Visual walkthrough — Direct vs indirect band gap materials

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Step 1 — Give the electron two addresses

WHAT. Inside a crystal an electron is not described by one number but by two: how much energy $E$ it has, and a second quantity called its crystal momentum, written $\hslash k$.

• $E$ = energy, measured in electron-volts (eV). Think "height on a hill."
• $k$ = the wavevector. It is a location along a horizontal axis. Think "east–west position."
• $\hslash$ (h-bar) = a fixed tiny number, Planck's constant divided by $2\pi$. It just converts $k$ into a momentum. So "$k$" and "$\hslash k$ = momentum" are the same axis, relabelled.

WHY two addresses? Because a rule of physics says both must balance in any jump the electron makes. One number is not enough to track a jump; we need the (energy, momentum) pair.

PICTURE. We draw a map whose up axis is energy and whose left–right axis is $k$. Every allowed electron state is a dot on this map.

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Step 2 — Draw the two bands as two valleys/hills

WHAT. The dots don't scatter randomly; they group into bands — smooth curves $E(k)$.

• The lower curve is the valence band: seats that are normally full. Its highest point is the valence band maximum, VBM.
• The upper curve is the conduction band: seats normally empty. Its lowest point is the conduction band minimum, CBM.
• The vertical gap between VBM and CBM is the band gap $E_g$ — a forbidden zone with no dots.

WHY these shapes? Near an extremum any smooth curve looks like a parabola (a $\cup$ or a $\cap$). The CBM is a valley ($\cup$); the VBM is a hill ($\cap$). That is all we need for the argument.

PICTURE. Two curves, a shaded forbidden band between them, and the two special points labelled.

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Step 3 — A falling electron must pay two bills

WHAT. When a conduction electron drops into an empty valence seat (a hole), it starts at $(k_c,E_c)$ and lands at $(k_v,E_v)$. Two conservation laws must hold at once.

Bill 1 — energy conservation: $\underset{\text{start energy}}{\underbrace{E_c}}-\underset{\text{end energy}}{\underbrace{E_v}}=\underset{\text{photon energy out}}{\underbrace{\hslash \omega }}$

Here $\hslash \omega$ ("h-bar omega") is the energy of the emitted light particle, a photon. The drop in height is carried away as light.

Bill 2 — crystal-momentum conservation: $\underset{\text{start momentum}}{\underbrace{\hslash k_c}}=\underset{\text{end momentum}}{\underbrace{\hslash k_v}}+\underset{\text{momentum the photon takes}}{\underbrace{q_{\text{photon}}}}$

$q_{\text{photon}}$ is the photon's own wavevector (momentum $÷\hslash$). This is the sideways bill.

WHY two bills? Energy tells us how far down; momentum tells us how far sideways. Both are conserved, so both must be paid by whatever leaves the crystal.

PICTURE. One arrow for the vertical drop (energy) and one for the horizontal shift (momentum), colour-coded.

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Step 4 — Weigh the photon's sideways power

WHAT. Now we measure how much sideways momentum a near-gap photon actually carries. A photon's wavevector is $q_{\text{photon}}=\frac{2\pi }{\lambda },$ where $\lambda$ is its wavelength. Near a typical gap, $\lambda \approx 1\mu \text{m}=10^{-6}\text{m}$, giving $q_{\text{photon}}\approx \frac{2\pi }{10^{-6}\text{m}}\approx 6\times 10^6{\text{m}}^{-1}.$

Compare that to the width of the map — how far $k$ can range before the pattern repeats. This half-width is the Brillouin-zone edge: $k_{\text{BZ}}\approx \frac{\pi }{a}\approx \frac{\pi }{5\times 10^{-10}\text{m}}\approx 6\times 10^9{\text{m}}^{-1},$ with $a$ the atomic spacing ($\sim 0.5\text{nm}$).

WHY compute a ratio? Because "big" and "small" only mean something relative to the jump we need. The natural yardstick for a sideways jump is the whole map width $k_{\text{BZ}}$.

$\frac{q_{\text{photon}}}{k_{\text{BZ}}}\approx \frac{6\times 10^6}{6\times 10^9}\approx 10^{-3}.$

PICTURE. The map with a giant $k_{\text{BZ}}$ ruler and a photon arrow so short it is barely a dot.

