Momentum operator in curvilinear coordinates

[ShayanJ]

Let's hear from Pauli:

The radial momentum operator defined through $\vec p_r f=\frac{\hbar}{ir}\frac{\partial}{\partial r}(rf)$ behaves in ordinary space in exactly the same way as the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$ in the half space. It is Hermitian, but its matrices cannot be diagonalised. However, the operator $p_r^2 f=-\hbar^2 \frac 1 r \frac{\partial^2}{\partial r^2}(rf)$ appearing in the Hamiltonian, can very well be diagonalized and has the eigenfunctions $\frac 1 r \sin{kr}$.

 

[DrDu]

Beyond all mathematical formalism it should be clear on physical grounds that there can't be a radial momentum operator:
It should generate a translation along r: $\psi(r)\to \psi(r-a)$. If the shift a is negative, part of the function will get shifted outside $[0,\infty]$ into the negative range. Hence this translation can't be unitary and its generator not self-adjoint.
On a closed interval [0, 1] this problem can be avoided as we can "fold back" that part which gets shifted out on one end on the other.

 

[vanhees71]

It's of course not $-\mathrm{i} \partial_r$ that represents a physical quantity as an operator, because it's of course not self-adjoint (not even Hermitean). Also the other operator discussed, which is made Hermitean by symmetrization (it's not even self-adjoint as pointed out before in #50) doesn't represent an observable, which one can easily interpret physically. This physical interpretation is possible by the symmetry analysis using the ray representations of the Galilei or proper orthochronous Poincare group underlying the derivation of the observable algebra, represented by the generators of the one-parameter subgroups.

The momentum operator is the generator of translations, and it's a (co-)vector given by
$$\vec{\hat{p}}=-\mathrm{i} \vec{\nabla}.$$
It's square is a scalar and as such invariant under isomorphisms. Thus one has to use the covariant expression
$$\Delta=\mathrm{div} \; \mathrm{grad}$$
in the nonrelativistic kinetic-energy operator
$$\frac{1}{2m} \hat{\vec{p}}^2=-\frac{1}{2m} \Delta$$
to get the correct Hamiltonian for the Schrödinger equation in position representation.

I guess, it was misleading to call the formal component $-\mathrm{i} \partial_r$ of the gradient in spherical coordinates $\hat{p}_r$, because it has not the meaning of a proper vector component. What's, however correct is
$$\vec{\nabla}=\vec{e}_r \partial_r + \vec{e}_{\vartheta} \frac{1}{r} \partial_{\vartheta} + \vec{e}_{\varphi} \frac{1}{r \sin \vartheta} \partial_{\varphi}.$$
Then you have to take into account the dependence of the basis vectors on position, when squaring the nabla operator.

That's all. It's just vector analysis in terms of (anholonomous!) orthonormal curvilinear coordinates, known already from the electrodynamics lecture! There's nothing specifically quantum mechanical here!

 

[DrDu]

Also the other operator discussed, which is made self-adjoint by symmetrization doesn't represent an observable, which one can easily interpret physically..

Dextercioby just provided all the mathematical details in post #50 why even this symmetrized operator isn't self-adjoint.

 

[vanhees71]

Dextercioby just provided all the mathematical details in post #50 why even this symmetrized operator isn't self-adjoint.

Good point; I've corrected my posting accordingly.

 

[vanhees71]

Ha!

I'd forgotten that (important) symmetrization trick too -- despite the fact that I've been crunching some quantum Runge-Lenz vector stuff in the hydrogen/kepler problem recently, where it's essential. (Sigh.)

So easy to overlook things like that...

The Runge-Lenz business is a bit cumbersome. I came to the conclusion that in this case it's easier to work in the position representation throughout (and use Mathematica to avoid boring technical calculations with derivatives:-)).

 

[strangerep]

The Runge-Lenz business is a bit cumbersome.

More than a "bit". My (by hand) computations of $[A_i, A_j]$ and ${\mathbf A}^2$ consume several pages of latex, even with lots of auxiliary utility formulas. This stuff must have been a massive PITA for people before computers were invented.

