The Schrödinger Equation and Atomic Wavefunctions

The hydrogen wavefunctions are not a historical curiosity confined to introductory courses — they are the literal computational starting point for essentially all of modern quantum chemistry. Density functional theory and Hartree–Fock calculations, used daily to predict reaction mechanisms, catalyst active sites, and drug-binding geometries, represent multi-electron molecular orbitals as combinations of basis functions built directly from hydrogen-like radial and angular forms (Slater-type and Gaussian-type orbitals are engineered to mimic the $e^{-r/na_0}$ decay and the $Y_\ell^m$ angular shapes derived here). Spectroscopists still use the Rydberg formula, an exact rearrangement of $E_n = -13.6\,\text{eV}/n^2$, to calibrate emission-line wavelengths from astrophysical sources and to identify hydrogen and hydrogenic ions in stellar atmospheres and interstellar gas by their characteristic Lyman, Balmer, and Paschen series. Rydberg atoms — hydrogen-like atoms deliberately excited to enormous $n$ (sometimes $n>100$) — are now a leading platform for quantum computing and quantum simulation, because their huge orbital radii ($r\propto n^2 a_0$) make their interactions with neighboring atoms and with microwave fields extraordinarily strong and easily controlled, a direct engineering exploitation of the $n$-dependence derived above. Closer to everyday inorganic chemistry, crystal field and ligand field theory (used to predict the colors and magnetism of transition-metal complexes) start by asking how the d-orbital angular shapes — pure solutions $Y_2^m$ of the same angular equation solved here — are split in energy by an octahedral or tetrahedral array of point charges; without the angular part of this derivation, there is no starting point for that entire branch of coordination chemistry. Even magnetic resonance imaging and NMR spectroscopy rely on the quantized $m_\ell$ (and its spin analogue $m_s$) structure that traces straight back to the single-valuedness argument in Step 2 of the first derivation.

Ask the tutor: How does adding a second electron break the hydrogen atom's exact solvability, and what approximations replace it?

Newtonian mechanics describes a particle by a trajectory $\mathbf{r}(t)$ obeying $F = ma$. Quantum mechanics describes a particle by a wavefunction $\psi(\mathbf{r},t)$ obeying the Schrödinger equation, a linear partial differential equation that plays the role $F=ma$ plays classically. For a system whose potential energy does not change with time — the hydrogen atom's proton sits still, to a superb approximation, while the electron's Coulomb attraction to it is fixed — the wavefunction can be written as a spatial part times a phase that oscillates in time, and the spatial part $\psi(\mathbf r)$ satisfies the time-independent Schrödinger equation,

This says: apply the Hamiltonian operator $\hat H$ (built from kinetic and potential energy) to the function $\psi$, and if $\psi$ is a valid solution you get the same function back, multiplied by a number $E$ — the energy. This is precisely the mathematics of a standing wave: only certain wavelengths (equivalently, certain curvatures) fit into a fixed boundary without destructive self-interference, and each allowed pattern has one energy. For hydrogen, $\hat H$ has a kinetic part depending on how sharply $\psi$ curves in space (the Laplacian $\nabla^2$, which measures the second derivative — the "concavity" — of $\psi$ in three dimensions) and a Coulomb potential part depending only on the distance $r$ from the electron to the proton:

Because the potential is spherically symmetric — it depends on $r$ alone, not on direction — it is natural to write $\nabla^2$ in spherical polar coordinates $(r,\theta,\phi)$. When this is done, the equation for $\psi(r,\theta,\phi)$ separates cleanly into a piece that depends only on $r$ and a piece that depends only on the angles $(\theta,\phi)$. This separability is the single most important structural fact in this article: it is why atomic orbitals factor into a radial part and an angular part, $\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^{m}(\theta,\phi)$, and why chemists can discuss "shape" (angular) and "size/number of nodes" (radial) as separate, combinable properties. Quantization itself is not an extra postulate bolted onto the equation — it is a mathematical consequence of demanding that $\psi$ be well-behaved: finite everywhere (no infinite probability density), single-valued (a wave cannot have two values at the same point), and normalizable ($\int|\psi|^2 d\tau = 1$, since the electron must be somewhere). Only for special values of energy does the radial equation produce a solution that decays properly as $r\to\infty$ instead of blowing up; only for special (integer) values of an angular separation constant does the angular equation produce a solution that is single-valued as $\phi$ increases by $2\pi$. Those special values are exactly the quantum numbers $n$, $\ell$, $m_\ell$.

