Physics 7440 Lecture Notes Week 8 Bravais lattices in at ...

Physics 7440 Lecture Notes Week 8

Bravais lattices in at most 2 dimensions

In each of 0-dimensional and 1-dimensional space there is just one type of Bravais lattice.

In two dimensions, there are five Bravais lattices. They are oblique, rectangular, centered
rectangular, hexagonal, and square.

The five fundamental two-dimensional Bravais lattices: 1 oblique, 2 rectangular, 3
centered rectangular, 4 hexagonal, and 5 square

Bravais lattices in 3 dimensions

14 conventional Bravais lattices, shown on the next page.

Copyright 2010 D. Reznik 60

PP C
monoclinic
triclinic

PC I F
orthorhombic

P I
tetragonal

P

rhombohedral

P

hexagonal

P (pcc) I (bcc) F (fcc)

cubic

Copyright 2010 D. Reznik 61

Wigner-Seitz cell

Wigner-Seitz cell is the smallest cell of the crystal structure with which one can generate
the lattice by translations of the type R= n1a1+ n2a2+ n3a3. Here is an example of the
Wigner-Seitz cell of the 2D triangular lattice.

The Wigner–Seitz cell around a lattice point is defined as the locus of points in space that
are closer to that lattice point than to any of the other lattice points.

The cell may be chosen by first picking a lattice point. Then, lines are drawn to all nearby
(closest) lattice points. At the midpoint of each line, another line is drawn normal to each
of the first set of lines.

In the case of a three-dimensional lattice, a perpendicular plane is drawn at the midpoint
of the lines between the lattice points. By using this method, the smallest area (or
volume) is enclosed in this way and is called the Wigner–Seitz primitive cell. All area (or
space) within the lattice will be filled by this type of primitive cell and will leave no gaps.

Non-Bravais lattices can be represented as a Bravais lattice with a basis.

For example a square lattice with atoms positioned both at the corners and in the middle
of the sides, can be represented as a square Bravais lattice with a 3-atom basis.

Conventional Cell

Many structures are typically describes in terms of a so-called conventional cell, which is
often larger than a primitive cell. For example the BCC lattice is usually described in
terms of a cubic unit cell. Such a unit cell fills space without any overlapping when
translated through some subset of the vectors of a Bravais lattice.

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Diffraction from crystals

For the purposes of diffraction, the crystal lattice can be viewed in terms of sets of
parallel planes, which can act in the same way as diffraction gratings when they are hit
with any propagaing wave that can penetrate inside. These waves can be electromagnetic
radiation or electrons, neutrons or other particles, which accoring to quantum mechanics
can propagate as waves.

The diffraction law is just the Bragg’s law: nλ=2dsinθ, where λ is the wavelength of the
incident radiation, d is the spacing between adjacent planes, and θ is the angle between
the incident beam propagation direction and the plane.

Using diffraction experiments one can measure the crystal structure by either scattering
x-rays or neutrons from single crystals or powder samples.

Consider a Fourier component of the incident radiation: A(x)=A0ei(kx-ωt) . Then |k|=2π/λ.

For this wave to de diffracted the Bragg condition must be satisfied: nλ=2dsin(θ), i.e.
|k|=nπ/dsinθ. Using the radiation of fixed wavelength and turning the sample and
changing θ. One could measure d-spacings between all sets of parallel planes, from which
the crystal structure can be deduced unambiguously. In fact it is sufficient to measure
diffracted intensity from a powder sample by changing θ only. Then each Bragg
reflection will appear at 2θ.

Note that the ω is always a function of k: this dispersion relation is often essential in
determining properties of the respective particles (photons, electrons, neutrons, etc.) But
in this case, it is irrelevant, thus as long as the wavelength (or DeBroglie wavelength) is
the same, the diffraction will occur at the same angle. For this reason, for example,
neutrons and x-rays can be used interchangeably to investigate the structure of materials
most of the time.