 

[vanhees71]

Yeah, it's going back to Pauli, who knew the Runge-Lenz vector from classical mechanics and thus was able to solve the hydrogen-atom problem in terms of matrix mechanics even before Schrödinger came up with his much more convenient wave-mechanics treatment. I'm always glad to live in a time where Computer algebra is available that helps us to avoid such cumbersome (but boring) calculations. Of course, it's worse doing it to get some familiarity with the operator algebra and Lie algebras and Lie groups.

 

[dextercioby]

Beyond all mathematical formalism it should be clear on physical grounds that there can't be a radial momentum operator:
It should generate a translation along r: $\psi(r)\to \psi(r-a)$. If the shift a is negative, part of the function will get shifted outside $[0,\infty]$ into the negative range. Hence this translation can't be unitary and its generator not self-adjoint.
On a closed interval [0, 1] this problem can be avoided as we can "fold back" that part which gets shifted out on one end on the other.

That's a nice insight. While I put all the gory details and shyan quoted Pauli's textbook (I sense Pauli quoting some original work of von Neumann), the "issue" of the so-called "radial angular momentum" had been solved without resorting to deficiency indices since 1973, see the original (behind a paywall, unfortunately) found here: http://scitation.aip.org/content/aapt/journal/ajp/41/8/10.1119/1.1987445

One can also read this: http://iopscience.iop.org/0143-0807/22/4/308/refs

 

[DrDu]

You might also enjoy reading Zhu, Klauder: Classical symptoms of quantum illnesses:
http://scitation.aip.org/content/aapt/journal/ajp/61/7/10.1119/1.17221

 

[dextercioby]

And I forgot to add the reference in the best QM textbook there is, section 6.2 of 1st volume of Galindo and Pascual, where my treatment is shortened and put in the full 3D context, where the deficiency indices differ.

 

[stevendaryl]

In case anybody else was as puzzled as I was, here's my resolution of a paradox involving the "momentum operator" p_r = -i \hbar (\partial_r + \frac{1}{r}).

What I'm about to say is presumably redundant with what's already been said, but is a less advanced argument.

On the one hand, we can prove that p_r has purely imaginary eigenvalues: Let \phi = \frac{1}{r} e^{-\lambda r}. Then p_r \phi = i \lambda \phi.

On the other hand, I thought I could prove that it was Hermitian. But that's a contradiction, because Hermitian operators have real eigenvalues. That was very puzzling. But then I saw that my proof had a mistake.

Here's the mistaken proof:

1. By definition, \partial_r^\dagger is that operator such that for all \phi and \psi: \int (\partial_r \psi)^* \phi \ r^2 dr d\Omega = \int \psi^* (\partial_r^\dagger \phi) \ r^2 dr d\Omega
2. We can rewrite the first integral: \int (\partial_r \psi)^* \phi \ r^2 dr d\Omega = \int (\partial_r \psi^*) \phi \ r^2 dr d\Omega
3. We can integrate by parts to get: \int (\partial_r \psi^*) \phi \ r^2 dr d\Omega = \psi^* \phi \ r^2 4 \pi|_{boundary} - \int \psi^* (\partial_r \phi \ r^2) dr d\Omega (where the first term on the right side of the equals is the boundary term).
4. The integral on the right side of the equals can be written as: - \int \psi^* (\partial_r \phi \ r^2) dr d\Omega= -\int \psi^* ((\partial_r + \frac{2}{r}) \phi)\ r^2 dr d\Omega
5. So, assuming that the boundary term is zero (WRONG!), we have: \int (\partial_r \psi)^* \phi \ r^2 dr d\Omega = - \int \psi^* ((\partial_r + \frac{2}{r}) \phi)\ r^2 dr d\Omega
6. From 5. and 1., we conclude \partial_r^\dagger = - (\partial_r + \frac{2}{r})
7. From 6., we conclude (-i \hbar (\partial_r + \frac{1}{r}))^\dagger = -i \hbar (\partial_r + \frac{1}{r})
8. So p_r^\dagger = p_r

The stupid error is to assume that the boundary term is zero. When you are dealing with cartesian coordinates, the boundary for an integral over x is -\infty, +\infty. If the wave function is square-integrable, then it has to vanish at both limits, so the boundary term goes to zero. In contrast, with the coordinate r,the boundary is 0, +\infty. The wave function has to vanish as r \rightarrow +\infty, but not as r \rightarrow 0.

