Phys 412 Solid State Physics

Lecturer: Réka Albert

What is a solid?• A material that keeps its shape

– Can be deformed by stress – Returns to original shape if it is not strained

too much

– Solid structure is defined by the atoms

Gas Liquid Crystalline Solid

What is Solid State Physics?• The body of knowledge about the fundamental

phenomena and classifications of solids • “fundamental phenomenon” = a characteristic

behavior exhibited by classes of solids• Examples:

– Ductile vs. brittle materials– Metals vs. Insulators– Superconductivity - discovered in 1911– Ferromagnetic materials

• The basic understanding of such “fundamental phenomena” has only occurred in the last 70 years - due to quantum mechanics

Phenomena and Principles in SSP• Mechanical

– Structures– Strength

• Thermal– Heat capacity– Heat conduction– Phase transitions

• Electrical– Insulators– Metals– Semiconductors– Superconductors

• Magnetic– Ferromagnetism

• Optical– Reflection, refraction– Colors

• Newton’s Laws

• Maxwell’s Equations

• Thermodynamics and Statistical Mechanics

• Quantum Mechanics– Schrodinger’s Equation– Pauli exclusion

principle

• Order and Symmetry

Course outline• Structures of solids: Kittel 1-5

crystal structurediffraction and reciprocal latticebinding atomic vibrations and elastic constantsthermal properties

• Electronic properties: Kittel 6-7free electron gasenergy bands – metals vs. insulatorssemiconductors (time permitting)

• Additional topics (time permitting)

Solid State Simulations• “bravais”: Crystal structure and x-ray diffraction• “laue”: Diffraction in perfect and imperfect

crystals• “born”: Lattice dynamics in one dimension• “debye”: Lattice dynamics and heat capacity• “drude”: Dynamics of the classical free electron

gas• “ising”: Ising model and ferromagnetism

Structure of Crystals (Kittel Ch. 1)

See many great sites like “Bob’s rock shop” with pictures and crystallography info: http://www.rockhounds.com/rockshop/xtal/index.html

A crystal is a repeated array of atoms.

Crystals • A crystal is a repeated array of atoms (or cells)

• Crystal = Lattice + Basis

CrystalLattice of points(Bravais Lattice) Basis of atoms

Possible crystal symmetries• Translation symmetry

• Point symmetries (rotation, reflection) always present

depend on basis

Characterizing the latticeEach lattice has translational symmetry - the atomic arrangementlooks the same when viewed from the lattice point at r or from thelattice point at

a1 , a2 , a3 - translation vectors (axes), only 2 needed in 2Dwhere u1 , u2 , u3 are arbitrary integers

There are multiple ways of choosing axes. Each choice determines a unit cell; the crystal is the repetition of these unit cells.

Primitive axes:– each point of the lattice can be described as– the parallelepiped defined by them has the smallest volume

The primitive translation vectors determine the primitive cell.There are many ways of choosing primitive axes, but there is always onelattice point (and as many atoms as there are in the basis) per primitive cell.

321 aaarr 321 uuu +++=′

321 aaa 321 uuu ++

Ex. Define a few possible sets of axes for the lattice below.Which are the primitive axes?What is the primitive cell?

Characterizing the basisA basis of atoms is attached to every lattice point, with every basis identicalCoordinates of atoms in the basis

a1 , a2 , a3 - lattice axes x j, yj , zj are between 0 and 1

Description of a crystal:1. What is the lattice?2. What choice of a1 , a2 , a3 do we wish to make?3. What is the basis?

321j aaar jjj zyx ++=

Two Dimensional Crystals

• Infinite number of possible crystals• Finite number of possible crystal types (Bravais lattices)• The entire infinite lattice is specified by the primitive

vectors a1 and a2 (also a3 in 3D)

a1

a2

φ

BasisLattice

Wallpaper patterns as crystals

Find the lattice and basis of these two wallpaper patterns.What symmetries do the patterns have?

Possible Two Dimensional Lattices

• Special angles φ = 90° and 60° lead to special crystal types• In addition to translations, these lattices are invariant under

rotations and/or reflections

• Ex: give a few examples of lattices with φ = 90°or φ = 60°:

a1

a2

φ

Possible Two Dimensional Lattices

These are the only possible special crystal types(Bravais lattices) in two dimensions.

a1

a2

Hexagonal Φ = 60, a1 = a26-fold rotation , reflections

a1

a2

φ

General oblique

a1

a2

Square4-fold rot., reflect.

a1

a2

Rectangular2-fold rot., reflect.

Centered Rectangular2-fold rot., reflect.

a1

a2

Ex.: Close packing of spheresWhat is the lattice corresponding to this arrangement?What symmetries does the lattice have?What axes can be defined for this lattice?What are the primitive axes?How many spheres are there in a cell? Hint: add spherefractions inside the cell.

Spheres inside the cell: 1/6+2/6+1/6+2/6=1.

60°

Crystalline layers with >1 atom basis

• Left - layers in the High Tc superconductors• Right - single layer of carbon graphite or

hexagonal BN (the two atoms are chemically different in BN, not in C )

Honeycomb Lattice(Graphite or BN layer)

a1

a2

CuO2 Square Lattice

a1

a2

Cu

O

Primitive Cell and Wigner-Seitz Cell

Wigner-Seitz Cell -- Unique

a1

a2

One possible Primitive Cell

Wigner Seitz Cell is most compact, highest symmetryprimitive cell possibleIt is defined by the neighboring lattice points: • connect a lattice point to all nearby lattice points• draw the perpendicular bisectors of lines • find enclosed polygon with smallest volume

a1

a2

Hexagonal Φ = 60, a1 = a26-fold rotation , reflections

a1

a2

Square4-fold rot., reflect.

a1

a2

Rectangular2-fold rot., reflect.

