1 Energy bands, holes and E–k diagrams

Run this chapter’s code in Google Colab (no installation).

Slides 3–9 (bands in metals and semiconductors, density of states, Fermi–Dirac function, holes) and 33–39 (periodic potential, Bloch waves, E–k diagrams, direct and indirect gaps, phonons).

1.1 Key relations

When $N$ atoms come together, each atomic level splits into $N$ closely spaced levels: an energy band (slide 3). In a metal the outermost band is only partly filled. In a semiconductor at 0 K the valence band (VB) is full, the conduction band (CB) is empty, and they are separated by the bandgap $E_g$ (slide 7).

An electron in a crystal of size $L$ can only take standing-wave modes, $kL=n\pi$ (slide 4). Counting them gives the density of states and, with the Fermi–Dirac occupation, the electron distribution (slides 5–6):

\[ g(E)=\frac{8\sqrt2\,\pi\, m^{*3/2}}{h^3}\,(E-E_c)^{1/2},\qquad f(E)=\frac{1}{1+\exp\!\left[(E-E_F)/k_BT\right]},\qquad n=\int g(E)f(E)\,dE . \]

At 0 K all states up to the Fermi energy $E_{F0}$ are full. For a metal with free-electron concentration $n$, $E_{F0}=\dfrac{h^2}{8m_e}\left(\dfrac{3n}{\pi}\right)^{2/3}$.

In the periodic potential $V(x)=V(x+ma)$ of the crystal the electron wavefunction is a Bloch wave $\psi_k(x)=U_k(x)e^{jkx}$ (slide 34). $\hbar k$ is the crystal momentum: an external force changes it as $F=d(\hbar k)/dt$ (slide 35). Plotting the allowed energies against $k$ gives the E–k diagram (slides 37–38). A photon can only be emitted efficiently if the CB minimum sits directly above the VB maximum (direct gap: GaAs, InP, GaN). In an indirect semiconductor (Si, Ge) a phonon must supply the missing momentum.

1.2 Where bands come from: the Kronig–Penney model

The simplest periodic potential is a row of thin barriers of strength $P$ spaced by $a$. Bloch’s theorem then gives a single equation for the allowed energies:

\[ \cos(ka)=\cos(\alpha a)+P\,\frac{\sin(\alpha a)}{\alpha a},\qquad \alpha=\frac{\sqrt{2m_eE}}{\hbar}. \]

The right-hand side can exceed 1 in magnitude, and then no real $k$ exists: that energy is a gap. Increase $P$ in the panel and watch the free-electron parabola break into bands, with gaps opening at $ka=\pm\pi$ (the edge of the Brillouin zone).

Show the Python

viewof kp = Inputs.form({
  P: Inputs.range([0, 12], {value: 4, step: 0.1, label: "barrier strength P"}),
  a: Inputs.range([0.2, 1.0], {value: 0.5, step: 0.01, label: "lattice period a (nm)"})
})

