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Model a proton as a uniformly charged thin spherical shell of radius $b$. Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem gives an electron potential energy $-C/b$ inside the shell and $-C/r$ outside. Relative to a point Coulomb potential, the perturbation is consequently $C(1/r-1/b)$ for $r<b$ and zero otherwise.

For the hydrogen atom ground state, time-independent perturbation theory gives the first-order energy correctionSince $E_{1s}=-C/(2a_0)$, the signed fractional change is $\Delta E/E_{1s}=-4b^2/(3a_0^2)+O((b/a_0{)}^3)$. The correction $\Delta E$ is positive: distributing the positive charge weakens the attraction near the origin and reduces the magnitude of the binding energy. Here $a_0$ is the Bohr radius and $b\ll a_0$ is the small-radius approximation.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

The unperturbed ground state is $|0,0⟩$ with energy $\hslash \omega$. Sincethe first-order energy correction vanishes and first-order nondegenerate perturbation theory gives

The linear confinement correction to Coulomb bound states uses $H^{\prime }=br$, and its first-order energy correction in a normalized Coulomb eigenstate is $b⟨r⟩$. Replace $a_0$ by $a_q$ in the hydrogen wavefunctions. The normalized spherical harmonics remove the angular integral, leavingFor $1s$, the integral of $r^3{e}^{-2r/a_q}$ yieldsFor $2s$, put $u=r/a_q$ and square the full radial polynomial, including its cross term:For $2p$, the same calculation givesConsequently the first two Coulomb levels become, to first order,Thus the second level splits into a singlet orbital level and a triply degenerate orbital level, with $E_{2s}-E_{2p}=b{a}_q+O(b^2)$. For the confining sign $b>0$, $2p$ lies lower than $2s$: its smaller mean radius gives the smaller positive correction. The expansion assumes these corrections and the induced couplings to other principal shells are small compared with the relevant Coulomb energy gaps; $b{a}_q^2/a\ll 1$ is the governing small parameter for fixed low $n$.

The expansion of the relativistic kinetic energy at fixed rest mass gives the perturbation $-p^4/(8m^3{c}^2)$. In the harmonic-oscillator ground state, $⟨p^4⟩=3(m\hslash \omega /2{)}^2$, proving the displayed first-order energy correction. Its small parameter is $\hslash \omega /(m{c}^2)$. This correction lowers the nonrelativistic energy and does not include other relativistic interactions.