Momentum operator in curvilinear coordinates

[DrDu]

I think that conclusion is a little strong, but I don't want to argue about it. However, I remember from studying quantum mechanics with periodic boundary conditions that if $x$ and $x+L$ represent the same physical location, then we replaced the usual commutation relation:

$[p, x] = -i \hbar$

by one that's appropriate for the ambiguity in $x$:

$[p, e^{\frac{2 \pi i x}{L}}] = \frac{2 \pi \hbar}{L} e^{\frac{2 \pi i x}{L}}$

The function $e^{\frac{2 \pi i x}{L}}$ is single-valued, even though $x$ is not.

We have used this (see e.g. Shyans post #13). The interpretation is however slightly different: x can be made single valued, e.g. taking mod_L. However this way it get's discontinuous and the commutation relations will only hold for exponentials $\exp(ikx)$ for $k=2\pi/L$. So this is at variance with eq. 4 which forms the starting point of the paper cited.

 

[vanhees71]

For example, in this paper, equation 3.2.4:
http://users.ece.gatech.edu/~alan/ECE6451/Lectures/StudentLectures/Brown_3p2_HydrogenAtom.pdf

This source clearly shows, how this is meant. Of course, this operator is NOT the operator representing the radial component of momentum in spherical coordinates. That's of course
$\hat{p}_r=-\mathrm{i} \vec{e}_r \cdot \vec{\nabla} =-\mathrm{i} \partial_r.$
Of course it is also true that
$\hat{p}_r^2 \psi(r) \neq -\Delta \psi(r),$
because the correct operator for the kinetic energy is given by the Laplace(-Beltrami) operator.

A naive derivation, aka "canonical quantization", is almost always wrong. You are lucky in Cartesian coordinates. The correct way to derive it is to use group-theoretical methods and write everything covariant. In this case that's the usual vector calculus, and the Laplace operator in spherical coordinates (applied to radial-symmetric wave functions) reads
$$\Delta \psi(r)=\frac{1}{r} \partial_r^2[r \psi(r)]=\frac{1}{r} \partial_r [\psi(r)+r \psi'(r)]=\frac{1}{r}[2 \psi'(r)+r \psi''(r)]=\psi''(r)+\frac{2}{r} \psi'(r).$$
The same holds true for the angular piece. It's correct expression can be written has $\hat{\vec{L}}^2/r^2$, where $\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}$.

Another nice example, where the canonical quantization fails is the quantization of the spinning top, which can be used as a (rough) model for the rotation bands of molecule spectra.

 

[stevendaryl]

This source clearly shows, how this is meant. Of course, this operator is NOT the operator representing the radial component of momentum in spherical coordinates. That's of course
$\hat{p}_r=-\mathrm{i} \vec{e}_r \cdot \vec{\nabla} =-\mathrm{i} \partial_r.$

The point that has been made in other papers about this definition of "radial momentum" is that it is not Hermitian:

$\hat{p}_r = \vec{e}_r \cdot \vec{p} = -i \hbar \frac{\partial}{\partial r}$
$(\hat{p}_r)^\dagger = \vec{p} \cdot \vec{e}_r = -i \hbar (\frac{2}{r} + \frac{\partial}{\partial r})$

So $\hat{p}_r$ defined that way isn't a Hermitian operator, and so, under the usual Rules of Quantum Mechanics, can't represent an observable.

The alternative definition of $p_r = -i \hbar (\frac{1}{r} + \frac{\partial}{\partial r})$ is Hermitian. (It is equal to $\frac{1}{2} (\hat{p}_r + (\hat{p}_r)^\dagger)$).

 

[DrDu]

The alternative definition of $p_r = -i \hbar (\frac{1}{r} + \frac{\partial}{\partial r})$ is Hermitian. (It is equal to $\frac{1}{2} (\hat{p}_r + (\hat{p}_r)^\dagger)$).

But it isn't self-adjoint, so it can't be an observable, either.

 

[stevendaryl]

The point that has been made in other papers about this definition of "radial momentum" is that it is not Hermitian:

$\hat{p}_r = \vec{e}_r \cdot \vec{p} = -i \hbar \frac{\partial}{\partial r}$
$(\hat{p}_r)^\dagger = \vec{p} \cdot \vec{e}_r = -i \hbar (\frac{2}{r} + \frac{\partial}{\partial r})$

So $\hat{p}_r$ defined that way isn't a Hermitian operator, and so, under the usual Rules of Quantum Mechanics, can't represent an observable.

The alternative definition of $p_r = -i \hbar (\frac{1}{r} + \frac{\partial}{\partial r})$ is Hermitian. (It is equal to $\frac{1}{2} (\hat{p}_r + (\hat{p}_r)^\dagger)$).

