---
title: The Schrödinger Equation in Three Dimensions and Hydrogen
module: The Quantum Hydrogen Atom
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  Extending the Schrödinger equation to three dimensions and separating it in
  spherical coordinates produces three ordinary differential equations, one per
  coordinate. Their boundary conditions generate the quantum numbers n, ℓ, and
  mℓ, quantize the angular momentum to √(ℓ(ℓ+1))ℏ with projections mℏ, and fix
  the bound-state energies of hydrogen at −Z²(13.6 eV)/n².
topics: [The Quantum Hydrogen Atom]
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 7 — Atomic Physics; §7-1 The Schrödinger Equation in Three Dimensions"
  - book: Tipler & Llewellyn
    ref: "§7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom"
draft: false
---

The one-dimensional Schrödinger equation of the previous module fixed a single
quantum number from a single boundary condition. A real atom sits in three
dimensions, and its electron feels a potential that depends only on the distance
to the nucleus. Both facts are structural. Three dimensions supply three
coordinates, each carrying its own boundary condition and therefore its own
quantum number; a central potential $V(r)$ makes the angular part of the problem
identical for every such potential, so it can be solved once and reused. The
result is the exact quantization of energy and angular momentum in hydrogen,
recovered from wave mechanics rather than postulated as in the
[Bohr model](/atomic-physics/early-models-and-old-quantum-theory/bohr-model-hydrogen).

## The equation in three dimensions

For a particle of mass $m$ moving in a potential $V(x,y,z)$, the
time-independent Schrödinger equation replaces the single second derivative of
the 1-D case with the Laplacian:[^tl-71]

$$
-\frac{\hbar^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2}
+ \frac{\partial^2\psi}{\partial y^2}
+ \frac{\partial^2\psi}{\partial z^2}\right)
+ V\psi = E\psi.
$$

The simplest three-dimensional bound problem is a particle in a cubical box:
$V=0$ inside $0 \le x,y,z \le L$ and infinite outside. The walls force $\psi$ to
vanish on each face, exactly as in one dimension applied to each axis
separately, so the wave function is a product of three sine standing waves and
the energy is a sum of three one-dimensional energies:

$$
E_{n_1 n_2 n_3} = \frac{\hbar^2 \pi^2}{2mL^2}\left(n_1^2 + n_2^2 + n_3^2\right),
\qquad n_1, n_2, n_3 = 1, 2, 3, \ldots
$$

The ground state $(1,1,1)$ has energy $3E_1$ with $E_1 = \hbar^2\pi^2/2mL^2$. The
first excited level is reached three different ways — $(2,1,1)$, $(1,2,1)$,
$(1,1,2)$ — all with energy $6E_1$. One energy shared by more than one wave
function is a **degeneracy**.

> **Definition (Degeneracy).** An energy level is _degenerate_ when two or more
> linearly independent wave functions share it. The number of such functions is
> the degree of degeneracy. Degeneracy is a consequence of symmetry: any change
> that lowers the symmetry of the potential splits the level.

$$
% caption: In a cubic well the first excited level is threefold degenerate;
% stretching the box to unequal side lengths removes the symmetry and splits it.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% cubic well ladder (left)
\draw[black] (0,0) -- (2.6,0);
\node[anchor=east, black] at (0,0) {$3E_1$};
\draw[black, thick] (0.2,1.8) -- (2.4,1.8);
\node[anchor=east, black] at (0,1.8) {$6E_1$};
\draw[black, thick] (0.2,3.4) -- (2.4,3.4);
\node[anchor=east, black] at (0,3.4) {$9E_1$};
\node[align=center, black!70] at (1.3,-0.7) {cubic well\\ $L_1 = L_2 = L_3$};
% split ladder (right)
\begin{scope}[xshift=5.6cm]
\draw[black] (0,0) -- (2.6,0);
\node[anchor=west, black] at (2.7,0) {ground};
\draw[acc, thick] (0.2,1.55) -- (2.4,1.55);
\draw[acc, thick] (0.2,1.8) -- (2.4,1.8);
\draw[acc, thick] (0.2,2.05) -- (2.4,2.05);
\node[anchor=west, acc] at (2.7,1.8) {split triplet};
\node[align=center, black!70] at (1.3,-0.7) {stretched well\\ $L_1 < L_2 < L_3$};
\end{scope}
\end{tikzpicture}
$$

