Solving the Radial Equation in Full

The separation of variables delivered the angular factors as spherical harmonics and left the radial factor $R_{n\ell }(r)$ defined by an ordinary differential equation that carries the Coulomb potential and the energy. That equation was quoted, not solved. Solving it is where the principal quantum number, the restriction $\ell \le n-1$, the $n^2$ degeneracy, and the exponential-times-polynomial form of the wave functions all originate. The method is the standard one for a Schrödinger eigenvalue problem: strip the asymptotic behaviour, expand the remainder in a power series, and force the series to terminate so the solution stays normalizable.

The radial equation as a one-dimensional problem

For a central Coulomb potential $V(r)=-kZ{e}^2/r$ with $k$ the Coulomb constant, the radial equation for $R(r)$ reads¹

with $\mu$ the electron–proton reduced mass. The first-derivative term is a nuisance. It disappears under the substitution

Substituting and multiplying by $r$ gives an equation identical in form to a one-dimensional Schrödinger equation for $u(r)$ on the half-line $r\ge 0$:

The barrier is why higher-$\ell$ states are held away from the nucleus. For $\ell =0$ the effective potential is the bare Coulomb well and $u$ may be finite and nonzero as $r\to 0$; for $\ell \ge 1$ the $1/r^2$ wall forces $u$ to zero there. The boundary conditions on $u$ are $u(0)=0$ (so that $R=u/r$ stays finite) and $u(r)\to 0$ as $r\to \infty$ (so that the state is bound and normalizable, ${\int }_0^{\infty }|u{|}^2\mathbf{d}r<\infty$).

Dimensionless form and asymptotics

Bound states have $E<0$. Set

where $a_0={\hslash }^2/\mu k{e}^2$ is the Bohr radius. In these variables every dimensioned quantity is absorbed and the equation becomes¹

The eigenvalue now lives entirely in the single number ${\rho }_0$; finding the allowed ${\rho }_0$ is finding the allowed energies. Two limits fix the shape of $u$ before any series is written.

• Large $\rho$. The $1/\rho$ and $1/{\rho }^2$ terms vanish, leaving ${\mathbf{d}}^{2}u/\mathbf{d}{\rho }^2\approx u$, with solutions $e^{\pm \rho }$. Normalizability discards $e^{+\rho }$, so $u\sim e^{-\rho }$ far out.
• Small $\rho$. The $1/{\rho }^2$ term dominates, leaving ${\mathbf{d}}^{2}u/\mathbf{d}{\rho }^2\approx \ell (\ell +1){\rho }^{-2}u$, whose solutions are ${\rho }^{\ell +1}$ and ${\rho }^{-\ell }$. Finiteness of $R=u/r$ discards ${\rho }^{-\ell }$, so $u\sim {\rho }^{\ell +1}$ near the origin.

Peeling both factors off, write the exact solution as their product with an unknown function $v(\rho )$:

The two extracted factors already satisfy the boundary conditions; $v$ must not spoil them, which means $v$ may grow no faster than a power of $\rho$.

The Frobenius series and its recursion

Differentiating $u={\rho }^{\ell +1}{e}^{-\rho }v$ twice and substituting reduces the equation to a single ODE for $v(\rho )$:¹²

Expand $v$ as a power series about the (regular singular) origin,

Insert the series, collect the coefficient of ${\rho }^j$, and set it to zero. Each power of $\rho$ gives one relation between consecutive coefficients — the recursion relation

One free constant $c_0$ (fixed later by normalization) generates the whole series. The recursion is the engine of the solution, and its large-$j$ behaviour decides everything.

