Past exam of the quantum condensed matter course of the University of Cambridge

Original paper, released PDF pages 493–498

Section A

1

Solution

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction Electrons of charge

- e

, the isotropic Drude model gives

v_{d} = - \frac{e τ}{m ^{*}} E = - μ_{e} E,

where

τ

is the momentum relaxation time.

The positive electron mobility

μ_{e} = ∣ v_{d} ∣/∣ E ∣

measures the ease of this drift in linear response. In the same approximation,

μ_{e} = \frac{e τ}{m ^{*}}, σ = n e μ_{e} .

Its units are

m^{2} V^{- 1} s^{- 1}

; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is

v = ℏ^{- 1} \nabla_{k} E

, while

ℏ \dot{k} = - e E

. Thus

(m^{*- 1})_{ij} = \frac{1}{ℏ ^{2}} \frac{\partial ^{2} E}{\partial k _{i} \partial k _{j}} .

For an isotropic parabolic minimum this reduces to

E = E_{c} + ℏ^{2} k^{2} / (2 m^{*})

and

m^{*} \dot{v} = - e E

. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

2

Solution

Work at zero temperature and fixed total carrier density. Write

n_{↑} = n /2 + δ

and

n_{↓} = n /2 - δ

. Let

g_{s} (E_{F})

be the single-spin density of states per unit volume in the unpolarised state. Since

d E_{s} / d n_{s} = E_{F s}

and

d E_{F s} / d n_{s} = 1/ g_{s}

, the linear kinetic-energy shifts cancel and the quadratic cost is

Δ E_{kin} = \frac{δ ^{2}}{g _{s} ( E _{F} )} + O (δ^{4}) .

The repulsive interaction instead decreases under spin polarization:

E_{int} = U (\frac{n ^{2}}{4} - δ^{2}), Δ E = [\frac{1}{g _{s} ( E _{F} )} - U] δ^{2} + O (δ^{4}) .

Hence the Stoner criterion for an opposite-spin contact repulsion is

U_{c} = \frac{1}{g _{s} ( E _{F} )} = \frac{2}{g _{tot} ( E _{F} )}, U > U_{c} destabilises the unpolarised state .

The two forms differ only in whether the stated density of states counts one spin or both. For the three-dimensional free electron model,

g_{tot} (E_{F}) = 3 n / (2 E_{F})

, so

U_{c} = \frac{4 E _{F}}{3 n} .

Equivalently, for fractional polarisation

ζ = (n_{↑} - n_{↓}) / n

, the kinetic-energy density is

\frac{3}{10} n E_{F} [(1 + ζ)^{5/3} + (1 - ζ)^{5/3}]

; its quadratic coefficient is

n E_{F} /3

, while the interaction contributes

- U n^{2} /4

. With number densities,

U

has units of energy times volume and the displayed interaction is an energy density; multiplying by the sample volume does not change the threshold.

3

Solution

An angle-resolved photoemission spectroscopy intensity map locates occupied electronic band structure through its bright dispersing ridges. For a nearly isotropic graphene cone,

E_{\pm} (q) = E_{D} \pm ℏ v_{F} ∣ q ∣, v_{F} = \frac{1}{ℏ} \frac{d E}{d k} .

Fit straight lines to the reasonably linear portions of the ridges, avoiding the broad central crossing, and extrapolate them to locate the Dirac point energy

E_{D}

. The PDF places the crossing near

k_{y} = 0

and plotted energy roughly

1.4

–

1.5 eV

. On a lower branch, a representative change from approximately

(∣ k_{y} ∣, E) = (0.10 \overset{˚}{A}^{- 1}, 0.9 eV)

towards

(0, 1.5 eV)

gives a slope near

6 eV \overset{˚}{A}

. Allowing for the ridge width gives about

5

–

7 eV \overset{˚}{A}

, hence

v_{F} ≃ (0.8 - 1.1) \times 1 0^{6} m s^{- 1}, approximately 1 0^{6} m s^{- 1} .

Here

1 eV \overset{˚}{A} /ℏ = 1.52 \times 1 0^{5} m s^{- 1}

To determine the carrier Fermi energy, identify the Fermi level from a calibrated photoemission cutoff, then measure its separation from the Dirac point:

E_{F}^{cone} = ∣ E_{F, plot} - E_{D} ∣ = ℏ v_{F} k_{F}, p_{F} = ℏ k_{F} .

The separation of the two Fermi crossings gives

2 k_{F}

for a cut through the cone centre. This determines the doping energy relative to the Dirac point; the absolute plotted energy origin is not itself a Fermi energy.

The reproduced panel has no marked Fermi level or stated energy-zero calibration. Its upper ridges end near

1.9 eV

, but that alone does not prove that the endpoint is a Fermi cutoff. If it is the intended cutoff, the estimate is

E_{F}^{cone} ≃ 0.4

–

0.5 eV

and

k_{F} \sim 0.06

–

0.08 \overset{˚}{A}^{- 1}

. If instead the plotted numbers are binding energies measured from an unstated Fermi zero, the inferred separation would be roughly

1.4

–

1.5 eV

. The supplied figure securely supports the slope estimate, but it does not distinguish these energy-reference conventions; an unconditional numerical Fermi energy requires the missing calibration.

Section B

4

a

Solution

Take

q = - e

as the signed electron charge. The Lorentz oscillator model is a driven damped bound charge:

m \ddot{u} + mγ \dot{u} + m ω_{0}^{2} u = q E (t) .

The first term is inertia, the second is a phenomenological frictional loss of coherent motion to other degrees of freedom, and the third is the linear restoring force from binding to the atom. The driving term is the electric force. The natural angular frequency

ω_{0}

represents the electronic transition; this classical oscillator describes its near-resonant linear response. The friction convention is

mγ \overset{u}{˙}

, so no factor two is hidden in

γ

b

Solution

Substitute harmonic amplitudes with the specified

e^{- iω t}

convention:

m (ω_{0}^{2} - ω^{2} - iγω) u_{0} = q E_{0} .

One electron contributes a dipole moment

p = q u_{0}

, so the electric polarization of a density

n

P_{0} = n q u_{0} = \frac{n q ^{2}}{m ( ω _{0}^{2} - ω ^{2} - iγω )} E_{0} .

Using

D = ϵ_{0} E + P = ϵ_{0} ϵ_{r} E

gives

ϵ_{r} (ω) = 1 + \frac{ω _{p}^{2}}{ω _{0}^{2} - ω ^{2} - iγω}, ω_{p}^{2} = \frac{n q ^{2}}{m ϵ _{0}} .

Thus the paper's dimensionless

ϵ_{ω}

is the relative permittivity, rather than the absolute permittivity

ϵ_{0} ϵ_{r}

. The electron's negative charge disappears from the electric susceptibility because both the force and dipole moment contain

q

c

Solution

Set

D = (ω_{0}^{2} - ω^{2})^{2} + γ^{2} ω^{2}

. The real part and imaginary part of the complex permittivity are

ϵ_{r}^{'} = 1 + \frac{ω _{p}^{2} ( ω _{0}^{2} - ω ^{2} )}{D}, ϵ_{r}^{''} = \frac{ω _{p}^{2} γω}{D} .