Ex. Find the Wigner-Seitz cell of the following lattices

Ex. How many atom fractions are contained in the BCC or FCC cell?

Primitive bcc cell

Regular rhombic dodecahedron

Primitive bcc cell

Ex. Consider a cube with unit side length.

Write down the primitive axes of the bcc cell using the unitvectors z,y,x ˆˆˆ

2/z

2/x

2/x−

2/z−2/y−

2/z2/y2/x +−=1a

2/z2/y2/x −+=2a

2/z2/y2/x −+=3a

Face Centered Cubic Lattice

X

y

z

a1

a3

a2

Wigner-Seitz CellOne Primitive Cell

Lattice Planes - Index System

• Define the plane by the reciprocals 1/n1, 1/n2, 1/n3

• Reduce to three integers with same ratio h,k,l• Plane is defined by h,k,l

n1 a1

Plane through the lattice n1 a1 , n2 a2, n3a3 n’s can be integers or rational fractions

n2 a2

n3 a3

Lattice planes in 2d crystals

• Infinite number of possible planes• Can be through lattice points or between lattice

points.• The lattice planes are independent of the basis.

a1

a2

φ

(22) (11)

(01)

(14)

(02)Basis

Schematic illustrations of lattice planes

• Each set of (h k) defines a family of parallel planes (e.g. planes that have the same intercept in different cells)

• Note that there always is a (h k) plane going through the origin!

a1

a2

φ

(14)

(11)

Equivalent lattice planes

• Distance between (h,k) planes: length of vector starting from cell origin and perpendicular to plane

• Low index planes: more lattice points, more widely spaced• High index planes: less lattice points, more closely spaced

• If the Miller indices contain a common divisor n, only every nth plane contains lattice points

a1

a2

φ

(01)(02)

(01) (02)…

Lattice

(100) plane parallel to yz plane; (110) plane parallel to z axis

Stacking hexagonal 2d layers to make close packed 3d crystal

• Can stack each layer in one of two ways, B or C above A

• Either way, each sphere has 12 equal neighbors• 6 in plane, 3 above, 3 below

A B C

Stacking hexagonal 2d layers to make hexagonal close packed (hcp) 3d crystal

• Stacking sequence: ABABAB • Hexagonal Bravais lattice, basis of 2 atoms

Stacking hexagonal 2d layers to make a face centered cubic (fcc) 3d crystal

• Stacking sequence: ABCABCABC • Leads to an fcc lattice• Basis of 1 atom

Cube

Note spheres ina line parallel to[110] direction incube

Face Centered Cubic Bravais LatticeTwo atoms (one Na, one Cl) per basisIn the conventional cubic lattice there are eight atoms per basis.

Ex. What are these eight atoms’ positions?

X

y

z

NaCl Structure

Simple Cubic Bravais LatticeTwo atoms per basis

X

y

z

CsCl Structure

a3

a2

a1

From http://www.ilpi.com/inorganic/structures/cscl/index.html

Diamond crystal structure

Face Centered Cubic Bravais LatticeTwo identical atoms per basisEx. What are the basis atoms’ positions? How does the diamondstructure differ from the NaCl structure?

Next Time

• Diffraction from crystals

• Reciprocal lattice

• Read Kittel Ch 2

Chapter II: Reciprocal lattice

Read chapter 2 of Kittel

How can we study crystal structure?• Need probe that can penetrate into crystal• X-rays, neutrons, (high energy electrons)

• X-rays discovered by Roentgen in 1895 - instant sensation round the world - view of his wife’s hand

• Neutrons (discovered in 1932) penetrate with almost no interaction with most materials

How can we study crystal structure?• X-rays scatter from the electrons - intensity proportional to

the density n(r) - Mainly the core electrons around the nucleus

• Similarly for high energy electrons• Neutrons scatter from the nuclei

(and electron magnetic moment)

• In all cases the scattering is periodic - the same in each cell of the crystal

• Diffraction is the constructive interference of the scattering from the very large number of cells of the crystal

The crystal can be viewed as made upof planes in different ways

• Low index planes: more lattice points, more widely spaced

• High index planes: less lattice points, more closely spaced

• Bragg model: incident waves are reflected specularlyfrom parallel planes

a1

a2

φ

Lattice

(01)

(14)

Bragg Scattering Law

• Condition for constructive interference:2d sin θ = n λ

• Maximum λ = 2d• Thus only waves with λ of order atomic size can

have Bragg scattering from a crystal

dθθ

d sin θ

λ

Single crystal diffraction

• Crystal must be oriented in all directions in 3D space using “Gonier Spectrometer”

• Observe scattering only at Bragg angles for a fixedwavelength x-ray or neutrons or …..

Rotate both sample and detector about axis

2θ

Alternative approach -energy dispersive diffraction

•For fixed angle θ , vary the energy (i.e., λ) to satisfy Bragg condition

•X-rays over broad energy range now available at synchrotrons

•Diffraction (Bragg scattering) from a single crystallite used toselect X-rays with desired wavelength

electrons

Photons - broadrange of energies

Single crystalmonochrometer

Experiment

Photons with selected energy

Scattered wave amplitude

• The Bragg law gives the condition for the constructive interference of waves scattered from lattice planes.

