Direct vs indirect band gap materials: Worked examples

This child page is a drill through every case the topic can hand you. We do not introduce a single new idea without first pinning it to a picture and plain words. If a symbol appears, it was earned on the parent note or built here.

Prerequisites worth a glance first: Band Theory Basics, Phonons and Lattice Vibrations, Optical Absorption in Semiconductors.

Symbols we will use (all earned here or on the parent)

Definition The cast of characters

$E$ — an electron's energy, in electron-volts (eV). One eV is the energy an electron gains crossing a 1-volt gap: a tiny, atom-sized unit of energy.
$k$ — crystal momentum direction/size (a wavevector). Think of it as which "lane" in the crystal the electron is travelling in. Measured in $m^{- 1}$ (per metre) because it is $2 π / wavelength$ .
$E_{g}$ — the band gap: the smallest energy jump between the top of the full (valence) band and the bottom of the empty (conduction) band.
$h ν$ — a photon's energy ( $h$ is Planck's constant, $ν$ its frequency). Light of frequency $ν$ delivers energy in lumps of size $h ν$ .
$ℏ$ — the reduced Planck constant, just $h / (2 π)$ . It is the conversion factor that turns a wavevector $k$ (a "per-metre" number) into a genuine momentum: momentum $= ℏ k$ . So $ℏ q_{p h}$ means "the photon's momentum."
$E_{p h}$ — a phonon's energy: a lump of lattice-vibration energy (a sound-quantum), typically a few hundredths of an eV.
$α$ — the absorption coefficient ( $cm^{- 1}$ ): how fast light dies out inside the material. Big $α$ = light absorbed in a thin slice.
$d$ — a depth (thickness) inside the material, measured from the illuminated surface, in cm or μm. It is the distance light has travelled into the slab.
$I$ , $I_{0}$ — the light intensity at depth $d$ inside the material ( $I$ ) and the incoming intensity at the surface ( $I_{0}$ ). The ratio $I / I_{0}$ is the fraction of light still surviving after travelling the distance $d$ .
$q_{p h}$ , $k_{B Z}$ — the photon's wavevector and the half-width of the Brillouin zone (the natural "size" of $k$ -space). We compare these to decide if a photon can bridge a $k$ -jump.

The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples below are each tagged with the cell they hit, and together they fill the whole table.

#	Case class	Concrete input	What it probes
A	Photon energy below gap, $h ν < E_{g}$	$h ν = 1.1$ eV on GaAs ( $E_{g} = 1.42$ )	Sign of $h ν - E_{g}$ is negative → no absorption
B	Photon energy at gap, $h ν = E_{g}$ (zero case)	$h ν = E_{g}$ exactly	$α \to 0$ limiting value at the edge
C	Photon energy above gap, $h ν > E_{g}$	direct vs indirect power law	Which curve shape → gap type
D	Momentum bookkeeping (any material)	$q_{p h}$ vs $k_{B Z}$	Why the jump must be vertical
E	Indirect two-branch (both signs of $\pm E_{p h}$ )	phonon-absorb vs phonon-emit	Split edge, low- $T$ behaviour
F	Degenerate / boundary case	Ge under pressure, near-equal valleys	When "direct or indirect?" is genuinely close
G	Real-world word problem	how thick a solar cell?	Turn $α$ into a device length
H	Exam twist	"bigger gap ⇒ indirect?" trap	Type vs magnitude are independent

Example 1 — Cell A: photon below the gap

Worked example Shine 1.1 eV light on GaAs (

$E_{g} = 1.42$ eV). Is it absorbed across the band gap? Forecast: Guess now — yes or no? And what does the formula $α \propto (h ν - E_{g})^{1/2}$ do when $h ν < E_{g}$ ?

Step 1. Compute $h ν - E_{g} = 1.1 - 1.42 = - 0.32$ eV. Why this step? The whole absorption law lives inside a square root of this quantity. Its sign decides everything.

Step 2. Put it in the direct law: $α \propto (h ν - E_{g})^{1/2} = (- 0.32)^{1/2}$ — the square root of a negative number is not a real absorption. Physically that means there is no final electron state to land in: the photon can't lift an electron a shorter distance than the gap. Why this step? A negative argument is the maths telling us "no allowed transition." The gap is a forbidden desert of energies.

Step 3. Conclusion: no band-to-band absorption. GaAs is transparent to this photon (it may still absorb by other means — free carriers, defects — but not across the gap).

Verify: Convert 1.1 eV to a wavelength: $λ = 1240/1.1 \approx 1127$ nm (near-infrared). GaAs's edge is at $1240/1.42 \approx 873$ nm. Since $1127 > 873$ , the light is "redder" (lower energy) than the edge → transparent. Sanity check passes.