For positive frequency the loss part is nonnegative. The real part starts at

1 + ω_{p}^{2} / ω_{0}^{2}

, rises below resonance, falls steeply through the resonance region, and approaches one from below at high frequency. The imaginary part is a positive absorption peak near

ω_{0}

, tending to zero on both sides.

For a nonmagnetic medium write its complex refractive index as

N = n_{r} + iκ

, with

N^{2} = ϵ_{r}

and optical extinction coefficient

κ

. In a transparent region with

κ ≪ n_{r}

and

ϵ_{r}^{'} > 0

n_{r} ≃ ϵ_{r}^{'}, κ ≃ \frac{ϵ _{r}^{''}}{2 ϵ _{r}^{'}}, α_{abs} = \frac{2 ωκ}{c} .

The real part controls phase velocity and optical dispersion; the loss part controls wave absorption and the cycle-averaged dissipated power density

ω ϵ_{0} ϵ_{r}^{''} ∣ E_{0} ∣^{2} /2

. Weak damping gives a narrow line but does not make wave absorption weak at its centre. The transparent-medium approximation applies away from the line.

Figure 1.
Single-resonance, multiple-resonance and free-electron optical responses
. Original dimensionless model plots. The first row shows one bound-electron resonance, the second shows three, and the third shows the Drude permittivity and its normal-incidence reflectance. Finite damping is retained; these are illustrative calculations, not material measurements. **Details**.

d

Solution

For independent electronic transitions, add the electric dipole moment responses:

ϵ_{r} (ω) = 1 + j \sum \frac{S _{j}}{ω _{j}^{2} - ω ^{2} - i γ _{j} ω}, S_{j} > 0.

The positive constants

S_{j}

include electron density and transition atomic oscillator strength. Each term adds an absorption peak to

ϵ_{r}^{''}

and a dispersive positive-to-negative contribution to

ϵ_{r}^{'} - 1

, as in the middle row of the figure.

Between well-separated absorption lines, where the medium is transparent, the refractive index generally increases with frequency:

d n_{r} / d ω > 0

, or

d n_{r} / d λ < 0

. This is normal dispersion. Near a resonance the real response turns down with increasing frequency, producing an interval of anomalous dispersion,

d n_{r} / d ω < 0

, accompanied by appreciable absorption. The absorptive and dispersive parts are linked by the Kramers-Kronig relations; the steep variation is not independent of the loss spectrum.

e

Solution

An unbound conduction Electron has no local restoring force, so set

ω_{0} = 0

while retaining inertia and momentum relaxation time

τ = 1/ γ

. This is the Drude model:

ϵ_{r} (ω) = 1 - \frac{ω _{p}^{2}}{ω ( ω + iγ )} .

Separating the response gives

ϵ_{r}^{'} = 1 - \frac{ω _{p}^{2}}{ω ^{2} + γ ^{2}}, ϵ_{r}^{''} = \frac{ω _{p}^{2} γ}{ω ( ω ^{2} + γ ^{2} )} .

For a good metal,

γ ≪ ω_{p}

. The real part is strongly negative at low frequency, crosses zero at

ω_{p}^{2} - γ^{2} ≃ ω_{p}

, and approaches one from below. The imaginary part diverges as

ω_{p}^{2} / (γω)

at low frequency, reflecting the finite dc electrical conductivity

σ_{0} = ϵ_{0} ω_{p}^{2} / γ

, then decays as

ω^{- 3}

. The low-frequency imaginary response describes conduction loss, not a finite-frequency bound-electron absorption line. Core-electron and interband responses can add a background in a real metal.

f

Solution

At normal incidence from vacuum, the Fresnel equations give the reflectance

R (ω) = \frac{N ( ω ) - 1}{N ( ω ) + 1}^{2}, N^{2} = ϵ_{r} .

A good metal has

R

close to one below its plasma frequency. At very low frequency the electromagnetic field penetrates only a skin depth; the local Drude limit gives

1 - R ≃ 8 ϵ_{0} ω / σ_{0}

. In the collisionless negative-permittivity range the transmitted field is evanescent and reflection remains strong. Near the plasma edge the reflectance drops sharply; above it

ϵ_{r}^{'} > 0

and propagating waves become possible. In the ideal free-electron high-frequency limit,

R \sim \frac{ω _{p}^{4}}{16 ω ^{4}} ⟶ 0.

Finite damping rounds the plasma edge, and interband absorption changes the detailed response of actual metals.

5

a

Solution

Choose one atomic orbital

ϕ_{A}

and

ϕ_{B}

on the two atoms and approximate a molecular orbital by the linear combination of atomic orbitals

ψ = c_{A} ϕ_{A} + c_{B} ϕ_{B} .

Projecting the Schrödinger equation onto the two orbitals gives a generalized eigenvalue problem

H c = ES c

, where

H_{ij} = ⟨ ϕ_{i} ∣ \hat{H} ∣ ϕ_{j} ⟩

and

S_{ij} = ⟨ ϕ_{i} ∣ ϕ_{j} ⟩

. In an orthonormal basis the matrix is

H = (ϵ_{A} t^{*} t ϵ_{B}), E_{\pm} = \frac{ϵ _{A} + ϵ _{B}}{2} \pm \frac{( ϵ _{A} - ϵ _{B} ) ^{2}}{4} + ∣ t ∣^{2} .

The coupling mixes the atomic states and produces a lower bonding molecular orbital and upper antibonding molecular orbital. For equivalent real orbitals with overlap

s

the normalized combinations are

(ϕ_{A} \pm ϕ_{B}) / 2 (1 \pm s)

; the precise energy ordering follows from the Hamiltonian matrix elements. This restricted orbital basis is an approximation to the molecule's wavefunction, rather than a claim that exact isolated-atom orbitals remain unchanged in the molecule.

b

Solution

A concrete example is samarium hexaboride,

SmB_{6}

: narrow, nearly localized

4 f

-derived states mix with a much more dispersive

5 d

-derived conduction band. Such band hybridization is the low-energy mechanism behind a bulk Kondo insulator gap. The two-band Hamiltonian is an effective illustration; correlations help set its narrow-band energy and renormalized mixing.^[ref]

c

Solution

Write the dispersive band as

E_{p} = A k^{2}

with

A > 0

and define

k_{c} = E_{fl} / A

. The uncoupled sketch is an upward-opening parabola and a horizontal flat band at positive energy. They cross on the sphere

k = k_{c}

in three dimensions, or at

\pm k_{c}

in a one-dimensional cut:

E_{p} (k_{c}) = E_{fl}, α = 0.

Without mixing the crossing is allowed. The left panel of the figure shows this case.

