Quantum chemistry — particle in a box revisited; H-atom solutions: Visual walkthrough

Step 1 — Draw the trap

WHAT. We put one particle (think: a tiny bead, or an electron) on a line and forbid it from leaving a region of length $L$ . Inside, nothing pushes on it. Outside, an infinitely strong wall pushes back.

WHY. We want the simplest possible situation where "being trapped" is the only physics. No gravity, no friction, no complicated forces — just walls. If quantization appears even here, it must come from confinement alone.

PICTURE. The floor of the box is flat (energy cost $= 0$ inside). The two side walls shoot straight up to infinity. We call the wall positions $x = 0$ (left) and $x = L$ (right).

Step 2 — The wave cannot touch the walls

WHAT. The particle is described by a wavefunction $ψ (x)$ : a curve whose squared height $∣ ψ (x) ∣^{2}$ tells you how likely the particle is to be found at $x$ . Because the walls are infinitely high, the particle is never outside, so $ψ = 0$ everywhere outside. A wave cannot jump, so it must already be $0$ at each wall.

WHY. These two pin-downs are the boundary conditions: $ψ (0) = 0 and ψ (L) = 0.$ Every restriction we ever get flows from these two equalities. Physically: the ends are tied down, exactly like a guitar string clamped at both ends.

PICTURE. The curve is forced to pass through the two red dots on the wall bases. Anything that doesn't hit both dots is not allowed.

Step 3 — What shape is the wave inside?

WHAT. Inside the box $V = 0$ , so the Schrödinger Equation $\hat{H} ψ = E ψ$ becomes $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} = E ψ .$ Let me name every piece:

$\frac{d ^{2} ψ}{d x ^{2}}$ — the curvature of the wave (how sharply it bends).
$ℏ$ — Planck's constant $h$ divided by $2 π$ ; sets the quantum scale.
$m$ — the particle's mass.
$E$ — the total energy we are trying to find.

Rearrange to put all constants on one side: $\frac{d ^{2} ψ}{d x ^{2}} = - k^{2} ψ, k^{2} = \frac{2 m E}{ℏ ^{2}} .$ Here $k$ is a new bundled constant — the wavenumber (how many wave-wiggles per unit length).

WHY this shape. The equation says: the curvature of $ψ$ is proportional to the negative of $ψ$ itself. Ask "what function bends back toward zero in proportion to its own height?" The answer is a sine or cosine. That is why trig enters — not by choice, but because only $sin$ and $cos$ satisfy "second derivative $= - k^{2} \times$ itself."

$ψ (x) = A sin (k x) + B cos (k x) .$

PICTURE. Steeper wiggle (bigger $k$ ) = more curvature = more bending back toward the axis.

Step 4 — The left wall kills the cosine

WHAT. Apply the first boundary condition $ψ (0) = 0$ : $ψ (0) = A sin (0) + B cos (0) = A \cdot 0 + B \cdot 1 = B .$ For this to equal $0$ we need $B = 0$ .

WHY. $cos (0) = 1 \neq = 0$ , so a cosine is already nonzero at the left wall — it cannot obey the pin-down at $x = 0$ . Only the sine, which is naturally zero there, survives. So

$ψ (x) = A sin (k x) .$

PICTURE. The teal cosine sits at full height on the left wall (violates the pin) and is deleted; the orange sine passes cleanly through the left dot and stays.

Step 5 — The right wall selects whole numbers (quantization!)

WHAT. Now apply the second boundary condition $ψ (L) = 0$ to $ψ (x) = A sin (k x)$ : $A sin (k L) = 0.$ We cannot set $A = 0$ (that erases the whole wave — no particle). So we must have $sin (k L) = 0.$

WHY — this is the heart of everything. A sine is zero only at $0, π, 2 π, 3 π, \dots$ — the integer multiples of $π$ . So $k L = nπ, n = 1, 2, 3, \dots \Rightarrow k = \frac{nπ}{L} .$

$n$ — a counting integer; it labels the allowed wave.
$k L = nπ$ — the box must hold a whole number of half-waves. A half-wave has length $π / k$ ; fit $n$ of them across $L$ .

$k$ can no longer be any value: it is forced onto a discrete ladder. This is quantization, born entirely from "both ends pinned."

PICTURE. $n = 1$ is one hump; $n = 2$ two humps; $n = 3$ three humps — each just barely returns to zero at the right wall. A "wave and a half" (dashed) misses the wall and is forbidden.