So the proof that p_r is symmetric (that is, \langle p_r \psi |\phi \rangle = \langle \psi|p_r \phi\rangle) only works in the case that \psi^* \phi r^2 goes to zero as r \rightarrow 0. That will be true if \psi and \phi are well-behaved at r=0, but square-integrability doesn't imply that. The function \psi = \frac{1}{r} e^{-\lambda r} is square-integrable, even though it is not well-behaved at r=0.

 

[DrDu]

That's what I desperately wanted to show to you! However, your terminology is still not right. To fully specify an operator you have to specify not only how it operates on functions, but also it's range of definition.
The operator p_r is symmetric or Hermitian on e.g. the differentiable functions which vanish at r=0. However, it is then not self-adjoint, as its adjoint operator is defined on all differentiable functions, irrespective of whether they vanish at r=0 or not.
Hermiticity is not enough to ensure reality of eigenvalues, the operator has to be self adjoint!

 

[stevendaryl]

The operator p_r is symmetric or Hermitian on e.g. the differentiable functions which vanish at r=0. However, it is then not self-adjoint

That's why I avoided the word "self-adjoint". The terminology I had seen (for example, here: http://en.wikipedia.org/wiki/Hermitian_adjoint) uses "Hermitian" and "self-adjoint" interchangeably, and uses the word "symmetric" to mean \langle A \psi|\phi\rangle = \langle \psi| A \phi\rangle.

 

[dextercioby]

But you see, quantum mechanics pretty much uses infinite-dimensional spaces everywhere, while "hermitean" in the language of mathematicians is outdated. The terminology put forward by Stone at the beginning of the 1930s won: symmetric, essentially self-adjoint and self-adjoint. "Hermitean" is only kept in the context of finite-dimensional vector spaces and usually is an adjective next to the substantive "matrix".

 

[Demystifier]

And I forgot to add the reference in the best QM textbook there is, section 6.2 of 1st volume of Galindo and Pascual, where my treatment is shortened and put in the full 3D context, where the deficiency indices differ.

I like that book too. :)

 

[vanhees71]

Now you leave me confused. Are you agreeing that $-\mathrm{i} \vec{\nabla}$ is only working in Cartesian coordinates? To me that doesn't make sense, because it's a vector operator and thus independent of the use coordinate system. Of course, spherical coordinates are singular along the polar axis, but this shouldn't do any harm to the properties of this operator, because then I need more charts, to cover the manifold, which in non-relativistic QT as well as in relativistic QFT is good old Euclidean $\mathbb{R}^3$. The momentum operator itself has also a well-defined coordinate-independent meaning, i.e., it's the generator of translations.

The Hilbert space for non-relativistic single-particle physics is the one and only separable Hilbert space, realized as $\ell^2$ in matrix and as $\mathrm{L}^2(\mathbb{R}^3,\mathbb{C}$ in wave mechanics. There the operator $-\mathrm{i} \vec{\nabla}$ is essentially self-adjoint, and the translations can continued to a unitary representation on the entire Hilbert space. Also in the physics community, it's well established to call operators Hermitean (math jargon symmetric) and self-adjoint (which is more than Hermitean taking into account the domains and co-domains properly). I also like Galindo/Pascual for such subtleties. Another good source is

Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
http://arxiv.org/abs/quant-ph/9907069

and

Bonneau, Guy, Faraut, Jacques: Self-adjoint extensions of operators and the teaching of quantum mechanics, Am. Jour. Phys. 69, 322, 2001
http://arxiv.org/abs/quant-ph/0103153

 

[DrDu]

The momentum operator generates translations along some cartesian directions, i.e., apx+bpy+cpz should be well defined for all a,b,c. This will hold true in whatever coordinate system you express it, specifically, you may use radial coordinates.
This doesn't mean that the appearing derivatives constitute self-adjoint operators on their own.

 

[vanhees71]

Of course not! $\partial_r$ is not even self-adjoint on the usual Hilbert space. The reasons were discussed at length in this thread (which got even featured ;-)).

 

[dextercioby]

Post#1:

I. The classical Hamiltonian for a free 3D-particle in Cartesian coordinates is H = p²_x + p²_x + p²_z (units such as m=1/2). You can compute the 3 fundamental PBs, then use the Dirac quantization scheme to get the 3 Born-Jordan CCRs then the Stone-von Neumann theorem leads you to the Schrödinger quantization of momenta in Cartesian position representation as p_x = - i ħ ∂/∂x, etc. which are essentially self-adjoint on S(R^3). Problem fully solved. A connection to the representation theory of the Heisenberg group can be made. It delivers the same conclusions.

