Particle in a Box, Atomic Shells, and the Birth of Energy Bands

This lesson connects three ideas that look separate but build on each other: confining an electron in a box, counting how many electrons fit into each atomic shell, and what happens once you don't confine electrons in one box but line up a whole periodic row of them — a crystal. The last piece is where band structure comes from.

The particle in a box

Start with an electron confined to a 1D region of width $L$, with infinite walls at $x=0$ and $x=L$. Inside the box the potential is zero, so the time-independent Schrödinger equation is

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi.$

The general solution is a mix of sine and cosine,

$\psi(x) = A\sin(kx) + B\cos(kx), \qquad k = \frac{\sqrt{2mE}}{\hbar}.$

The walls force $\psi=0$ at $x=0$ and $x=L$, since the electron can never be found outside the box. Plugging in $x=0$ kills the cosine term ($B=0$). Plugging in $x=L$ requires

$\sin(kL) = 0 \implies k_n = \frac{n\pi}{L}, \quad n = 1,2,3,\dots$

So only discrete wavevectors — and therefore discrete energies — are allowed:

$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad \psi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right).$

The shapes follow directly from this formula. $\psi_1$ is a single smooth arch — half of a sine wave, no interior nodes. $\psi_2$ has one node in the middle, so it looks like an S-shaped wiggle: a hump up, then a hump down. $\psi_3$ has two interior nodes, giving three lobes that alternate sign.

Squaring gives the probability density $|\psi_n(x)|^2$, and squaring erases the sign information — every lobe becomes a positive-valued bump. That's why $|\psi_1|^2$ is one bump, $|\psi_2|^2$ is two bumps, and $|\psi_3|^2$ is three bumps of similar height, all sitting between the same two walls. The number of bumps in $|\psi_n|^2$ always equals $n$; the number of nodes in $\psi_n$ always equals $n-1$. This is the general rule for any bound state: higher energy means more nodes, more oscillation, more localized probability lobes.

Quantum numbers and how many states each shell holds

For an electron bound in three dimensions (like in an atom), the orbital angular momentum quantum number $\ell$ labels the shape of the state, and for each $\ell$ there are $2\ell+1$ orientations (magnetic quantum number $m_\ell$), each of which holds 2 electrons once you include spin. So the number of states in a subshell is $2(2\ell+1)$:

• $s$: $\ell = 0 \to 2$ states
• $p$: $\ell = 1 \to 6$ states
• $d$: $\ell = 2 \to 10$ states
• $f$: $\ell = 3 \to 14$ states

This counting matters later: it's the same logic — count the distinct standing-wave states, then fill them two electrons at a time — that you use to fill up energy levels in a solid. The particle in a box gave you the idea of discrete quantized levels; this gives you the bookkeeping for how many electrons occupy each one before you're forced into the next level up.

Free electron gas: why $k$-space needs three dimensions

Now put many free electrons in a 3D box of volume $V = L_x L_y L_z$ instead of a 1D line. The same boundary-condition logic applies in each direction independently, so the wavefunction separates and each direction gets its own quantum number:

$k_x = \frac{n_x\pi}{L_x}, \quad k_y = \frac{n_y\pi}{L_y}, \quad k_z = \frac{n_z\pi}{L_z}, \qquad n_x,n_y,n_z = 1,2,3,\dots$

$E = \frac{\hbar^2}{2m}\left(k_x^2+k_y^2+k_z^2\right).$

It's tempting to think $k$ is just "the wavelength," since $k = 2\pi/\lambda$ in 1D — so why would you need three of them? The resolution is that $k$ is really a vector, $\vec{k} = (k_x,k_y,k_z)$. Its magnitude sets the wavelength, but its direction sets which way the wave is traveling. A plane wave in 3D can propagate along any direction, not just left or right, so you need all three components to fully specify the state. Each allowed triplet $(k_x,k_y,k_z)$ is a single point sitting in an abstract 3D space — $k$-space — and as you add more electrons to the box, you fill up more and more of these points, starting from the origin outward, because lower $|\vec{k}|$ means lower energy.

At absolute zero, filling points from the center outward and stopping once you've placed all the electrons produces a filled sphere in $k$-space, bounded by the Fermi wavevector $k_F$. Everything inside the sphere is occupied, everything outside is empty.

From a free electron to an electron in a crystal

Free electrons only feel a flat, zero potential. A real electron in a crystal feels the periodic pull of the ion cores, repeating every lattice spacing. The Kronig-Penney model captures this with the simplest periodic potential possible: alternating flat wells and flat barriers, period $(a+d)$ — a well of width $a$ where $V=0$, followed by a barrier of width $d$ where $V=V_0$.

Because the potential is periodic, Bloch's theorem says the solution can't just be any wavefunction — it has to satisfy

$\psi(x + a+d) = e^{ik(a+d)}\psi(x),$

where $k$ is the Bloch wavevector, the crystal analog of the free-electron $k$. Inside a well, the electron behaves like a free particle with

$\alpha = \frac{\sqrt{2mE}}{\hbar}, \qquad \psi(x) = Ae^{i\alpha x} + Be^{-i\alpha x}.$

Inside a barrier, if $E<V_0$, the wavefunction decays/grows exponentially with

$\beta = \frac{\sqrt{2m(V_0-E)}}{\hbar}, \qquad \psi(x) = Ce^{\beta x} + De^{-\beta x}.$

Matching $\psi$ and $\psi'$ at each interface, and using the Bloch condition to relate the well region to the barrier region one period over, gives four linear equations in $A,B,C,D$. A nontrivial solution exists only if a certain determinant vanishes. Taking the standard limit of thin, tall barriers (barrier width $d\to 0$, height $V_0\to\infty$, with the product $P = \dfrac{mV_0 a d}{\hbar^2}$ held fixed as a measure of barrier strength) collapses that determinant condition down to the compact form:

$\cos[k(a+d)] = P\,\frac{\sin(\alpha a)}{\alpha a} + \cos(\alpha a).$

Why gaps in energy appear

This equation is the whole story. The left-hand side, $\cos[k(a+d)]$, can only ever take values between $-1$ and $1$, because it's a cosine of a real number. The right-hand side depends only on the energy $E$ (through $\alpha = \sqrt{2mE}/\hbar$) and the barrier strength $P$ — it has nothing to do with $k$.

So pick an energy $E$, compute the right-hand side. If it lands between $-1$ and $1$, you can find a real $k$ that solves the equation — this energy is allowed, and the electron can exist as a propagating Bloch wave with that $k$. If the right-hand side falls outside $[-1,1]$, there is no real $k$ that works. The only way to formally satisfy the equation is for $k$ to become complex, which means the wave doesn't propagate — it decays. That energy is forbidden. This is exactly what "some $E$ will not equal $\alpha k$" is pointing at: not every energy corresponds to a valid propagating state in the crystal, only the ones where the right-hand side stays bounded.

Sweeping $E$ continuously, the right-hand side oscillates in and out of the $[-1,1]$ window, carving energy into an alternating sequence of allowed bands and forbidden gaps.

At low energy, near $\alpha a \to 0$, the right-hand side stays close to $\cos[k(a+d)]$'s free-particle behavior, so the lowest band looks almost like the ordinary free-electron parabola $E = \hbar^2k^2/2m$. As $E$ grows, the periodic potential increasingly distorts this, and gaps open up at the points where the right-hand side crosses outside $[-1,1]$. Each allowed band, once you sort them by energy, plays the same role that $n=1,2,3,\dots$ played for the particle in a box: a band index labeling successive groups of allowed energies, separated now not by simple quantization but by genuine forbidden gaps carved out by the crystal's periodicity.