Let’s see more precisely under which conditions there will be a reflection of a wave from
a crystal lattice. Let A0ei(kx-ωt) be the amplitude the incident radiation and A0’ei(k’x-ωt) be
the amplitude of the scattered radiation. We now see what the difference between k and
k’ is:

k’ -k’
θ
Copyright 2010 D. Reznik
k
Q=k-k63’

|Q|=|k-k’|=2|k|sinθ=2sin(θ)nπ/dsin(θ)=2nπ/d. The direction of Q is normal to the surfaces
of the parallel planes. Note that that |Q| is independent of either λ or θ, i.e. it is an
intrinsic property of the crystal.

Every set of equally-spaced parallel planes can be uniquely defined by a vector of
magnitude 2π/d and direction normal to the planes. These vectors form a lattice in k-
space (called that way because 2π/d has the same units as k), which is called reciprocal
lattice. The nodes of the lattice are called reciprocal lattice vectors, and for this reason k-
space is also called the reciprocal space.

Now let’s write down the general experssion for the elastic scattering cross section of a
plane wave from a crystal lattice using Fermi’s golden rule:

dσ ~ |< kf |U(r) | ki >|2 δ (εi −ε ), where <r|k>= eik⋅r
dΩ f

Then we get:

dσ eiQ⋅rU (r)dr

∫ ∫ ∫dΩ
~ |< k f |U(r) | ki >|2 δ (εi −ε )= eik⋅rU (r)eik '⋅rdr = ei(k−k ')⋅rU (r)dr =
f
r
r r

Notice that the scattering cross section is the function of Q only. Thus it can be written as
dσ .
dQ

The important result is that the scattering cross section for elastic scattering is
proportional to the Fourier transform of the scattering potential. We can repersent U(r) as:

∫U(r)= U geig⋅rd 3g ; and then dσ ~UQ
dQ

Now consider a periodic potential in 3D:
U(r)=U(r+ l1a1+ l2a2+ l3a3), where a1, a2, a3 are the unit vectors of a Bravais lattice.
Note that in a general case, a1, a2, a3 are not orthogonal.

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∫ ∫ ∫U(r)= U geig⋅rd 3g = e e dig⋅r ig⋅(l1a1+l2a2 +l3a3 )
U e d g =ig⋅( r+l1a1+l2a2 +l3a3 ) 3 U 3g ; This means that
g g

Ug = U e ,ig⋅(l1a1+l2a2 +l3a3 ) i.e. Ug= U n1,n2,n3 δ(2πL-g.(l1a1+ l2a2+ l3a3)), where L is some integer.
g

In other words, Ug=0 unless g satisfies: g.(l1a1+ l2a2+ l3a3)=2πN for some integers

l1, l2, l3.

What we need to do now is to figure out how to find these values of g.

Rather than deriving them step-by step, it is easier to write down the solution and then
show that it is a correct one.

g=(n1b1+ n2b2+ n3b3), where b1 = 2π (a2 × a3) b2 = 2π (a3 × a1) b3 = 2π (a1 × a2 ) , and
a1 ⋅ (a2 × a3 ) a1 ⋅ (a2 × a3) a1 ⋅ (a2 × a3 )

b1, b2, b3 are integers.

It is easy to see that if we plug this expression for g.(l1a1+ l2a2+ l3a3)=2πN

This set of vectors forms a lattice in k-space, which is called the reciprocal lattice.

Notation: We shall denote the reciprocal lattice vectors by the their indices: (n1, n2, n3),
and the direct lattice vectors by (l1, l2, l3). (n1, n2, n3). (l1, l2, l3)= n1l1+ n2l2+ n3l3
Notice that ai.bj= 2πδij.
Struture factor and atomic form factor.

∫U(r)= U geig⋅r

∫ ∑Fourrier transform of U: eiQ⋅rU (r)dr = U δn1,n2,n3 (Q − g)n1,n2,n3 , where

r n1n2n3

∫U =Nn1,n2,n3 eiQ⋅rU (r)dr , where N is the total number of unit cells. (see problem set 8),

unit
cell

where N is the total number of unit cells, which approaches infinity.