So that actually is the starting point of the disputed paper. By definition, if $\hat{A}$ is an operator, then $\hat{A}^\dagger$ is defined by:

$\langle \psi | \hat{A}^\dagger | \phi \rangle = \langle \phi | \hat{A} | \psi \rangle^*$

For the Hilbert space of square-integrable functions in 3-dimensional space, this implies:

$\int \psi^* (\hat{A}^\dagger \phi) \sqrt{g} dr d\theta d\phi = \int (\hat{A} \psi)^* \phi \sqrt{g} dr d\theta d\phi$

where $g$ is the determinant of the metric, which is just $g = r^4 sin^2(\theta)$ in the usual spherical coordinates in flat 3-space. It follows that:

$(-i \hbar \partial_r)^\dagger = -i \hbar (\partial_r + \frac{\partial_r \sqrt{g}}{\sqrt{g}}) = -i \hbar(\partial_r + \frac{2}{r})$

So, really, that's the point of the article, which is that unless $(\partial_j g) = 0$ (which is only true in general for Cartesian coordinates $x_j$ in flat space), then $-i \hbar \partial_j$ will not be a Hermitian operator, so it can't represent an observable.

 

[stevendaryl]

But it isn't self-adjoint, so it can't be an observable, either.

Are you using "self-adjoint" to mean $\hat{A}^\dagger = \hat{A}$? Then it certainly is self-adjoint.

By definition of adjoint with the usual Hilbert space of square-integrable functions on 3-space,

$\int \psi^* ((\partial_r)^\dagger \phi) \ r^2 sin^2(\theta) dr d\theta d\phi = \int (\partial_r \psi)^* \phi \ r^2 sin^2(\theta) dr d\theta d\phi$
$= \int \partial_r (\psi^* \phi \ r^2 sin^2(\theta)) dr d\theta d\phi - \int \psi^* (\partial_r \phi \ r^2 sin^2(\theta)) dr d\theta d\phi$

The first integral goes to zero, and the second one is equal to

$- \int \psi^* (\partial_r \phi + \frac{2}{r}) \ r^2 sin^2(\theta) dr d\theta d\phi$

(because you have to take a derivative of $r^2 sin^2(theta)$ as well as $\phi$)

So the conclusion (as I pointed out in another post) is that $(\partial_r)^\dagger = - (\partial_r + \frac{2}{r})$.
This implies that $(-i \hbar(\partial_r + \frac{1}{r}))^\dagger = -i \hbar(\partial_r + \frac{1}{r})$. So it's self-adjoint.

 

[stevendaryl]

The real point of the article, now that I've thought about it a little more, is that the adjoint of a differential operator involves the metric:

In curvilinear coordinates $x^a$, with corresponding metric $g_{a b}$ and determinant $g$,

$\partial_a^\dagger = - (\partial_a + \frac{1}{\sqrt{g}} \partial_a \sqrt{g})$

So $-i \hbar \partial_a$ is not self-adjoint, except in the special case of Cartesian coordinates in flat space.

 

[DrDu]

The point is not about $p_r$ being symmetric but about the ranges of definition of $p_r$ and $p_r^\dagger$ not to coincide.

 

[stevendaryl]

The point is not about $p_r$ being symmetric but about the ranges of definition of $p_r$ and $p_r^\dagger$ not to coincide.

I'm not sure I understand what that sentence means. What does it mean for $p_r$ to be symmetric?

 

[DrDu]

Hermitian and symmetric are synonymous.

$= \int \partial_r (\psi^* \phi \ r^2 sin^2(\theta)) dr d\theta d\phi - \int \psi^* (\partial_r \phi \ r^2 sin^2(\theta)) dr d\theta d\phi$

The first integral goes to zero,

Edit: Ups, that was not what I wanted to write. Take the function $\phi=r^{-1} \exp(-r)$. It is square integrable and has a well defined first derivative. Hence you could include it into the range of definition of $p_r$. However, then your can't include it into the range of definition of $p_r^\dagger$ as $\langle \phi |p_r \phi\rangle$ diverges. But if you don't include it into the range of definition of $p_r$, you can include it into the range of definition of $p_r^\dagger$. Hence you can't make the ranges of $p_r$ and $p_r^\dagger$ to coincice. Hence $p_r$ can't be self adjoint.
This is formalized using the deficiency index of the operator which is (1,0) for $p_r$. For an operator to have a self-adjoint extensions, the two indices have to be equal.

 

[ShayanJ]

Again, an introduction to quantum theory by Levin, this time the last part of section 10.1.
He says that the operator $\hat{P}_{rad}(r)=\hat e_r \cdot \hat{ \vec P}(\vec r)$ is not hermitian and can't be the radial momentum operator but its symmetrization $\hat{P}_{rad}(r)=\frac 1 2[(\frac{\vec r}{r})\cdot \hat{ \vec P}(\vec r) +\hat{ \vec P}(\vec r)\cdot (\frac{\vec r}{r})]$ leads to the correct Hermitian form. Now because we have $\hat{ \vec P}(\vec r)\cdot (\frac{\vec r}{r})=-\frac{2 i \hbar }{r}$ , we'll get $\hat{P}_{rad}(r)=-i\hbar (\frac{\partial}{\partial r}+\frac 1 r)$.

