---
title: The Schrödinger Equation in Three Dimensions
module: Central Potentials
moduleNumber: 7
lessonNumber: 1
order: 701
summary: >
  A central potential depends only on the distance from a force center, so the
  three-dimensional Schrödinger equation separates in spherical coordinates. The
  angular factor is a spherical harmonic; the radial factor obeys a
  one-dimensional equation with an effective potential whose centrifugal barrier
  depends on the angular-momentum quantum number. The free particle and the
  spherical box fix the two limiting cases through the spherical Bessel functions.
topics: [Central Potentials]
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 4 — Quantum Mechanics in Three Dimensions; §4.1 The Schrödinger Equation in Spherical Coordinates"
  - book: Shankar
    ref: "Ch. 12 — Rotational Invariance and Angular Momentum; §12.6 The Radial Equation"
  - book: Cohen-Tannoudji
    ref: "Ch. VII — Particle in a Central Potential; §A The Radial Equation"
draft: false
---

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](/quantum-mechanics/angular-momentum/orbital-angular-momentum-and-spherical-harmonics),
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

$$
-\frac{\hbar^2}{2\mu}\,\nabla^2 \psi(\vec r) + V(\vec r)\,\psi(\vec r)
  = E\,\psi(\vec r).
$$

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
$(r, \theta, \varphi)$. The Laplacian in these coordinates is

$$
\nabla^2 = \frac{1}{r^2}\,\frac{\partial}{\partial r}\!\left(r^2\,\frac{\partial}{\partial r}\right)
  + \frac{1}{r^2\sin\theta}\,\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial}{\partial\theta}\right)
  + \frac{1}{r^2\sin^2\theta}\,\frac{\partial^2}{\partial\varphi^2}.
$$

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

$$
\nabla^2 = \frac{1}{r^2}\,\frac{\partial}{\partial r}\!\left(r^2\,\frac{\partial}{\partial r}\right)
  - \frac{L^2}{\hbar^2 r^2},
$$

turns the Schrödinger equation into

$$
-\frac{\hbar^2}{2\mu}\,\frac{1}{r^2}\,\frac{\partial}{\partial r}\!\left(r^2\,\frac{\partial\psi}{\partial r}\right)
  + \frac{L^2}{2\mu r^2}\,\psi
  + V(r)\,\psi = E\,\psi.
$$

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 $Y_\ell^m(\theta,\varphi)$:

$$
\psi_{n\ell m}(r,\theta,\varphi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\varphi),
\qquad
L^2\,Y_\ell^m = \hbar^2\,\ell(\ell+1)\,Y_\ell^m.
$$

$$
% caption: A central-potential eigenstate factors into a radial amplitude that
% carries the energy dependence and an angular harmonic fixed by the two
% angular-momentum quantum numbers.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0,
  bx/.style={draw, minimum width=30mm, minimum height=14mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[bx, draw=acc, text=acc, thick] (P) at (0,0) {full state\\$\psi_{n\ell m}$};
\node[bx] (R) at (5.2,1.4) {radial part\\$R_{n\ell}(r)$};
\node[bx] (Y) at (5.2,-1.4) {angular part\\$Y_\ell^m$};
\draw[->, acc, thick] (P) -- (R);
\draw[->, acc, thick] (P) -- (Y);
\node[anchor=west, font=\scriptsize] at (7.1,1.4) {depends on $V(r)$, $E$};
\node[anchor=west, font=\scriptsize] at (7.1,-1.4) {same for every $V(r)$};
\end{tikzpicture}
$$

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

$$
-\frac{\hbar^2}{2\mu}\,\frac{1}{r^2}\,\frac{\d}{\d r}\!\left(r^2\,\frac{\d R}{\d r}\right)
  + \left[V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2}\right]R = E\,R.
$$

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

$$
u(r) = r\,R(r),
\qquad\text{equivalently}\qquad
R(r) = \frac{u(r)}{r}.
$$

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

> **Theorem (Radial equation).** With $u = rR$, the radial factor of a
> central-potential eigenstate obeys
> $$
> -\frac{\hbar^2}{2\mu}\,\frac{\d^2 u}{\d r^2}
>   + V_{\text{eff}}(r)\,u = E\,u,
> \qquad
> V_{\text{eff}}(r) = V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2}.
> $$
> This is identical in form to the one-dimensional Schrödinger equation on the
> half-line $r \in [0,\infty)$, with an **effective potential** equal to the
> true potential plus a repulsive **centrifugal barrier**.