II. A problem for you: consider the classical 3D particle in spherical coordinates. Write its canonical Hamiltonian, then its 3 fundamental PBs, then use the Dirac quantization scheme to derive a “spherical equivalent” of the known 3 Born-Jordan CCRs, then use the Stone-von Neumann theorem and obtain the quantization of the 3 “spherical” momenta in the 'position representation' as differential operators. Compare to case I. Do you get contradictions? If so, where?

 

[dextercioby]

Post#2:

I see a simple geometry problem: In R^3 (realistic QM models always assume the freedom of infinite motion and also infinite time evolution), one's free to use any coordinates he likes to describe the infinite motion. There's an impressive list displayed in one of the classics: the 2 volume book of Morse and Feshbach.

Then what does a freedom of reparametrization of R^3 mean for QM? Well, essentially a chain of isomorphisms of Hilbert spaces. If R^3 is described in Cartesian coordinates, one has a simple identification of coordinates and (canonical) momenta via the Newtonian/Hamiltonian dynamics + Dirac "canonical quantization". One obtains the Born-Jordan CCRs and their representation on L^(R^3). But if R^3 is described in spherical coordinates (which are almost mandatory for a neat resolution of the quantum dynamics of the Hydrogen atom, for example), then the classical Hamiltonian momenta p_r, p_theta and p_phi don't get a Dirac "canonical quantization" (as you may have discovered in Post#1). We say that the 3 components of spherical momenta are not quantum observables, cause there's no (essentially) self-adjoint operator to describe them, whether the particle is in a potential field or free. The Hilbert space isomorphic to L^2(R^3) would then be L^((0,∞), r^2 dr) ⊗ L^2(S^2) and the only sensible observable in this space is necessarily the Hamiltonian, which, thanks to the nice work by T. Kato pioneered by F. Rellich, is always shown to be self-adjoint. Then you move to cylinder coordinates. Then to parabolic ones, then to ellipsoidal ones, etc. to check if the triplet of quantum (canonical) momenta are observables or not. Fortunately, the Hamiltonian is self-adjoint every time (its spectral equation will always be reduced to a Sturm-Liouville ODE). A thing which is actually cool, because the (time-independent) Hamiltonian through its spectral equation always provides us with a complete system of (generalized) states of feasible values of energy, but not necessarily of momentum. But, despite this considerations I outlined, a true description of quantization should be done from a pure diff-geom. treatment of the classical dynamics (-> Arnold, -> Marsden) which is to be quantized: I refer to what's known as geometric quantization which originated within the context of the Groenewold-van Hove no-go theorem,then with the work of Moyal and reached maturity through the works of Souriau and Kostant.

On the other hand, and here I’m trying to build a connection with what vH71 said several times, the representation theory of the Galilei group or the (restricted) Poincare group which he mentions always considers the Galilean space-time or the Minkowski space-time as being in a Cartesian spatial representation/parametrization. That is our classical and quantum fields are always functions of x,y,z,t, not of r, phi, z for example. I can only suspect/guess that, if one goes to GR with its simplest space-time (the Schwarzschild one), one should be forced to use a spherical (or other type of) "spatial" parametrization of the "spatial" 3D submanifold, so that the Weyl-Wigner-Bargmann-type representation theory (which one would like to carry over from QM and QFT into a QFT on a curved space-time) would necessarily consider (projective) representations of the symmetry groups of this GR space-time in which a Cartesian parametrization of the “spatial” 3D submanifold would no longer be possible. Would this be doable? I don't know... A somehow related problem, I guess, can be seen as trying to put a "classical" Dirac spinor in a curved space-time, thing which is possible (up to technicalities as described by Wald in his famous chapter 13) only because a physical curved space-time is locally flat, thus the known R^3 and its particular aforementioned Cartesian parametrization "creeps" into the “curvy” picture precisely to accommodate the Dirac spinors.

 

[strangerep]

Dex, I'm very interested to explore this stuff, but maybe it should go in a new thread (referencing this one)?