P.S.
Dude, this book by Levin is really nice!

EDIT: oops...stevendaryl has already mentioned this in post #38!
So why the argument?

 

[stevendaryl]

Hermitian and symmetric are synonymous.

What if $\phi=r^{-(1.1)} \exp(-r)$and $\psi=1/\exp(-r)$? Both functions are square integrable and both are in the range of definition of $p_r$, yet, the first integral won't vanish.

An operator is only Hermitian or not relative to a set of elements of the Hilbert space. That has nothing particularly to do with $p_r$. With the usual 1-D momentum operator, $p_x = -i \hbar \partial_x$, the proof that it is Hermitian uses:

$\int \partial_x (\psi^* \phi) dx = 0$

The proof goes:

1. $\int \partial_x (\psi^* \phi) dx = 0$
2. But also, $\int \partial_x (\psi^* \phi) dx = \int (\partial_x \psi^*) \phi dx + \int \psi^* (\partial_x \phi) dx$
3. Therefore, $\int (\partial_x \psi^*) \phi dx = - \int \psi^* (\partial_x \phi) dx$
4. By definition of $\dagger$, it follows that $\partial_x^\dagger = - \partial_x$
5. Therefore, $(-i \hbar \partial_x)^\dagger = -i \hbar \partial_x$

If equation 1 doesn't hold, then the proof that $-i \hbar \partial_x$ is Hermitian isn't valid.

 

[DrDu]

This is some standard stuff which I am a bit rusty. The reference of coice is the book by Glazman and Akhiezer.
The following article is also very instructive:
http://arxiv.org/abs/quant-ph/0103153

 

[strangerep]

[...] its symmetrization [...] leads to the correct Hermitian form.

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...

 

[dextercioby]

DrDu is right, of course. Let's discuss both operators for the so-called "radial momentum" in spherical coordinates. More precisely, let's set the problem properly from a mathematical viewpoint: Let's take the "complicated" one first:

$$p_r =: -i\hbar \left(\frac{\partial}{\partial r} + \frac {1}{r} \right)$$

which we want to study in the Hilbert space $\mathcal{H} = \mathcal{L}^2 \left((0,\infty), r^2 dr\right)$.

A careful analysis shows that this operator in well defined in the Hilbert space under study, it has a maximum domain dense everywhere (in the Hilbert space) made up of absolutely continuous functions on any finite sub-interval of $(0,\infty)$ which go to 0 in the limit r→∞ and moreover go to 0 when r→0 with the additional property that the partial derivative (actually we neglect spherical angles, so we can use safely use d instead of ∂) also belongs to $\mathcal{H}$, so it has a unique adjoint: $p_{r}^{\dagger}$ (whose domain we needn't calculate, see below).

Now let's use what von Neumann invented as far back as 1929: Assume the adjoint $p_{r}^{\dagger}$ is a (closed) extension of $p_r$. Thus, for any $\psi (r) \subset D_{p_{r}^{\dagger}}$ we can consider the deficiency indices equation in the form:

$$p_{r}^{\dagger} \psi_{\pm} (r) = \pm i \frac{\hbar}{d} \psi_{\pm} (r)$$ , with d having the dimension of length and d>0 (it's a constant to account for the length dimension in the derivative + fraction in the LHS)

This is a genuine differential equation, since we assume that $p_{r}^{\dagger}$ acts in the same way as $p_r$.

The ODE:

$$-i\hbar \frac{d\psi_{\pm} (r)}{dr} - i\hbar \frac{\psi_{\pm} (r)}{r} = \pm i \frac{\hbar}{d} \psi_{\pm} (r)$$ has the solution (family of solutions):

$$\psi_{+} (r) = \frac{C_{+}}{r} e^{ -\frac{r}{d}}$$ and

$$\psi_{-} (r) = C_{-} r {} {} e^{ +\frac{r}{d}}$$

The second solution is not in the Hilbert space $\mathcal{H}$, while the first solution is. [The first solution is in the domain of $p_{r}^{\dagger}$, but not in the domain of $p_{r}$ - thus a hint that $p_r$ can't be self-ajoint]. So the v-N deficiency indices are (1,0), in other words, $p_r$ is a closed symmetric operator with no self-adjoint extension on its maximum domain of definition.
Thus DrDu is right.

For the simpler (so-called 'unsymmetrized' version) $p_r = -i \hbar \frac{\partial}{\partial r}$ one is again invited to do the simpler maths outlined above (it's already done in a different Hilbert space in the quoted article by Bonneaux and al. which in turn cite the book by Akhiezer & Glazman as their source). He/she will reach the same conclusion bolded above.

So you see, even if you naively symmetrize something, it won't generate self-adjoint operators, hence won't generate observables. In other words, the "radial momentum" is not an observable in QM, it turns out that only the Schrödinger/Dirac Hamiltonian is.

 

[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?