The Scalar Relativistic Coulomb Problem

The Klein–Gordon equation in an attractive electrostatic Coulomb field has an exactly soluble bound-state problem. Its spectrum depends on orbital angular momentum as well as the principal quantum number, even though the particle has no spin. The familiar formula also assumes a specific boundary at the singular source. Square integrability alone does not always select that boundary.

Required background. External Potentials fixes electrostatic coupling; the Klein–Gordon Inner Product fixes normalization. Separation in spherical harmonics and terminating confluent hypergeometric series are assumed. Helpful background. Klein–Gordon to Schrödinger provides the low-energy check.

Radial equation for an electrostatic point source

Take an infinitely heavy prescribed source, $\mathbf A=0$, and

$$V(r)=q\Phi(r)=-\frac{a\hbar c}{r},\qquad a>0.$$

For charge $-e$ in the field of charge $+Ze$, $a=Z\alpha$ with $\alpha=e^2/(4\pi\epsilon_0\hbar c)$. This is a time-component vector potential, not a scalar mass modification. Ignore source size, recoil, radiation, and other interactions. Write

$$\phi(t,\mathbf x)=e^{-iEt/\hbar} Y_{\ell m_\ell}(\theta,\varphi)\frac{u(r)}r.$$

Normalize the angular factor by $\int |Y_{\ell m_\ell}|^2d\Omega=1$. Using $\nabla^2(Yu/r)=Y[u''-\ell(\ell+1)u/r^2]/r$ gives

$$\begin{aligned} u''+\biggl[& \frac{E^2-m^2c^4}{\hbar^2c^2} +\frac{2Ea}{\hbar cr}\\ &+\frac{a^2-\ell(\ell+1)}{r^2} \biggr]u=0. \end{aligned}$$

The attractive $1/r$ coefficient is proportional to $E$, and the electrostatic square adds an attractive inverse-square term. Both features follow from vector coupling.

The origin boundary is part of the model

For a given partial wave assume strictly $a<\ell+\tfrac12$ and define

$$\lambda=\sqrt{(\ell+\tfrac12)^2-a^2}>0.$$

Set $s=\tfrac12+\lambda$. The two local radial behaviors are $u_\pm\sim r^{1/2\pm\lambda}$. Here we choose the less-singular branch

$$u(r)\sim C r^{1/2+\lambda},$$

excluding the coefficient of the other independent behavior. This agrees with the Friedrichs form boundary for the semibounded inverse-square radial expression $-d^2/dr^2+(\lambda^2-\tfrac14)/r^2$. The full stationary KG problem still has energy-dependent coefficients; it is not an ordinary energy-independent Schrödinger Hamiltonian.

For $\ell=0$ and $0<a<1/2$, both radial powers vanish at zero, both are square integrable in $dr$ there, and both give finite KG charge. Thus neither $u(0)=0$ nor finite norm selects the textbook branch. The selected spatial radial amplitude $u/r\sim r^{-1/2+\lambda}$ is mildly divergent in this s wave. “Regular” means the declared less-singular branch, not a bounded spatial amplitude.

A point-source model including all partial waves needs $a<1/2$ for this construction. A higher partial wave’s weaker bound does not repair the s-wave problem. At $\lambda=0$ a logarithmic second solution appears; imaginary $\lambda$ requires a different boundary analysis. Finite source size and other short-distance interactions can change the boundary data and spectrum (Dereziński and Richard, 2017; Burgess et al., 2017).

Termination of the radial series

For a bound mode $|E|<mc^2$, define

$$\begin{aligned} \kappa&=\frac{\sqrt{m^2c^4-E^2}}{\hbar c},\\ \rho&=2\kappa r,\\ \nu&=\frac{aE}{\sqrt{m^2c^4-E^2}}. \end{aligned}$$

The origin and large-distance factors suggest $u=\rho^s e^{-\rho/2}F(\rho)$. Substitution gives

$$\rho F''+(2s-\rho)F'+(\nu-s)F=0.$$

The chosen origin solution is proportional to $\,{}_1F_1(s-\nu;2s;\rho)$. Generically it contains an $e^\rho$ term at large positive $\rho$, overwhelming the prefactor’s decay. For this origin branch it becomes a polynomial precisely when

$$s-\nu=-n_r,\qquad n_r=0,1,2,\ldots.$$

Then $F$ is proportional to $L_{n_r}^{2\lambda}(\rho)$. The unsquared condition is $\nu=n_r+s>0$; since $a>0$, it selects $E>0$. With $N=n_r+\tfrac12+\lambda$, solving gives

$$E_{n_r\ell}=mc^2\left(1+\frac{a^2}{N^2}\right)^{-1/2}.$$

Do not append a negative sign after squaring the condition. Those values do not solve the same selected attractive-potential problem. Charge conjugation changes the potential coupling as well as the frequency.

