The hydrogen atom is the single most important exactly solvable problem in all of physics. One proton, one electron, held together by the electrostatic pull between them — and from that bare minimum the Schrödinger equation conjures the entire architecture of chemistry: the shells and subshells, the shapes we draw as orbitals, and ultimately the layout of the periodic table itself. Solve this one atom and you understand why the second row of the table has eight elements, why the transition metals arrive when they do, why any of it looks the way it does.

This page does one thing: it takes the electron sitting in the proton's electric field and works out what quantum mechanics allows. The answer is not a smear of possibilities but a beautifully structured catalogue of allowed states, each pinned down by three whole numbers — the quantum numbers. Everything else on the page is the story of where those three numbers come from and what they mean.

The electron in the Coulomb well

The proton, charge +e, sits at the origin. The electron, charge -e, feels the attractive electrostatic potential energy given by Coulomb's law — it depends only on the distance r to the proton:

This is a well: deep and steep near the proton, flattening toward zero far away. An electron with negative total energy is bound — trapped in the well, unable to escape to infinity — and it is those bound states we want. Plug this V(r) into the time-independent Schrödinger equation for the electron's wavefunction \psi(\mathbf{r}):

Crucially, V depends on r alone — not on which direction you look. The potential is spherically symmetric. That single fact is the master key that unlocks the whole problem.

Spherical symmetry ⇒ separate the equation

When a problem has spherical symmetry, you should stop using x, y, z and switch to spherical coordinates (r, \theta, \varphi) — the natural language of a round problem. In these coordinates the Laplacian \nabla^2 splits cleanly into a piece that acts only on r and a piece (the angular part) that acts only on the angles \theta, \varphi. That split lets us separate variables: we look for solutions that are a product of a radial function and an angular function,

The angular factors Y_\ell^m are exactly the spherical harmonics — the very same functions that describe every spherically symmetric problem, from the vibrations of a drum-sphere to the multipoles of a planet's gravity. They are handed to us ready-made by the maths, and their two labels \ell and m come along for free. All that is left is to solve a single ordinary differential equation for the radial function R_{n\ell}(r), which knows about the Coulomb well. Solving it — demanding that \psi stay finite and normalisable — is what quantises the energy and produces the third label, n.

Three quantum numbers

The demand that \psi be a well-behaved, single-valued, normalisable function does not permit just any solution. It forces three integers, nested inside one another like Russian dolls:

Read the nesting carefully: pick n, and that limits how large \ell may be; pick \ell, and that limits how large |m| may be. A single state of the hydrogen atom is specified by naming all three: (n, \ell, m).

The energy depends only on n

Here is the small miracle. Even though a state carries three labels, solving the radial equation reveals that its energy depends on n alone — the two angular numbers \ell and m drop out entirely:

This is exactly the Bohr formula — the same −13.6 eV/n² that Bohr guessed in 1913 from quantised orbits — but now it falls out honestly from the full Schrödinger equation, with no ad-hoc postulates. The ground state n=1 lies deepest at -13.6\ \text{eV}; the levels crowd toward E = 0 as n \to \infty, where the electron finally breaks free and the atom is ionised.

Because many different (\ell, m) states share the same energy, level n is degenerate. Count them: for each \ell there are 2\ell+1 values of m, and \ell runs from 0 to n-1, so

Each electron also carries a spin that can point two ways, so the number of distinct electron states in shell n is 2n^2 — the 2, 8, 18, 32, \dots capacities that give the periodic table its rows.

Orbitals: shape from ℓ, radial nodes from n and ℓ

The wavefunction \psi itself is not directly observable; what is real is |\psi|^2, the probability density for finding the electron. Because |\psi|^2 = |R_{n\ell}|^2\,|Y_\ell^m|^2, the angular factor |Y_\ell^m|^2 fixes the shape, and it is entirely governed by \ell. Physicists give the values of \ell letters for historical reasons:

• \ell = 0 — an s orbital: perfectly spherical (no angular dependence at all);
• \ell = 1 — a p orbital: two lobes, a dumbbell pointing along an axis;
• \ell = 2 — a d orbital: the familiar four-lobed cloverleaf shapes;
• and onward, \ell = 3 is f, and so on.

The radial factor R_{n\ell}(r) controls how the electron density is spread out in distance. The most useful way to see it is the radial probability distribution, P(r) = r^2\,|R_{n\ell}(r)|^2 — the probability of finding the electron in a thin shell at radius r (the r^2 is the growing surface area of that shell). A key feature is its radial nodes — radii where P(r) touches zero, spherical surfaces on which the electron is never found. Their number follows a tidy rule:

So the 1s (n=1,\ell=0) has none, the 2s has one, the 2p has none, the 3s has two. Switch the orbital below and watch the peaks — and the nodes between them — appear and shift outward:

Worked examples

Example 1 — how many subshells and orbitals in a shell? Take n = 3. The allowed values of \ell run 0, 1, 2 — that is n = 3 subshells (3s, 3p, 3d). Counting orbitals, each \ell contributes 2\ell + 1:

Nine spatial orbitals — matching the general n^2 — and 2n^2 = 18 electron states once spin is included.

Example 2 — the energy of a level. What is the energy of the n = 3 level, and how much energy frees an electron sitting there? From E_n = -13.6/n^2:

Ionising from n=3 means lifting the electron to E = 0, which costs +1.51\ \text{eV}. (From the ground state it would cost the full 13.6\ \text{eV} — the ionisation energy of hydrogen.)

Example 3 — counting radial nodes. How many radial nodes does the 4d orbital have? Here n = 4 and, since it is a d orbital, \ell = 2. So

One spherical surface where the electron is never found. Notice that 2p (4-2... no — 2-1-1) and 3d (3-2-1) both have zero radial nodes: the highest-\ell orbital in any shell is always node-free.