempts thus far. $\endgroup$ Apr 27, 2022 at 9:40
  • 2
    $\begingroup$ I answered a similar question here, but I am not aware of the subtleties @HTNW is refering to (@HTNW : can you provide some references you found on that ?). You can also take a look at the answers below this other question $\endgroup$ Apr 27, 2022 at 12:56
  • 1
    $\begingroup$ @Nobody-Knows-I-am-a-Dog You do not want $Q:H\to H$ because there's not enough "space" in the output to put the necessary information. Physics trumps mathematics when you're setting up a problem. (Unless you pick a basis and break $Q$ into components.) I refer to this note on Wikipedia about tensoring things onto $L^p$ spaces; but I remember seeing a suggestion that the technical stuff collapses if you assume $V$ is a Hilbert space? I'm sure "spectral analysis" generalizes to $Q:H\to V\otimes H$ (though I wouldn't know). $\endgroup$
    – HTNW
    Apr 27, 2022 at 18:55
  • 2
    $\begingroup$ @Nobody-Knows-I-am-a-Dog You can at least "obviously" generalize eigenvalues and eigenvectors to the equation $Q|\psi\rangle=\mathbf r\otimes|\psi\rangle.$ Don't know how to extend this to the whole spectrum, but that's why we typically just pick a basis and fall back to the components of $Q.$ $\endgroup$
    – HTNW
    Apr 29, 2022 at 19:44

1 Answer 1

1
$\begingroup$

What you want is impossible - there is no "abstract position operator" because quantization can change with coordinates. Famously, quantization is not a functor, and in fact choosing different coordinates for your classical system can produce different quantized systems or fail altogether, see also this answer of mine.

Furthermore, starting with something like $L^2(V)$ is really the wrong way to go about trying to abstract quantum mechanics: The fundamental structure of quantum mechanics is the algebra of observables and the Hilbert space its represented on, not some measure space $(V,\mu)$. And when we construct the algebra of observables from the algebra of classical functions on phase space, we pick the functions $x_i$ and $p_i$ of some particular Darboux coordinate system as our "position" and "momentum" operators - there is no notion of "coordinate transformation" there that we could preserve, since due to Gronewold-van Howe any non-linear expressions in $x_i$ and $p_i$ will not necessarily be preserved after quantization - that is, different coordinates $Q_i(x,p)$ and $P_i(x,p)$ will not necessarily fulfill $\hat{Q}_i(x,p) = Q_i(\hat{x},\hat{p})$, even after resolving ordering ambiguities in the second expression, so we do not preserve the "freedom of coordinate choice" in the course of quantization.

The most "abstract" definition of the position and momentum operators is via the Stone-von Neumann theorem: We start with the abstract algebra of $n$ position and momentum operators with Lie bracket $[x_i,p_j] = \delta_{ij}$ and then ask what its possible representations are - and the Stone-von Neumann theorem says there is, up to isomorphism, only a single irreducible representation, namely that on $L^2(\mathbb{R}^n)$ with $x_i$ as multiplication in the $i$-th variable and $p_i$ as differentiation. So we really need not worry about some more general construction of the quantum mechanical Hilbert space: The Stone-von Neumann theorem guarantees we can always choose our familiar $L^2(\mathbb{R}^n)$.

Since you mention Minkowski space: quantum mechanics of single relativistic particles is a hack, a set of tools that are useful but ultimately inconsistent, and the correct relativistic quantum theory is quantum field theory, where you do not have position operators in the traditional sense (see Reeh-Schlieder theorem, Newton-Wigner localization).

$\endgroup$
1
  • $\begingroup$ Much food for thought for weeks. THNKS. Happy. But unhappy with QFT. We plug in the reals and their topology instead of obtaining it as emerging from a lower level principle. Unhappy due to some supposedly "fishy" math parts it is said to contain. Unhappy due to my quite limited understanding of it. Your post hit the nail on the head and is VERY helpful. $\endgroup$ May 11, 2022 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.