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In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

Put $R=f{e}^{-\gamma r}$. ThenAfter substitution and cancellation of the ${\gamma }^{2}f$ terms, the equation becomesFor the power series $f={\sum }_{n=0}^{\infty }{a}_n{r}^n$, the coefficient of $r^{n-1}$ isThus

If the series does not terminate, then $a_n/a_{n-1}\sim 2\gamma /n$, so $f$ acquires the large-$r$ behaviour $e^{2\gamma r}$ and $R$ grows like $e^{\gamma r}$. Normalizability therefore requires termination, which occurs whenfor some positive integer $N$. HenceThe ground state has $N=1$, so the hydrogen ground-state energy isThis is the series-termination quantization of the Coulomb radial equation.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.