Step 6 — Turn allowed $k$ into allowed energy

WHAT. In Step 3 we defined $k^{2} = \frac{2 m E}{ℏ ^{2}}$ , i.e. $E = \frac{ℏ ^{2} k ^{2}}{2 m}$ . Substitute the allowed $k = \frac{nπ}{L}$ : $E_{n} = \frac{ℏ ^{2}}{2 m} (\frac{nπ}{L})^{2} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} .$ Using $ℏ = \frac{h}{2 π}$ so $ℏ^{2} = \frac{h ^{2}}{4 π ^{2}}$ , the $π^{2}$ cancels:

WHY. We wanted energies, and $k$ was just a stepping stone. Because only special $k$ -values are allowed, only special $E$ -values are allowed — the allowed $k$ -ladder becomes an allowed $E$ -ladder.

PICTURE. The energy ladder: rungs at heights $1, 4, 9, 16$ (in units of $\frac{h ^{2}}{8 m L ^{2}}$ ), so gaps grow $3, 5, 7, \dots$ . The lowest rung sits above zero — the Zero-point Energy.

Step 7 — Edge case: why $n = 0$ (and negative $n$ ) are banned

WHAT. Check the degenerate values of $n$ our formula seems to allow.

$n = 0$ : then $k = 0$ and $ψ (x) = A sin (0) = 0$ everywhere. A wave that is zero everywhere means the particle is nowhere — there is no particle. Rejected.
Negative $n$ : $sin (\frac{- nπ x}{L}) = - sin (\frac{nπ x}{L})$ . The minus sign is absorbed into $A$ , so it is the same physical state, not a new one. No new information.

WHY. A physical state needs a nonzero, normalizable $ψ$ . $n = 0$ fails "nonzero"; $n < 0$ is a duplicate. So the ladder truly starts at $n = 1$ , which carries the irreducible zero-point energy $E_{1} = \frac{h ^{2}}{8 m L ^{2}} > 0$ . A confined quantum particle can never be perfectly still.

PICTURE. The flat $n = 0$ line (no wave, crossed out) versus the surviving $n = 1$ hump sitting at the lowest real rung.

Step 8 — Same trick, harder geometry: the H-atom in one glance

WHAT. Swap the flat box floor for the Coulomb well $V (r) = - \frac{e ^{2}}{4 π ε _{0} r}$ and go to 3D. Because $V$ depends only on distance $r$ , the wave splits into a radial piece and an angular piece: $ψ (r, θ, ϕ) = R_{n ℓ} (r) Y_{ℓ}^{m} (θ, ϕ) .$

WHY the same story works. Each direction still gets a "must fit / must be single-valued" rule — the 3D echo of "both ends pinned":

The angle $ϕ$ must return to itself after a full turn ( $2 π$ ): $e^{im (ϕ + 2 π)} = e^{im ϕ} \Rightarrow m_{ℓ} \in Z$ . Same "whole number fits" logic as the box.
The radial wave must die at $r \to \infty$ (electron stays bound). That pin-down quantizes the energy: $E_{n} = - \frac{13.6 eV}{n ^{2}} .$
$n^{2}$ in the denominator now (attraction weakens far out), and $E < 0$ because the electron is bound — below the free-electron reference of $0$ .

PICTURE. Left: the box's flat floor with rising rungs. Right: the Coulomb funnel with rungs crowding up toward zero. Same machine, mirror-image ladder.

Recall Which rule made each quantum number?

$n$ (size/energy) ::: radial wave must decay as $r \to \infty$ . $ℓ$ (shape) ::: angular equation must stay finite at the poles. $m_{ℓ}$ (orientation) ::: $ϕ$ -wave must be single-valued after $2 π$ .

The shapes these produce are the Atomic Orbitals and Shapes, labelled by Quantum Numbers, and they replace the older orbit picture in Bohr Model vs Quantum Mechanics.

The one-picture summary

Everything above compressed: pin both ends → only whole-number half-waves fit → $k$ is a ladder → $E \propto n^{2}$ is a ladder → the lowest rung floats above zero. The atom is the same, bent into a funnel.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a jump-rope held by two kids, one at each end, both hands never moving. If you shake it, you can make one big loop, or two loops, or three — but never "two and a half" loops, because the ends are held still and a half-loop would need a moving hand. Each allowed number of loops is a different "note," and a bigger number of loops takes more effort — that's more energy. That's the whole quantum box: the two held hands are the walls, the loops are the wave patterns $n = 1, 2, 3$ , and "you can't have half a loop" is why energy comes in separated steps instead of a smooth slope. You can't even hold it perfectly flat and call it a state — a flat rope is no wave at all, so the calmest real pattern is the single loop, which still costs a little energy (the zero-point energy). The hydrogen atom is the same rope trick done on a curved, funnel-shaped floor: instead of the energy steps climbing forever, they bunch together as they climb toward the top of the funnel, where the electron would finally break free.