A Hardy-type inequality and some spectral characterizations for the Dirac–Coulomb operator

Abstract

We prove a sharp Hardy-type inequality for the Dirac operator. We exploit this inequality to obtain spectral properties of the Dirac operator perturbed with Hermitian matrix-valued potentials ${\mathbf {V}}$ of Coulomb type: we characterise its eigenvalues in terms of the Birman–Schwinger principle and we bound its discrete spectrum from below, showing that the ground-state energy is reached if and only if ${\mathbf {V}}$ verifies some rigidity conditions. In the particular case of an electrostatic potential, these imply that ${\mathbf {V}}$ is the Coulomb potential.

Appendix A. Partial wave subspaces

In this appendix, we recall the partial wave subspaces associated to the Dirac equation. We sketch here this topic, referring to [26, Sect. 4.6] for further details.

Let $Y^l_n$ be the spherical harmonics. They are defined for $n = 0, 1, 2, \ldots$, and $l =-n,-n + 1,\ldots , n,$ and they satisfy $\Delta _{\mathbb {S}^2} Y^l_n= n(n + 1)Y^l_n$, where $\Delta _{\mathbb {S}^2}$ denotes the usual spherical Laplacian. Moreover, $Y^l_n$ form a complete orthonormal set in $L^2(\mathbb {S}^2)$. For $j = 1/2, 3/2, 5/2, \ldots ,$ and $m_j = -j,-j + 1, \ldots , j$, set

$$\begin{aligned} \begin{aligned} \psi ^{m_j}_{j-1/2}&:= \frac{1}{\sqrt{2j}} \left( \begin{array}{c} \sqrt{j+m_j}\,Y^{m_j-1/2}_{j-1/2}\\ \sqrt{j-m_j}\,Y^{m_j+1/2}_{j-1/2}\\ \end{array}\right) ,\\ \psi ^{m_j}_{j+1/2}&:=\frac{1}{\sqrt{2j+2}} \left( \begin{array}{c} \sqrt{j+1-m_j}\,Y^{m_j-1/2}_{j+1/2}\\ -\sqrt{j+1+m_j}\,Y^{m_j+1/2}_{j+1/2}\\ \end{array}\right) ; \end{aligned} \end{aligned}$$

then $\psi ^{m_j}_{j\pm 1/2}$ form a complete orthonormal set in $L^2(\mathbb {S}^2)^2$. For $k_j:=\pm (j+1/2)$ we set

$$\begin{aligned} \Phi ^+_{m_j,\pm (j+1/2)}:= \left( \begin{array}{c} i\,\quad \psi ^{m_j}_{j\pm 1/2}\\ 0 \end{array}\right) , \quad \Phi ^-_{m_j,\pm (j+1/2)}:= \left( \begin{array}{c} 0\\ \psi ^{m_j}_{j\mp 1/2} \end{array}\right) . \end{aligned}$$

Then, the set $\lbrace \Phi ^+_{m_j,k_j},\Phi ^-_{m_j,k_j} \rbrace _{j,k_j,m_j}$ is a complete orthonormal basis of $L^2(\mathbb {S}^2)^4$ and

$$\begin{aligned} (1+2{{\mathbf {S}}}\cdot L)\Phi _{m_j,k_j}=-k_j\beta \Phi _{m_j,k_j}, \end{aligned}$$

(A.1)

where the spin angular momentum operator${{\mathbf {S}}}$ and the orbital angular momentumL are defined as

$$\begin{aligned} {\mathbf {S}}= \frac{1}{2}\left( \begin{array}{cc} \sigma &{} 0\\ 0 &{} \sigma \end{array} \right) \quad \text {and}\quad L:=-ix\wedge \nabla . \end{aligned}$$

(A.2)

So, we can write

$$\begin{aligned} \psi (x)= \sum _{j,k_j,m_j} \frac{1}{r}\left( f^+_{m_j,k_j}(r)\Phi ^+_{m_j,k_j}({{\hat{x}}})+ f^-_{m_j,k_j}(r)\Phi ^-_{m_j,k_j}({{\hat{x}}})\right) \end{aligned}$$

(A.3)

and, by definition,

$$\begin{aligned} \int _{{{{\mathbb {R}}}}^3}|\psi |^2\,dx= \sum _{j,k_j,m_j} \int _0^{+\infty }|f^+_{m_j,k_j}(r)|^2+ |f^-_{m_j,k_j}(r)|^2\,dr. \end{aligned}$$

Thanks to [26, Eq. 4.109] and (A.3), we have that

$$\begin{aligned} \begin{aligned} \int _{{{{\mathbb {R}}}}^3} \frac{|\psi |^2}{|x|}\,dx&= \sum _{j,k_j,m_j} \int _0^{+\infty } \frac{1}{r} \left( |f^+_{m_j,k_j}(r)|^2+ |f^-_{m_j,k_j}(r)|^2\right) dr,\\ \int _{{{{\mathbb {R}}}}^3} \frac{|(1+2{{\mathbf {S}}}\cdot L)\psi |^2}{|x|}\,dx&= \sum _{j,k_j,m_j} \int _0^{+\infty } \frac{k_j^2}{r} \left( |f^+_{m_j,k_j}(r)|^2+ |f^-_{m_j,k_j}(r)|^2\right) dr. \end{aligned} \end{aligned}$$

(A.4)

From (A.4), we directly deduce that

$$\begin{aligned} \int _{{{{\mathbb {R}}}}^3} \frac{|\psi |^2}{|x|}\,dx \le \int _{{{{\mathbb {R}}}}^3} \frac{|(1+2{{\mathbf {S}}}\cdot L)\psi |^2}{|x|}\,dx, \end{aligned}$$

(A.5)

and that (A.5) is attained if and only if $f_{m_j,k_j}^\pm =0$ for $k_j\ne \pm 1$, or equivalently $j\ne 1/2$.