Figure 1.
Avoided crossings and Fermi-volume redistribution in a two-band model
. Original model calculations. The first three panels use an exactly flat second band and show zero, weak and strong hybridisation in units of its energy. The fourth uses a finite heavy-band curvature and fixed total electron density; two Fermi radii separate before the inner pocket vanishes. It distinguishes the general two-surface result from the eventual single-surface limit. **Details**.

For dimensional clarity

A = ℏ^{2} / (2 m_{ℓ})

in the usual band effective mass convention. The printed product

m_{p} ℏ^{2} k^{2}

is consistent only if

m_{p}

is a curvature coefficient with inverse-mass units; it cannot simultaneously denote the effective mass itself. Using

A

avoids relying on that notation.

d

Solution

At nonzero band hybridization, the crossing becomes an avoided crossing. The separation at

k_{c}

2 α

. If

α ≪ E_{fl}

, appreciable mixing occurs only where

∣ A k^{2} - E_{fl} ∣ ≲ α

; away from that region each branch retains nearly its original orbital character. For

α \sim E_{fl}

, the avoided crossing is broad and both branches are substantially reshaped, as shown in the figure.

Use

E_{-}

for the lower branch and

E_{+}

for the upper one. For

α > 0

E_{-} (k) < E_{fl} < E_{+} (k), min_{k} E_{+} (k) = E_{+} (0) = \frac{E _{fl} + E _{fl}^{2} + 4 α ^{2}}{2} .

The lower branch increases towards

E_{fl}

from below. Thus an indirect gap exists in the exactly flat-band model. In the unbounded-parabola limit its lower bound is

Δ_{ind} = \frac{E _{fl}^{2} + 4 α ^{2} - E _{fl}}{2} ≃ \frac{α ^{2}}{E _{fl}} (α ≪ E_{fl}) .

A finite Brillouin zone gives a finite maximum of

E_{-}

below

E_{fl}

and hence a slightly larger indirect gap. This hybridization gap differs from the local direct gap

2 α

The material can therefore be a band insulator when the lower band is full and the upper band empty. With two spin-degenerate orbitals per unit cell, the required filling is

2 electrons per unit cell, i.e. half of the combined four-state capacity .

With spin degeneracy removed, replace two by one for each independent spin sector. Other fillings are metallic in this noninteracting model.

e

Solution

At each wavevector the orbital-basis Hamiltonian is

H (k) = (E_{p} (k) α α E_{fl}) .

A phase choice makes the momentum-independent mixing real and nonnegative. The characteristic equation is

(E_{p} - E) (E_{fl} - E) - α^{2} = 0,

E_{\pm} (k) = \frac{1}{2} [E_{p} + E_{fl} \pm (E_{p} - E_{fl})^{2} + 4 α^{2}] .

For

α > 0

, an eigenvector obeys

c_{fl} / c_{p} = (E_{\pm} - E_{p}) / α

. At the bare crossing, the normalized states are

(∣ p ⟩ \pm ∣ fl ⟩) / 2

with energies

E_{fl} \pm α

. The lower state's parabolic-orbital weight is

∣ c_{p, -} ∣^{2} = \frac{1}{2} [1 - \frac{E _{p} - E _{fl}}{( E _{p} - E _{fl} ) ^{2} + 4 α ^{2}}],

which tends to one below the crossing and zero above it. This makes the exchange of band character in the sketches explicit.

f

Solution

Let the common uncoupled Fermi wavevector be

k_{0}

. Equal carrier densities imply equal Fermi surface volumes, and hence equal radii for the spherical spin-degenerate electron pockets. The conserved total density is

n_{tot} = \frac{2 k _{0}^{3}}{3 π ^{2}} .

If hybridisation leaves only one occupied electron-like band, state counting gives

k_{F} = 2^{1/3} k_{0}, p_{F} = 2^{1/3} p_{0}, \frac{Δ p _{F}}{p _{0}} = 2^{1/3} - 1 ≃ 0.260.

This is the familiar single-large-Fermi-surface answer. It requires the other hybridised band to be empty.

For finite heavy-band mass, that conclusion does not follow from an arbitrary nonzero coupling. At sufficiently small

α

, the upper band still has an occupied pocket, so there are two Fermi momenta, not one. The exact fixed-density statement is Fermi-volume conservation in two hybridized electron bands:

k_{F, -}^{3} + k_{F, +}^{3} = 2 k_{0}^{3} .

For an explicit weak-coupling check, write

E_{p} = A k^{2}

E_{h} = C + B k^{2}

, with

0 < B ≪ A

, and enforce the uncoupled equality at

k_{0}

C = (A - B) k_{0}^{2}

μ_{0} = A k_{0}^{2}

. Put

μ = μ_{0} + δ μ

and

y = k_{F}^{2} - k_{0}^{2}

. The Fermi-surface equation is

(δ μ - A y) (δ μ - B y) - α^{2} = 0.

At fixed density

δ μ = O (α^{2})

, giving

k_{F, \mp} = k_{0} \pm \frac{α}{2 k _{0} A B} + O (α^{2}) .

(5)

For arbitrary coupling, put

q_{\pm} = k_{F, \mp}^{2}

. The roots of the same characteristic equation, before discarding unoccupied pockets, are

q_{\pm} = \frac{( A + B ) μ - A C \pm [( A - B ) μ - A C ] ^{2} + 4 A B α ^{2}}{2 A B} .

(6)

Choose the chemical potential so that

[max (q_{+}, 0)]^{3/2} + [max (q_{-}, 0)]^{3/2} = 2 k_{0}^{3}

. This specifies the Fermi momentum of each pocket for any supplied curvatures and coupling.

The transition to one occupied pocket occurs at

μ = E_{+} (0)

. Set

q_{*} = 2^{2/3} k_{0}^{2}

; at the transition the two squared radii are

q_{*}

and zero. Their sum and product give

μ_{c} = \frac{A ( C + B q _{*} )}{A + B}, α_{c}^{2} = μ_{c} (μ_{c} - C) = \frac{A B ( C + B q _{*} ) ( A q _{*} - C )}{( A + B ) ^{2}} .

(7)

For

0 < α < α_{c}

there are two pockets; for

α \geq α_{c}

the only nonzero radius is

2^{1/3} k_{0}

. The outer radius grows continuously and the inner one shrinks. The nearly flat band limit can make depopulation occur at a small coupling, but the mass ratio or a single-surface assumption is needed to turn the question into a unique finite-coupling numerical change.

6

a

Solution

Count wavevectors in a large volume with periodic boundary conditions. One spin-degenerate spherical band supplies

\frac{N ( k )}{V} = \frac{2}{( 2 π ) ^{3}} \frac{4 π k ^{3}}{3} = \frac{k ^{3}}{3 π ^{2}}, \frac{1}{V} \frac{d N}{d k} = \frac{k ^{2}}{π ^{2}} .

For the conduction band,

k = 2 m_{e}^{*} (ϵ - ϵ_{c}) /ℏ

and

d k / d ϵ = m_{e}^{*} / (ℏ^{2} k)

. Transforming the state count to energy gives

g_{e} (ϵ) = \frac{1}{2 π ^{2}} (\frac{2 m _{e}^{*}}{ℏ ^{2}})^{3/2} ϵ - ϵ_{c}, ϵ \geq ϵ_{c} .

