Hydrogen Wave Functions and Orbitals

The previous lesson extracted the quantum numbers and energies of hydrogen from the separated Schrödinger equation. This lesson uses the wave functions themselves. Each one is a product of a radial factor and an angular factor, and its square is a probability density in three-dimensional space. The radial factor answers how far from the nucleus?; the angular factor answers in what direction? Together they replace the Bohr orbit with a diffuse cloud whose shape depends on $ℓ$ and $m$ .

The complete wave function

Assembling the three separated factors, the hydrogen wave function is

ψ_{n ℓ m} (r, θ, ϕ) = C_{n ℓ m} R_{n ℓ} (r) f_{ℓ m} (θ) g_{m} (ϕ) = C_{n ℓ m} R_{n ℓ} (r) Y_{ℓ m} (θ, ϕ),

with $C_{n ℓ m}$ fixed by normalization. The energy depends only on $n$ , but the wave function depends on all three quantum numbers, one from each coordinate. For any $n$ there are $n$ values of $ℓ$ and $2 ℓ + 1$ values of $m$ , so most energy levels carry several distinct wave functions — the degeneracy of the inverse-square force.

The radial functions for the three lowest shells, in the hydrogenic form with Bohr radius $a_{0} = ℏ^{2} / μ k e^{2}$ , are:¹

$n$	$ℓ$	$R_{n ℓ} (r)$
$1$	$0$	$2 a_{0}^{- 3/2} e^{- r / a_{0}}$
$2$	$0$	$\frac{1}{2 2} a_{0}^{- 3/2} (2 - \frac{r}{a _{0}}) e^{- r /2 a_{0}}$
$2$	$1$	$\frac{1}{2 6} a_{0}^{- 3/2} \frac{r}{a _{0}} e^{- r /2 a_{0}}$
$3$	$0$	$\propto (1 - \frac{2 r}{3 a _{0}} + \frac{2 r ^{2}}{27 a _{0}^{2}}) e^{- r /3 a_{0}}$
$3$	$1$	$\propto \frac{r}{a _{0}} (1 - \frac{r}{6 a _{0}}) e^{- r /3 a_{0}}$
$3$	$2$	$\propto \frac{r ^{2}}{a _{0}^{2}} e^{- r /3 a_{0}}$

Two structural facts read off the table. The exponential decay rate is set by $n$ alone through $e^{- r / n a_{0}}$ , so higher shells extend farther. Near the origin $R_{n ℓ} \propto r^{ℓ}$ , so only $ℓ = 0$ states are nonzero at the nucleus; the higher the $ℓ$ , the more the centrifugal barrier pushes the electron out.

Probability density and the radial distribution

Born's rule makes $∣ ψ ∣^{2} d τ$ the probability of finding the electron in the volume element $d τ$ . In spherical coordinates the volume element is

d τ = r^{2} sin θ d r d θ d ϕ .

Two different questions have two different answers. The probability density $∣ ψ ∣^{2}$ is the chance per unit volume at a point. The radial probability $P (r) d r$ is the chance of finding the electron anywhere in the thin spherical shell between $r$ and $r + d r$ , obtained by multiplying the density by the shell volume $4 π r^{2} d r$ (for a spherically symmetric state):

P (r) = 4 π r^{2} ∣ ψ ∣^{2} \propto r^{2} ∣ R_{n ℓ} (r) ∣^{2} .

The radial probability weights the density by the shell volume 4πr²dr, which vanishes at the origin and grows outward.

The ground state

For $n = 1$ , both $ℓ$ and $m$ are zero, the Laguerre polynomial is $1$ , and

ψ_{100} = C_{100} e^{- Z r / a_{0}}, C_{100} = \frac{1}{π} (\frac{Z}{a _{0}})^{3/2} .