Anyway, the crucial difference between the Souriau approach and the usual approach is that (iiuc) Souriau concentrates on quantization applied to the symmetries of the solution space of a particular (classical) equation of motion. The background space (i.e., phase space) is just a mathematical artifact. What matters is the solution subspace thereof, and hence the symmetries of that subspace. The background coordinate system is, in the end, unphysical.

 

[vanhees71]

I had to look up the Groenewold-van Hove theorem ;-). That's very interesting and easy to prove. It only shows once more on a more formal level that what's called "canonical quantization" is at best a heuristical tool. As I said before, the observable algebra should be determined on physics grounds, and the most elegant and direct way are the symmetry considerations as, e.g., worked out in Ballentine's and Weinberg's (the latter of course also for relativistic QFT) textbooks. I don't know about "geometric quantization". Do you have a good review article on this approach? In some sense, of course, the symmetry approach is also geometric in some sense. This idea goes back to Riemann and most prominently Felix Klein in the 19th century, where they investigated the symmetry properties of geometries and the possibility to reconstruct geometries from and classify them according to their symmetry groups. This culminated in a "hype" about what then was called "the theory of invariants", and from a theoretical-physics point of view this was one of the most fruitful developments in mathematics since the discovery of calculus (Noether's theorem is one highlight, the use of group theory in QT by Weyl, van der Waerden from the mathematical and by Wigner, Bargmann et al from the physical side are others).

Concerning quantum theory in curved general-relativistic space time be warned! It's very complicated and not fully understood as far as I know, and I'm not an expert on it. This doesn't only hold for the quantization of the gravitational interaction which is very puzzling for decades now but even for the more simple problem to do relativistic QFT in a general-relativistic background spacetime (i.e., spacetime stays "unquantized" as in non-relativistic and special relativistic QT). There is quite some literature about this topic. The most simple spaces are not of the Schwarzschild type (I don't know whether there exists something about QFT with a Schwarzschild background spacetime) but the most symmetric Robertson-Walker-Friedmann-Lemaitre spacetimes. A good source on this is the famous review article by B. deWitt:

DeWitt, Bryce S.: Quantum field theory in curved spacetime, Phys. Rept. 19, 295–357, 1975
http://dx.doi.org/10.1016/0370-1573(75)90051-4

I'm sure there are newer papers on the subject but, as I said, I'm not an expert in these issues.

 

[stevendaryl]

There is something (actually, quite a few things) that I don't understand about the business of operators being self-adjoint. In curved space (consider just two-dimensions, for simplicity), what I would think would be the position space representation of the inner product of two wave functions \psi and \phi is:

\langle \psi | \phi \rangle = \int \psi^* \phi \sqrt{g} dx dy

where g is the determinant of the metric tensor. In this case, we have, with p_x = -i \hbar \partial_x,

\langle p_x \psi | \phi \rangle = \langle \psi | {P}_x \phi \rangle + ST

where {P}_x is the operator defined by {P}_x f = -i \hbar \frac{1}{\sqrt{g}} \partial_x (\sqrt{g} f) = (p_x - i \hbar \frac{1}{\sqrt{g}} \partial_x \sqrt{g}) f, and where ST is the "surface term": \int ({P}_x (D_x (\psi^* \phi))) \sqrt{g} dx dy

So p_x is only symmetric if ST = 0 and g is constant.

(Note: An identity that can be used is that: \frac{1}{\sqrt g} \partial_x \sqrt{g} = \Gamma^i_{i x}, where \Gamma is the connection coefficients (implicit summation over the dummy index i) So the operator P_x can actually be written in the form: P_x f = (p_x - i \hbar \Gamma^i_{i x}) f, which seems like sort of a covariant derivative, except that since f is a scalar, there's no difference between partial derivatives and covariant derivatives.)
​

So if an operator being symmetric is a necessary (but maybe not sufficient) condition for being an observable, then the usual momentum operator isn't an observable in curved space. What does that mean?

 

[strangerep]

\langle \psi | \phi \rangle = \int \psi^* \phi \sqrt{g} dx dy

where g is the determinant of the metric tensor. In this case, we have, with p_x = -i \hbar \partial_x,

In the earlier spherical-polar conundrum, the resolution involving recognizing the the radial momentum should be written (classically) as something like $\hat e_r \cdot \nabla$, and then symmetrized.