For the valence band use

∣ d k / d ϵ ∣ = m_{h}^{*} / (ℏ^{2} k)

and

k = 2 m_{h}^{*} (ϵ_{v} - ϵ) /ℏ

, obtaining

g_{h} (ϵ) = \frac{1}{2 π ^{2}} (\frac{2 m _{h}^{*}}{ℏ ^{2}})^{3/2} ϵ_{v} - ϵ, ϵ \leq ϵ_{v} .

Each density of states is zero beyond its indicated band edge. The factor two counts spin; an extra valley degeneracy was not assumed in these stipulated isotropic bands.

b

Solution

In the nondegenerate conduction band, the Fermi-Dirac distribution reduces to its Boltzmann distribution tail. Thus

n = \int_{ϵ_{c}}^{\infty} g_{e} (ϵ) e^{- (ϵ - μ) / (k_{B} T)} d ϵ = N_{c} e^{- (ϵ_{c} - μ) / (k_{B} T)}, N_{c} = 2 (\frac{m _{e}^{*} k _{B} T}{2 π ℏ ^{2}})^{3/2} .

Here the substitution

x = (ϵ - ϵ_{c}) / (k_{B} T)

leaves

\int_{0}^{\infty} x^{1/2} e^{- x} d x = π /2

. For electron holes use the empty-state probability

1 - f (ϵ) ≃ e^{(ϵ - μ) / (k_{B} T)}

, not the conduction-band Boltzmann factor. This gives

p = N_{v} e^{- (μ - ϵ_{v}) / (k_{B} T)}, N_{v} = 2 (\frac{m _{h}^{*} k _{B} T}{2 π ℏ ^{2}})^{3/2} .

The chemical potential cancels in the product. With

E_{g} = ϵ_{c} - ϵ_{v}

n p = N_{c} N_{v} e^{- E_{g} / (k_{B} T)} = \frac{1}{2} (m_{e}^{*} m_{h}^{*})^{3/2} (\frac{k _{B} T}{π ℏ ^{2}})^{3} e^{- E_{g} / (k_{B} T)} = n_{i}^{2} .

This semiconductor mass-action law holds in thermal equilibrium for nondegenerate carriers, including a doped semiconductor. Only in an intrinsic semiconductor must

n = p = n_{i}

. The factors

N_{c}, N_{v}

are the effective density of states; the factor

1/2

in the final expression follows from the two spin factors already included in (a).

c

Solution

Ordinary doped semiconductor devices need a regime where their donor impurities or acceptor impurities supply mobile charge carriers while intrinsic thermal excitation remains sufficiently small.

At low temperature, donor impurities or acceptor impurities may no longer be ionized. Dopant freeze-out reduces the carrier density, raises electrical resistance and changes the intended carrier injection and p-n junction response. At high temperature the intrinsic carrier concentration grows as

n_{i} \propto T^{3/2} e^{- E_{g} / (2 k_{B} T)}

when band effective masses and band gap are treated as constant. Once it becomes comparable with the net doping density, it overwhelms the intended electron-hole imbalance; reverse leakage and dissipation increase and useful p/n control is lost.

The usual operating window lies between freeze-out and intrinsic conduction, with mobility and thermal management placing further practical limits. This is a statement about conventional doped-device operation, not a fundamental prohibition on engineered cryogenic or wide-bandgap devices.

d

Solution

An ideal diffusion-dominated p-n junction obeys the Shockley diode equation

I (V, T) = I_{s} (T) [e^{e V / (η k_{B} T)} - 1],

with diode ideality factor

η = 1

in the simplest model. Forward bias gives exponential current growth; reverse bias gives a nearly constant

- I_{s}

until diode reverse breakdown, outside this ideal expression.

For a long abrupt p-n junction in low carrier injection, the diode saturation current is

I_{s} = e A n_{i}^{2} (\frac{D _{p}}{L _{p} N _{D}} + \frac{D _{n}}{L _{n} N _{A}}) = C (T) n_{i}^{2} .

Here

A

is area,

N_{D}, N_{A}

are ionized dopant densities, and

D, L

are minority-carrier diffusion coefficients and diffusion lengths in the appropriate regions. At each temperature, extract

I_{s}

either from the reverse-saturation plateau below breakdown or from a forward exponential fit using its measured diode ideality factor.

Extracting a semiconductor band gap from diode saturation current requires correcting for the transport prefactor:

lo g \frac{I _{s} ( T )}{C ( T ) T ^{3}} = constant - \frac{E _{g}}{k _{B} T} .

A plot against

1/ T

has slope

- E_{g} / k_{B}

when the gap and effective masses are nearly temperature-independent. Alternatively divide the measured

I_{s}

by a known

C (T)

to recover the full

n_{i}^{2} (T)

. The prefactor cannot generally be assumed constant over an arbitrarily broad range. Junction-generation or recombination-dominated current can scale as

n_{i}

instead, so the diffusion regime should be identified from the current-voltage behaviour.

Figure 1.
Ideal diode characteristic and extraction of a semiconductor activation energy
. Original model sketches. The normalized current-voltage curve shows forward exponential growth and reverse saturation. The activation panel plots a model proportional to $n_{i}^{2} / T^{3}$ for a fixed illustrative gap; it shows the slope to infer after correcting transport prefactors, not measured device data. **Details**.

e

Solution

A current-carrying sample in a transverse magnetic field develops a transverse electric field because the Lorentz force deflects its charge carriers. In steady state this Hall voltage balances the transverse deflection. With Hall coefficient

R_{H} = E_{y} / (j_{x} B_{z})

, one isotropic electron band gives

R_{H} = - 1/ (n e)

and one hole band gives

R_{H} = + 1/ (p e)

When Electrons and electron holes both conduct, the weak-field two-carrier result is

R_{H} = \frac{p μ _{h}^{2} - n μ _{e}^{2}}{e ( p μ _{h} + n μ _{e} ) ^{2}}, σ = e (n μ_{e} + p μ_{h}) .

The two-carrier Hall coefficient weights each species by mobility squared. Its sign changes when

p μ_{h}^{2} = n μ_{e}^{2} .

Temperature changes dopant ionization, intrinsic electron-hole generation, and the two charge carrier mobilities, so it can move the system across this condition. For example, a p-type semiconductor can be hole-dominated in its extrinsic regime but become negative in its intrinsic regime if

n ≃ p

and

μ_{e} > μ_{h}

. A sign change is possible, not mandatory: in many n-type semiconductors Electrons dominate at both temperatures. The density majority alone does not determine the Hall sign in a two-carrier material.