Termination and quantization

If the series never stops, the ratio of successive coefficients settles to

which is the coefficient ratio of the Taylor series of $e^{2\rho }$. An infinite $v$ therefore grows like $e^{2\rho }$, making $u\sim {\rho }^{\ell +1}{e}^{+\rho }$ — exponentially divergent, not normalizable. The only escape is termination: some $c_{j+1}$ must vanish while $c_j\ne 0$, after which every later coefficient is zero and $v$ is a polynomial. From the recursion, the numerator vanishes at $j=j_{\max }$ when

Define the integer on the left as the principal quantum number

Because $j_{\max }\ge 0$, the polynomial degree is $j_{\max }=n-\ell -1\ge 0$, which is precisely the restriction

The energy depends on $n$ alone. The angular momentum $\ell$ entered the radial equation through the centrifugal term and left through the termination count $n-\ell -1$, but it cancelled out of the eigenvalue. That cancellation is special to the $1/r$ potential and is examined as a symmetry in the next lesson. The Rydberg energy scales as $Z^2$ because the length scale contracts as $a_0/Z$ while the potential deepens as $Z$; the two combine to $Z^2$.

The associated Laguerre polynomials

The terminating $v(\rho )$ is, up to normalization, an associated Laguerre polynomial. In the convention of Griffiths & Schroeter,¹

a polynomial of degree $q=n-\ell -1$. The lower index is the polynomial degree, the upper index $2\ell +1$ is set by the centrifugal factor. The first few are

Assembling $R=u/r={\rho }^{\ell }{e}^{-\rho }v/\kappa$ and restoring $\rho =Zr/n{a}_0$ (so $2\rho =2Zr/n{a}_0$) gives the normalized hydrogenic radial function

The three ingredients are visible: a power $r^{\ell }$ from the centrifugal asymptote, an exponential $e^{-Zr/n{a}_0}$ from the large-$r$ asymptote whose range $n{a}_0/Z$ grows with $n$, and a degree-$(n-\ell -1)$ polynomial supplying the oscillations. The overall sign is convention.

Counting nodes and states

The polynomial degree $n-\ell -1$ equals the number of positive roots of $L_{n-\ell -1}^{2\ell +1}$, and each root is a value of $r$ where $R_{n\ell }$ changes sign — a radial node. The full node count of ${\psi }_{n\ell m}$ splits cleanly:

• radial nodes: $n-\ell -1$, the spheres where $R_{n\ell }(r)=0$;
• angular nodes: $\ell$, the nodal cones and planes of $Y_{\ell m}$;
• total nodes: $(n-\ell -1)+\ell =n-1$, independent of $\ell$.

A state of fixed $n$ always has $n-1$ nodal surfaces, redistributed between radial and angular as $\ell$ runs from $0$ to $n-1$. The circular state $\ell =n-1$ has no radial nodes and a single-hump $R$; the penetrating state $\ell =0$ puts all $n-1$ nodes in the radial direction.

The degeneracy of the level $E_n$ counts the distinct $(n,\ell ,m)$ triples at fixed $n$. For each $\ell$ there are $2\ell +1$ values of $m$, so

doubling to $2n^2$ once electron spin is included. The $\ell$-sum being exactly $n^2$ is the arithmetic face of the accidental degeneracy.

Hydrogenic scaling with nuclear charge

Every result carries $Z$ in a fixed pattern, because the Coulomb problem has a single length and a single energy scale that absorb it. Rescaling $r\to Zr$ maps the $Z$-charged problem onto hydrogen:

• length: the wave function depends on $r$ only through $Zr/a_0$, so all radii contract as $a_0/Z$; the most probable radius of the ground state is $a_0/Z$;
• energy: $E_n=-Z^2\mathrm{Ry}/n^2$, quadratic in $Z$;
• expectation values: $⟨r⟩\propto n^2{a}_0/Z$ and $⟨1/r⟩\propto Z/n^2{a}_0$, the subject of a later lesson.

The scaling makes every one-electron ion — ${\mathrm{He}}^{+}$ ($Z=2$), ${\mathrm{Li}}^{2+}$ ($Z=3$), and the highly charged ions of laboratory spectroscopy — a rescaled copy of hydrogen, with the same wave functions and a spectrum stretched by $Z^2$. The exact solvability ends the moment a second electron is added, or the $1/r$ potential is perturbed by relativistic and fine-structure corrections, both of which are computed by feeding these exact wave functions into perturbation theory.