• We need a deeper analysis to determine the scattering intensity from the basis of atoms, i.e. from the spatial distribution of electrons within each cell.

• We will use the periodicity of the electron number densityn(r) to perform Fourier analysis.

• We end up with a second lattice associated with the crystal – reciprocal lattice

Periodic Functions and Fourier Analysis

• Any periodic function can be expressed in terms of its periodic Fourier components (harmonics).

• Example of density n(x) in 1D crystal:

n(x) = n0 + Σp>0[Cp cos (2π p x/a) + Sp sin (2π p x/a) ]

• Easier expression:n(x) = Σp npexp( i 2π p x/a)

(easier because exp( a + b) = exp( a ) exp( b) )

• Expression for Fourier Components:np = a-1 ∫

0a dx n(x) exp( - i 2π p x/a)

Periodic functions and Fourier Analysis • Define vector position r = (x,y) (2D), r = (x,y,z) (3D).• Fourier analysis

f(r) = ΣG fG exp( i G ⋅ r) where the G’s are vectors, i.e.,

exp( i G ⋅ r) = exp( i (Gx x + Gy y + Gz z) )

• A periodic function satisfies f(r) = f(r + T) where T is any translationT(n1,n2,…) = n1 a1 + n2 a2 (+ n3 a3 in 3D),

where the n’s are integers • Thus

f(r + T) = ΣG fG exp( i G . r) exp( i G ⋅ T) = f( r )⇒ exp( i G ⋅ T) = 1 ⇒ G ⋅ T = 2π x integer

Reciprocal Lattice• The reciprocal lattice is the set of vectors G in Fourier space

that satisfy the requirement G ⋅ T = 2π x integer for any translation T(n1,n2,…) = n1 a1 + n2 a2 (+ n3 a3 in 3D)

• How to find the G’s ??• Define vectors bi by

bi ⋅ aj = 2π δij , where δii = 1, δij = 0 if i ≠ j

• If we define the vectorsG(m1,m2,…) = m1 b1 + m2 b2 (+ m3 b3 in 3D),

where the m’s are integers, then clearly G ⋅ T = 2π x integer for any T

Reciprocal Lattice and Translations• Note: Reciprocal lattice is defined only by the vectors

G(m1,m2,…) = m1 b1 + m2 b2 (+ m3 b3 in 3D),where the m’s are integers and

bi ⋅ aj = 2π δij , where δii = 1, δij = 0 if i ≠ j

• The only information about the actual basis of atoms is in the quantitative values of the Fourier components fG in the Fourier analysis

f(r) = ΣG fG exp( i G ⋅ r)

• Inversion:fG = Vcell

-1∫cell dr f(r) exp(- i G ⋅ r)

Reciprocal Lattice and Fourier Analysis in 1D

• In 1D, b = 2 π /a, b and a parallel• Periodic function f(x):

f(x) = Σp fp exp( i 2π p x/a)= Σp fp exp( i p b), p = integer

• The set of all integers x b are the reciprocal lattice

a

b

Real & Reciprocal lattices in 2 D

• Two lattices associated with crystal lattice• b1 perpendicular to a2 , b2 perpendicular to a1

• Wigner-Seitz cell of reciprocal lattice called the “First Brillouin Zone” or just “Brillouin Zone”

a1

a2b2

b1

b2

b1

Wigner-Seitz Cell

Brillouin Zone

a1

a2

Ex. What is the relationship between b1 and b2, if a1>a2?

Reciprocal Lattice in 3D

• The primitive vectors of the reciprocal lattice are defined by the vectors bi that satisfy

bi ⋅ aj = 2π δij , where δii = 1, δij = 0 if i ≠ j

• How to find the b’s?

• Note: b1 is orthogonal to a2 and a3, etc.• In 3D, this is found by noting that (a2 x a3 ) is orthogonal

to a2 and a3

• Also volume of primitive cell V = |a1 ⋅ (a2 x a3 )|• Then bi = (2π / V ) (aj x ak ),

where (i, j, k) = (1,2,3), (2,3,1) or (3,1,2)

2/z2/y2/x +−=1a

2/z2/y2/x −+=2a

222 /z/y/x ++−=3a

kji aab ×=Vπ2

Ex. Write the primitive vectorsof the reciprocal lattice in termsof z,y,x ˆˆˆ

Real and reciprocal lattice(recall Bravais exercises)

• the reciprocal vector G= h b1 + k b2 + l b3 is perpendicular to the real lattice plane with index (h k l)

• the distance between two consecutive (h k l) planes is

• See also Problem 2.1 in Kittel

Gπndhkl

2=

Scattering and Fourier Analysis

• The in and out waves have the form: exp( i kin. r - i ωt) and exp( i kout. r - i ωt)

• If the in wave drives the electron density, which then radiates waves, the outgoing amplitude is proportional to:

F= ∫space dr n(r) exp(i (kin - kout )⋅ r)

d

λkin

kout

θ

Scattering and Fourier Analysis

• Define ∆k = kout - kint• Then we know from Fourier analysis that

F = ∫space dr n(r) exp(- i ∆k . r) = N cell V cell nG

only if ∆k = G, where G = recip. lat. vector• Otherwise integral vanishes ⇒ no diffraction• nG = V cell

-1∫cell dr n(r) exp(- i G ⋅ r)