Show the Python

kpOut = {
  const hbar = 1.054571817e-34, me = 9.1093837e-31, q = 1.602176634e-19, a = kp.a*1e-9;
  const E0 = hbar**2/(2*me*a*a)/q;              // eV, energy for alpha*a = 1
  const pts = [], free = [], bands = [];
  let inBand = false, start = 0;
  for (let i = 1; i <= 6000; i++) {
    const x = 4*Math.PI*i/6000, E = E0*x*x;     // x = alpha*a
    const rhs = Math.cos(x) + kp.P*Math.sin(x)/x;
    if (Math.abs(rhs) <= 1) {
      const ka = Math.acos(rhs);
      pts.push({k: ka/Math.PI, E}); pts.push({k: -ka/Math.PI, E});
      if (!inBand) { inBand = true; start = E; }
    } else if (inBand) { inBand = false; bands.push([start, E]); }
    let kf = x % (2*Math.PI); if (kf > Math.PI) kf -= 2*Math.PI;
    free.push({k: kf/Math.PI, E});
  }
  if (inBand) bands.push([start, E0*16*Math.PI**2]);
  const gaps = bands.slice(1).map((b, i) => b[0] - bands[i][1]);
  return {pts, free, bands, gaps};
}
Plot.plot({
  width: Math.min(width, 1050), height: 420, x: {label: "ka/π (reduced zone)", domain: [-1, 1]},
  y: {label: "E (eV)", grid: true},
  marks: [
    Plot.dot(kpOut.free, {x: "k", y: "E", r: 0.6, fill: "#52514e", fillOpacity: 0.35}),
    Plot.dot(kpOut.pts, {x: "k", y: "E", r: 1.3, fill: "#2a78d6"}),
    Plot.tip(kpOut.pts, Plot.pointer({x: "k", y: "E", title: d => `ka/π = ${d.k.toFixed(3)}\nE = ${d.E.toFixed(2)} eV`}))
  ]
})

Show the Python

html`<div class="result-box">Blue: allowed states in the crystal. Grey: free-electron parabola folded into the first zone (P = 0).<br>
Band gaps: ${kpOut.gaps.slice(0, 3).map((g, i) => `E<sub>g${i+1}</sub> = <b>${g.toFixed(2)} eV</b>`).join(" · ")}</div>`

Show the Python

a, P = 0.5e-9, 4.0
x = np.linspace(1e-4, 4*np.pi, 8000)               # alpha*a
E = (sl.hbar*x/a)**2/(2*sl.me)/sl.qe
rhs = np.cos(x) + P*np.sin(x)/x
ok = np.abs(rhs) <= 1
ka = np.where(ok, np.arccos(np.clip(rhs, -1, 1)), np.nan)
band = np.floor(x/np.pi)                          # band index -> extended-zone unfolding
k_ext = np.where(band % 2 == 0, ka + band*np.pi, (band + 1)*np.pi - ka)

fig, ax = plt.subplots(1, 2, figsize=(9, 4.2), sharey=True)
ax[0].plot(rhs, E); ax[0].axvspan(-1, 1, color=sl.PALETTE[0], alpha=0.08)
ax[0].set(xlabel="cos(αa) + P sin(αa)/(αa)", ylabel="E (eV)", xlim=(-3, 5))
ax[1].plot(k_ext/np.pi, E, label="crystal"); ax[1].plot(x/np.pi, E, ls="--", color=sl.INK2, lw=1.2, label="free electron")
for m in (1, 2, 3): ax[1].axvline(m, color=sl.GRID, lw=1)
ax[1].set(xlabel="ka/π (extended zone)"); ax[1].legend(fontsize=8)
plt.tight_layout(); plt.show()

Figure 1.1: Kronig–Penney model (a = 0.5 nm, P = 4). Left: the right-hand side of the dispersion relation; energies where it leaves the band [−1, 1] are forbidden. Right: the resulting E–k diagram in the extended zone, compared with a free electron.

1.3 Metals: the Fermi energy

Slides 3–6 picture a metal as a partly filled band. With one free electron per atom the free-electron formula already gives the right order of magnitude for $E_{F0}$, and the average electron energy at 0 K is $\tfrac35E_{F0}$:

Show the Python

metals = {"Na": 2.65e28, "Al": 18.1e28, "Cu": 8.47e28, "Ag": 5.86e28, "Au": 5.90e28}
EF = {m: sl.h**2/(8*sl.me)*(3*n/np.pi)**(2/3)/sl.qe for m, n in metals.items()}
pd.DataFrame({"n (10^28 m^-3)": {m: n/1e28 for m, n in metals.items()},
              "E_F0 (eV)": EF, "mean energy 3/5 E_F0 (eV)": {m: 0.6*e for m, e in EF.items()},
              "Fermi speed (10^6 m/s)": {m: np.sqrt(2*e*sl.qe/sl.me)/1e6 for m, e in EF.items()}}).round(2)

