The Schrödinger Equation in Three Dimensions

Every force that points toward or away from a single center and whose strength depends only on the distance to that center derives from a potential $V(r)$ with no angular dependence. The Coulomb attraction of a nucleus, the isotropic oscillator, and the idealized spherical box all belong to this class. Rotational invariance of $V(r)$ makes the Hamiltonian commute with every component of orbital angular momentum, and that symmetry is what lets the three-dimensional stationary-state problem collapse into a one-dimensional radial equation. The angular dependence is fixed once and for all by the spherical harmonics; only the radial motion carries the information specific to a given $V(r)$.

The stationary-state equation in three dimensions

The time-independent Schrödinger equation for a particle of mass $\mu$ in a potential $V(\vec{r})$ is

The mass is written $\mu$ rather than $m$ to keep it distinct from the magnetic quantum number and, in a two-body problem such as hydrogen, to signal that the relevant mass is the reduced mass of the pair. For a central potential$V(\vec{r})=V(r)$, with $r=|\vec{r}|$, the natural coordinates are spherical $(r,\theta ,\phi )$. The Laplacian in these coordinates is

The two angular terms are precisely $-L^2/({\hslash }^2{r}^2)$, where $L^2$ is the total orbital-angular-momentum operator whose spectrum was found by diagonalizing it on the sphere. Substituting the compact form,

turns the Schrödinger equation into

Only the $L^2$ term carries angular derivatives, and it acts on the sphere at fixed $r$. This is the structural reason separation succeeds.

Separation of variables

Because $[H,L^2]=0$ and $[H,L_z]=0$ for a central potential, the Hamiltonian shares an eigenbasis with the compatible pair $\{L^2,L_z\}$. Seek a product solution in which the angular factor is already an eigenfunction of that pair, the spherical harmonic $Y_{\ell }^m(\theta ,\phi )$:

Applying $L^2{Y}_{\ell }^m={\hslash }^2\ell (\ell +1)Y_{\ell }^m$ and dividing out the common factor $Y_{\ell }^m$ leaves an ordinary differential equation for $R_{n\ell }(r)$ alone:

The three-dimensional problem has been reduced to a family of radial problems, one for each value of $\ell$. The magnetic quantum number $m$ has dropped out entirely: the radial motion cannot depend on the orientation of the angular momentum, only on its magnitude. Every energy level obtained from this equation is therefore at least $(2\ell +1)$-fold degenerate, once for each allowed $m$.

The radial equation and the effective potential

The first-derivative term is awkward. It is removed by the substitution

Direct differentiation gives $\frac{1}{r^2}\frac{\mathbf{d}}{\mathbf{d}r}(r^2{R}^{\prime })=\frac{1}{r}{u}^{\prime \prime }$, so the radial equation becomes

The reduction is exact, not an approximation. Everything known about one-dimensional wave mechanics — nodes, curvature, the relation between confinement and energy — transfers directly, with the single change that the domain is a half-line and the effective potential carries the $\ell$-dependent barrier.

The centrifugal term ${\hslash }^2\ell (\ell +1)/2\mu r^2$ is the quantum image of the classical angular-momentum barrier $L^2/2\mu r^2$ that keeps an orbiting particle away from the center. It is repulsive (positive) and grows without bound as $r\to 0$ for every $\ell \ge 1$, so a particle with angular momentum is expelled from the origin. For $\ell =0$ the barrier vanishes and the particle can reach $r=0$.

Boundary conditions and behavior at the origin

Two conditions select the acceptable radial solutions.

• Regularity at the origin. Normalizability of $\psi$ requires ${\int }_0^{\infty }|R{|}^2{r}^2\mathbf{d}r<\infty$, and in terms of $u=rR$ this reads ${\int }_0^{\infty }|u{|}^2\mathbf{d}r<\infty$, exactly the one-dimensional condition. A finite $R(0)$ demands $u(0)=0$; the wavefunction cannot diverge at the center.
• Decay at infinity. For a bound state ($E<0$ relative to the potential's asymptotic value) $u$ must vanish as $r\to \infty$ so the state is square-integrable.

The small-$r$ behavior is fixed by the centrifugal term, which dominates $V(r)$ whenever $V$ is less singular than $r^{-2}$. Near the origin the radial equation reduces to

Trying $u\sim r^s$ gives the indicial equation $s(s-1)=\ell (\ell +1)$, with roots $s=\ell +1$ and $s=-\ell$. The second is discarded because it makes $R=u/r\sim r^{-\ell -1}$ diverge (and violates $u(0)=0$ for $\ell \ge 1$; the $\ell =0$ case is excluded separately because $R\sim 1/r$ produces a delta-function source in ${\nabla }^2$). The physical solution therefore behaves as

Higher angular momentum suppresses the wavefunction more strongly near the center, the wave-mechanical statement that the centrifugal barrier keeps such a particle out.

