Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

Question

In Nolting's QM book (Theoretical Physics 7), in the chapter on central potentials, a radial momentum operator $\hat{p}_r$ is defined as \begin{equation} \hat{p_r} = -i \hbar \Big( \frac{\partial}{\partial r} + \frac{1}{r}\Big). \end{equation} In two following problems, we try to find the conditions satisfied by wave functions that make $\hat{p_r}$ hermitian. We can show that these conditions are

\begin{equation} \lim_{r\to\infty} r\psi(r) = 0 \; \; \; , \; \; \; \lim_{r\to 0} r \psi(r) = 0. \end{equation}

In the other problem, we attempt to solve the eigenvalue problem of $\hat{p_r}$, but the eigenfunctions are of the form

\begin{equation} \psi(r) = \frac{1}{r} e^{i \lambda r / \hbar}, \end{equation} which doesn't satisfy the second condition for $\hat{p_r}$ to be hermitian. The argument is that this is why $\hat{p_r}$ is not a good observable. But doesn't that violate the spectral theorem? As far as I understand it, the spectral theorem says that if $\hat{p_r}$ is hermitian, then there exists an orthonormal basis $\textbf{for the space in which $\hat{p_r}$ is hermitian}$, consisting of eigenvectors of $\hat{p_r}$. But here we can't find ny eigenvector of $\hat{p_r}$ in the space where it is hermitian.

Note: I understand this could qualify as an MSE question, but it's regarding physics and has a physical implication, so I posted it here.

Answer 1

The operator $$\hat p_r \ (...) = -i \hbar \frac{1}{r}\partial_r \left[\ r \ (...)\right] $$ is obvoiusly hermitean/symmetric on $\mathit L^2(\mathbb R_+ , r^2 dr)$:

$$\int_0^\infty \ r^2 \ dr \ \psi(r)^* \ \hat p_r \ \phi(r) = \int_0^\infty \ dr \ \psi(r)^* \ (-i \hbar) r \partial_r \left[\ r \ \phi(r)\right] = \\ \int_0^\infty \ dr \ \phi(r) \ \left[(-i \hbar) \ r \partial_r \ r \ \psi(r)\right]^* $$

for functions in $\mathit L^2$ bounded at $r=0$. For functions with a power law at $r=0$ we have

$$ r^{-1} \partial_r \left(\ r \ r^\alpha\right) =(\alpha +1 ) \ r^{\alpha-1} $$

$$ 2\alpha>-1 : \int_0 dr \ r^2 \ r^{2\alpha-2} <\infty $$

So we see that for functions in the domain of $\hat p$ with a singularity ar $r=0$, $\psi=r^{-1/2+\epsilon}$, the domain of the adjoint operator includes singularities up to $r^{-3/2+\epsilon}$.

Since domains of $\hat p_r$ and $\hat p_r^*$ are not identical, the operator is hermitean, but not selfadjoint.

This fact is normal for the generator of the translation group on a Hilbert space over a domain with a boundary. There is the freedom, to specify phase changes for a wave, reflected at the boundary.

Answer 2

Addressing a point which is not covered in the answer by @Roland F and the two pertinent comments to it:

The OP is conflating spectral theory and the general theory of operators in Quantum Mechanics which both serve as the mathematical foundation in its standard (textbook) formulation. The two topics are related in the following way: a definition of an operator (call it $A$) is given by specifying a triplet (operator, domain, Hilbert space) $\left(A, D(A)\subseteq \mathcal{H}, \mathcal{H}\right)$ and "an action" of $A$ on elements of its domain. This action defines not only the maximal domain of the operator, but also the range of the operator $\mathcal{R}(A)$, which is also a subset of the Hilbert space. This is, however, very general, but weirdly in QM very restrictive!

The spectral theory applied to Quantum Mechanics wants to solve the following equation:

$$Af = \lambda f, \text{for} f\in\ D(A)\cap\mathcal{R}(A), \lambda\in\mathbb{R}.$$

The OP discovers that the radial momentum operator when defined on the appropriate Hilbert space $L^2 (\mathbb{R}_+, r^2 dr)$ is Hermitean, or what mathematicians call symmetric. Then he attempts to solve its spectral equation and finds out that this is impossible, the "function-type" solution he found is not member of the Hilbert space, let alone the very particular subset formed by intersecting the maximal domain and the range.

So $p_r$ doesn't qualify as a "good observable" for two reasons: if we trust the textbook statement that "a good observable" is represented by a self-adjoint operator (hence its spectrum is a subset of $\mathbb{R}$), then there's no way to deem $p_r$ self-adjoint. In mathematics we say that $p_r$ is a symmetric operator on its maximal domain with von Neumann deficiency indices $(0,\aleph_0)$ and its spectrum is the whole set of complex numbers. These mathematical attributes of $p_r$ don't go well at all with the rest of the axioms of the textbook formulation, the Born one (there's no guarantee that if I devise a virtually infinite mechanism to measure this particular momentum, the forcibly real value shown by the measurement apparatus is the true momentum value), then the von Neumann one (after the measurement takes place, the physical state to which the system "collapses" is the Hilbert space eigenvector of the operator for the particular value being measured).

Bottom line: yes, $p_r$ is flawed and its study is interesting only as a nice application of functional analysis in physics, but fortunately we need to square it in the Hamiltonian of systems with spherical symmetry, so that the Hamiltonians obtained are "good observables", or at least "better observables".