## Spherical coordinates and separation

For the hydrogen atom the proton is treated as fixed and the electron moves in
the Coulomb potential

$$
V(r) = -\frac{kZe^2}{r},
$$

where $Z$ is the atomic number ($Z=1$ for hydrogen, $Z=2$ for singly ionized
helium). Nuclear motion is folded in exactly by replacing the electron mass $m$
with the reduced mass $\mu = m/(1 + m/M_N)$.[^tl-71] Because $V$ depends only on
$r = (x^2 + y^2 + z^2)^{1/2}$, the natural coordinates are spherical:
$x = r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$, $z = r\cos\theta$.

$$
% caption: Spherical coordinates: r measures distance from the nucleus, the
% polar angle opens from the z axis, and the azimuth turns around it.
\begin{tikzpicture}[scale=1.15, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,0) -- (3.2,0) node[right, black!70] {$y$};
\draw[->, black] (0,0) -- (0,3.0) node[above, black!70] {$z$};
\draw[->, black] (0,0) -- (-1.7,-1.3) node[below left, black!70] {$x$};
% point P and radius
\coordinate (P) at (1.9,2.0);
\draw[acc, very thick, ->] (0,0) -- (P);
\node[acc, anchor=south west] at (P) {P};
\node[acc, anchor=west] at (1.45,1.15) {$r$};
% projection to xy plane
\coordinate (Q) at (1.9,0.55);
\draw[black, dashed] (P) -- (Q);
\draw[black, dashed] (0,0) -- (Q);
% polar angle arc from z-axis
\draw[black, thick] (0,1.05) arc (90:47:1.05);
\node[black] at (0.55,1.3) {polar};
% azimuth arc in plane
\draw[black, thick] (0.85,0) arc (0:16:0.85);
\node[black] at (1.15,0.18) {azimuth};
\end{tikzpicture}
$$

The transformed equation is formidable but standard,[^tl-72]

$$
-\frac{\hbar^2}{2\mu}\left[\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial\psi}{\partial r}\right)
+ \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)
+ \frac{1}{r^2\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2}\right]
+ V(r)\psi = E\psi.
$$

The solution proceeds by **separation of variables**: seek product solutions

$$
\psi(r,\theta,\phi) = R(r)\,f(\theta)\,g(\phi).
$$

Substituting and dividing turns the one partial differential equation into three
ordinary ones. The separation works because each derivative touches only its own
factor, and the radial terms can be collected on one side of an equation whose
other side depends only on the angles. Two independent variables set equal force
both sides to a constant.

- **Azimuthal equation.** The $\phi$-dependence separates first, giving
  $g(\phi) = e^{im\phi}$. Single-valuedness, $g(\phi + 2\pi) = g(\phi)$, forces
  $m$ to be an integer, positive, negative, or zero.
- **Polar equation.** The $\theta$-dependence yields the associated Legendre
  functions $f_{\ell m}(\theta)$. Finiteness at $\theta = 0$ and $\theta = \pi$
  forces $\ell$ to be a non-negative integer and restricts $|m| \le \ell$.
- **Radial equation.** The $r$-dependence carries the potential and produces the
  principal quantum number $n$ and the energy.

The product of the two angular factors is the **spherical harmonic**
$Y_{\ell m}(\theta,\phi) = f_{\ell m}(\theta)\,g_m(\phi)$, the same family of
functions for _every_ central potential, since the angular equations never
reference $V(r)$.

| $\ell$ | $m$ | $Y_{\ell m}(\theta,\phi)$ |
| --- | --- | --- |
| $0$ | $0$ | $\sqrt{1/4\pi}$ |
| $1$ | $0$ | $\sqrt{3/4\pi}\,\cos\theta$ |
| $1$ | $\pm 1$ | $\mp\sqrt{3/8\pi}\,\sin\theta\,e^{\pm i\phi}$ |
| $2$ | $0$ | $\sqrt{5/16\pi}\,(3\cos^2\theta - 1)$ |
| $2$ | $\pm 1$ | $\mp\sqrt{15/8\pi}\,\sin\theta\cos\theta\,e^{\pm i\phi}$ |
| $2$ | $\pm 2$ | $\sqrt{15/32\pi}\,\sin^2\theta\,e^{\pm 2i\phi}$ |