2021

Original paper, released PDF pages 433–437

Section A

1

Solution

For state counting in a Bloch band, take a primitive cubic lattice of spacing

a

and impose periodic boundary conditions on lengths

L_{i} = N_{i} a

. The allowed wavevectors are

k_{i} = 2 π l_{i} / L_{i}

. The first Brillouin zone has side

2 π / a

, so it contains

N_{i}

distinct values in each direction and hence

N_{x} N_{y} N_{z} = N

wavevectors. Each orbital-derived energy band has one state per wavevector per spin orientation. Therefore

states per band in the whole crystal = 2 N, states per atom per band = 2.

The phrase “per atom” attached to

2 N

in the PDF is a counting error:

2 N

is the total, not the per-atom count.

By the Pauli exclusion principle, a monovalent crystal supplies

N

valence Electrons and half-fills a spin-degenerate band; its Fermi level lies within allowed states, giving a metal in the independent-electron picture. Two electrons per atom can fill one band. If a band gap separates it from all empty bands, the crystal is a band insulator. An even valence count only permits insulation: overlapping bands can still produce a metal. Conversely, electron interactions can produce a Mott insulator, and symmetry breaking can permit insulation outside this simple spin-degenerate electronic band structure criterion. The relevant count in a crystal with a basis is per primitive unit cell.

2

Solution

In a simple metal, conduction Electrons can move collectively in response to an incident electromagnetic wave. The Drude model gives, with time dependence

e^{- iω t}

ϵ_{r} (ω) = 1 - \frac{ω _{p}^{2}}{ω ( ω + iγ )}, ω_{p}^{2} = \frac{n e ^{2}}{ϵ _{0} m ^{*}} .

In the low-loss range

γ ≪ ω < ω_{p}

, the relative permittivity is approximately

1 - ω_{p}^{2} / ω^{2} < 0

. The wavevector inside is predominantly imaginary, so the field is evanescent and penetrates only a short distance. Surface currents reradiate the field; little energy enters a good conductor. The normal-incidence Fresnel equations give

R = \frac{ϵ _{r} - 1}{ϵ _{r} + 1}^{2} ≃ 1 (ω < ω_{p}, γ small) .

Many ordinary metals have their plasma frequency in the ultraviolet, placing visible light below this reflection edge. For

ω > ω_{p}

the free-carrier contribution permits propagation; at

ω ≫ ω_{p}

ϵ_{r} \to 1

and

R \sim ω_{p}^{4} / (16 ω^{4}) \to 0

. The electrons' inertial response becomes too small to screen the rapidly oscillating field.

Ultraviolet transparency is a simple-metal model prediction, not a universal property of real metals. Electronic transitions between bands and bound-electron absorption can remain strong, shift the plasma edge and prevent transparency; transmission also depends on thickness and attenuation. The robust explanation is the change from a reflective negative-permittivity range to a possible propagating range above the plasma edge.

3

Solution

Let adjacent equal masses

m

be separated by

a

, with alternating strong and weak spring constants

K

and

κ

K ≫ κ > 0

. The pattern repeats after

2 a

, even though the masses are identical. Thus the first Brillouin zone is

- \frac{π}{2 a} \leq q \leq \frac{π}{2 a} .

Choose the strong bond inside each two-atom primitive unit cell. With displacement amplitudes

(u, v) e^{i (2 na q - ω t)}

, the normal mode equations are

m ω^{2} (u v) = (K + κ - K - κ e^{2 i q a} - K - κ e^{- 2 i q a} K + κ) (u v) .

The alternating-spring chain dispersion relation is consequently

ω_{\pm}^{2} (q) = \frac{K + κ \pm ( K - κ ) ^{2} + 4 K κ cos ^{2} ( q a )}{m} .

The lower acoustic phonon branch has in-phase motion within the strongly bound pair and

ω_{-} (0) = 0

. Its small-

q

slope is the speed of sound,

c_{s} = a \frac{2 K κ}{m ( K + κ )} ≃ a \frac{2 κ}{m} .

The upper optical phonon branch has opposite motion within each pair, with

ω_{+} (0) = 2 (K + κ) / m

. Here “optical” names the gapped branch; equal neutral masses need not couple to light through an electric dipole.

In the strong-bond limit of the alternating-spring chain, the pairs act almost as rigid dimers in the low-frequency branch, while their internal stretch gives a high, nearly flat branch:

ω_{-}^{2} ≃ \frac{2 κ}{m} sin^{2} (q a), ω_{+}^{2} ≃ \frac{2 K}{m} + \frac{2 κ}{m} cos^{2} (q a) .

(5)

At the zone edge the two frequencies are

2 κ / m

and

2 K / m

, separated by a large forbidden frequency interval. Neighbouring cells are out of phase there. Both branches have horizontal tangents at the edge.

Figure 1.
Phonon branches and displacement patterns of a strongly dimerized equal-mass chain
. Original calculation for $κ / K = 0.06$ . The horizontal coordinate uses the two-atom-cell Brillouin-zone boundary. Strong solid bonds and weak dashed bonds identify each dimer. Arrows show relative displacements at the zone centre and edge; their overall sign and amplitude are arbitrary. **Details**.

Section B

4

a

Solution

For a Hamiltonian with a lattice-periodic potential

V (r + R) = V (r)

, its simultaneous energy eigenstates and lattice-translation eigenstates can be chosen in the Bloch wave form

ψ_{n k} (r) = e^{i k \cdot r} u_{n k} (r), u_{n k} (r + R) = u_{n k} (r) .

Equivalently,

ψ_{n k} (r + R) = e^{i k \cdot R} ψ_{n k} (r)

. The index

n

labels an energy band; crystal momentum

ℏ k

is defined modulo a reciprocal lattice vector. These are the content of Bloch's theorem.

b

Solution

Use one orthonormalized localized atomic orbital

∣ j ⟩

at each position

ja

. In the nearest-neighbour tight-binding model, translation invariance gives the on-site energy and real hopping integral

E_{0} = ⟨ j ∣ H ∣ j ⟩, t = ⟨ j ∣ H ∣ j + 1 ⟩ = ⟨ j + 1∣ H ∣ j ⟩ .

The Bloch state is

∣ k ⟩ = \frac{1}{N} \sum_{j} e^{ikja} ∣ j ⟩ .

Since

H ∣ j ⟩ = E_{0} ∣ j ⟩ + t ∣ j - 1 ⟩ + t ∣ j + 1 ⟩

, shifting the two summation indices gives

H ∣ k ⟩ = [E_{0} + t e^{ika} + t e^{- ika}] ∣ k ⟩, E (k) = E_{0} + 2 t cos (ka) .

The hopping integral is an energy, and its sign sets whether

k = 0

is a minimum or maximum.

E_{0}

is the diagonal matrix element in the crystal Hamiltonian, which can differ from an isolated-atom energy.