Conserved normalization and weak coupling

The mode’s KG norm is

$$Q=\int_0^\infty dr\, \frac{E+a\hbar c/r}{mc^2}|u(r)|^2.$$

For these positive-energy states the weight is positive. Fix the constant by $Q=1$, not by $\int|u|^2dr=1$ unless the conversion is explicitly included. Let $n=n_r+\ell+1\geq\ell+1$. At fixed $n,\ell$,

$$\begin{aligned} \frac{E_{n\ell}}{mc^2} ={}&1-\frac{a^2}{2n^2}\\ &+a^4\left[\frac{3}{8n^4} -\frac{1}{2n^3(\ell+\tfrac12)}\right]+O(a^6). \end{aligned}$$

The leading binding energy is the Schrödinger Coulomb value. The next term lifts its accidental degeneracy in $\ell$, while rotations retain degeneracy in $m_\ell$. There is no spin-orbit splitting because the particle has spin zero. This is therefore not the Dirac hydrogen spectrum. Real charged spin-zero bound systems also need recoil, finite-size, and interaction corrections omitted by this model.

The singularity threshold is not a pair threshold

For the selected ground state,

$$E_{1s}=mc^2\sqrt{\frac{1+\sqrt{1-4a^2}}2}.$$

As $a\to\tfrac12^-$, it tends to $mc^2/\sqrt2$. The state does not reach the negative continuum in this limit. Loss of a real origin exponent diagnoses the singular point-source problem, not a universal pair-production threshold. A physical instability analysis requires a specified source profile and field dynamics. Pair-Creation Thresholds separates invariant kinematics from production rates.

Exercises

1. Derive the ground-state expression from $n_r=\ell=0$.

Solution

Put $\lambda=\sqrt{1/4-a^2}$ and $s=1/2+\lambda$. Since $s^2+a^2=s$, the spectrum gives $E^2/(m^2c^4)=s^2/(s^2+a^2)=s$. The positive root is the displayed expression.

2. Check the $a^4$ term using the Schrödinger expectation of $-\mathbf P^4/(8m^3c^2)$. Use $\varepsilon_n=-mc^2a^2/(2n^2)$, $\langle V\rangle=2\varepsilon_n$, and $\langle V^2\rangle=m^2c^4a^4/[n^3(\ell+\tfrac12)]$.

Solution

On the Schrödinger state, $\mathbf P^2\psi=2m(\varepsilon_n-V)\psi$. Taking its squared norm yields

$$\begin{aligned} \langle\mathbf P^4\rangle &=4m^2\langle(\varepsilon_n-V)^2\rangle\\ &=m^4c^4a^4\left[ \frac4{n^3(\ell+\tfrac12)}-\frac3{n^4}\right]. \end{aligned}$$

Multiplication by $-1/(8m^3c^2)$ reproduces the exact expansion’s correction. This expectation-value identity does not require commuting $\mathbf P^2$ with $V$ as operators.

3. For $\ell=0$, $0<a<1/2$, verify that finite charge near the origin does not exclude either radial behavior.

Solution

The dominant charge weight is proportional to $1/r$, so $|u_\pm|^2/r\sim r^{\pm2\lambda}$. Both integrals converge because $0<\lambda<1/2$. The less-singular choice is an additional boundary specification, not a consequence of charge integrability.

References

• F. Bastianelli, Relativistic Quantum Mechanics, University of Bologna lecture notes, 2023–2024, pp. 23–24 — scalar Coulomb reduction and spectrum.
• C. P. Burgess, P. Hayman, M. Rummel, M. Williams, and L. Zalavári, “Point-particle effective field theory II: relativistic effects and Coulomb/inverse-square competition,” Journal of High Energy Physics 07, 072, 2017, doi:10.1007/JHEP07(2017)072 — short-distance source data in relativistic Coulomb systems.
• J. Dereziński and S. Richard, “On Schrödinger Operators with Inverse Square Potentials on the Half-Line,” Annales Henri Poincaré 18, 869–928, 2017, doi:10.1007/s00023-016-0520-7 — boundary extensions of the inverse-square radial expression.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — electrostatic scalar bound states.