The free particle in spherical coordinates

Setting $V(r)=0$ isolates the effect of the centrifugal barrier alone. Write $k^2=2\mu E/{\hslash }^2$ with $E>0$; the radial equation for $R$ becomes

In the dimensionless variable $\rho =kr$ this is the spherical Bessel equation. Its two independent solutions are the spherical Bessel function of the first kind $j_{\ell }(\rho )$ and the spherical Neumann function $n_{\ell }(\rho )$,

The lowest few, in closed form:

Their behavior at the two ends decides which is physical. Near the origin

so $j_{\ell }$ carries the regular $r^{\ell }$ behavior demanded above while $n_{\ell }$ diverges. A free particle filling all space therefore uses $R_{\ell }(r)=j_{\ell }(kr)$ only; the Neumann functions reappear when the origin is excluded, as in scattering off a hard sphere. Both oscillate and decay as $1/\rho$ at large argument,

the phase shift $\ell \pi /2$ recording the angular momentum. The energy spectrum is continuous, $E={\hslash }^2{k}^2/2\mu$ for any $k>0$, exactly as for a plane wave; the spherical form simply reorganizes the same free-particle states into simultaneous eigenstates of $\{H,L^2,L_z\}$ instead of momentum.

The infinite spherical well

Confining the particle inside a sphere of radius $a$ with $V(r)=0$ for $r<a$ and $V=\infty$ for $r>a$ turns the continuous free spectrum into a discrete one. Inside, the regular solution is $R(r)=j_{\ell }(kr)$; the wall forces $R(a)=0$, hence

The zeros of $j_{\ell }$ are not evenly spaced and, except for $\ell =0$, are not known in closed form. Write ${\beta }_{n\ell }$ for the $n$-th positive zero of $j_{\ell }$. The quantization condition $ka={\beta }_{n\ell }$ fixes the allowed energies

For $\ell =0$ the barrier is absent and $u=rR=A\sin (kr)$, giving the familiar one-dimensional condition $\sin (ka)=0$, i.e. ${\beta }_{n0}=n\pi$ and

the infinite-square-well spectrum on the half-line. The first few zeros for higher $\ell$ interleave with these:

Because ${\beta }_{1\ell }$ increases with $\ell$, the ordering of levels by energy is $1s<1p<1d<2s<1f<2p<\dots$, and the accidental coincidences that make hydrogen so special are absent: here the energy depends on both $n$ and $\ell$, and each level is only $(2\ell +1)$-fold degenerate from the $m$-sum, with no extra $\ell$-degeneracy.

The radial nodes and the meaning of the radial quantum number

The label $n$ counts the radial nodes plus one, in the same way the principal number does for the infinite line. A state $R_{n\ell }$ has $n-1$ nodes in the open interval $0<r<a$ (not counting the forced zero at $r=a$), and each added node raises the curvature of $u$ and thus the energy, exactly the one-dimensional relationship between node count and energy. The angular quantum number contributes a separate, angular node structure through $Y_{\ell }^m$: the wavefunction has $\ell -|m|$ nodal cones in $\theta$ and $|m|$ nodal planes in $\phi$. Total node count is the sum, and it grows with both excitation of the radial motion and of the rotation.

The three ingredients of a central-potential spectrum

Every problem in this module — the free particle just solved, the hydrogen atom, and the isotropic oscillator — is built from the same three pieces, and the labor is entirely in the third.

• The angular harmonic $Y_{\ell }^m$ is universal: it is the same function for every central potential, carrying the $(2\ell +1)$-fold degeneracy and the parity $(-1{)}^{\ell }$ regardless of $V(r)$.
• The centrifugal barrier ${\hslash }^2\ell (\ell +1)/2\mu r^2$ is also universal in form, and it sets the $r^{\ell +1}$ behavior of $u$ at the origin for every problem.
• The true potential $V(r)$ is the only ingredient that changes, and it alone determines the radial functions, the energy spectrum, and whether the degeneracy exceeds the mandatory $2\ell +1$.

The last point is the thread of the next two lessons. A generic $V(r)$ produces levels that depend on both the radial count and $\ell$, as the spherical box does. The Coulomb and oscillator potentials produce an extra, accidental degeneracy — energies that depend on a single combined quantum number — and that degeneracy is the fingerprint of a symmetry larger than rotation.¹²³