In your case, maybe one should start from $\hat e_x(z) \cdot \nabla \, \Big|_z$, where $z$ denotes a point on the manifold $M$, and the $\hat e_x(z)$ is a unit vector in the tangent space at $z$, i.e., $TM_z$. So when one moves from point to point on $M$, one must also adjust $\hat e_x$ accordingly, and one must work in the 2nd tangent space $TTM$, iiuc. (In high-falutin' language, one must choose a ``horizontal/vertical'' decomposition of $TTM$ which is independent of local coordinates -- which is essentially what's happening here, afaict.)

But I could be wrong.

 

[vanhees71]

The point is to formulate your observables in a manifestly covariant way. In non-relativistic quantum theory in the position representation ("wave mechanics") this boils down to express everything in terms of the classical differential operators grad, curl, and div or in terms of the nabla calculus. This includes constructs like $-\mathrm{i} \vec{r} \times \vec{\nabla}$ for orbital angular momentum. Then there is no problem with writing down the Hamiltonian in terms of the operator algebra. Sometimes operator-ordering problem occur. I think then one has to more or less guess the right form of the Hamiltonian, relying on more or less handwaving concepts like Weyl ordering. In QFT often normal ordering helps to avoid some complications with renormalization parts of Feynman diagrams but at the same time it can be problematic concerning, e.g., gauge symmetries. The symmetries are the most helpful properties in finding first the operator algebra and then the correct Hamiltonian for a given problem. Last but not least one should always be aware that the final judgement about theories and models will always be experiments and observation. I think there is no generally valid scheme to guess the Hamiltonian from the classical analogue of a given problem, and sometimes there's no classical analogue at hand (e.g., Bose-Einstein condensation, superconductivity and superfluidity, etc.).

 

[TrickyDicky]

There is something (actually, quite a few things) that I don't understand about the business of operators being self-adjoint. In curved space (consider just two-dimensions, for simplicity), what I would think would be the position space representation of the inner product of two wave functions \psi and \phi is:

\langle \psi | \phi \rangle = \int \psi^* \phi \sqrt{g} dx dy

where g is the determinant of the metric tensor. In this case, we have, with p_x = -i \hbar \partial_x,

\langle p_x \psi | \phi \rangle = \langle \psi | {P}_x \phi \rangle + ST

where {P}_x is the operator defined by {P}_x f = -i \hbar \frac{1}{\sqrt{g}} \partial_x (\sqrt{g} f) = (p_x - i \hbar \frac{1}{\sqrt{g}} \partial_x \sqrt{g}) f, and where ST is the "surface term": \int ({P}_x (D_x (\psi^* \phi))) \sqrt{g} dx dy

So p_x is only symmetric if ST = 0 and g is constant.

(Note: An identity that can be used is that: \frac{1}{\sqrt g} \partial_x \sqrt{g} = \Gamma^i_{i x}, where \Gamma is the connection coefficients (implicit summation over the dummy index i) So the operator P_x can actually be written in the form: P_x f = (p_x - i \hbar \Gamma^i_{i x}) f, which seems like sort of a covariant derivative, except that since f is a scalar, there's no difference between partial derivatives and covariant derivatives.)
​

So if an operator being symmetric is a necessary (but maybe not sufficient) condition for being an observable, then the usual momentum operator isn't an observable in curved space. What does that mean?

The point is to formulate your observables in a manifestly covariant way. In non-relativistic quantum theory in the position representation ("wave mechanics") this boils down to express everything in terms of the classical differential operators grad, curl, and div or in terms of the nabla calculus.

This is how I think of this(let me know if it is not correct). Whether expressing operators in curvilinear coordinates or in curved space one must keep them linear in QM by fiat, so in the case of operators acting on scalar functions like the Hamiltonian or kinetic energy one can use for example the Laplace-Beltrami linear operator in curved spaces but for vectorial operators like momentum QM's ban on nonlinear operators only leaves open the option vanhees refers to: "express everything in terms of the classical differential operators grad, curl, and div", no curved covariant derivative as in stevendaryl's example.