The orthogonal tight-binding approximation matters. If raw neighbouring orbitals retain a nonzero orbital overlap

s

in their normalization, projection instead gives a generalized eigenvalue problem and the nonorthogonal nearest-neighbour tight-binding band

E (k) = [E_{0} + 2 h cos (ka)] / [1 + 2 s cos (ka)]

. The stated cosine band follows after orthonormalization and nearest-neighbour truncation, or after neglecting that overlap metric while retaining the hopping it motivates.

c

Solution

A site in a primitive cubic lattice has six nearest neighbours, at

\pm a \hat{x}

\pm a \hat{y}

and

\pm a \hat{z}

. Repeating the Bloch state calculation with equal hopping integrals

t

along all six bonds gives

E (k) = E_{0} + 2 t [cos (k_{x} a) + cos (k_{y} a) + cos (k_{z} a)] .

The cosine sum ranges from

- 3

3

within the first Brillouin zone. Therefore

E_{m i n} = E_{0} - 6∣ t ∣, E_{m a x} = E_{0} + 6∣ t ∣, W = E_{m a x} - E_{m i n} = 12∣ t ∣.

For

t < 0

the minimum is at the zone centre; for

t > 0

the maximum is there.

d

Solution

For

∣ k_{i} a ∣ ≪ 1

, expand each cosine to second order:

E (k) = E_{0} + 6 t - t a^{2} (k_{x}^{2} + k_{y}^{2} + k_{z}^{2}) + O (t a^{4} k^{4}) .

To this order a fixed energy fixes

k_{x}^{2} + k_{y}^{2} + k_{z}^{2}

, so the constant-energy surfaces are spheres. The band effective mass tensor satisfies

(m^{*- 1})_{ij} = \frac{1}{ℏ ^{2}} \frac{\partial ^{2} E}{\partial k _{i} \partial k _{j}} = - \frac{2 t a ^{2}}{ℏ ^{2}} δ_{ij}, m^{*} = - \frac{ℏ ^{2}}{2 t a ^{2}} .

For bonding hopping

t < 0

this is a positive electron mass at the band bottom. If

t > 0

k = 0

is a maximum with negative electronic mass, or positive electron hole mass

ℏ^{2} / (2 t a^{2})

. The spherical shape is a quadratic approximation; cubic anisotropy appears in the fourth-order terms.

e

Solution

Taking

E_{1} > 0

so that the zone centre is a minimum, the effective mass for exponentially decaying tight-binding hopping is

m^{*} (a) = \frac{ℏ ^{2}}{4 E _{1} a ^{2}} e^{a / a_{1}} .

Its logarithmic derivative is

\frac{1}{m ^{*}} \frac{d m ^{*}}{d a} = \frac{1}{a _{1}} - \frac{2}{a} .

The curve decreases for

a < 2 a_{1}

, has a unique minimum at

a = 2 a_{1}

, and increases for

a > 2 a_{1}

m_{m i n}^{*} = \frac{ℏ ^{2} exp ( 2 )}{16 E _{1} a _{1}^{2}} .

The formal limits are an

a^{- 2}

divergence as

a \to 0

and an exponential divergence as

a \to \infty

. The small-separation extrapolation eventually invalidates the localized-orbital tight-binding model; the sketch describes the specified formula.

Figure 1.
Effective mass versus atomic spacing for exponentially decaying nearest-neighbour hopping
. Original plot of the specified tight-binding model, normalized by its minimum. The minimum occurs at $a = 2 a_{1}$ , where the logarithmic derivative of the hopping magnitude balances the explicit $a^{2}$ band-curvature factor. **Details**.

f

Solution

The electronic bandwidth is

W (a) = 24 E_{1} e^{- a / a_{1}}

. With

a

reduced from

a_{1}

0.99 a_{1}

\frac{Δ W}{W} = \frac{W ( 0.99 a _{1} )}{W ( a _{1} )} - 1 = e^{0.01} - 1 ≃ 0.01005.

Thus the energy band widens by approximately 1.005 percent, or 1 percent to first order. Compression increases the orbital coupling.

5

a

Solution

For a degenerate Fermi gas, the Pauli exclusion principle restricts thermally changed occupations to an energy shell of width

k_{B} T

around the Fermi energy. The number of active Electrons is of order

g (E_{F}) k_{B} T \sim N k_{B} T / E_{F}

; each carries an excitation energy of order

k_{B} T

. Thus the thermal energy grows as

N (k_{B} T)^{2} / E_{F}

and the electronic heat capacity is proportional to

T

More precisely, the fixed-particle-number Sommerfeld expansion gives

U (T) - U (0) = \frac{π ^{2}}{6} g (E_{F}) (k_{B} T)^{2} + O (T^{4}), C_{e} = \frac{π ^{2}}{3} g (E_{F}) k_{B}^{2} T + O (T^{3}) .

Here

g

is the total spin-summed density of states for the sample. For a three-dimensional parabolic band,

g (E_{F}) = 3 N / (2 E_{F})

, yielding

C_{e} = A \frac{N k _{B}^{2} T}{E _{F}}, A = \frac{π ^{2}}{2}, k_{B} T ≪ E_{F} .

The linear law concerns the electronic contribution; the low-temperature lattice heat capacity usually adds a

T^{3}

term.

b

Solution

Let

n = N / V

and

c_{e} = C_{e} / V

be the carrier density and electronic heat capacity per unit volume. In isotropic kinetic theory, the angular average

⟨ v_{x}^{2} ⟩ = v_{F}^{2} /3

gives the electronic thermal conductivity

K_{e} = \frac{1}{3} c_{e} v_{F} ℓ_{Q} = \frac{1}{3} c_{e} v_{F}^{2} τ_{Q},

where

ℓ_{Q} = v_{F} τ_{Q}

is the mean free path associated with the heat-current relaxation time. With

v_{F}^{2} = 2 E_{F} / m^{*}

K_{e} = \frac{2 A}{3} \frac{n k _{B}^{2} T τ _{Q}}{m ^{*}} .

The Drude model gives the electrical conductivity

σ = n e^{2} τ_{σ} / m^{*}

, where

τ_{σ}

is the momentum relaxation time. Therefore

\frac{K _{e}}{σ T} = \frac{2 A}{3} (\frac{k _{B}}{e})^{2} \frac{τ _{Q}}{τ _{σ}} .

When the same approximately elastic collisions relax heat and charge currents,

τ_{Q} = τ_{σ}

. Using

A = π^{2} /2

gives the Wiedemann-Franz law

K_{e} = L_{0} σ T, L_{0} = \frac{π ^{2}}{3} (\frac{k _{B}}{e})^{2} ≃ 2.44 \times 1 0^{- 8} W Ω K^{- 2} .

L_{0}

is the Lorenz number. Degenerate electrons move at the Fermi velocity, not the classical thermal velocity; using classical equipartition would give the wrong coefficient.

c

Solution

The Wiedemann-Franz law applies to a normal degenerate metal when the same itinerant Electrons carry heat and charge, scattering is approximately elastic over the thermal shell, and the transport probability varies slowly across the energy interval

k_{B} T

around the Fermi level. The comparison uses the electronic thermal conductivity

K_{e}

, with any lattice contribution removed.

It works particularly well as

T \to 0

when elastic impurity or defect scattering dominates. It also often holds approximately for

Θ_{D} ≪ T ≪ T_{F}

, where thermally populated phonons act as a nearly elastic classical scattering field. Room temperature need not lie above the Debye temperature for every metal.