Table 1.1: Free-electron Fermi energy of common metals (electron concentrations from tables).

	n (10^28 m^-3)	E_F0 (eV)	mean energy 3/5 E_F0 (eV)	Fermi speed (10^6 m/s)
Na	2.65	3.24	1.94	1.07
Al	18.10	11.67	7.00	2.03
Cu	8.47	7.03	4.22	1.57
Ag	5.86	5.50	3.30	1.39
Au	5.90	5.53	3.32	1.39

Show the Python

E = np.linspace(0, 9, 2000)
g = sl.dos_3d(E, 1.0)
fig, ax = plt.subplots(1, 2, figsize=(9, 3.8))
for T, c in ((300, sl.PALETTE[0]), (3000, sl.PALETTE[1])):
    ax[0].plot(E, sl.fermi_dirac(E, EF["Cu"], T), color=c, label=f"f(E), {T} K")
    ax[1].plot(E, g*sl.fermi_dirac(E, EF["Cu"], T)/1e28, color=c, label=f"{T} K")
ax[1].plot(E, g/1e28, ls="--", color=sl.INK2, lw=1.2, label="g(E)")
ax[0].set(xlabel="E (eV)", ylabel="f(E)"); ax[0].legend(fontsize=8)
ax[1].set(xlabel="E (eV)", ylabel="10²⁸ states m⁻³ eV⁻¹"); ax[1].legend(fontsize=8)
plt.tight_layout(); plt.show()

Figure 1.2: Copper (E_F0 = 7.0 eV): density of states g(E), Fermi–Dirac function f(E), and the electron distribution g(E)f(E) at 300 K and at an (unphysical) 3000 K. Only electrons within a few k_BT of E_F can change state, which is why the electronic heat capacity of metals is so small.

1.4 Holes (slides 8–9)

A photon with $h\nu>E_g$ breaks a bond and creates an electron–hole pair (EHP). The empty bond, the hole, moves because neighbouring valence electrons tunnel into it. It behaves like a positive particle with its own effective mass $m_h^*$. When a wandering CB electron meets a hole they recombine: in a direct-gap material the energy is released as a photon. This is the LED.

The photogeneration threshold is $\lambda_g=hc/E_g$:

Show the Python

pd.DataFrame({m: {"E_g (eV)": p["Eg"], "λ_g (nm)": sl.wavelength_from_eV(p["Eg"])*1e9,
                  "gap": "direct" if p["direct"] else "indirect"} for m, p in sl.MATERIALS.items()}).T

Table 1.2: Bandgap and cut-off wavelength λ_g = hc/E_g of the semiconductors of Table 3.1.

	E_g (eV)	λ_g (nm)	gap
Ge	0.66	1878.548461	indirect
Si	1.11	1116.974761	indirect
InP	1.35	918.40147	direct
GaAs	1.42	873.128158	direct
AlAs	2.17	571.355753	indirect

1.5 Direct and indirect gaps, phonons (slides 37–39)

Near a band edge the E–k diagram is a parabola, $E=E_c+\hbar^2(k-k_0)^2/2m_e^*$. The figure contrasts GaAs (CB minimum at $k=0$) with Si (CB minimum near the zone edge along [100], $k_0\approx0.85\cdot2\pi/a$).