## Quantization of angular momentum

The angular equations do more than supply functions; they diagonalize angular
momentum. Rewriting the classical energy of a particle in a central field in
terms of the radial momentum $p_r$ and the angular momentum $L$,

$$
\frac{p_r^2}{2\mu} + \frac{L^2}{2\mu r^2} + V(r) = E,
$$

and promoting $p_r^2$ and $L^2$ to operators reproduces the two kinetic terms of
the spherical Schrödinger equation. The operator $(L^2)_{\text{op}}$ is exactly
the angular part, and it acts on the spherical harmonics as an eigenvalue
equation:[^tl-72]

$$
(L^2)_{\text{op}}\,Y_{\ell m}(\theta,\phi) = \ell(\ell+1)\hbar^2\,Y_{\ell m}(\theta,\phi).
$$

For every central potential the magnitude of the angular momentum is therefore
quantized, and its projection on the $z$ axis is quantized independently:

> **Theorem (Angular-momentum quantization).** In any potential $V = V(r)$,
> $$
> |\mathbf{L}| = \sqrt{\ell(\ell+1)}\,\hbar, \qquad \ell = 0, 1, 2, \ldots
> $$
> $$
> L_z = m\hbar, \qquad m = 0, \pm 1, \pm 2, \ldots, \pm\ell.
> $$
> The **orbital quantum number** $\ell$ fixes the length; the **magnetic
> quantum number** $m$ fixes the projection.

Two features are unusual. First, $|\mathbf{L}| = \sqrt{\ell(\ell+1)}\,\hbar$ is
strictly larger than the maximum projection $\ell\hbar$, so $\mathbf{L}$ can
never lie along the $z$ axis. The two follow from the uncertainty principle for
angular momentum: no two components of $\mathbf{L}$ can be known at once (except
when $\mathbf{L} = 0$), so a fully aligned vector, which would fix all three
components, is forbidden. Second, for a given $\ell$ there are exactly
$2\ell + 1$ allowed values of $m$, one orientation per integer projection. This
is **space quantization**: $\mathbf{L}$ points only along the discrete set of
cones whose axis is $z$.

$$
% caption: The vector model for ℓ = 2: L has length √6 ℏ and lies on one of five
% cones, one per allowed projection Lz from 2ℏ down to −2ℏ.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,-3.1) -- (0,3.3) node[right, black!70] {$z$};
% five allowed Lz levels (length sqrt6 ~ 2.449)
% angles: cos = m/sqrt6 ; m = 2,1,0,-1,-2
% m=2
\draw[acc, thick, ->] (0,0) -- ({2.449*sin(35.26)},{2.449*cos(35.26)});
\draw[black, dashed] (0,{2.449*cos(35.26)}) -- ({2.449*sin(35.26)},{2.449*cos(35.26)});
\node[acc, anchor=east] at (-0.15,2.0) {$L_z = 2$};
% m=1
\draw[black, thick, ->] (0,0) -- ({2.449*sin(65.9)},{2.449*cos(65.9)});
\draw[black, dashed] (0,{2.449*cos(65.9)}) -- ({2.449*sin(65.9)},{2.449*cos(65.9)});
\node[black, anchor=east] at (-0.15,1.0) {$L_z = 1$};
% m=0
\draw[black, thick, ->] (0,0) -- (2.449,0);
\node[black, anchor=south] at (1.9,0.05) {$L_z = 0$};
% m=-1
\draw[black, thick, ->] (0,0) -- ({2.449*sin(114.1)},{2.449*cos(114.1)});
% m=-2
\draw[black, thick, ->] (0,0) -- ({2.449*sin(144.7)},{2.449*cos(144.7)});
\node[black, anchor=west] at (0.1,-2.35) {equal length};
\end{tikzpicture}
$$

### Worked example: allowed orientations for ℓ = 2

For $\ell = 2$ the projections are $L_z = m\hbar$ with $m = 0, \pm 1, \pm 2$, so
$L_z \in \{2\hbar, \hbar, 0, -\hbar, -2\hbar\}$. The magnitude is

$$
|\mathbf{L}| = \sqrt{\ell(\ell+1)}\,\hbar = \sqrt{6}\,\hbar \approx 2.45\hbar.
$$