The set of reciprocal lattice vectors determines thepossible x-ray reflections

d

λ∆k = Gkin

kout

θ

Elastic Scattering

• For elastic scattering (energy the same for in and out waves)

| kin | = | kout |, or kin2 = kout

2 = ( kin + G)2

• Then one arrives at the condition for diffraction: 2 | kin. G | = G2

d

λ∆k = Gkin

kout

θ

Ewald Construction

kout = kin + G| 2 kin⋅ G | =2 | kin | | G | cos (90º + θ)= 2 | kin | | G | sin θ

lπ2=• ∆ka3

kin

koutG

2θ 90º−θ

• Laue equations:

kπ2=• ∆ka2

hπ2=• ∆ka1

Equivalent to Bragg Condition

• From last slide, since G2 = | G |2 :

| G | = 2 | kin | sin θ• But | kin | = 2π/λ, and | G | = n (2π/d), where d =

spacing between planes (see Kittel prob. 2.1)• ⇒ Bragg condition 2d sin θ = n λ

kin

koutG

2θ

Geometric Construction of Diffraction Conditions

• Consequence of condition | 2 kin ⋅ G | = G2

• | kin ⋅ G/2 | = (G/2)2

• The vector kin (also kout) lies along the perpendicular bisecting plane of a G vector

• One example is shown

b2

kin

b1

kout

G

Diffraction and the Brillouin Zone

• Brillouin Zone formed byperpendicular bisectors of G vectors

• Consequence:No diffraction for any kinside the first Brillouin Zone

• Special role of Brillouin Zone (Wigner-Seitz cell of reciprocal lattice) as opposed to any other primitive cell

b2

kin

Brillouin Zone

b1

kout

G

Comparison of diffraction from different lattices

• The Bragg condition can also be written | G | = 2 | kin | sin θ

⇒ sin θ = (λ /4π) | G | • Thus the ratios of the sines of the angles for

diffraction are given by:sin θ1 / sin θ2 = | G1 | / | G2 |

• Each type of lattice has characteristic ratios the positions of diffraction peaks as a function of sin θ

• Simple scaling with λ

Experimental Powder Pattern

• Diffraction peaks at angles satisfying the Bragg condition

Differences for imperfect powderaverages

Reciprocal Lattice units

http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/teaching.html

Fourier analysis of the basis • The intensity of the diffraction at each G is

proportional to the square of the scattering amplitude

F = N ∫cell dr n(r) exp(- i G . r)= NSG

• SG – structure factor• Regard the crystal density n(r) as a sum of atomic-

like densities natom (r - Ri), centered at point Rin(r) = ∑ all i natom i ( r - Ri)

• Then also SG = ∑ i in cell ∫space dr natom i (r - Ri) exp(- i G ⋅r)

Cell

One atom per cell and Form Factor• Then one can set Ri = 0 and SG is the Fourier

transform of one atom densityf = ∫space dr natom (r) exp(- i G ⋅ r)

• Called Form Factor• In the limit of point-like atoms f=Z

natom (r) |r|

|G|

f Values of |G| for aparticular crystal

More than one atom per cell

• SG = ∑ i in cell ∫space dr natom i ( r - Ri) exp(- i G ⋅ r)

= ∑ i in cell exp(- i G ⋅ Ri) ∫space dr natom i ( r - Ri) exp(- i G ⋅ (r - Ri) )

= ∑ i in cell exp(- i G ⋅ Ri) ∫space dr natom i ( r) exp(- i G ⋅ r)

= ∑ i in cell exp(- i G ⋅ Ri) fG atom i

• Interpretation: Form factor fG atom i and phase factorexp(- i G . Ri) for each atom in unit cell

Structure factor and atomic form factor

• The amplitude of the scattered electromagnetic wave is

• The structure factor of the base is

• The atomic form factor

GG NSF =

∑ ⋅−=cell in i

i atomGG )iexp(fS iRG

)iexp()r(n dVffcell unit

jj atom

Gj rG ⋅−== ∫

“Pure” Structure factor

• Often the basis contains more than one atom that is same element, e.g., diamond structure

• Then fG atom i = fG atom is the same andSG = ∑ i in cell exp(- i G ⋅ Ri) SG

atom i

= fG atom ∑ i in cell exp(- i G ⋅ Ri)

• Define “pure” structure factorS0

G = (1/n) ∑ i in cell exp(- i G ⋅ Ri)

where n = number of atoms in cell

• Then SG = n S0G fG atom

Body Centered Cubic viewed as Simple Cubic with 2 points per cell

S0G = (1/2) ∑ i =1,2 exp(- i G ⋅ Ri)

= (1/2) ( 1 + exp(- i G ⋅ R2) = (1/2) exp(- i G ⋅ R2/2) [exp( i G ⋅ R2/2) + exp(- i G ⋅ R2/2) ] = exp(- i G ⋅ R2/2) cos ( G ⋅ R2/2)

Result: If G = (v1 v2 v3) 2π/a|S0

G | = 1 if sum of integers is even| S0

G | = 0 if sum is odd

Same as we found before! FCC reciprocal lattice

aa1

a3

a2

Points at R1 = (0,0,0)R2 = (1,1,1) a/2

Face Centered Cubic viewed as Simple Cubic with 4 points per cell

aa1

a2

a3

Points at (0,0,0) ; (1,1,0) a/2 ; (1,0,1) a/2 ; (0,1,1) a/2

S0G = (1/4) ∑ i =1,4 exp(- i G . Ri)

Result:

If G = (v1 v2 v3) 2π/athen

S0G = 1 if all integers

are odd or all are even

S0G = 0 otherwise

Same as we found before! BCC reciprocal lattice

Structure factor for diamond

• Ex: diamond structureS0

G = (1/2) ∑ i =1,2 exp(- i G . Ri)

• R1 = + (1/8, 1/8, 1/8)aR2 = - (1/8, 1/8, 1/8)a

Binding in Crystals (Kittel Ch. 3)

A crystal is a repeated array of atoms

Why do they form?