 

[vanhees71]

This I don't understand. The operator $\vec{\nabla}$ is linear by definition. In, e.g., spherical coordinates, it's
$$\vec{\nabla} \psi=\vec{e}_r \partial_r \psi+\frac{\vec{e}_{\vartheta}}{r} \partial_{\vartheta} \psi + \frac{\vec{e}_{\varphi}}{r \sin \vartheta} \partial_{\varphi} \psi.$$

 

[TrickyDicky]

This I don't understand. The operator $\vec{\nabla}$ is linear by definition. In, e.g., spherical coordinates, it's
$$\vec{\nabla} \psi=\vec{e}_r \partial_r \psi+\frac{\vec{e}_{\vartheta}}{r} \partial_{\vartheta} \psi + \frac{\vec{e}_{\varphi}}{r \sin \vartheta} \partial_{\varphi} \psi.$$

Sure, that was my point too. I was trying to address stevendaryl's question: "then the usual momentum operator isn't an observable in curved space. What does that mean?"
So you just wrote the gradient in spherical coordinates, with of course a non-coordinate(nonholonomic) basis as espherical curvilinear coordinates are orthonormal like you also pointed out in a previous post and the basis dependence on position is canceled out, but you cannot do that for true spatial curvature as stevendaryl noticed. Then I'm basically saying again what all of you have already been repeating in this thread: one cannot then single out the radial component and declare it an operator on its own, for one thing the noncoordinate basis makes it not viable already before geting into most of the considerations that have been commented in the thread about self-adjointness, hermiticity, etc(wich are all correct of course). If one tries to use a coordinate basis one must introduce nonlinearity in the operator that is not possible in QM by definition .
Then I made a distinction between the gradient operator which is vector valued used for $\hat p$ and the scalar-valued Laplacian operator used for $\hat H$, in the latter case, the component operators issue doesn't come up of course and we don't have any problem applying it to any manifold, i.e. its generalization the Laplace-Beltrami op.) maintains the linearity when applied to curved spaces like for instance spherical harmonics problems can be fitted into when constrained to $S^2$. So it seems it all comes down to the vectorial nature of the momentum operator and that the authors of the paper linked in the OP ignored the basis-dependence intrinsic to QM dynamics introduced by Planck constant h.

 

[vanhees71]

I guess, here is some confusion. Of course, the momentum opertor in Euclidean space is always the same, no matter in which curvilinear coordinates you express the operator $\vec{\nabla}$.

Another thing is when you have a curved, i.e., non-Euclidean, space. Then the symmetries of this space tell you, whether there is one or more than one momentum operator. There's a momentum operator, if the space admits one or several translation(s), and the generator of the translation is a momentum operator.

 

[TrickyDicky]

I guess, here is some confusion.

Can you quote the part you find confused and specify how it is wrong? I'm not saying anything disagreeing with the rest of your post that I'm aware of.

 

[vanhees71]

In the thread it looked as if different things have been mixed up. On the one hand there is the gradient operator in Euclidean space, expressed in curvilinear coordinates. This is a covariant operator and as such it doesn't make a difference whether I describe it in terms of Cartesian or any curvilinear coordinates. It's almost trivial that the momentum operator doesn't depend on the coordinates you use to express it.

On the other hand there are non-Euclidean spaces, where you have to figure out whether there's a momentum operator at all or not. If there is one, it must be possible to express it in terms of any (local) coordinate system, i.e., it's generally (diffeomorphic) covariant.

 

[TrickyDicky]

In the thread it looked as if different things have been mixed up. On the one hand there is the gradient operator in Euclidean space, expressed in curvilinear coordinates. This is a covariant operator and as such it doesn't make a difference whether I describe it in terms of Cartesian or any curvilinear coordinates. It's almost trivial that the momentum operator doesn't depend on the coordinates you use to express it.

Yes, when using nonholonomic basis where i.e. $\hat p_x$ is a bona fide operator but the key issue is that it does depend on the coordinates when using a coordinate-basis.

On the other hand there are non-Euclidean spaces, where you have to figure out whether there's a momentum operator at all or not. If there is one, it must be possible to express it in terms of any (local) coordinate system, i.e., it's generally (diffeomorphic) covariant.

And it seems there is no such momentum operator in those spaces, while there is a Hamiltonian operator. Note that this doesn't arise in curved spacetime where by virtue of the equivalence principle one can always use orthonormal frames locally.