The smallest angle between $\mathbf{L}$ and the $z$ axis occurs at the largest
projection, $m = \ell = 2$:

$$
\cos\theta_{\min} = \frac{L_z}{|\mathbf{L}|} = \frac{2}{\sqrt{6}} = 0.816
\quad\Longrightarrow\quad \theta_{\min} = 35.3^\circ.
$$

The vector cannot reach $\theta = 0$, confirming that $\mathbf{L}$ never aligns
with the axis.[^tl-72]

## The radial equation and energy quantization

The angular constant $\ell(\ell+1)$ feeds into the radial equation, which for
the Coulomb potential reads

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

The bracket is an **effective potential**
$V_{\text{eff}}(r) = \hbar^2\ell(\ell+1)/2\mu r^2 - kZe^2/r$: the attractive
Coulomb term plus a repulsive centrifugal barrier that grows with $\ell$.
Bound states have $E < 0$; the well confines the electron, and only discrete
energies give normalizable $R(r)$.

$$
% caption: The Coulomb well plus the centrifugal barrier. Negative total energy
% (E < 0) gives bound, quantized states; positive energy is unbound.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (5.6,0) node[right, black!70] {$r$};
\draw[->, black] (0,-2.6) -- (0,2.3) node[above, black!70] {$V$};
% Coulomb-like well: -1/r shifted, plus small centrifugal rise near origin
\draw[acc, very thick] (0.35,2.0) .. controls (0.6,-1.0) and (1.1,-2.3) .. (1.6,-2.05)
  .. controls (2.6,-1.6) and (4.0,-0.7) .. (5.4,-0.35);
% E<0 bound level
\draw[black!70, dashed] (0.75,-1.25) -- (4.2,-1.25);
\node[black!70, anchor=west] at (4.2,-1.25) {$E < 0$ bound};
% E>0 unbound
\draw[black!70, dashed] (0,0.9) -- (5.2,0.9);
\node[black!70, anchor=west] at (4.0,1.15) {$E > 0$ free};
\node[acc, anchor=west] at (2.7,-1.95) {$V_{eff}(r)$};
\end{tikzpicture}
$$

Solving the radial equation for hydrogen gives energies that depend only on the
principal quantum number:[^tl-72]

$$
E_n = -\left(\frac{kZe^2}{\hbar}\right)^2\frac{\mu}{2n^2}
= -\frac{Z^2 E_1}{n^2},
\qquad E_1 = \tfrac{1}{2}\!\left(\frac{ke^2}{\hbar}\right)^2\!\mu \approx 13.6~\text{eV},
$$

with $n = 1, 2, 3, \ldots$ and the restriction $n > \ell$. These are the Bohr
energies exactly. The radial functions have the form
$R_{n\ell}(r) = A_{n\ell}\,e^{-r/na_0}\,r^{\ell}\,l_{n\ell}(r/a_0)$, where the
$l_{n\ell}$ are Laguerre polynomials and $a_0 = \hbar^2/(\mu ke^2)$ is the Bohr
radius.

The three quantum numbers and their ranges close the section:

$$
n = 1, 2, 3, \ldots, \qquad
\ell = 0, 1, 2, \ldots, n-1, \qquad
m = -\ell, -\ell+1, \ldots, \ell-1, \ell.
$$

That $E_n$ depends on $n$ alone, and not on $\ell$, is special to the
inverse-square force. In classical terms, the energy of an orbit in a $1/r^2$
field depends only on the semimajor axis, not the eccentricity; the largest
$\ell = n-1$ is the nearly circular orbit and small $\ell$ the eccentric one.
For any non-Coulomb central force the degeneracy in $\ell$ is lifted, a fact
that shapes the [periodic table](/atomic-physics/many-electron-atoms/periodic-table-atomic-spectra).

