06. Angular momentum and hydrogen orbitals

Angular momentum and hydrogen orbitals

Actually see what 1s, 2p and 3d look like, and work out which part of the picture n, l and m each control.

What you will see

The probability cloud developing dot by dot — each dot a possible outcome of a position measurement
n sets the size and the radial node spheres, l sets the shape, m sets the orientation
Slicing a 3d orbital open reveals that the cloverleaf is hollow in the middle

Assumed background

Spherical coordinates
Stationary states and eigenvalue problems (module 03)
Basic ideas about angular momentum (module 05 helps but is not required)

Hydrogen was the first real victory of quantum mechanics: a two-particle three-dimensional problem solved completely and exactly, with spectral lines matching experiment to several decimal places.

It is also where chemistry starts — the shape of the periodic table, the directionality of covalent bonds and the geometry of molecules are all written into the lobes below.

Hydrogen orbitals: actually seeing what |ψ(r)|² looks like

Each dot is one possible outcome of a position measurement, and their density is the probability density. Change the quantum numbers and watch the lobes, the nodes and the sheer size respond to n, l and m.

Loading 3D scene…

Quantum numbers

Principal quantum number n3

Angular quantum number l2

Magnetic quantum number m0

Current: 3d_{z^2} · l < n, |m| ≤ l

Representation

Point-cloud opacity0.72

The dumbbells and cloverleaves of chemistry class are the *real* spherical harmonics. For the complex ones, |ψ|² is symmetric about z and shows no azimuthal structure at all. Both are correct — they are just different choices of basis.

Structure and clipping

Radial node spheres (0)Radius ringsClipping plane

Radial probability P(r) = r²|R(r)|²

The dashed verticals are radial nodes. Watch how fast the peak moves outward with n — that is what it means for an atom to "grow".

\psi_{320}(r,\theta,\varphi) = R_{32}(r)\,Y_{2}^{0}(\theta,\varphi)

E = −13.6/3² eV = -1.512 eV

Radial nodes 0Angular nodes 2Total nodes 2Sampled points 0 / 26,000

lobes with ψ > 0
lobes with ψ < 0
radial node spheres

What to look for

Start with 1s → 2s → 3s. All three are spherical, but the radius grows fast and node spheres appear inside. That is where the "size" of an atom comes from.
Hold n = 2 and take l from 0 to 1: the sphere becomes a dumbbell. Now change m and the dumbbell swings to a new direction — that is all p_z, p_x and p_y are.
Choose 3d with m = ±2, switch on the clipping plane and drag it. From outside it is a cloverleaf; only a cut reveals that the middle is empty.
Switch from real to complex orbitals: the shape immediately becomes a ring, symmetric about z. Both describe the same set of eigenstates in different bases — chemists prefer the real ones, angular-momentum algebra prefers the complex ones.
Switch on the node spheres and count them: there are exactly n − l − 1.

✕ "The electron orbits the nucleus"

A stationary state has no trajectory. The cloud is not a path the electron traced out; it is "the probability that a position measurement lands here". The density does not change with time at all.

✕ "Orbitals have a definite boundary"

|ψ|² is non-zero beyond any finite radius; it merely decays exponentially. The "edge" you see is where the sampled points thin out, not a wall.

✕ "Nodes are places the electron cannot cross"

The density is zero on a nodal surface, but that is not a barrier. Asking how the electron "gets across" a node already presupposes a trajectory, and there is none.

Think it through

Why does the hydrogen energy depend only on n and not on l or m? (Hint: this extra degeneracy is peculiar to the Coulomb potential; other central forces do not have it.)
2p_z and 2p_x have the same energy and differ only by a rotation. Does "the electron is in 2p_z" have any absolute meaning?
The total node count is always n − 1. Recall that the nth state of a one-dimensional well has n nodes — how are the two facts related?

This is an interactive 3D scene (hydrogen-orbital) and needs JavaScript and WebGL to run.

Three quantum numbers, three jobs

Any central-force stationary state separates:

\psi_{n\ell m}(r,\theta,\varphi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\varphi).

Each quantum number then owns one aspect of the picture, and you can verify each in the scene:

Number	Range	What it controls
$n$	$1,2,3,\ldots$	Size ( $\langle r\rangle \sim n^2 a_0$ ) and energy; $n-\ell-1$ radial node spheres
$\ell$	$0 \ldots n-1$	Shape: sphere, dumbbell, cloverleaf…; $\ell$ angular nodes
$m$	$-\ell \ldots \ell$	Orientation of the lobes, or (for complex orbitals) the phase circulation about $z$

The total node count is always $n-1$ — the three-dimensional version of “the nth state of a one-dimensional well has n nodes”.

Energy depends only on n: an unusual degeneracy

E_n = -\frac{13.6\ \mathrm{eV}}{n^2}.