Once the equation is solved, $\psi$ itself is not observable and carries no direct physical meaning as an amplitude of "stuff." What is physically meaningful is $|\psi|^2$, the Born probability density: $|\psi(\mathbf r)|^2\,d\tau$ is the probability of finding the electron in the infinitesimal volume element $d\tau$ around point $\mathbf r$, given the atom is in that particular quantum state. An "orbital" drawn as a cloud or a boundary surface is a contour of $|\psi|^2$ (or of $\psi$ itself, showing where it is positive versus negative, which matters for bonding overlap) — not a picture of a path the electron travels.

The full time-independent Schrödinger equation for the hydrogen atom, written out with the Laplacian in spherical coordinates, is

where:

• $\hbar = h/2\pi$ is the reduced Planck constant,
• $m_e$ is the electron's (reduced) mass,
• $e$ is the elementary charge, $\varepsilon_0$ the vacuum permittivity,
• $r,\theta,\phi$ are spherical coordinates centered on the proton,
• $E$ is the total energy (an eigenvalue, i.e. a number determined by the equation, not chosen freely).

Reading it: the bracketed term is the kinetic energy operator's action on $\psi$, decomposed into a radial curvature piece and an angular curvature piece (the operator inside the last two terms in braces is, up to a constant, the square of the angular momentum operator $\hat L^2$). The $-e^2/4\pi\varepsilon_0 r$ term is just Coulomb's law for potential energy between charges $-e$ and $+e$ separated by $r$. The whole equation says: (kinetic energy operator + potential energy operator) acting on $\psi$ reproduces $E\psi$ — total mechanical energy is conserved and quantized. Separating $\psi(r,\theta,\phi) = R(r)\,\Theta(\theta)\,\Phi(\phi)$ and dividing through by $\psi$ turns this single partial differential equation into three ordinary differential equations linked by two separation constants. Demanding physically acceptable solutions (finite, single-valued, normalizable) forces those constants — and hence the energy — to be quantized:

where:

• $n$ is the principal quantum number, arising from the radial equation's requirement that $R(r)\to 0$ as $r\to\infty$,
• all the fundamental constants combine into a single number, $13.6\ \text{eV}$ (or $2.18\times10^{-18}\,\text{J}$), so that $E_n = -13.6\,\text{eV}/n^2$.

Reading it: this is the exact Bohr formula, but now derived from a wave equation rather than assumed from planetary orbits. Energy depends only on $n$ for hydrogen (a one-electron atom) — a special "accidental" degeneracy of the pure Coulomb potential — so all orbitals with the same $n$ (2s and 2p, or 3s, 3p, 3d) share exactly the same energy in isolated hydrogen, a degeneracy that will be lifted in multi-electron atoms by electron-electron repulsion (foreshadowing why 2s lies below 2p once you add more electrons). The angular equation, independent of the potential's exact radial form, produces the quantized total orbital angular momentum and its projection along an axis:

where:

• $\ell = 0,1,2,\dots,n-1$ is the angular momentum (azimuthal) quantum number, labeled s, p, d, f,
• $m_\ell = -\ell,\dots,+\ell$ is the magnetic quantum number, the projection of angular momentum onto a chosen (z) axis,
• $Y_\ell^{m}(\theta,\phi)$ are the spherical harmonics, universal angular functions independent of which central potential you use.

Reading it: angular momentum magnitude and orientation are quantized before you even know the specific potential (Coulomb, harmonic, whatever) — this is a general feature of any spherically symmetric problem, which is why the same $Y_\ell^m$ shapes (spherical s, dumbbell p, cloverleaf d) recur for every central-force problem in chemistry and are inherited unchanged when you build multi-electron atoms or even nuclear shell models.