U =n1,n2,n3 is called a structure factor (Common notation: Sn1,n2,n3 ) if each atom scatters like
a delta function in space, (which is true for the case of nuclear scattering of neutrons). If

there is only 1 atom per unit cell, then Un1,n2,n3 is the same for all n. Otherwise, it is not the

∑same. In this case Un1,n2,n3 = b eign1n2n3 ⋅ri , where sites i all belong to the unit cell and bi is
i

i

proportional to the scattering cross section of a single atom that is sitting at site i. bi
would be the same for the same atoms and different for different ones.

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Un1,n2,n3 = is called a form factor if there is one atom per unit cell and the scattering
potential for each atom is not a delta function in space (as is in the case of x-ray
scattering or magnetic neutron scattering where the scattering is done by the atomic
electrons bound to the nucleus). The form factor is a Fourier transform of the potential
created by a single isolated atom. It would be different for different kinds of atoms. In the
case of a lattice with a basis, the structure factor would contain the atomic form factors
instead of the bis. Note that bi is independent of the Miller indices n1,n2,n3, but the form
factor does depend on the Miller indices.

Properties of reciprocal lattice vectors

1. Each reciprocal lattice vector is normal to a set of lattice planes of the direct lattice.

Proof: Consider some reciprocal lattice vector: (n1, n2, n3) and some direct lattice vector
(l1, l2, l3). Then we can generate a lattice plane consisting of lattice sites (l1’, l2’, l3’) =(l1-
mn3, l2-mn3, l3+m(n1+n2) for all integer values of m.

These sites include (l1, l2, l3) and (l1’, l2’, l3’) . (n1, n2, n3) = (n1, n2, n3). (l1, l2, l3); I.e. they
lie in a plane normal to (n1, n2, n3).

In fact reciprocal lattice vectors are used to index sets of parallel lattice planes using
Miller indices, which are (n1, n2, n3) values for the shortest reciprocal lattice vector
normal to the plane. They are normally denoted as h,k,l.

Intercepts with the crystal axes of a lattice plane are inversely proportional to the Miller
indices of the plane.

An equivalent way to construct Miller indices is as follows:

Find the intercepts on the axes in terms of the lattice constants a1,a2,a3.

Take reciprocals of these numbers and then reduce to three integers having the same
ratio, usually the smallest three integers. The result, (hkl) is called the index of the plane.

2. |g| for the smallest reciprocal lattice vector normal to the set of parallel planes is 2π/d
where d is the spacing of the lattice planes.

Proof: if n is a unit vector normal to the plane with Miller indeces (hkl), the interplanar
spacing is n.a1/h. But n=g/|g|. Thus d=g.a1/h|g|=2π/|g|.

3. The volume of the unit cell of the reciprocal lattice is inversely proportional to the
volume of a unit cell of the direct lattice.

The proof is straightforward and is given on the top of page 12 in Ziman.

4. The direct lattice is the reciprocal of its own reciprocal lattice.

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See Homework 8 problem 1b.

5. The Wigner-Seitz cell of the reciprocal lattice is called the first Brillouin zone (BZ).

In 3D a Bragg plane is a plane is reciprocal space such that the vector normal to this
plane that connects it with the origin is ½ of a reciprocal lattice vector. I.e. it is connected
to its mirror reflection by a full reciprocal lattice vector. In 1D the equivalent of Bragg
planes are points: (2n+1)π/a

An equivalent definition of the 1st Brillouin zone, which also is easy to extend to 2nd, 3rd,
etc, zone is:

The 1st BZ is the set of points in k-space that can be reached from the origin without
crossing any Bragg plane.

Then the 2nd BZ is the set of points in k-space that can be reached from the origin by
crossing only 1 Bragg plane.

And the nnd BZ is the set of points in k-space that can be reached from the origin by
crossing only (n+1) Bragg plane.

Bloch’s theorem in 3D

It is straightforward to extend our derivation of Bloch’s theorem to 3D. One just has to
replace x with the 3D vercor r, have all translations be vectors and have all ks to be
vectors as well. In addition any multiplications and additions of these must be replaced
with dot products and vector additions.

Then Bloch’s theorem for an electron in a potential that is periodic in 3D can be written
down as:

<r|ψ>= eik'⋅ru(r) where u(r)=u(r+ l1a1+ l2a2+ l3a3).