$2s$ and $2p$ look nothing alike yet have identical energy. This is not true for central forces in general — in a many-electron atom $2s$ sits below $2p$ .

The Coulomb potential is special because it has an extra conserved quantity: the Laplace–Runge–Lenz vector. Classically that is “the major axis of the ellipse does not precess”; quantum mechanically it enlarges the symmetry group from $SO(3)$ to $SO(4)$ and lifts the degeneracy from $2\ell+1$ to $n^2$ .

Real versus complex orbitals

The toggle in the scene matters.

Complex spherical harmonics $Y_\ell^m$ are eigenstates of $\hat L_z$ with definite angular-momentum projection $m$ . But their $|\psi|^2$ is symmetric about $z$ and draws a doughnut with no azimuthal structure at all.

Real orbitals are linear combinations of $m$ and $-m$ :

p_x = \tfrac{1}{\sqrt2}\left(Y_1^{-1}-Y_1^{1}\right),\qquad p_y = \tfrac{\ii}{\sqrt2}\left(Y_1^{-1}+Y_1^{1}\right).

They are no longer eigenstates of $\hat L_z$ , but they have the same energy, so they are perfectly legitimate stationary states. Every dumbbell and cloverleaf in a chemistry course is one of these, because bonding cares about direction.

Both pictures describe the same set of states in different bases. Use complex ones for angular-momentum algebra, real ones for chemical bonds.

The radial distribution: why atoms grow

Do not read $|R_{n\ell}(r)|^2$ as “the probability of finding the electron at radius r”. The area of a spherical shell grows as $r^2$ , so the correct radial density is

P(r) = r^2\left|R_{n\ell}(r)\right|^2 .

For $1s$ the peak of $P(r)$ sits exactly at $r=a_0$ — the Bohr radius reappears as the most probable radius rather than an actual orbit.

The $P(r)$ inset in the control panel marks the radial nodes with dashed lines. Change $n$ and watch how quickly the peak moves outward.

The radial equation and the effective potentialadvanced

Set $u(r) = rR(r)$ and the radial equation takes a one-dimensional form:

-\frac{\hbar^2}{2m}u'' + \left[\underbrace{-\frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2\ell(\ell+1)}{2mr^2}}_{V_{\text{eff}}(r)}\right]u = Eu .

The second term is the centrifugal barrier. For $\ell>0$ it diverges near the origin and pushes the wavefunction away from it. That is why only $s$ orbitals ( $\ell=0$ ) are non-zero at the nucleus — and hence why only $s$ electrons feel the contact interaction with nuclear spin (hyperfine structure, the Fermi contact term in NMR).

Compare $3s$ and $3d$ in the scene: the first still has noticeable density near the nucleus, the second is completely hollow there.

Common misreadings

Key formulas

Separation of variables
$\psi_{n\ell m} = R_{n\ell}(r)Y_\ell^m(\theta,\varphi)$
True for any central force
Hydrogen levels
$E_n = -13.6\,\mathrm{eV}/n^2$
Depends on n alone — special to the Coulomb potential
Radial probability
$P(r) = r^2|R_{n\ell}(r)|^2$
Do not forget the r² volume factor
Counting nodes
$(n-\ell-1)\ \text{radial} + \ell\ \text{angular} = n-1$
Same origin as the one-dimensional node theorem

Think it through

In hydrogen $3s$ , $3p$ and $3d$ are degenerate; in sodium $3s < 3p < 3d$ . Explain the ordering using “an s electron penetrates inside the screening of the inner electrons”.
Take n from 1 to 5 in the scene and the camera pulls back automatically. Without that, $1s$ would be almost invisible on the scale of $5g$ — estimate $\langle r\rangle_{n=5}/\langle r\rangle_{n=1}$ .
$2p_z$ and $2p_x$ differ only by a rotation and have the same energy. Does “the electron is in $2p_z$ ” have any absolute meaning? When does that statement become meaningful?

Angular momentum and hydrogen orbitals

What you will see

Assumed background

Hydrogen orbitals: actually seeing what |ψ(r)|² looks like

Quantum numbers

Representation

Structure and clipping

Radial probability P(r) = r²|R(r)|²

Three quantum numbers, three jobs

Energy depends only on n: an unusual degeneracy

The radial distribution: why atoms grow

Common misreadings

Key formulas

Think it through

Go deeper · matching textbook sections

Having finished this module

Quantum numbers

Representation

Structure and clipping

Radial probability P(r) = r²|R(r)|²

Three quantum numbers, three jobs

Energy depends only on n: an unusual degeneracy

The radial distribution: why atoms grow

Common misreadings

∎Key formulas

Think it through

Go deeper · matching textbook sections

Having finished this module

Key formulas