Example 1 — Foundational: confirm normalization fixes the constant N in the 1s wavefunction. The (unnormalized) 1s radial function found above is $R_{10}(r) = Ne^{-r/a_0}$, and the full wavefunction (since $\ell=0$ has the angularly uniform $Y_0^0 = 1/\sqrt{4\pi}$) is $\psi_{100} = \dfrac{N}{\sqrt{4\pi}}e^{-r/a_0}$. Normalization requires the total probability of finding the electron somewhere in all space to equal exactly 1:

The angular integral over $\theta,\phi$ of $|Y_0^0|^2$ gives exactly 1 (spherical harmonics are separately normalized), leaving

Using the standard integral $\int_0^\infty r^2 e^{-2r/a_0}dr = \dfrac{2a_0^3}{8} = \dfrac{a_0^3}{4}$ (from $\int_0^\infty x^ne^{-\alpha x}dx = n!/\alpha^{n+1}$ with $n=2,\ \alpha = 2/a_0$):

So the correctly normalized 1s wavefunction is $\psi_{1s} = \dfrac{1}{\sqrt{\pi a_0^3}}\,e^{-r/a_0}$. As a limiting-case check: at $r=0$, $\psi_{1s}$ is finite and maximal ($1/\sqrt{\pi a_0^3}$, not infinite), and as $r\to\infty$, $\psi_{1s}\to 0$ — both required for a normalizable bound state, and both properties this exponential form correctly reproduces.

Example 2 — Applied: where is the electron most likely to be found in the 1s state, and how does this differ from |ψ|² itself? This is the classic trap for students conflating $|\psi|^2$ (probability density, per unit volume) with the radial distribution function $P(r) = 4\pi r^2 |R(r)|^2$ (probability per unit radial shell thickness, integrated over all angles). For 1s hydrogen,

Although $|\psi|^2$ itself is maximal at $r=0$ (the electron is most dense right at the nucleus), $P(r)$ vanishes at $r=0$ because the volume of a shell of radius $r\to 0$ shrinks to zero even faster than the density grows. Maximizing $P(r)$: set $dP/dr = 0$,

The non-trivial root is $r = a_0$. So the most probable distance from the nucleus in the ground state is exactly one Bohr radius — reproducing the "size" of Bohr's original orbit — but now understood as the peak of a continuous probability distribution, not the radius of a physical circular path. As a limiting check, note $P(r)\to 0$ both as $r\to 0$ (shell volume vanishes) and as $r\to\infty$ (exponential decay dominates any polynomial growth), consistent with a single, well-defined maximum in between.

Example 3 — Exam/competition-level: compare the number of radial nodes and total nodes for the 4d and 5p hydrogen orbitals, and use this to predict their relative extents. The general node-counting rules, which follow directly from the degree of the Laguerre polynomial in $R_{n\ell}(r)$ and from the angular nodal planes in $Y_\ell^m$, are:

For 4d: $n=4,\ \ell=2$, so radial nodes $= 4-2-1 = 1$, angular nodes $=2$, total $=3$. For 5p: $n=5,\ \ell=1$, so radial nodes $=5-1-1=3$, angular nodes $=1$, total $=4$. Both obey total nodes $=n-1$ exactly (3 and 4 respectively), confirming internal consistency of the counting rule. Physically, radial nodes are spherical shells where $R_{n\ell}(r)=0$ (the wavefunction changes sign as you move outward), while angular nodes are planes or cones through the nucleus where $Y_\ell^{m}=0$. Because 5p has one more total node than 4d, and because a higher principal quantum number $n$ generically pushes the outermost probability maximum of the radial distribution to larger $r$ (the exponential decay constant scales as $1/n$, and the polynomial prefactor's degree grows with $n$), 5p is the more diffuse, larger-average-radius orbital of the two, despite 4d having a higher angular momentum. This is exactly the kind of comparison that underlies the (n+ℓ) ordering rule used to build up the periodic table, foreshadowing why electron configurations fill 4s before 3d in multi-electron atoms even though hydrogen alone would never distinguish them by energy.