Example 2 — Cell B: photon exactly at the gap (the zero/limit case)

Worked example What is

$α$ right at the band edge, $h ν = E_{g}$ , for a direct material? Forecast: Will $α$ jump up like a step, or rise gently from zero?

Step 1. Set $h ν = E_{g}$ , so $h ν - E_{g} = 0$ . Why this step? This is the boundary between "no absorption" (Cell A) and "absorption" (Cell C). Limits live here.

Step 2. Direct law: $α \propto (h ν - E_{g})^{1/2} = 0^{1/2} = 0$ . So $α \to 0$ exactly at the edge and rises as a square-root curve just above it. Look at Figure s01 — the blue direct curve leaves the axis vertically at first (infinite slope of $x$ at $0$ ) then bends over. Why this step? The joint density of states — the number of electron pairs available to make the jump — vanishes at the edge and grows like $energy above edge$ , just like a 3D free particle.

Figure s01 — what to look at: two normalised absorption curves just above the edge

h ν = E_{g}

(dashed grey line). Focus on how each curve leaves the axis: the blue direct curve shoots up almost vertically (square-root, infinite starting slope), while the orange indirect curve creeps along the axis (parabola, zero starting slope). That steep-vs-flat departure is the visual fingerprint of gap type.

Step 3. For an indirect material the law is $(h ν - E_{g})^{2}$ , which at $0$ is also $0$ but leaves the axis flat (zero slope) — the orange curve in Figure s01 hugs the axis far longer. That is why indirect absorption is weak near the edge.

Verify: Let $x = h ν - E_{g}$ (the energy above the edge, our earlier "argument"). Slopes at the edge: direct $\frac{d}{d x} x^{1/2} = \frac{1}{2} x^{- 1/2} \to \infty$ as $x \to 0^{+}$ (steep). Indirect $\frac{d}{d x} x^{2} = 2 x \to 0$ as $x \to 0^{+}$ (flat). The steep-vs-flat contrast is the whole story, confirmed.

Example 3 — Cell C: photon above the gap, identify the type

Worked example Absorption data: plotting

$α^{1/2}$ vs $h ν$ gives a straight line hitting zero at $1.12$ eV. Direct or indirect? Name the material. Forecast: Which power law, when you take a square root of $α$ , straightens out?

Step 1. If $α \propto (h ν - E_{g})^{2}$ then $α^{1/2} \propto (h ν - E_{g})$ — a straight line in $α^{1/2}$ vs $h ν$ . Why this step? Taking the same root as the inverse of the power linearises the plot; the power that straightens tells you the exponent.

Step 2. Since $α^{1/2}$ (not $α^{2}$ ) is linear, the exponent is $2$ → indirect gap. The intercept gives $E_{g} = 1.12$ eV. Why this step? The straight line crosses zero exactly where $h ν = E_{g}$ (absorption begins).

Step 3. $E_{g} \approx 1.12$ eV, indirect → this is silicon. (Contrast Example from parent where $α^{2}$ linear at 1.42 eV → direct GaAs.)

Verify: Which root linearises which law? Direct $α \propto x^{1/2} \Rightarrow α^{2} \propto x$ (linear in $α^{2}$ ). Indirect $α \propto x^{2} \Rightarrow α^{1/2} \propto x$ (linear in $α^{1/2}$ ). Here $α^{1/2}$ linear ⇒ indirect. Consistent. See also Silicon Solar Cells.

Example 4 — Cell D: momentum bookkeeping (any material)

Worked example For GaP, an indirect emission at

$λ = 550$ nm (green), lattice constant $a = 0.545$ nm. Show the photon cannot supply the $k$ -jump; estimate what fraction it does cover. Forecast: Order of magnitude of $q_{p h} / k_{B Z}$ — bigger than $0.1$ , or tiny?

Step 1. Photon wavevector $q_{p h} = 2 π / λ = 2 π / (550 \times 1 0^{- 9}) \approx 1.14 \times 1 0^{7} m^{- 1}$ ; its momentum is $ℏ q_{p h}$ (recall $ℏ$ turns a wavevector into a momentum). Why this step? We compare the photon's wavevector to the zone size; multiplying both by the same constant $ℏ$ would not change the ratio, so we work with $q_{p h}$ directly.

Step 2. Zone half-width $k_{B Z} = π / a = π / (0.545 \times 1 0^{- 9}) \approx 5.76 \times 1 0^{9} m^{- 1}$ . Why this step? This is the natural "distance across $k$ -space" — the scale of an indirect $k$ -jump.

