Institutt for fysikk, NTNU TFY4340 Mesoscopic Physics Spring 2010

Lecture notes, week 3 (January 19)

SOLID STATE PHYSICS UPDATE

Crystal structure A Bravais lattice is formed by all possible lattice vectors

R = n_{1}a_{1} + n_{2}a_{2} + n_{3}a_{3},

where n_{i} = 0, ±1, ±2, …, and a_{1}, a_{2}, and a_{3} are primitive vectors. A primitive unit cell is spanned by the primitive vectors:

The primitive cell has volume

V_{3} = |(a_{1} × a_{2}) · a_{3}|.

(Subscript 3 indicates a 3–dimensional volume.) In a 2–dimensional system (2D), the ”volume” (i.e., area) of the primitive cell is

V_{2} = A = |a_{1} × a_{2}|.

Example: 2D triangular lattice.

Primitive vectors:

a_{1} = ax̂
a_{2} = \frac{a}{2}x̂ + \sqrt{a^{2} − (a/2)^{2}}ŷ = \frac{a}{2}x̂ + \frac{\sqrt{3}a}{2}ŷ

Area of primitive cell:

V_{2} = A = |a_{1} × a_{2}| = |\frac{\sqrt{3}a^{2}}{2} ẑ| = \frac{\sqrt{3}a^{2}}{2} .

Example: 3D face–centered cubic (FCC) lattice. Primitive vectors:

a_{1} = \frac{a}{2} (x̂ + ŷ)
a_{2} = \frac{a}{2} (x̂ + ẑ)
a_{3} = \frac{a}{2} (ŷ + ẑ)

Primitive cell volume:

V_{3} = | (a_{1} × a_{2}) · a_{3}| = \frac{a^{3}}{8} |(−ŷ − ẑ) · (ŷ + ẑ)| = \frac{a^{3}}{8} | − 2| = \frac{a^{3}}{4} .

This is the volume pr lattice point in the FCC Bravais lattice. A Wigner–Seitz cell is the region of space around a given lattice point P which is closer to P than to all the other lattice points. Example: 2D triangular lattice. In other words, the 2D triangular lattice has a hexagonal Wigner–Seitz cell.

The reciprocal lattice of a Bravais lattice consists of all the (wave–)vectors K that correspond to plane waves

e^{iK·r}

with the periodicity of the Bravais lattice, i.e.,

e^{iK·(r+R)} = e^{iK·r},

which implies that K · R = 0, ±2π, ±4π, …. Hence,

K = m_{1}b_{1} + m_{2}b_{2} + m_{3}b_{3},

with integer–valued m_{i} and the following relations between the primitive reciprocal lattice vectors b_{i} and the primitive real space lattice vectors a_{i}:

b_{1} = \frac{2π}{V} (a_{2} × a_{3})
b_{2} = \frac{2π}{V} (a_{3} × a_{1})
b_{3} = \frac{2π}{V} (a_{1} × a_{2})

Here, V = V_{3} = | (a_{1} × a_{2}) · a_{3}|, as above. One also has the relation b_{i} · a_{j} = 2πδ_{ij} between primitive reciprocal and direct lattice vectors. Here, δ_{ij} is the Kronecker delta, i.e., δ_{ij} = 1 if i = j and δ_{ij} = 0 if i ≠ j.

Brillouin zones: The Wigner–Seitz cell in reciprocal space, around the origin K = 0, is by definition the first Brillouin zone (1BZ). [More generally, the n-th BZ corresponds to the region of K–space with K = 0 as the n-th nearest neighbour.]

We will mainly discuss experiments that involve transport of electrons in semiconductors like silicon, Si, and gallium arsenide, GaAs. The crystal structure of Si is the diamond lattice, i.e., two FCC sublattices, one centered at the origin, and the other shifted an amount (a/4, a/4, a/4) with respect to the first one. Gallium arsenide has the zincblende lattice structure, with an FCC Ga lattice centered at (0, 0, 0) and an FCC As lattice centered at (a/4, a/4, a/4). The reciprocal lattice of the FCC lattice is the body–centered cubic (BCC) lattice.

Electrons in solid state crystals

Several factors complicate the quantum–mechanical description of electrons in a crystal:

• the periodic potential, V (r + R) = V (r), due to the ionic lattice

• there are many electrons, interacting with each other (Coulomb repulsion)

• the lattice is not static, but vibrating, even at zero temperature (T = 0) (the ground state of a harmonic oscillator has energy E = h̄ω/2)

• the crystal is not perfect, there are impurities and other lattice defects

• the temperature in a real system is not zero • the electron spin couples to its orbital motion, so–called spin–orbit coupling

• and so on

We start by making the most drastic simplifications and neglect all the complications listed above and assume that we have free, non–interacting electrons moving around in our crystal. I.e., the potential is simply a constant, and we may as well put V = 0. However, we retain the fact that electrons are fermions, so that a given single–particle quantum state may at most be occupied by a single electron. This is the Pauli exclusion principle. A ”single–particle quantum state” is a solution ψ of the (time independent) single–particle Schrödinger equation (SE)

[−\frac{h̄^{2}}{2m}∇^{2} + V (r)] ψ(r) = Eψ(r).

For simplicity, we let our ”constant potential crystal” be a cube with volume V = L^{3}. Concerning boundary conditions, we may choose between (a) vanishing wave function, ψ = 0, at the edges of the cube, or (b) periodic boundary conditions (PBC), ψ(x + L, y, z) = ψ(x, y, z) etc. With PBC, the eigensolutions are plane waves,

ψ(x, y, z) = C e^{ik·r},

with wave vectors

k = k_{x}x̂ + k_{y}ŷ + k_{z}ẑ = \frac{2π}{L} (n_{1}x̂ + n_{2}ŷ + n_{3}ẑ) .

Once more, the n_{i} are integer–valued.

Spin degeneracy

Electrons have spin S, with S = |S| = \sqrt{s(s + 1)}h̄ = \sqrt{3/4}h̄ and S_{z} = m_{s}h̄ = ±h̄/2. Thus, we have 2 possible spin states for each ”orbital state” ψ(r): ξ(+) for a ”spin up” electron and ξ(−) for a ”spin down” electron.

Energy eigenvalus Inserting the plane wave solution into the SE yields the familiar free–electron energy eigenvalues

E(k) = \frac{h̄^{2}k^{2}}{2m} .

The function E(k) is also known as the dispersion relation. If the system is sufficiently small, the energy spectrum is discrete, with a separation between the energy levels proportional to 1/L^{3}. In a macroscopic system, with a large value of L, we have an approximately continuous energy spectrum.

Density of states

Having determined the dispersion relation E(k), we are able to calculate the density of states (DOS), both in k–space and as a function of the energy E. Carrying on with the free–electron model, we see that there is one allowed value of the wave vector k in a cube of volume (2π/L)^{3} in k–space. Taking spin degeneracy into account, this means that there are 2 allowed states in this volume. Hence, the DOS in k–space is

D(k) = \frac{2}{(2π/L)^{3}} = \frac{V}{4π^{3}} .

Within a sphere of radius k, the number of allowed states is then

N(k) = D(k) · V_{k} = \frac{V}{4π^{3}} · \frac{4}{3}πk^{3} = \frac{V}{3π^{2}} k^{3}.

Hence, since k = \sqrt{2mE/h̄^{2}}, the number of allowed states with energy less than E is

N(E) = \frac{V}{3π^{2}} · (\frac{2mE}{h̄^{2}} )^{3/2} .