At intermediate temperatures, inelastic small-angle electron-phonon scattering can relax heat much more efficiently than charge. Then

τ_{Q} \neq = τ_{σ}

and

K_{e} / (σ T)

can be smaller than

L_{0}

. Strong energy dependence of scattering, additional neutral heat carriers, or a transition out of the normal metallic state also invalidates the simple derivation. An observed deviation alone does not identify which assumption failed.

d

Solution

For one conduction electron per atom, let

n_{a}

be the atom density. State counting in a Bloch band gives the Fermi wavevector

k_{F} = (3 π^{2} n_{a})^{1/3}

. The Debye model preserves the three vibrational modes per atom by a spherical Debye wavevector cutoff

q_{D} = (6 π^{2} n_{a})^{1/3}

, so

q_{D} = 2^{1/3} k_{F} ≃ 1.26 k_{F} .

Both wavevectors are of order the inverse atomic spacing. The numerical factor uses the spherical Debye approximation; a real zone boundary and the largest-frequency mode depend on direction and branch. Because

q_{D} < 2 k_{F}

, short-wavelength acoustic phonons can scatter Fermi surface electrons through substantial angles. For nearly elastic scattering within a spherical Fermi surface,

q = 2 k_{F} sin (θ /2), 1 - cos θ = \frac{q ^{2}}{2 k _{F}^{2}} .

Near room temperature, when a substantial fraction of the phonon spectrum is occupied, electron-phonon scattering is often the main source of electrical resistance. For

T ≫ Θ_{D}

, the phonon population and scattering rate are approximately proportional to

T

, giving

ρ \propto T

σ \propto T^{- 1}

and roughly constant

K_{e}

when the Wiedemann-Franz law holds.

For small-angle electron-phonon transport at low temperature, thermally active acoustic modes instead have

q_{T} \sim \frac{k _{B} T}{ℏ c _{s}} = q_{D} \frac{T}{Θ _{D}} ≪ k_{F} .

They predominantly scatter forward. The factor

1 - cos θ

makes such collisions inefficient at relaxing an electrical current, while a phonon energy of order

k_{B} T

can still substantially change an electron's thermal excitation. Heat-current relaxation therefore need not share the charge-current relaxation time. In the clean three-dimensional acoustic-phonon model this gives

τ_{Q}^{- 1} \propto T^{3}

but

τ_{σ}^{- 1} \propto T^{5}

, so the phonon-limited electrical resistivity scales as

T^{5}

and

K_{e} \propto T^{- 2}

. These laws describe the regime before residual elastic scattering takes over; at sufficiently low temperature impurities give nearly constant residual resistivity and

K_{e} \propto T

, restoring the usual Lorenz number.^[ref]

Phonons also transport heat themselves, with

K_{ph} ≃ c_{ph} c_{s} ℓ_{ph} /3

. At low temperature their population decreases and their phonon mean free path can become boundary- or defect-limited. At higher temperature Umklapp scattering can relax the phonon heat current. Such lattice conduction is an additional contribution, not the quantity constrained by the electronic Wiedemann-Franz law.

6

a

Solution

Write

ε_{s} = ϵ_{0} ϵ

and take

e > 0

α > 0

, so the p side is

x < 0

. In the depletion approximation, mobile Electrons and electron holes are negligible compared with the fixed ionized dopants. Their charge density is

ρ (x) = e [n_{D}^{+} (x) - n_{A}^{-} (x)] ≃ e αx .

Applying Poisson equation gives

\frac{d ^{2} V}{d x ^{2}} = - \frac{ρ ( x )}{ε _{s}} = - \frac{e αx}{ϵ _{0} ϵ} .

Outside the depletion region, mobile carriers approximately neutralize the dopants, so

ρ ≃ 0

and

V^{''} ≃ 0

. Full dopant ionization alone does not remove the mobile-carrier charge: the result also requires the depletion approximation.

b

Solution

At equilibrium, majority Electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.

Since electronic band energies shift by

- e V (x)

, both the conduction band and valence band undergo band bending down from p to n by

e V_{b}

at zero applied bias. Their separation remains the same gap. In the nondegenerate thermal equilibrium limit the built-in voltage can also be expressed using neutral-edge densities as

e V_{b} = k_{B} T lo g (n_{n} / n_{p}) = k_{B} T lo g (n_{n} p_{p} / n_{i}^{2})

Take the n-contact potential and its electrochemical potential as reference. Applying

V_{ext}

to the p side changes the total depletion-layer drop to

U = V_{b} - V_{ext}, E_{c, p} - E_{c, n} = E_{v, p} - E_{v, n} = e U .

Thus

V_{ext} < 0

is reverse bias: it increases the barrier and widens the depletion region. The bands have horizontal tangents at the neutral edges in this approximation and a smooth cubic bend within the linearly graded layer.

A biased junction is not in global thermal equilibrium. Its contact electron electrochemical potentials obey

μ_{p} - μ_{n} = - e V_{ext} .

Under reverse bias the p-contact level is higher by

e ∣ V_{ext} ∣

. A single horizontal equilibrium chemical potential cannot describe this state; electron and hole quasi-Fermi levels describe the nonequilibrium populations. The sketch marks the contact levels. Detailed spatial quasi-Fermi profiles require transport and recombination information beyond the electrostatic doping profile.

Figure 1.
Band bending, electric field and capacitance of a linearly graded junction
. Original depletion-model sketches with the p region on the left. The top panels compare equilibrium with reverse bias; under bias only the contact electrochemical levels are specified. The bottom panels show the negative electric field and the normalized positive differential capacitance. Energy and length scales are illustrative. **Details**.

c

Solution

At equilibrium the diffusion currents and drift currents cancel separately for each carrier species. Electron diffusion is from n to p, while the electric field drives electrons from p to n; hole diffusion is from p to n, while hole drift is from n to p. The net current is zero, although these opposing microscopic fluxes are present. In conventional-current signs, both majority diffusion contributions point p to n and the balancing drift contributions point n to p.

Forward bias,

V_{ext} > 0

, reduces the barrier and enhances majority-carrier diffusion and minority carrier injection. The ideal low-injection Shockley diode equation gives

J = J_{s} [e^{e V_{ext} / (k_{B} T)} - 1],

where positive current is p to n. For reverse bias, majority diffusion is suppressed and the field sweeps thermally supplied minority carriers across the junction, giving approximately

J = - J_{s}

. This saturation is limited by carrier supply rather than a current proportional to the field. Depletion-layer generation, recombination and eventual diode reverse breakdown modify the ideal result.

d

Solution

Let the neutral edges be

- x_{p}

and

x_{n}

and take the exterior electric field to vanish. Integrating

d E_{x} / d x = ρ / ε_{s} = e αx / ε_{s}

between the edges gives

x_{n}^{2} = x_{p}^{2}

, so

x_{n} = x_{p} = w /2

. This is also the condition of equal and opposite integrated depletion charges.