## Counting states and the level diagram

For a given $n$ there are $n$ values of $\ell$, and for each $\ell$ there are
$2\ell + 1$ values of $m$. The total number of spatial states with energy $E_n$
is

$$
\sum_{\ell=0}^{n-1}(2\ell + 1) = n^2,
$$

which doubles to $2n^2$ once electron spin is included in the next lessons. The
$m$-degeneracy reflects the absence of any preferred direction in space; a
[magnetic field](/atomic-physics/many-electron-atoms/periodic-table-atomic-spectra)
supplies one and splits the levels.

$$
% caption: The (n, ℓ, mℓ) count for n = 2: the 2s subshell holds one orbital,
% the 2p subshell three, for four spatial states — 2n² = 8 with spin.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  box/.style={draw, minimum width=17mm, minimum height=9mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc, text=acc, thick] (root) at (0,2.3) {$n = 2$};
\node[box] (s) at (-2.5,0.6) {s subshell};
\node[box] (p) at (2.5,0.6) {p subshell};
\draw[black, ->] (root) -- (s);
\draw[black, ->] (root) -- (p);
\node[box, minimum width=13mm] (s0) at (-2.5,-1.2) {2s\\ 1 orbital};
\node[box, minimum width=13mm] (p1) at (0.6,-1.2) {2p};
\node[box, minimum width=13mm] (p2) at (2.5,-1.2) {2p};
\node[box, minimum width=13mm] (p3) at (4.4,-1.2) {2p};
\draw[black, ->] (s) -- (s0);
\draw[black, ->] (p) -- (p1);
\draw[black, ->] (p) -- (p2);
\draw[black, ->] (p) -- (p3);
\node[black] at (2.5,-2.15) {3 orbitals};
\end{tikzpicture}
$$

States are named by the value of $n$ followed by a letter code for $\ell$: **S**
for $\ell=0$, **P** for $\ell=1$, **D** for $\ell=2$, **F** for $\ell=3$, then
alphabetically. The codes descend from the spectroscopists' _sharp_,
_principal_, _diffuse_, and _fundamental_ line series. Radiative transitions
between levels are not arbitrary; conservation of angular momentum, together with
the photon's intrinsic spin of $1\hbar$, restricts them by the **selection
rules**

$$
\Delta\ell = \pm 1, \qquad \Delta m = 0, \pm 1.
$$

$$
% caption: The hydrogen level ladder, energies −13.6/n² eV converging to the
% ionization limit; the Lyman and Balmer transitions obey Δℓ = ±1.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% ionization limit
\draw[black, thick] (0,4.0) -- (5.2,4.0);
\node[black!70, anchor=west] at (5.2,4.0) {ionization ($E = 0$)};
% levels
\draw[black!70, thick] (0,3.55) -- (5.0,3.55); \node[anchor=east, black!70] at (0,3.55) {$n=4$};
\draw[black!70, thick] (0,3.0) -- (5.0,3.0);   \node[anchor=east, black!70] at (0,3.0) {$n=3$};
\draw[black!70, thick] (0,1.7) -- (5.0,1.7);   \node[anchor=east, black!70] at (0,1.7) {$n=2$};
\draw[black!70, thick] (0,-1.4) -- (5.0,-1.4); \node[anchor=east, black!70] at (0,-1.4) {$n=1$};
% Lyman transitions to n=1
\draw[acc, ->] (1.0,1.7) -- (1.0,-1.35);
\draw[acc, ->] (1.6,3.0) -- (1.6,-1.35);
\node[acc, anchor=north] at (1.0,-1.45) {Lyman};
% Balmer transitions to n=2
\draw[black, dashed, ->] (3.4,3.0) -- (3.4,1.75);
\draw[black, dashed, ->] (4.1,3.55) -- (4.1,1.75);
\node[black, anchor=south] at (3.75,3.6) {Balmer};
\end{tikzpicture}
$$

The wave mechanics of a central potential thus reproduces the Bohr spectrum
while adding structure the old model lacked: a spread of angular-momentum states
$\ell$ at each energy, a genuine spatial probability cloud instead of an orbit,
and selection rules that decide which lines appear. The
[shapes of those clouds](/atomic-physics/quantum-hydrogen-atom/hydrogen-wave-functions)
are the subject of the next lesson.

[^tl-71]: **Tipler & Llewellyn**, §7-1 — the three-dimensional Schrödinger equation, the cubic-box degeneracy, and the transformation to spherical coordinates with the reduced-mass substitution.
[^tl-72]: **Tipler & Llewellyn**, §7-2 — separation of variables, the $L^2$ eigenvalue equation and space quantization, the radial equation with the effective potential, and the hydrogen energies $E_n = -Z^2 E_1/n^2$.