Show the Python

k = np.linspace(-1, 1, 400)                              # in units of 2π/a
fig, ax = plt.subplots(1, 2, figsize=(9, 3.8), sharey=True)
for a_, (name, k0, Eg, mc) in zip(ax, (("GaAs", 0.0, 1.42, 0.067), ("Si", 0.85, 1.11, 0.26))):
    a_m = sl.MATERIALS[name]["a"]
    kk = k*2*np.pi/a_m
    a_.plot(k, Eg + (sl.hbar*(kk - k0*2*np.pi/a_m))**2/(2*mc*sl.me)/sl.qe, label="CB")
    a_.plot(k, -(sl.hbar*kk)**2/(2*0.5*sl.me)/sl.qe, label="VB")
    a_.annotate("", xy=(k0, Eg), xytext=(0, 0), arrowprops=dict(arrowstyle="->", color=sl.PALETTE[7], lw=1.8))
    a_.set(title=name, xlabel="k (2π/a)", ylim=(-1.2, 3)); a_.legend(fontsize=8, loc="lower right")
ax[0].set_ylabel("E (eV)"); plt.tight_layout(); plt.show()

k_photon = 2*np.pi/sl.wavelength_from_eV(1.11)
print(f"photon wavevector at Eg(Si): {k_photon:.2e} m^-1;  zone edge π/a = {np.pi/sl.MATERIALS['Si']['a']:.2e} m^-1")
print(f"ratio: {k_photon/(np.pi/sl.MATERIALS['Si']['a']):.1e}  -> photons carry almost no crystal momentum")
v_s = 9000                                               # longitudinal sound speed in Si (m/s)
print(f"acoustic phonon at k0: E = ħ v_s k0 ≈ {sl.hbar*v_s*0.85*2*np.pi/sl.MATERIALS['Si']['a']/sl.qe*1e3:.0f} meV (linear estimate; real Si phonons at X: 19–57 meV)")

Figure 1.3: Parabolic E–k sketches. GaAs: a vertical (direct) transition emits a photon. Si: the CB minimum is displaced by k₀, so recombination needs a phonon to carry momentum ħk₀. On this scale the photon wavevector would be a vertical arrow (about 10⁻³ of the axis).

photon wavevector at Eg(Si): 5.63e+06 m^-1;  zone edge π/a = 5.78e+09 m^-1
ratio: 9.7e-04  -> photons carry almost no crystal momentum
acoustic phonon at k0: E = ħ v_s k0 ≈ 58 meV (linear estimate; real Si phonons at X: 19–57 meV)

A photon cannot supply the momentum $\hbar k_0$, but a lattice vibration (phonon, slide 39) can: it has a large $k$ and a small energy (tens of meV). Three-body processes are slow, so radiative recombination in Si takes milliseconds, while non-radiative recombination through defects (slide 38c) takes microseconds. Si therefore cannot be used for LEDs.

Going deeper

Effective mass from curvature. $1/m^*=\hbar^{-2}\,d^2E/dk^2$. Compute it at the bottom of the first two Kronig–Penney bands. Why is it negative at the top of a band? This is the origin of the hole.
Tight binding. For $N$ coupled atoms the levels spread over a band of width $4t$ (hopping $t$). Diagonalise an $N\times N$ tridiagonal matrix with numpy.linalg.eigvalsh and watch the levels fill a band as $N$ grows (slide 3b).
Real band structures. Use the empirical pseudopotential method (a 30-line Python script with the Cohen–Bergstresser form factors) to compute the E–k diagram of Si and GaAs along L–Γ–X and find $E_g$ and $k_0$.
Strained and alloyed Ge becomes a direct-gap semiconductor (GeSn alloys): a route to light sources on silicon.

Complements

S. O. Kasap, Optoelectronics and Photonics, ch. 3; Principles of Electronic Materials and Devices, ch. 4–5.
C. Kittel, Introduction to Solid State Physics: chapters on free-electron gas and energy bands (Kronig–Penney).
M. L. Cohen and J. R. Chelikowsky, Electronic Structure and Optical Properties of Semiconductors (pseudopotentials).

1.6 Try it yourself

In the live panel, at which $P$ does the first gap reach 1.1 eV (the Si gap) for $a=0.543$ nm? Is the model realistic?
Show that the number of states in one Kronig–Penney band equals the number of unit cells (count $k=2\pi m/Na$).
Compute $E_{F0}$ for Mg (two free electrons per atom, $n_\text{atoms}=4.3\times10^{28}$ m$^{-3}$).
How large must the phonon energy be to assist an indirect transition in Ge, where the CB minimum is at the L point, $k_0=\sqrt3\,\pi/a$? Use a sound speed of 5000 m/s.