What are characteristic bonding mechanisms?

Do particular mechanisms lead to particular types of crystal structures?

Binding of atoms to form crystals

• Binding is due to the electrons

Ultimate description is quantum mechanicsQuantum states of electrons change as atoms are

brought together • Leads to solid crystal structures of the nuclei

• Can understand basic bonding mechanics from simple quantum arguments

Full quantitative understanding now possible - more later in course

Basic types of binding

• Van der Waals

• Ionic

• Metallic

• Covalent

• (Hydrogen)

Closed-Shell Binding

Metallic BindingCovalent Binding

Ionic Binding

Van der Waals Bonding• Attraction because electrons can interact and be correlated

even if they are on well-separated atoms

• Consider closed shell “inert”that does not form strong chemical bonds

• Isolated closed shell atom - electron distributedsymmetrically around the atom

• What happens if two atoms come together?

+

-

Van der Waals Bonding• What happens if two closed shell atoms come together?

• Electrons on one atom are attracted to the other nucleus, but repelled by the other electrons

• Energy reduced - a net attraction - because the electrons can correlate to reduce the repulsion

+

-

+

-

+- +-

Van der Waals Bonding• Quantum Effect: Electron on each atom is like a fluctuating

dipole - uncertainty principle

• Dipole on atom 1 creates electric field E on atom 2 proportional to 1/R3

• E generates dipole D on atom 2: D = α E where α = polarizability

• The interaction of the two dipoles is proportional to

• Always attractive

+

-

+

- R

6R1~E D

Rare Gas Solids

• Attractive energy ~ • The analysis breaks down at short distance where the

wavefunctions overlapShort distance repulsion - Due to exclusion principle

• Final forms for interaction between two atoms

(Lennard-Jones)

(exponential)

A, B, ρ empirical parameters

6R1

126~)(RB

RARE +−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

ρRexp B

RA~)R(E 6

Total Energy of Crystal

Distance Between Atoms

Ene

rgie

s of C

ryst

alThe general shape applies for any type of binding

Equilibrium Lattice Constant• Ecrystal= N Ecell and Vcrystal = N Vcell

• Ecell is given by all pairwise interactions of an “origin” atom

where R is the nearest neighbor distance, ρi is the distance to atom i in units of R, and

• Also

dimensionless sums • Minimum is for dE/dR = 0

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∑ ∑

i i ii RRNRE

612

2)(ρσ

ρσε

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= ∑ ∑

i i ii RRNRE

661212 112)(ρ

σρ

σε

B ,A ≡≡ 126 44 εσεσ

Rare Gas Solids

• Atoms nearly spherical

• Short-range non-directional attraction and repulsion

⇒ Close packed structures HCP or FCC• For the fcc structure

in very good agreement with the nearest-neighbor distance in Ne, Ar, Kr, Xe

4539214113188121612

. .i ii i

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∑∑ ρρ

( ) 0910 00 .RRR

dRdE

=⇒==σ

Ionic Solids• Much stronger binding than Van der Waals attractive

energy ~ 1/R

• 1. Pay energy to form ions

• 2. Gain energy to bring ions together.

• Is there a net attraction?

Na Cl Na+ Cl-

Na+ Cl- Na+Cl-

Na+ Cl-Na+ Cl-

Na+ Cl-Na+ Cl-

cohesive energy-

ionization energy of Na+ - electron affinity of Cl-

Ionic Solids• Attractive (electrostatic) interaction ~ • attraction for opposite charge, repulsion for same charge• long range• Result: Attractive energy defined to be - αq2/R, • α is the Madelung constant (depends on structure) • q= charge, R = nearest neigh. dist. in crystal

• Repulsion similar to closed shell systems (exponential works best)

• Final form Ecell(R) = - αq2/R + zλ exp(-R/ρ)

• (z = number of nearest neighbors, λ = parameter)

ijRq2±

Ionic Solids

• Madelung constantα

• sum over each ion i , the sign depends on the charge of I• If reference ion negative, + will apply to positive ions• Care must be taken for the series to converge.

• General Method: Ewald sum given in Kittel appendix• Convergent sums can be found by summing over neutral

shells of neighbors

Values of αNaCl structure 1.748CsCl struc. (bcc) 1.763ZnS structure 1.638

ii ρα 1∑±=

• Ex. Calculate the Madelung constant of a 1D line of ions of alternating signs with ion spacing R. The reference ion is negative

• Hint: ln (1+x) = x-x2/2+x3/3-x4/4

ii ρα 1∑±=

Face Centered Cubic Bravais Lattice

NaCl Structure

Favored for ionic crystals with large size differenceClose packed negative ions with small positive ions

Simple Cubic Bravais Lattice

X

y

z

CsCl Structure

a3

a2

a1

From http://www.ilpi.com/inorganic/structures/cscl/index.html

Favored for ionic crystals with small size difference

Covalent crystalsCovalent bond

- usually formed from two electrons, one from each atom - the spins of the atoms are antiparallel- electrons partly localized in the region between the twoatoms

• The covalent bond has strong directional properties, so thecrystals formed by covalent bonding tend to be less packed.• The strength of the covalent bond can be comparable to

ionic bonds• There is a continuous range of crystals between the ionic

and covalent limits.