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 $\hbar^2\ell(\ell+1)/2\mu r^2$ 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
$r = 0$.

$$
% caption: Effective potential $V_\mathrm{eff}$ for an attractive well: the
% centrifugal barrier raises and steepens the inner wall as the angular-momentum
% quantum number grows, pushing the classically allowed region outward.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,-2.6) -- (0,3.0) node[anchor=south, black] {$V$};
\draw[->, black] (0,0) -- (6.4,0) node[anchor=east, black] {$r$};
% attractive base potential V(r) ~ -1/r  (l=0 curve)
\draw[black, very thick, domain=0.42:6.2, samples=120, variable=\x]
  plot ({\x}, {-1.6/\x});
\node[black, anchor=west] at (4.3,-0.55) {$l = 0$};
% l=1: -1/r + b/r^2
\draw[acc, very thick, domain=0.62:6.2, samples=140, variable=\x]
  plot ({\x}, {-1.6/\x + 0.9/(\x*\x)});
\node[acc, anchor=west] at (2.9,0.45) {$l = 1$};
% l=2: -1/r + 3b/r^2
\draw[black, very thick, densely dashed, domain=0.95:6.2, samples=140, variable=\x]
  plot ({\x}, {-1.6/\x + 2.7/(\x*\x)});
\node[black, anchor=west] at (2.2,1.35) {$l = 2$};
\end{tikzpicture}
$$

### Boundary conditions and behavior at the origin

Two conditions select the acceptable radial solutions.

- **Regularity at the origin.** Normalizability of $\psi$ requires
  $\int_0^\infty |R|^2\,r^2\,\d r < \infty$, and in terms of $u = rR$ this reads
  $\int_0^\infty |u|^2\,\d r < \infty$, 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

$$
\frac{\d^2 u}{\d r^2} \approx \frac{\ell(\ell+1)}{r^2}\,u.
$$

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

$$
u(r) \sim r^{\ell+1},
\qquad
R(r) \sim r^{\ell}
\qquad (r \to 0).
$$

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

> **Definition (Radial normalization).** The full three-dimensional state
> $\psi_{n\ell m} = R_{n\ell}\,Y_\ell^m$ is normalized when the angular harmonic
> is normalized on the sphere and
> $$
> \int_0^\infty |R_{n\ell}(r)|^2\,r^2\,\d r
>   = \int_0^\infty |u_{n\ell}(r)|^2\,\d r = 1.
> $$
> The factor $r^2$ is the radial part of the volume element
> $\d V = r^2\sin\theta\,\d r\,\d\theta\,\d\varphi$; folding it into $u = rR$
> makes the radial problem look one-dimensional.

## The free particle in spherical coordinates

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

$$
\frac{1}{r^2}\,\frac{\d}{\d r}\!\left(r^2\,\frac{\d R}{\d r}\right)
  + \left[k^2 - \frac{\ell(\ell+1)}{r^2}\right]R = 0.
$$

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 $n_\ell(\rho)$,

$$
j_\ell(\rho) = (-\rho)^\ell\!\left(\frac{1}{\rho}\,\frac{\d}{\d\rho}\right)^{\!\ell}\frac{\sin\rho}{\rho},
\qquad
n_\ell(\rho) = -(-\rho)^\ell\!\left(\frac{1}{\rho}\,\frac{\d}{\d\rho}\right)^{\!\ell}\frac{\cos\rho}{\rho}.
$$