# Energy bands, holes and E–k diagrams {#sec-bands} [![Open in Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/{{< var github_user >}}/{{< var repo >}}/blob/main/{{< var subdir >}}/notebooks/bands.ipynb){.colab-link}  Run this chapter's code in Google Colab (no installation). ```{python} #| include: false import numpy as np import matplotlib.pyplot as plt import pandas as pd import semilib as sl sl.use_style() ``` *Slides 3–9 (bands in metals and semiconductors, density of states, Fermi–Dirac function, holes) and 33–39 (periodic potential, Bloch waves, E–k diagrams, direct and indirect gaps, phonons).* ## Key relations When $N$ atoms come together, each atomic level splits into $N$ closely spaced levels: an **energy band** (slide 3). In a metal the outermost band is only partly filled. In a semiconductor at 0 K the **valence band** (VB) is full, the **conduction band** (CB) is empty, and they are separated by the **bandgap** $E_g$ (slide 7). An electron in a crystal of size $L$ can only take standing-wave modes, $kL=n\pi$ (slide 4). Counting them gives the **density of states** and, with the Fermi–Dirac occupation, the electron distribution (slides 5–6): $$ g(E)=\frac{8\sqrt2\,\pi\, m^{*3/2}}{h^3}\,(E-E_c)^{1/2},\qquad f(E)=\frac{1}{1+\exp\!\left[(E-E_F)/k_BT\right]},\qquad n=\int g(E)f(E)\,dE . $$ At 0 K all states up to the **Fermi energy** $E_{F0}$ are full. For a metal with free-electron concentration $n$, $E_{F0}=\dfrac{h^2}{8m_e}\left(\dfrac{3n}{\pi}\right)^{2/3}$. In the periodic potential $V(x)=V(x+ma)$ of the crystal the electron wavefunction is a **Bloch wave** $\psi_k(x)=U_k(x)e^{jkx}$ (slide 34). $\hbar k$ is the **crystal momentum**: an external force changes it as $F=d(\hbar k)/dt$ (slide 35). Plotting the allowed energies against $k$ gives the **E–k diagram** (slides 37–38). A photon can only be emitted efficiently if the CB minimum sits directly above the VB maximum (**direct gap**: GaAs, InP, GaN). In an **indirect** semiconductor (Si, Ge) a phonon must supply the missing momentum. ## Where bands come from: the Kronig–Penney model {#sec-kp} The simplest periodic potential is a row of thin barriers of strength $P$ spaced by $a$. Bloch's theorem then gives a single equation for the allowed energies: $$ \cos(ka)=\cos(\alpha a)+P\,\frac{\sin(\alpha a)}{\alpha a},\qquad \alpha=\frac{\sqrt{2m_eE}}{\hbar}. $$ The right-hand side can exceed 1 in magnitude, and then no real $k$ exists: that energy is a **gap**. Increase $P$ in the panel and watch the free-electron parabola break into bands, with gaps opening at $ka=\pm\pi$ (the edge of the Brillouin zone). ```{ojs} //| echo: false viewof kp = Inputs.form({ P: Inputs.range([0, 12], {value: 4, step: 0.1, label: "barrier strength P"}), a: Inputs.range([0.2, 1.0], {value: 0.5, step: 0.01, label: "lattice period a (nm)"}) }) ``` ```{ojs} //| echo: false kpOut = { const hbar = 1.054571817e-34, me = 9.1093837e-31, q = 1.602176634e-19, a = kp.a*1e-9; const