A second integration with

E_{x} (w /2) = 0

yields

E_{x} (x) = \frac{e α}{2 ε _{s}} (x^{2} - \frac{w ^{2}}{4}), ∣ x ∣ \leq \frac{w}{2} .

It is negative inside the layer, pointing from n to p, and zero at both edges. Its largest magnitude occurs at the metallurgical junction:

x = 0, ∣ E_{x} ∣_{m a x} = \frac{e α w ^{2}}{8 ε _{s}} .

For reference, choosing

V (- w /2) = 0

gives the cubic electric potential

V (x) = \frac{e α}{2 ε _{s}} (\frac{w ^{2} x}{4} - \frac{x ^{3}}{3} + \frac{w ^{3}}{12}) .

e

Solution

The electric potential difference is the negative integral of the field:

V_{b} - V_{ext} = - \int_{- w /2}^{w /2} E_{x} (x) d x = \frac{e α}{2 ε _{s}} \int_{- w /2}^{w /2} (\frac{w ^{2}}{4} - x^{2}) d x = \frac{e α w ^{3}}{12 ε _{s}} .

Consequently

w (V_{ext}) = [\frac{12 ϵ _{0} ϵ ( V _{b} - V _{ext} )}{e α}]^{1/3}, w (0) = (\frac{12 ϵ _{0} ϵ V _{b}}{e α})^{1/3} .

The linearly graded junction depletion width has this cube-root dependence because of the linearly growing space charge: the charge on one side scales as

w^{2}

, the field as

w^{2}

, and its potential integral as

w^{3}

. The formula assumes

V_{b} - V_{ext} > 0

and that the depletion approximation remains applicable.

f

Solution

The positive charge per unit interface area on the n side and negative charge on the p side are

Q_{n} = \int_{0}^{w /2} e αx d x = \frac{e α w ^{2}}{8}, Q_{p} = - Q_{n} .

Use the signed p-side charge

Q = Q_{p}

because

V_{ext}

is the p-side voltage relative to n. Differentiating the width relation gives

\frac{d w}{d V _{ext}} = - \frac{w}{3 ( V _{b} - V _{ext} )}, C = \frac{d Q _{p}}{d V _{ext}} = \frac{e α w ^{2}}{12 ( V _{b} - V _{ext} )} .

Using part (e), the positive junction capacitance per unit area is

C = \frac{ε _{s}}{w} = [\frac{e α ε _{s}^{2}}{12 ( V _{b} - V _{ext} )}]^{1/3} .

It decreases with increasing reverse-bias magnitude as

(V_{b} + ∣ V_{ext} ∣)^{- 1/3}

. Equivalently,

C^{- 3} = \frac{12 ( V _{b} - V _{ext} )}{e α ε _{s}^{2}},

so a linear

C^{- 3}

versus voltage plot diagnoses this ideal linear grading and can determine

α

For this linearly graded junction capacitance, the charge convention matters: if “stored charge” means the positive magnitude

Q_{n}

, then

d Q_{n} / d V_{ext} = - ε_{s} / w

; the positive measured capacitance is

- d Q_{n} / d V_{ext} = d Q_{n} / d V_{R}

for

V_{R} = - V_{ext}

. The total net depletion charge is zero and is not the stored terminal charge. At substantial forward bias, minority carrier storage and the failure of the sharp depletion approximation prevent extrapolation of the apparent divergence at

V_{ext} = V_{b}

ψ_{n k} (r) = e^{i k \cdot r} u_{n k} (r), u_{n k} (r + R) = u_{n k} (r) .

Here

R

is any lattice translation and

n

labels an energy band. Under that translation the wavefunction gains phase

e^{i k \cdot R}

. The crystal momentum

ℏ k

labels this translation eigenvalue and is defined modulo

ℏ G

for a reciprocal lattice vector

G

The mechanical momentum is different. In zero magnetic field its operator is

- i ℏ\nabla

, which acts as

- i ℏ\nabla ψ_{n k} = ℏ k ψ_{n k} + e^{i k \cdot r} (- i ℏ\nabla u_{n k}) .

A general Bloch state is therefore not a mechanical-momentum eigenstate. Expanding its periodic factor gives plane-wave components with momenta

ℏ (k + G)

, with probabilities determined by their Fourier coefficients. For a local Hamiltonian

p^{2} / (2 m_{e}) + V (r)

, its mean momentum satisfies

⟨ p ⟩ = m_{e} v_{g} = \frac{m _{e}}{ℏ} \nabla_{k} E_{n} (k),

not generally

ℏ k

. The lattice exchanges actual momentum with the electron; crystal-momentum conservation in scattering is instead modulo reciprocal-lattice vectors, as in Umklapp scattering.

2

Solution

Use Thomas-Fermi screening, treating a homogeneous three-dimensional conductor in static linear response. For a positive test charge the screened electric potential and bare Coulomb potential are

V (r) = \frac{Q}{4 π ε _{b} r} e^{- r / λ_{TF}}, V_{bare} (r) = \frac{Q}{4 π ε _{b} r} .

Here

ε_{b}

is the background permittivity before the mobile carriers screen; it is

ϵ_{0}

in the simplest free-electron model. Both curves have the same leading

1/ r

divergence near the origin, while the screened curve lies below the bare one and decays much faster.

A positive potential attracts Electrons. If

g (E_{F})

is their total spin-summed density of states per volume and energy,

δ n = e g (E_{F}) V

and

ρ_{ind} = - e^{2} g (E_{F}) V

. Inserting this charge into Poisson equation gives the Thomas-Fermi screening length

λ_{TF} = \frac{ε _{b}}{e ^{2} g ( E _{F} )} .

A larger density of states lets the conductor rearrange more charge for the same potential change and shortens the screening length. For a parabolic free electron model,

g (E_{F}) = 3 n / (2 E_{F})

, hence

λ_{TF}^{2} = 2 ε_{b} E_{F} / (3 n e^{2})

Figure 1.
Bare and Thomas-Fermi-screened potentials of a positive test charge
. Original dimensionless model curves on the same axes, normalized by $Q / (4 π ε_{b} λ_{TF})$ . The near-origin divergence is clipped for display. The local screening approximation omits microscopic nonlocal effects. **Details**.

The exponential form is the intended local screening model. A microscopic zero-temperature metal also has Friedel oscillations, so the exponential alone is not an exact description at every length scale.

3

Solution

The Moiré pattern sets a superlattice period

L = 50 a = 12.5 nm, G = \frac{2 π}{L} .

For the stipulated scalar one-dimensional model, right- and left-moving states have

E (k) = ℏ v ∣ k ∣

. The

n

th Bragg crossing satisfies

E (k) = E (k - n G)

and hence

k = n G /2

. The Bragg gap ladder for a linear dispersion therefore has centres

E_{n} ≃ \frac{nπ ℏ v}{L}, Δ E ≃ \frac{π ℏ v}{L} = \frac{h v}{2 L} ≃ 0.165 eV \approx 0.17 eV .