ZnS and Diamond structure

ZnS Structure with Face Centered Cubic Bravais Lattice

C, Si, Ge form diamond structure with only one type of atom

• Favored if there isstrong directional covalent bonding

• Each atom has 4neighbors in tetrahedron

• Explained by simple bonding pictures andfull electronic calculations

(110) plane in ZnS crystalzig-zag Zn-S chains of atoms

(diamond if the two atoms are the same)

X

y

z

(110) plane in diamond structure crystal

Calculated valence electron density in a (110) plane in a Si crystal

(Cover of Physics Today, 1970)

Metallic binding

• Tends to be non-directional because electrons are spread out

• The valence electrons of the atom are free to move• Typically leads to close packed structures• See Kittel Table 3 - almost all metals are FCC,

HCP, or BCC

A B C

Hydrogen Bonds• H is a special case• If it is ionized it is just a single proton (unlike all other atoms

in the periodic table)

• A proton can always be attracted to regions of high electron density - i.e., it can cause extra binding because it attracts electrons

Example: Water

• (Does not happen with other atoms because of the repulsion of the core electrons)

O-2 O-2

Proton attracting secondwater molecule

Atomic and Ionic Radii

• Atoms and ions have typical sizes• Governed by cores which are filled shells and do not change

much in different solids

• Somewhat arbitrary, but chosen so that sum of radii is nearest neighbor distance

• Tables in KittelNa+ Cl-

Summary: binding of crystals

• Four primary types of binding: van der Waals, ionic, covalent, metallic

• Typical structures:– Close packed for rare gases, metals– Simple two-atom-per-cell structures with large Madelung

constants for ionic crystals– Open structures with few neighbors, directional bonds for

covalent binding

Phonons I - Crystal Vibrations(Kittel Ch. 4)

• Positions of atoms in their perfect lattice positions are given by:R0(n1, n2, n3) = n1

0 x + n20 y + n3

0 z

For simplicity here we consider only one atom per cell and assume an orthogonal coordinate system

For convenience let ni = (ni10, ni2

0, ni30 ) denote atom i which has

position R0i

• The displacement of atom i can be written

∆Ri = ui x + vi y + wi z=(n1i- n1i0) x + (n2i- n2i

0) y+ (n3i- n3i0) z

Displacements of Atoms

z

x

y

Energy & Force due to Displacements• The energy of the crystal changes if the atoms are displaced. • The change in energy can be written as a function of the positions

of all the atoms:E(R1, R2, R3, …)=E(R1

0 +∆R1, R20 +∆R2, ..)

• There are no linear terms if we expand about the equilibrium positions – equilibrium defined by dE/d R (R=R0 )=0

• To lowest order in the displacements the energy is quadratic -Hooke’s law - harmonic limit

E = E0 + (1/2) Σi j ∆Ri . Di j .∆Rj + ….

Energy & Force due to Displacements• The general expression for force on atom s is

Fs = - dE/d Rs

• From the harmonic expression the force is given byFs = - Σ j Ds j .∆Rj

• The D’s are called force constants - the ratio of force on atom s to displacement of atom j - the generalization of the force constant of a spring

• There are no forces at the equilibrium positions.• The force is due to the displacement of atoms i and the lowest

order terms are linear in the displacements• Note that Ds s exists and its sign is negative! • What matters is the distance between Ri and Rs

Linear chain• Consider atoms in a line restricted to move along the line

• Fs = - Σ j Ds j . uj

• Consider the case of only nearest neighbor interactions.• Fs = - (Ds s-1. us-1+ Ds s. us +Ds s-1. us-1)

• Or, in analogy with elastic springs, assume that force depends on the relative displacements

• Fs = - Σi C (us - us+i)

• Fs = -C [(us - us+1) + [(us - us-1) ]= C [ us+1 + us-1 - 2 us]

i ui=∆Ri

a

Oscillations of linear chain

• Newton’s Law:M d2 us / dt2 = Fs = C [ us+1 + us-1 - 2 us]

• Time dependence: Let us(t) = us exp(-iωt) ) (also sin or cos is correct but not as elegant) Then

M ω2 us = C [ us+1 + us-1 - 2 us]

• How to solve? Looks complicated - an infinite number of coupled oscillators!

sa us

a

Oscillations of linear chain

• Since the equation is the same at each s, the solution must have the same form at each s differing only by a phase factor. This is most easily written

us = u exp(ik (s a) )• Then

M ω2 u = C [exp(ik a) + exp(- ik a) - 2 ] uor

ω2 = (C/ M ) [2 cos(ka) - 2]

us

New representation

Oscillations of linear chain

• A more convenient form isω2 = ( C / M ) [2 cos(ka) - 2]

= 4 ( C / M ) sin2(ka/2) (using cos(x) = cos2 (x/2) - sin2(x/2) = 1 - 2 sin2(x/2))

• Finally: ω = 2 ( C / M ) 1/2 | sin (ka/2) |

ui

Oscillations of a linear chain• We have solved the infinite set of coupled oscillators!

• The solution is an infinite set of independent oscillators, eachlabeled by k (wavevector) and having a frequency

ωk = 2 ( C / M ) 1/2 | sin (ka/2) |

• The relation ωk as a function of k is called the dispersion curve

0 2π/aπ/a

ωk

Brillouin Zone• Consider k ranging over all reciprocal space.