E0 = hbar**2/(2*me*a*a)/q; // eV, energy for alpha*a = 1 const pts = [], free = [], bands = []; let inBand = false, start = 0; for (let i = 1; i <= 6000; i++) { const x = 4*Math.PI*i/6000, E = E0*x*x; // x = alpha*a const rhs = Math.cos(x) + kp.P*Math.sin(x)/x; if (Math.abs(rhs) <= 1) { const ka = Math.acos(rhs); pts.push({k: ka/Math.PI, E}); pts.push({k: -ka/Math.PI, E}); if (!inBand) { inBand = true; start = E; } } else if (inBand) { inBand = false; bands.push([start, E]); } let kf = x % (2*Math.PI); if (kf > Math.PI) kf -= 2*Math.PI; free.push({k: kf/Math.PI, E}); } if (inBand) bands.push([start, E0*16*Math.PI**2]); const gaps = bands.slice(1).map((b, i) => b[0] - bands[i][1]); return {pts, free, bands, gaps}; } Plot.plot({ width: Math.min(width, 1050), height: 420, x: {label: "ka/π (reduced zone)", domain: [-1, 1]}, y: {label: "E (eV)", grid: true}, marks: [ Plot.dot(kpOut.free, {x: "k", y: "E", r: 0.6, fill: "#52514e", fillOpacity: 0.35}), Plot.dot(kpOut.pts, {x: "k", y: "E", r: 1.3, fill: "#2a78d6"}), Plot.tip(kpOut.pts, Plot.pointer({x: "k", y: "E", title: d => `ka/π = ${d.k.toFixed(3)}\nE = ${d.E.toFixed(2)} eV`})) ] }) ``` ```{ojs} //| echo: false html`<div class="result-box">Blue: allowed states in the crystal. Grey: free-electron parabola folded into the first zone (P = 0).<br> Band gaps: ${kpOut.gaps.slice(0, 3).map((g, i) => `E<sub>g${i+1}</sub> = <b>${g.toFixed(2)} eV</b>`).join(" · ")}</div>` ``` ```{python} #| label: fig-kp #| fig-cap: "Kronig–Penney model (a = 0.5 nm, P = 4). Left: the right-hand side of the dispersion relation; energies where it leaves the band [−1, 1] are forbidden. Right: the resulting E–k diagram in the extended zone, compared with a free electron." a, P = 0.5e-9, 4.0 x = np.linspace(1e-4, 4*np.pi, 8000) # alpha*a E = (sl.hbar*x/a)**2/(2*sl.me)/sl.qe rhs = np.cos(x) + P*np.sin(x)/x ok = np.abs(rhs) <= 1 ka = np.where(ok, np.arccos(np.clip(rhs, -1, 1)), np.nan) band = np.floor(x/np.pi) # band index -> extended-zone unfolding k_ext = np.where(band % 2 == 0, ka + band*np.pi, (band + 1)*np.pi - ka) fig, ax = plt.subplots(1, 2, figsize=(9, 4.2), sharey=True) ax[0].plot(rhs, E); ax[0].axvspan(-1, 1, color=sl.PALETTE[0], alpha=0.08) ax[0].set(xlabel="cos(αa) + P sin(αa)/(αa)", ylabel="E (eV)", xlim=(-3, 5)) ax[1].plot(k_ext/np.pi, E, label="crystal"); ax[1].plot(x/np.pi, E, ls="--", color=sl.INK2, lw=1.2, label="free electron") for m in (1, 2, 3): ax[1].axvline(m, color=sl.GRID, lw=1) ax[1].set(xlabel="ka/π (extended zone)"); ax[1].legend(fontsize=8) plt.tight_layout(); plt.show() ``` ## Metals: the Fermi energy {#sec-metals} Slides 3–6 picture a metal as a partly filled band. With one free electron per atom the free-electron formula already gives the right order of magnitude for $E_{F0}$, and the average electron energy at 0 K is $\tfrac35E_{F0}$: ```{python} #| label: tbl-metals #| tbl-cap: "Free-electron Fermi energy of common metals (electron concentrations from