Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 1

Question 1

We use the given vector identity, with A = a_{2}, B = a_{3}, and C = a_{3} × a_{1}:

V Ṽ = V | (b_{1} × b_{2}) · b_{3}|
= V (\frac{2π}{V} )^{3} | ((a_{2} × a_{3}) × (a_{3} × a_{1})) · (a_{1} × a_{2}) |
= V (\frac{2π}{V} )^{3} | ((a_{2} · (a_{3} × a_{1})) a_{3} − (a_{3} · (a_{3} × a_{1})) a_{2}) · (a_{1} × a_{2}) |
= V (\frac{2π}{V} )^{3} |V a_{3} · (a_{1} × a_{2}) |
= V (\frac{2π}{V} )^{3} |V · V |
= (2π)^{3}

Here, we used the fact that a_{3} · (a_{3} × a_{1}) = 0.

Question 2

We have two primitive vectors in direct space, a_{i} = (a_{ix}, a_{iy}) (i = 1, 2), and two in reciprocal space, b_{i} = (b_{ix}, b_{iy}) (i = 1, 2), and four equations relating them, b_{i} · a_{j} = 2πδ_{ij} (i, j = 1, 2). These four equations are exactly what we need in order to determine the four unknowns b_{ix}, b_{iy} (i = 1, 2). A little algebra yields

b_{1x} = \frac{2πa_{2y}}{a_{1x}a_{2y} − a_{1y}a_{2x}}
b_{1y} = \frac{2πa_{2x}}{a_{1y}a_{2x} − a_{1x}a_{2y}}
b_{2x} = \frac{2πa_{1y}}{a_{2x}a_{1y} − a_{2y}a_{1x}}
b_{2y} = \frac{2πa_{1x}}{a_{2y}a_{1x} − a_{2x}a_{1y}}

With a regular hexagonal lattice, with nearest neighbour distance a, we have

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

which yields

b_{1x} = \frac{2π}{3a}
b_{1y} = −\frac{2π}{\sqrt{3}a}
b_{2x} = \frac{2π}{3a}
b_{2y} = \frac{2π}{\sqrt{3}a}

In other words, the reciprocal lattice is triangular, with primitive lattice vectors

b_{1} = \frac{2π}{3a} (x̂ − \sqrt{3}ŷ)
b_{2} = \frac{2π}{3a} (x̂ + \sqrt{3}ŷ)

with magnitude |b_{1}| = |b_{2}| = b = 4π/3a. The first Brillouin zone is the area of k–space around the lattice point K = 0 that is closer to the origin than to any other point of the reciprocal lattice, i.e., the so–called Wigner–Seitz cell. With a triangular reciprocal lattice, the 1BZ becomes hexagonal:

The area of the primitive cell in direct space is

A = |a_{1} × a_{2}| = |a_{1x}a_{2y} − a_{1y}a_{2x}| = \frac{3\sqrt{3}a^{2}}{2} ,

and the corresponding area in reciprocal space is

Ã = |b_{1} × b_{2}| = |b_{1x}b_{2y} − b_{1y}b_{2x}| = \frac{8π^{2}}{3\sqrt{3}a^{2}} .

The product of these two is AÃ = 4π^{2}, as it should be.

Question 3

Solution of the Schrödinger equation (SE) in 2D, with periodic boundary conditions (PBC), yields plane–wave solutions

ψ(x, y) ∼ e^{ik·r}

with allowed values of the wave vector,

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

Hence, there is 1 allowed value of k in an area (2π/L)^{2} in k–space, and, because of spin degeneracy g_{s} = 2, there are 2 allowed states within this k–space area. As in 3D, the DOS in k–space is constant,

D_{2}(k) = \frac{2}{(2π/L)^{2}} = \frac{A}{2π^{2}} .

The number of states with absolute value of the wave vector smaller than a given k is

N_{2}(k) = πk^{2} · \frac{A}{2π^{2}} = \frac{Ak^{2}}{2π} ,

and since E(k) = h̄^{2}k^{2}/2m, i.e., k = \sqrt{2mE/h̄^{2}}, the number of states with energy less than E is

N_{2}(E) = \frac{A}{2π} · \frac{2mE}{h̄^{2}} = \frac{mA}{πh̄^{2}}E.

Therefore, the 2D DOS is

D_{2}(E) = \frac{dN_{2}}{dE} = \frac{mA}{πh̄^{2}},

a constant, independent of the energy.

In 1D, a similar way of reasoning yields

D_{1}(k) = \frac{2}{2π/L} = \frac{L}{π}
N_{1}(k) = 2k · \frac{L}{π} = \frac{2Lk}{π}
N_{1}(E) = \frac{2L}{π} · \sqrt{2mE/h̄^{2}} = \frac{2L\sqrt{2m}}{πh̄}\sqrt{E}
D_{1}(E) = \frac{dN_{1}}{dE} = \frac{L\sqrt{2m}}{πh̄}E^{−1/2}

To sum up: The dimensionality of the system determines the energy dependence of the DOS. A qualitative sketch of the DOS in 1, 2, and 3 dimensions is given in the figure below:

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