The expression for ωk is periodic

ωk = 2 (C/ M ) ½ | sin (ka/2)|

0 2π/aπ/a

ωk

-2π/a −π/a

Brillouin Zone

• All the information is in the first Brillouin Zone - the rest is repeated with periodicity 2π/a - that is, the frequencies are the same for ωk and ωk+G where G is any reciprocal lattice vector G = integer times 2π/a

• What does this mean?

Meaning of periodicity in reciprocal space

• In fact the motion of atoms with wavevector k is identical to the motion with wavevector k + G

• All independent vibrations are described by k inside BZ

sin (ka/2) with k ~ 2π/3 sin ( (k + 2π/a) a/2)

ui

Group velocity of vibration wave• The wave us = u exp(ik (s a) - iω t) is a traveling wave• Phase velocity vφ = ω / k • Group velocity vk = d ωk / dk = slope of ωk vs k

ωk = 2 ( C / M ) 1/2 sin (ka/2) so

vk = a ( C / M ) 1/2 cos (ka/2)

0-π/a π/a

ωk

vk = v sound

vk = 0 at BZ boundary

What is significance of zero Group velocity at BZ Boundary?

• Fundamentally different from elastic wave in a continuum• Since ωk is periodic in k it must have vk = d ωk / dk = 0

somewhere!• Occurs at BZ boundary because ωk must be symmetric about the

points on the boundary

0-π/a π/a

ωk

vk = 0 at BZ boundary

What is significance of zero group velocity at BZ Boundary?

• Example of Bragg Diffraction!• Any wave (vibrations or other waves) is diffracted if k is on a BZ

boundary• us = u exp(ik (s a) ) = u exp (±isπ) = u(-1)s

• Leads to standing wave with group velocity = 0

Meaning of periodicity in reciprocal space -- II

• This is a general result valid in all crystals in all dimensions• The vibrations are an example of excitations. The atoms are not in

their lowest energy positions but are vibrating. • The excitations are labeled by a wavevector k and are periodic

functions of k in reciprocal space. • All the excitations are counted if one considers only k inside the

Brillouin zone (BZ). The excitations for k outside the BZ are identical to those inside and are not independent excitations.

Diffraction and the Brillouin Zone

• Brillouin Zone formed byperpendicular bisectors of G vectors

• Special Role of Brillouin Zone (Wigner-Seitz cell of recip. lat.) as opposed to any other primitive cell

• No diffraction for any kinside the first Brillouin Zone

• Now we see that there are no independent excitationsoutside of the first Brilluin Zone

b2

k2

Brillouin Zone

b1

k1 G

Sound Velocity• In the long wavelength (small k) limit the atomic

vibration wave us = u exp(ik (s a) - iω t) is an elastic wave

• Atoms act like a continuum for ka << 1• ωk = ( C / M ) 1/2 ka• Sound velocity

vsound = a ( C / M ) 1/2

0-π/a π/a

ωk

vk = v sound

• N independent oscillators, each labeled by k (wavevector) and having a frequency

ωk = 2 ( C / M ) 1/2 | sin (ka/2) |• Leading to a wave us = u cos (ksa - ω t) • If end atoms are fixed at us = 0, possible wavelenghts

k=πn/(N-1)a, N values <=π/a• If periodic boundary conditions uN+s = us

k=+-2πn/Na, N values<= π/aThese discrete choices for waves are called the normal modes ofcrystal excitations. The normal modes serve as a basis for describing arbitrarilycomplex excitations.

Normal modes of a finite set of oscillators

Oscillations in higher dimensions

• For k in x direction each atom in the planes perpendicular to x moves the same:

us = u exp(ik (s a) - iω t) • For motion in x direction, same as linear chain

ω = 2 ( C / M ) 1/2 | sin (ka/2) | • longitudinal wave

ui

a

Oscillations in higher dimensions

• Transverse motion: k in x direction; motion vs in y directionvs = v exp(ik (s a) - iω t)

• Central forces give no restoring force! Unstable!• Need other forces - non-central or second neighbor

vi

a

Oscillations in higher dimensions

• Transverse motion: k in x direction; motion vs in y directionvs = v exp(ik (s a) - iω t)

• Second neighbor forcesω2 = (1/2)( C / M ) [4 cos(ka) - 4]

• The end result is the same!

vi

a

4 neighbors

Geometric factor = cos2(p/4)

Two atoms per cell - Linear chain• To illustrate the effect of having two different atoms per cell,

consider the simplest case atoms in a line with nearest neighbor forces only

• Now we must calculate force and acceleration of each of the atoms in the cell

Fs1 = C [ us-1

2 + us2 - 2 us

1] = M1 d2 us1 / dt2

andFs

2 = C[ us+11 + us

1 - 2 us2] = M2 d2 us

2 / dt2

Cell s us1

a

us2

Note subscripts

Oscillations with two atoms per cell• Since the equation is the same for each cell s, the solution must

have the same form at each s differing only by a phase factor. This is most easily written

us1 = u1 exp(ik (s a) - iω t )

us2 = u2 exp(ik (s a) - iω t )

• Inserting in Newton’s equations gives the coupled equations-M1 ω2 u1 = C[(exp(-ik a) + 1) u2 - 2 u1]

and -M2 ω2 u2 = C [(exp( ik a) + 1) u1 - 2 u2]

2 C- M1 ω2 - C(exp(-ik a) + 1)

- C(exp( ik a) + 1) 2 C- M2 ω2= 0

Oscillations with two atoms per cell

• Exercise: Find the simplest form of the equation connecting ωand k

• Use cos(x) = cos2 (x/2) - sin2(x/2) = 1 - 2 sin2(x/2))

• How many dispersion relations (branches) does this correspond to?