The density $∣ ψ_{100} ∣^{2} \propto e^{- 2 Z r / a_{0}}$ is maximal at the origin and falls monotonically. The radial distribution multiplies this by $r^{2}$ , which vanishes at the origin, so $P (r)$ starts at zero, rises, and peaks. Setting $d P / d r = 0$ gives the most probable radius

r_{mp} = \frac{a _{0}}{Z},

the Bohr radius for hydrogen.¹ The two curves say complementary things: the electron is most densely found at the nucleus but most likely found in the shell at $a_{0}$ , because there is far more shell area out there.

For the ground state, |ψ|² peaks at the nucleus while the radial probability P(r) = r²|ψ|² peaks at the Bohr radius a₀.

Excited states

For $n = 2$ the possibilities are $ℓ = 0$ (the 2s state, spherically symmetric) and $ℓ = 1$ (the three 2p states):¹

ψ_{200} = C_{200} (2 - \frac{Z r}{a _{0}}) e^{- Z r /2 a_{0}}, ψ_{210} = C_{210} \frac{Z r}{a _{0}} e^{- Z r /2 a_{0}} cos θ,

with $ψ_{21, \pm 1} \propto (Z r / a_{0}) e^{- Z r /2 a_{0}} sin θ e^{\pm i ϕ}$ . The 2p radial distribution peaks at the second Bohr orbit, $r_{max} = 2^{2} a_{0} = 4 a_{0}$ . The 2s distribution has two maxima: a large one near $4 a_{0}$ and a small subsidiary bump close to the nucleus, the signature of the extra factor near the origin. This inner bump lets an $ℓ = 0$ electron penetrate the core of a heavier atom, which is decisive for the ordering of subshells.

Radial distributions for 1s, 2s, and 2p. The 2p peak sits at the second Bohr orbit 4a₀; the 2s curve adds a small inner bump near the nucleus.

Radial nodes

The number of times $R_{n ℓ} (r)$ crosses zero (for $r > 0$ ) is $n - ℓ - 1$ , the number of radial nodes. The 1s and 2p functions have none; the 2s function has one, at $r = 2 a_{0}$ , where the factor $(2 - r / a_{0})$ vanishes. Nodes carry no probability, and each one separates regions of opposite sign in the wave function.

The radial functions R₂₀ and R₂₁. R₂₀ (2s) crosses zero once at 2a₀; R₂₁ (2p) starts at zero, rises, and decays without a node.

Angular shapes and orbitals

The angular factor $∣ Y_{ℓ m} (θ, ϕ) ∣^{2}$ fixes the directional distribution, and it depends on $ℓ$ and $m$ but never on the radial part.¹

$ℓ = 0$ (s). $Y_{00}$ is constant, so the density is spherically symmetric — a round cloud.
$ℓ = 1$ , $m = 0$ (p). The density is $\propto cos^{2} θ$ , largest along the $z$ axis and zero in the equatorial plane: two lobes, a dumbbell.
$ℓ = 1$ , $m = \pm 1$ . The density is $\propto sin^{2} θ$ , largest in the equatorial plane: a ring, or toroid, around the $z$ axis.
$ℓ = 2$ (d). Four-lobed cloverleaf patterns and their variants.

Angular probability shapes: the s orbital is a sphere, the p orbital a two-lobed dumbbell, the d orbital a four-lobed cloverleaf.

These angular shapes are the orbitals of chemistry. The directional lobes of the p and d clouds determine the geometry of the bonds an atom can form. A more literal picture plots the density as a dot cloud, dense where $∣ ψ ∣^{2}$ is large.

A dot-density rendering of the ground-state cloud: each dot marks a sampled electron position, densest at the nucleus and thinning outward.

With the spatial wave functions in hand, the description is still incomplete: the electron carries an intrinsic angular momentum the Schrödinger equation never produced. That spin adds the fourth quantum number and splits the spectral lines.

Tipler & Llewellyn, §7-3 — the complete wave function $ψ_{n ℓ m} = C_{n ℓ m} R_{n ℓ} Y_{ℓ m}$ , the ground-state normalization and most-probable radius $a_{0} / Z$ , the $n = 2$ radial distributions with the 2p peak at $4 a_{0}$ , and the angular ( $s$ / $p$ / $d$ ) probability shapes. ↩ ↩² ↩³ ↩⁴