Now electronic dispersions can be “reduced” to the first Brillouin zone by translating
each k-vector by a reciprocal lattice vector that would put the result into the first BZ, and
make the corresponding change in u(r.

One can also extend our calculations of |ψk> and ε(k) (now k is a vector) in a similar
manner. If the periodic potential is treated as weak, one obtains the same expression as in
1D:

∑|ψk>=|k>+ U neαt |k+qn>, but qn now enumerates all reciprocal lattice vectors. It is

n≠0 ε0 −ε0
k+qn k

∑easy to see that this works, since in this notation, we can write U(r)= U eiqn⋅r
qn

n

Copyright 2010 D. Reznik 67

∑Similarly, ε 0 Un2
εk= k + U0+ ε0 −ε0 ; Analogously to 1D, this perturbation expansion
k
n≠0 k +q n

does not work if ε0 ≈ ε 0 n , which is satisfied near the Bragg planes, i.e. boundaries
k k +q

between Brillouin zones. By the same logic, energy gaps should appear at the zone

boundaries that would separate different electronic bands.

k k-g
Γ Γ

g

The above figure illustrates this point. Here if k is on the zone boundary, there exists k-g

that is also on the zone boundary such that |k|=|k-g|, which is he necessary and sufficient

condition for ε0 = ε 0 n .
k k +q

We now have a general prescription for calculating the energies of each Bloch state,
which we can follow up to a point analytically, and which defines the framework for
numerical calculations for more complicated/realistic cases.

But we do not have a prescription for how to fill these states with particles and, calculate,
for example, the Fermi energy as a function of particle density. In order to do that we
need to figure out how to integrate over Bloch states, or, in other works how to figure out
what is the number of states (which can be either occupied or empty) in a finite region of
k-space.

Counting states in 3D

Let’s once again impose periodic boundary conditions. In 3D this is somewhat more
complicated:

<r+L1a1|ψ>=<r|ψ>, <r+L2a2|ψ>=<r|ψ>, <r+L3a3|ψ>=<r|ψ>, where L1, L2, L3, are large
integers. Note that it is no longer possible to make this equivalent to a real geometrical
construct, like a circle for 1D. However, it is very useful a formal trick, since it makes it
easier to work with these boundary conditions than with particles in a box. This is
because solutions for particles in a box are always standing waves, but propagating waves
are possible using periodic boundary conditions.

For a Bloch state of wavevector k, these conditions imply that
eik⋅(L1⋅a1) = eik⋅(L2⋅a2 ) = eik⋅(L3⋅a3) =1, which can only be achieved by having k of the form

k= m1 b1 + m2 b2 m3 b3 , where m1, m2, m3 are integers, and b1, b2, b3 are reciprocal lattice
L1 L2 L3

vectors.

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Thus allowed values of k are obtained by dividing the components of reciprocal lattice
vectors along b1, b2, b3 into L1, L2, L3 parts respectively. In this way the k-vectors sit on a
mesh in reciprocal space. Notice that if we fill the first Brillouin zone with states, then

-L1/2<m1<L1/2, -L2/2<m2<L2/2, -L3/2<m3<L3/2.

The total number of states in this case will be L1* L2* L3=N – the total number of unit
cells in our crystal.

We had the same result in 1D (see homework 7).

Note that the volume of a zone is 8π3/vc, where vc is the volume of a unit cell in real
space given by a1 ⋅ (a2 × a3 ) .

If there are N unit cells, vc=V/N, where V is the volume of the crystal. Thus the volume
of k-space per allowed k-vector is 8π3/Nvc=8π3/V, i.e. there are V/8π3 allowed k-vectors

per unit volume of reciprocal space.

In the limit of a large crystal, we can replace the sum over k-states by an integral:

∑ → V 3 ∫∫∫ d 3k
8π
k

Note that when we calculate properties of solids per unit volume, (e.g. specific heat), V
drops out, and such properties depend on the structure and the composition of the unit
cell only.

Copyright 2010 D. Reznik 69