Step 3. Ratio $q_{p h} / k_{B Z} \approx 1.14 \times 1 0^{7} /5.76 \times 1 0^{9} \approx 2.0 \times 1 0^{- 3}$ . Why this step? The photon covers only $0.2%$ of the jump — the phonon must supply the other $99.8%$ . See Phonons and Lattice Vibrations.

Verify: Units: $(m^{- 1}) / (m^{- 1})$ = dimensionless ✓. Magnitude $\sim 1 0^{- 3}$ matches the parent's generic estimate ✓.

Example 5 — Cell E: indirect two branches (both signs of $\pm E_{p h}$ )

Worked example Silicon indirect edge

$E_{g} = 1.12$ eV, phonon energy $E_{p h} = 0.058$ eV. Where are the phonon-absorption and phonon-emission thresholds, and which dominates at very low temperature? Forecast: Two thresholds — will they sit above or below $1.12$ eV, and which survives when the crystal is cold?

Step 1. Indirect law splits: $α \propto (h ν - E_{g} \pm E_{p h})^{2}$ . The two thresholds are where each bracket hits zero. Why this step? The $\pm$ means the crystal can absorb a phonon (adding $E_{p h}$ of help, so the photon needs less) or emit one (photon must supply extra).

Step 2. Phonon-absorption branch: $h ν = E_{g} - E_{p h} = 1.12 - 0.058 = 1.062$ eV (photon aided by an existing phonon → lower threshold). Phonon-emission branch: $h ν = E_{g} + E_{p h} = 1.12 + 0.058 = 1.178$ eV. Figure s02 shows the double edge. Why this step? Sign $-$ lowers the needed photon energy; sign $+$ raises it. Both signs are real, physical cases — Cell E demands we show both.

Figure s02 — what to look at: the silicon indirect edge (blue) is the sum of two dashed sub-edges — a green phonon-absorption branch turning on early at

1.062

eV and a red phonon-emission branch turning on later at

1.178

eV. Notice the true gap

E_{g} = 1.12

eV (dotted grey) sits exactly halfway between the two coloured thresholds. Watch the green branch: it needs a pre-existing phonon, so it fades as the crystal cools.

Step 3. At very low $T$ almost no phonons exist to be absorbed, so the absorption branch (which needs a pre-existing phonon) fades — only the emission branch ( $1.178$ eV) survives. Why this step? You can only absorb a phonon that is there; cooling empties the phonon reservoir.

Verify: Absorption threshold $1.062$ eV, emission threshold $1.178$ eV; gap between them $= 2 E_{p h} = 0.116$ eV. Check $1.178 - 1.062 = 0.116$ ✓. Midpoint $= (1.062 + 1.178) /2 = 1.12 = E_{g}$ ✓ (the two branches straddle the true gap symmetrically).

Example 6 — Cell F: the degenerate / boundary case

Worked example Germanium's lowest conduction valley (at

$L$ , indirect) is only $\sim 0.14$ eV below its direct valley (at $Γ$ ). If strain raises the $L$ valley by $0.14$ eV, what happens? Forecast: Does Ge stay indirect, or can it be tuned to the boundary and beyond?

Step 1. Indirect gap uses the $L$ valley: $E_{g}^{in d} \approx 0.66$ eV. Direct gap uses the $Γ$ valley: $E_{g}^{d i r} \approx 0.66 + 0.14 = 0.80$ eV. Why this step? The smaller of the two decides the true gap and its type. Here $L$ (indirect) wins by $0.14$ eV.

Step 2. Raise the $L$ valley by exactly $0.14$ eV (tensile strain). Now $E_{L} = E_{Γ}$ — the material sits on the boundary: direct and indirect thresholds coincide at $0.80$ eV. This is the degenerate case — "direct or indirect?" has no unique answer; both channels open together. Why this step? The boundary is not a contradiction — it is a tunable knob (strained Ge is engineered this way for on-chip lasers).

Step 3. Push a hair further (strain $> 0.14$ eV of lift): $Γ$ becomes the lowest valley → Ge turns direct and can lase. Why this step? We test what happens just past the boundary to show the transition is not a single fragile point but a genuine crossover into direct behaviour — that is exactly the regime device engineers exploit.

Verify: Boundary condition $E_{Γ} - E_{L} = 0.14 - 0.14 = 0$ ✓. Beyond it $E_{Γ} - E_{L} < 0$ ⇒ direct. Note gap shrinks to $0.66$ eV worth of indirect but the direct channel opens at $0.80$ eV → type flips, magnitude changes independently (foreshadows Cell H).