The lowest few, in closed form:

$$
j_0(\rho) = \frac{\sin\rho}{\rho},
\qquad
j_1(\rho) = \frac{\sin\rho}{\rho^2} - \frac{\cos\rho}{\rho},
\qquad
j_2(\rho) = \left(\frac{3}{\rho^3} - \frac{1}{\rho}\right)\sin\rho - \frac{3}{\rho^2}\cos\rho,
$$
$$
n_0(\rho) = -\frac{\cos\rho}{\rho},
\qquad
n_1(\rho) = -\frac{\cos\rho}{\rho^2} - \frac{\sin\rho}{\rho}.
$$

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

$$
j_\ell(\rho) \xrightarrow[\rho\to 0]{} \frac{\rho^\ell}{(2\ell+1)!!},
\qquad
n_\ell(\rho) \xrightarrow[\rho\to 0]{} -\frac{(2\ell-1)!!}{\rho^{\ell+1}},
$$

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,

$$
j_\ell(\rho) \xrightarrow[\rho\to\infty]{} \frac{1}{\rho}\sin\!\left(\rho - \frac{\ell\pi}{2}\right),
\qquad
n_\ell(\rho) \xrightarrow[\rho\to\infty]{} -\frac{1}{\rho}\cos\!\left(\rho - \frac{\ell\pi}{2}\right),
$$

the phase shift $\ell\pi/2$ recording the angular momentum. The energy spectrum
is continuous, $E = \hbar^2 k^2/2\mu$ 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.

$$
% caption: Spherical Bessel functions of the first kind $j_\ell(\rho)$: each
% starts flat as a power $\rho^\ell$ near the origin, peaks later for larger
% $\ell$ as the centrifugal barrier delays the first oscillation, then decays
% like $1/\rho$.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,0) -- (9.2,0) node[anchor=north, black] {$x$};
\draw[->, black] (0,-0.75) -- (0,3.1) node[anchor=south, black] {$j_l(x)$};
\draw[black] (0,2.4) -- (-0.12,2.4) node[anchor=east, black!70, font=\scriptsize] {1};
% j0 = sin r / r, scaled *2.4
\draw[acc, very thick, domain=0.001:9.0, samples=200, variable=\x]
  plot ({\x}, {2.4*sin(\x r)/\x});
\node[acc, anchor=south] at (0.7,2.45) {$j_0$};
% j1 = sin/r^2 - cos/r, scaled *2.4
\draw[black, very thick, densely dashed, domain=0.05:9.0, samples=220, variable=\x]
  plot ({\x}, {2.4*(sin(\x r)/(\x*\x) - cos(\x r)/\x)});
\node[black, anchor=south] at (2.05,1.55) {$j_1$};
% j2, scaled *2.4
\draw[black, very thick, densely dotted, domain=0.15:9.0, samples=240, variable=\x]
  plot ({\x}, {2.4*((3/(\x*\x*\x) - 1/\x)*sin(\x r) - 3/(\x*\x)*cos(\x r))});
\node[black, anchor=south] at (3.55,1.15) {$j_2$};
\end{tikzpicture}
$$

## 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

$$
j_\ell(ka) = 0.
$$

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

$$
E_{n\ell} = \frac{\hbar^2 \beta_{n\ell}^2}{2\mu a^2},
\qquad n = 1, 2, 3, \dots,
\quad \ell = 0, 1, 2, \dots.
$$

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

$$
E_{n0} = \frac{n^2\pi^2\hbar^2}{2\mu a^2},
$$

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

| $\ell$ | first zero $\beta_{1\ell}$ | second zero $\beta_{2\ell}$ | spectroscopic label |
| --- | --- | --- | --- |
| $0$ | $\pi \approx 3.142$ | $2\pi \approx 6.283$ | $1s,\ 2s$ |
| $1$ | $4.493$ | $7.725$ | $1p,\ 2p$ |
| $2$ | $5.763$ | $9.095$ | $1d,\ 2d$ |
| $3$ | $6.988$ | $10.417$ | $1f,\ 2f$ |

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.

