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

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

$$$$\lim_{r\to\infty} r\psi(r) = 0 \; \; \; , \; \; \; \lim_{r\to 0} r \psi(r) = 0.$$$$

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

$$$$\psi(r) = \frac{1}{r} e^{i \lambda r / \hbar},$$$$ which doesn't satisfy the second condition for $$\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.

• What is "the space in which $P_r$ is hermitian"? What is a "good observable"? Despite that: Strictly speaking, the spectral theorem applies to self-adjoint operators, not hermitian (yes, there is a difference). Commented Feb 19 at 17:57
• Related: physics.stackexchange.com/q/9349/2451 , physics.stackexchange.com/q/661586/2451 and links therein. Commented Feb 19 at 17:59
• @TobiasFünke The space referred to is the space of functions which satisfy the conditions, on the functions, mentioned in the question. A good observable is an observable for which you can construct an experiment which measures its possible values. Can you perhaps allude to the distinction between hermitian and self adjoint and which one of those we use in QM?
– EM_1
Commented Feb 19 at 20:33
• @CosmasZachos I've seen it but I'm not sure I understand it. It says the operator has "deficiency indices" and I don't know what that is. My understanding is that the operator has a singularity at the origin, but I'm not sure that's what the answer is referring to. Also, the conditions found above require the limits to vanish, not the functions themselves at the indicated points, so I'm not sure the singularity is the problem.
– EM_1
Commented Feb 19 at 21:33

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.

• I think this answer is incomplete. It is also true that on $\mathcal{D} = C_c^\infty(\mathbb{R})$, the momentum operator is not self-adjoint. The domain of the adjoint is $H^1(\mathbb{R})$. This however does not mean that this $p$ is physically unviable. The condition you must check is whether or not $p$ is essentially self adjoint on the given domain.
– Prox
Commented Feb 20 at 15:42
• And this discussion about the boundary condition at the origin is the physical interpretation of the various boundary conditions that you can add that would make it essentially self-adjoint.
– Prox
Commented Feb 20 at 15:44

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".