Example 7 — Cell G: real-world word problem (device thickness)

Worked example A

silicon cell has $α \approx 100 cm^{- 1}$ at $800$ nm; a GaAs cell has $α \approx 1 0^{4} cm^{- 1}$ . How thick must each be to absorb $\sim 63%$ of that light? Forecast: Which needs to be hundreds of microns, which just a micron?

Step 1. Light decays as $I = I_{0} e^{- α d}$ (Beer–Lambert), where $d$ is the depth travelled into the slab, $I_{0}$ is the intensity entering the surface, and $I$ is what survives at depth $d$ . Absorbing $63%$ means $I / I_{0} = e^{- 1}$ , i.e. $α d = 1$ , so $d = 1/ α$ . Why this step? $1/ α$ is the natural "absorption depth" — the slab thickness in which the light dims by a factor $e$ . See Optical Absorption in Semiconductors.

Step 2. Silicon: $d = 1/ (100 cm^{- 1}) = 0.01 cm = 100 μ m$ . Why this step? Indirect ⇒ small $α$ ⇒ big $d$ . Silicon wafers really are $\sim 100$ s of microns.

Step 3. GaAs: $d = 1/ (1 0^{4} cm^{- 1}) = 1 0^{- 4} cm = 1 μ m$ . Why this step? Direct ⇒ large $α$ ⇒ tiny $d$ . Matches the parent's "GaAs absorbs in ~1 μm."

Verify: Ratio of thicknesses $= α_{G a A s} / α_{S i} = 1 0^{4} /100 = 100$ . Predicted $d$ ratio $= 100 μ m /1 μ m = 100$ ✓. Units: $1/ cm^{- 1} = cm$ ✓.

Example 8 — Cell H: the exam twist (type vs magnitude)

Worked example Rank these by band-gap

magnitude and separately by type: Si (1.12 eV, indirect), GaAs (1.42, direct), GaN (3.4, direct), AlAs (2.16, indirect). "Does a bigger gap mean indirect?" Forecast: Will magnitude order and type order line up, or scramble?

Step 1. By magnitude: Si $1.12 <$ GaAs $1.42 <$ AlAs $2.16 <$ GaN $3.4$ . Why this step? Magnitude is just the number $E_{g}$ — set by how far apart the bands are.

Step 2. By type: direct = {GaAs $1.42$ , GaN $3.4$ }; indirect = {Si $1.12$ , AlAs $2.16$ }. Type is set by whether the extrema share $k$ , not by the size of $E_{g}$ . Why this step? Notice direct appears at both a small gap (GaAs) and the largest gap (GaN); indirect at both smallest (Si) and a mid gap (AlAs). No correlation.

Step 3. Answer the twist: No — a bigger gap does not imply indirect. The counterexample is decisive: the largest-gap material here, GaN at $3.4$ eV, is direct, while a smaller-gap one, AlAs at $2.16$ eV, is indirect. So the two properties are independent axes — magnitude is set by how far apart the bands sit in energy, type by where in $k$ -space their extrema line up. You must always check the $E$ – $k$ positions, never guess type from $E_{g}$ alone. Why this step? A single clean counterexample (max-gap member is direct) is enough to demolish the "bigger ⇒ indirect" rule, and stating the two independent physical causes prevents the learner from re-inventing the trap.

Verify: Test the claim "bigger gap ⇒ indirect" on the max-gap member: GaN is direct ⇒ claim is false by counterexample ✓. (This is exactly the parent's third steel-manned mistake.)

Recall Recap of the matrix

Below-gap ( $h ν < E_{g}$ ) ::: Negative argument in the → no band-to-band absorption; material transparent. At-gap zero case ( $h ν = E_{g}$ ) ::: $α = 0$ ; direct leaves the axis steeply (), indirect flatly ( $()^{2}$ ). Above-gap type ID ::: $α^{2}$ linear ⇒ direct; $α^{1/2}$ linear ⇒ indirect; intercept $= E_{g}$ . Momentum ( $q_{p h} / k_{B Z}$ ) ::: $\sim 1 0^{- 3}$ → photon can't bridge the jump → phonon needed for indirect. Two branches ( $\pm E_{p h}$ ) ::: Absorption edge at $E_{g} - E_{p h}$ , emission at $E_{g} + E_{p h}$ ; only emission survives cold. Boundary case ::: When two valleys coincide in energy, direct and indirect thresholds meet; tunable (strained Ge). Device thickness ::: $d = 1/ α$ ; Si $\sim 100 μ$ m, GaAs $\sim 1 μ$ m. Type vs magnitude ::: Independent — GaN wide+direct, AlAs wide+indirect.

Related applications: LEDs and Laser Diodes, Recombination Mechanisms.