$$
% caption: Lowest levels of the infinite spherical well, ordered by the zeros of
% the spherical Bessel functions; left labels give the dimensionless energy
% $2\mu a^2 E/\hbar^2 = \beta_{n\ell}^2$. Unlike hydrogen, states of different
% $\ell$ at the same radial count sit at different energies.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% vertical energy axis
\draw[->, black] (0,0) -- (0,6.3) node[anchor=south, black] {$E$};
% levels at heights proportional to beta^2: 1s 9.87, 1p 20.19, 1d 33.21, 2s 39.48, 1f 48.83, 2p 59.68
% scale by /9.5 and shift
\def\lev#1#2#3{\draw[acc, very thick] (0.6,#1) -- (2.6,#1); \node[anchor=west, font=\scriptsize] at (2.75,#1) {#2}; \node[anchor=east, black, font=\scriptsize] at (0.5,#1) {#3};}
\lev{1.04}{1s}{9.9}
\lev{2.13}{1p}{20.2}
\lev{3.50}{1d}{33.2}
\lev{4.16}{2s}{39.5}
\lev{5.14}{1f}{48.8}
\lev{6.00}{2p}{59.7}
\end{tikzpicture}
$$

## 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 $\varphi$. Total node count is the sum, and it grows with both
excitation of the radial motion and of the rotation.

$$
% caption: Radial functions $u_{n0}(r) = rR_{n0}$ for the l = 0 states of the
% spherical box: the ground state has no interior node, the first excited state
% one, matching the sine standing waves of the one-dimensional well.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes: r from 0 to a=pi (map a to x=6)
\draw[->, black] (0,0) -- (6.7,0) node[anchor=north, black] {$r$};
\draw[black] (0,-2.2) -- (0,2.2) node[anchor=south, black, font=\scriptsize] {$u(r)$};
\draw[black] (6.0,0.08) -- (6.0,-0.08) node[anchor=north, black!70, font=\scriptsize] {$a$};
% u_10 = sin(pi r / a): one hump. x in [0,6] maps r in [0,a], arg = pi * x/6
\draw[acc, very thick, domain=0:6, samples=120, variable=\x]
  plot ({\x}, {1.9*sin(180*\x/6)});
\node[acc, anchor=south] at (3.0,1.95) {$u_{10}$ (0 nodes)};
% u_20 = sin(2 pi r / a): one interior node at r=a/2 (x=3)
\draw[black, very thick, densely dashed, domain=0:6, samples=140, variable=\x]
  plot ({\x}, {1.55*sin(360*\x/6)});
\node[black, anchor=north] at (4.5,-1.6) {$u_{20}$ (1 node)};
\fill[black] (3.0,0) circle (2.0pt);
\end{tikzpicture}
$$

## The three ingredients of a central-potential spectrum

Every problem in this module — the free particle just solved, the
[hydrogen atom](/quantum-mechanics/central-potentials/the-hydrogen-atom), and the
[isotropic oscillator](/quantum-mechanics/central-potentials/the-isotropic-oscillator-and-hidden-symmetry) —
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 $\hbar^2\ell(\ell+1)/2\mu r^2$** 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.[^gs-radial][^shankar-radial][^ct-central]

[^gs-radial]: Griffiths & Schroeter, _Introduction to Quantum Mechanics_, 3rd ed. (Cambridge, 2018), §4.1 — separation in spherical coordinates, the substitution $u = rR$, the effective potential with the centrifugal barrier, and the free particle / infinite spherical well via the spherical Bessel functions. Publisher: https://doi.org/10.1017/9781316995433
[^shankar-radial]: Shankar, _Principles of Quantum Mechanics_, 2nd ed. (Springer, 1994), Ch. 12 §12.6 — the radial equation, boundary conditions at the origin ($u \sim r^{\ell+1}$), and the free-particle solutions in spherical coordinates. Publisher: https://doi.org/10.1007/978-1-4757-0576-8
[^ct-central]: Cohen-Tannoudji, Diu & Laloë, _Quantum Mechanics_, Vol. I (Wiley, 1977), Ch. VII §A — the general theory of a particle in a central potential and the structure of the radial equation.