tables)." metals = {"Na": 2.65e28, "Al": 18.1e28, "Cu": 8.47e28, "Ag": 5.86e28, "Au": 5.90e28} EF = {m: sl.h**2/(8*sl.me)*(3*n/np.pi)**(2/3)/sl.qe for m, n in metals.items()} pd.DataFrame({"n (10^28 m^-3)": {m: n/1e28 for m, n in metals.items()}, "E_F0 (eV)": EF, "mean energy 3/5 E_F0 (eV)": {m: 0.6*e for m, e in EF.items()}, "Fermi speed (10^6 m/s)": {m: np.sqrt(2*e*sl.qe/sl.me)/1e6 for m, e in EF.items()}}).round(2) ``` ```{python} #| label: fig-metal #| fig-cap: "Copper (E_F0 = 7.0 eV): density of states g(E), Fermi–Dirac function f(E), and the electron distribution g(E)f(E) at 300 K and at an (unphysical) 3000 K. Only electrons within a few k_BT of E_F can change state, which is why the electronic heat capacity of metals is so small." E = np.linspace(0, 9, 2000) g = sl.dos_3d(E, 1.0) fig, ax = plt.subplots(1, 2, figsize=(9, 3.8)) for T, c in ((300, sl.PALETTE[0]), (3000, sl.PALETTE[1])): ax[0].plot(E, sl.fermi_dirac(E, EF["Cu"], T), color=c, label=f"f(E), {T} K") ax[1].plot(E, g*sl.fermi_dirac(E, EF["Cu"], T)/1e28, color=c, label=f"{T} K") ax[1].plot(E, g/1e28, ls="--", color=sl.INK2, lw=1.2, label="g(E)") ax[0].set(xlabel="E (eV)", ylabel="f(E)"); ax[0].legend(fontsize=8) ax[1].set(xlabel="E (eV)", ylabel="10²⁸ states m⁻³ eV⁻¹"); ax[1].legend(fontsize=8) plt.tight_layout(); plt.show() ``` ## Holes (slides 8–9) A photon with $h\nu>E_g$ breaks a bond and creates an **electron–hole pair** (EHP). The empty bond, the **hole**, moves because neighbouring valence electrons tunnel into it. It behaves like a positive particle with its own effective mass $m_h^*$. When a wandering CB electron meets a hole they **recombine**: in a direct-gap material the energy is released as a photon. This is the LED. The photogeneration threshold is $\lambda_g=hc/E_g$: ```{python} #| label: tbl-cutoff #| tbl-cap: "Bandgap and cut-off wavelength λ_g = hc/E_g of the semiconductors of Table 3.1." pd.DataFrame({m: {"E_g (eV)": p["Eg"], "λ_g (nm)": sl.wavelength_from_eV(p["Eg"])*1e9, "gap": "direct" if p["direct"] else "indirect"} for m, p in sl.MATERIALS.items()}).T ``` ## Direct and indirect gaps, phonons (slides 37–39) {#sec-direct} Near a band edge the E–k diagram is a parabola, $E=E_c+\hbar^2(k-k_0)^2/2m_e^*$. The figure contrasts GaAs (CB minimum at $k=0$) with Si (CB minimum near the zone edge along [100], $k_0\approx0.85\cdot2\pi/a$). ```{python} #| label: fig-ek #| fig-cap: "Parabolic E–k sketches. GaAs: a vertical (direct) transition emits a photon. Si: the CB minimum is displaced by k₀, so recombination needs a phonon to carry momentum ħk₀. On this scale the photon wavevector would be a vertical arrow (about 10⁻³ of the axis)." k = np.linspace(-1, 1, 400) # in units of 2π/a fig, ax = plt.subplots(1, 2, figsize=(9, 3.8), sharey=True) for a_, (name, k0, Eg, mc) in zip(ax, (("GaAs", 0.0, 1.42, 0.067), ("Si", 0.85, 1.11, 0.26))): a_m = sl.MATERIALS[name]["a"] kk = k*2*np.pi/a_m a_.plot(k, Eg + (sl.hbar*(kk - k0*2*np.pi/a_m))**2/(2*mc*sl.me)/sl.qe, label="CB") a_.plot(k, -(sl.hbar*kk)**2/(2*0.5*sl.me)/sl.qe, label="VB") a_.annotate("", xy=(k0, Eg), xytext=(0, 0), arrowprops=dict(arrowstyle="->", color=sl.PALETTE[7], lw=1.8)) a_.set(title=name, xlabel="k (2π/a)", ylim=(-1.2, 3)); a_.legend(fontsize=8, loc="lower right") ax[0].set_ylabel("E (eV)"); plt.tight_layout(); plt.show() k_photon = 2*np.pi/sl.wavelength_from_eV(1.11) print(f"photon wavevector at Eg(Si): {k_photon:.2e} m^-1; zone edge π/a = {np.pi/sl.MATERIALS['Si']['a']:.2e} m^-1") print(f"ratio: {k_photon/(np.pi/sl.MATERIALS['Si']['a']):.1e} -> photons carry almost no crystal momentum") v_s = 9000 # longitudinal sound speed in Si (m/s) print(f"acoustic phonon at k0: E = ħ v_s k0 ≈ {sl.hbar*v_s*0.85*2*np.pi/sl.MATERIALS['Si']['a']/sl.qe*1e3:.0f} meV (linear estimate; real Si phonons at X: 19–57 meV)") ``` A photon cannot supply the momentum $\hbar k_0$, but a lattice vibration (phonon, slide 39) can: it has a large $k$ and a small energy (tens of meV). Three-body processes are slow, so radiative recombination in Si takes milliseconds, while non-radiative recombination through defects (slide 38c) takes microseconds. Si therefore cannot be used for LEDs. ::: {.callout-tip collapse="true"} ## Going deeper - **Effective mass from curvature.** $1/m^*=\hbar^{-2}\,d^2E/dk^2$. Compute it at the bottom of the first two Kronig–Penney bands. Why is it *negative* at the top of a band? This is the origin of the hole. - **Tight binding.** For $N$ coupled atoms the levels spread over a band of width $4t$ (hopping $t$). Diagonalise an $N\times N$ tridiagonal matrix with `numpy.linalg.eigvalsh` and watch the levels fill a band as $N$ grows (slide 3b). - **Real band structures.** Use the empirical pseudopotential method (a 30-line Python script with the Cohen–Bergstresser form factors) to compute the E–k diagram of Si and GaAs along L–Γ–X and find $E_g$ and $k_0$. - **Strained and alloyed Ge** becomes a direct-gap semiconductor (GeSn alloys): a route to light sources on silicon. ::: ::: {.callout-note collapse="true"} ## Complements - S. O. Kasap, *Optoelectronics and Photonics*, ch. 3; *Principles of Electronic Materials and Devices*, ch. 4–5. - C. Kittel, *Introduction to Solid State Physics*: chapters on free-electron gas and energy bands (Kronig–Penney). - M. L. Cohen and J. R. Chelikowsky, *Electronic Structure and Optical Properties of Semiconductors* (pseudopotentials). ::: ## Try it yourself 1. In the live panel, at which $P$ does the first gap reach 1.1 eV (the Si gap) for $a=0.543$ nm? Is the model realistic? 2. Show that the number of states in one Kronig–Penney band equals the number of unit cells (count $k=2\pi m/Na$). 3. Compute $E_{F0}$ for Mg (two free electrons per atom, $n_\text{atoms}=4.3\times10^{28}$ m$^{-3}$). 4. How large must the phonon energy be to assist an indirect transition in Ge, where the CB minimum is at the L point, $k_0=\sqrt3\,\pi/a$? Use a sound speed of 5000 m/s.