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Step 5 — The direct case: a clean vertical drop

WHAT. If the CBM sits directly above the VBM (both at the same $k$, usually $k=0$, the $\Gamma$ point), the electron's required jump is already purely vertical.

WHY it works. Bill 2 asks for zero sideways shift ($k_c=k_v$), and a photon gladly supplies zero sideways momentum. One particle — the photon — pays both bills. Every attempt succeeds, so recombination is fast, radiative, bright. These are the LEDs and Laser Diodes materials (GaAs, GaN, InP).

PICTURE. VBM and CBM stacked; a single straight photon arrow drops the electron.

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Step 6 — The indirect case: the ledge is offset

WHAT. In silicon and germanium the CBM sits off to the side — at a different $k$ from the VBM. Now the electron's jump needs a real sideways step $\Delta k=k_c-k_v$, and $\Delta k$ is a large fraction of the map width. The photon (Step 4) cannot supply it.

WHY a photon fails. A vertical photon arrow misses the seat. To land, the electron must ALSO shift sideways — and only a phonon (a quantum of lattice vibration, from Phonons and Lattice Vibrations) carries big momentum with tiny energy. So Bill 2 becomes $\hslash k_c=\hslash k_v+q_{\text{photon}}\pm q_{\text{phonon}},$ where $\pm q_{\text{phonon}}$ is the sideways momentum the vibration adds ($+$) or removes ($-$). The energy bill gains a matching small correction $\pm E_{\text{phonon}}$.

PICTURE. Offset ledge; a diagonal path made of a vertical photon leg plus a horizontal phonon leg.

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Step 7 — Degenerate and edge cases (never skip these)

WHAT / WHY, case by case:

• $\Delta k=0$ exactly (perfectly direct). Photon alone works; no phonon term. Strongest emission.
• $\Delta k$ tiny but nonzero. Still needs a phonon in principle, but the probability is not zero — "quasi-direct." Real crystals are never mathematically perfect.
• Photon momentum set to zero ($q_{\text{photon}}\to 0$). The rule $k_c\approx k_v$ becomes exact. Our whole conclusion is the small-$q$ limit, so this is the clean idealization, not a contradiction.
• Phonon absorption vs emission ($\pm$). At low temperature few phonons exist to absorb, so only the emission ($-E_{ph}$) branch survives, and the absorption edge shifts. This is why indirect absorption edges are temperature-dependent and appear as two sub-edges.
• Gap magnitude is irrelevant here. Whether $E_g$ is $0.7$ eV (Ge) or $3.4$ eV (GaN) does not decide the type; only the horizontal offset of the extrema does. Big-and-direct (GaN) and big-and-indirect (AlAs) both exist.

PICTURE. Four mini-panels showing: exact-direct, tiny-offset, low-$T$ single branch, and big-gap-either-type.

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The one-picture summary

Everything above collapses into one comparison: vertical arrow succeeds (direct); diagonal arrow needs a phonon crutch (indirect).

Recall Feynman: the whole walkthrough in plain words

Picture a giant board game. Up means "how much energy an electron has"; left–right means "where it is in momentum." An electron sitting up high wants to fall into an empty seat down low, and when it falls it must hand off its bill to something that leaves the crystal. Falling down is easy — a particle of light (photon) happily carries away that energy. But the light particle has almost no sideways push, about one-thousandth of the board's width. So light can only drop an electron straight down. If the empty seat is directly below (a direct material like GaAs), the drop is straight and light flies out — the thing glows. If the seat is shoved off to the side (an indirect material like silicon), a straight drop misses. The electron then needs a second helper — a lattice shiver called a phonon — to nudge it sideways at the exact same moment. Getting both helpers at once is rare, so the electron usually just leaks its energy away as heat. That is the entire secret of why GaAs makes lasers and silicon makes solar cells but lousy light bulbs — and it never depended on how big the gap was, only on whether the seat was straight down or off to the side.

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Recall

On the $E$–$k$ map, what does an optical (photon-only) transition always look like? ::: A vertical arrow, because $k_c\approx k_v$. Why is the photon's momentum negligible on this map? ::: $q_{\text{photon}}/k_{\text{BZ}}\approx 10^{-3}$ — about a thousandth of the Brillouin-zone width. What supplies the missing sideways momentum in an indirect jump? ::: A phonon, which carries large $k$ but tiny energy. Does a larger $E_g$ imply indirect? ::: No — gap magnitude and gap type are independent (GaN is wide and direct).