= 02 C- M1 ω2 - C(exp(-ik a) + 1)

- C(exp( ik a) + 1) 2 C- M2 ω2

Oscillations with two atoms per cell• Solution

0 2π/aπ/a

ωk“Acoustic”

“Optic”

“Gap” frequencies at whichno vibrations can occur

Oscillations with two atoms per cell

• Limits:• k ~ 0

Acoustic: ω2 = (1/2) (C/ (M1 + M2) ) k2 a2

Optic: ω2 = 2 C[(1 / M1 ) + (1/M2) ] = 2 C/µ• k = π/a

Acoustic: ω2 = 2 C/ Mlarge Optic: ω2 = 2 C/ Msmall

0 2π/aπ/a

ωk“Acoustic”

“Optic”“Gap” frequencies at which

no vibrations can occur

Acoustic -Total Mass

Optic -Reduced Mass

Modes for k near 0

• Optic at k = 0 - opposed motion - larger displacement of smaller mass

us1

a

us2

• Acoustic at k near 0 - motion of cell as a whole

us1

a

us2

Modes for k at BZ boundary

• Optic at k = π/a - motion of smaller massus

1

a

us2= 0

• Each type of atom moves in opposite directions in adjacent cells• Leads to two modes, each with only one type of atom moving • Acoustic at k = π/a - motion of larger mass

us1= 0

a

us2

Atom 2 does not movebecause there are no forces on it!

Oscillations in 3 dimension with N atoms per cell

• Result

0 2π/aπ/a

ωk

3 Acoustic modesEach has ω ~ k at small k

3 (N -1) Optic Modes

Quantization of Vibration waves• Each independent harmonic oscillator has quantized energies:

en = (n + 1/2) hν = (n + 1/2) hω• We can use this here because we have shown that vibrations in a

crystal are independent waves, each labeled by k (and index for the type of mode - 3N indices in a 3 dimen. crystal with N atoms per cell)

• Since the energy of an oscillator is 1/2 kinetic and 1/2 potential, the mean square displacement is given by(1/2) M ω2 u2 = (1/2) (n + 1/2) hωwhere M and u are appropriate to the particular mode(e.g. total mass for acoustic modes, reduced mass for optic modes , ….)

Quantization of Vibration waves• Quanta are called phonons• Each phonon carries energy hω• For each independent oscillator (i.e., for each independent wave in

a crystal), there can be any integer number of phonons • These can be viewed as particles• They can be detected experimentally as creation or destruction of

quantized particles• Later we will see they can transport energy just like a gas of

ordinary particles (like molecules in a gas).

Inelastic Scattering and Fourier Analysis

• The in and out waves have the form: exp( i kin. r - i ωint) and exp( i kout. r - i ωoutt)

• For elastic scattering we found that diffraction occurs only for kin - kout = G

• For inelastic scattering the lattice planes are vibrating and the phonon supplies wavevectorkphonon and frequency ωphonon

d

λkin

kout

Inelastic Scattering and Fourier Analysis

• Result:• Inelastic diffraction occurs for

kin - kout = G ± kphononωin - ωout = ± ωphonon or Εn - Εout = ± hωphonon

kin ωin koutωout

kphononωphonon

Quantum Mechanics

Experimental Measurements of Dispersion Curves

• Dispersion curves ω as a function of k are measured by inelastic diffraction

• If the atoms are vibrating then diffraction can occur with energy loss or gain by scattering particle

• In principle, can use any particle - neutrons from a reactor, X-rays from a synchrotron, He atoms which scatter from surfaces, …...

Experimental Measurements of Dispersion Curves

• Neutrons are most useful for vibrationsFor λ ~ atomic size, energies ~ vibration energiesBUT requires very large crystals (weak scattering)

• X-ray - only recently has it been possible to have enough resolution (meV resolution with KeV X-rays!)

• “Triple Axis” - rotation of sample and two monochrometers

Neutrons or X-rays with broad range of energies Single crystal

monchrometer

Sample

selected energy in

Single crystalmonchrometer

Detector

selected energy out

Experimental Measurements of Dispersion Curves

• Alternate approach for NeutronsUse neutrons from a sudden burst, e.g., at the new “spallation”source being built at Oak Ridge

• Measure in and out energies by “time of flight”

Burst of neutrons at measuredtime (broad range of energies)

Sample

Mechanical chopperselects velocity, i.e.,energy in

Detector

Timing at detectorselects energy out

More on Phonons as Particles• Quanta are called phonons, each with energy hω• k can be interpreted as “momentum”• What does this mean?

NOT really momentum - a phonon does not change the total momentum of the crystalBut k is “conserved” almost like real momentum - when a phonon is scattered it transfers “k” plus any reciprocal lattice vector, i.e.,

∑ kbefore = ∑ kafter + G• Example : scattering of particles

kin = kout + G ± kphononwhere + means a phonon is created, - means a phonon is destroyed

Summary• Normal modes of harmonic crystal:

Independent oscillators labeled by wavevector k and having frequency ωk

• The relation ωk as a function of k is called a dispersion curve - 3N curves for N atoms/cell in 3 dimensions

• Quantized energies (n + 1/2) h ωk

• Can be viewed as particles that can be created or destroyed - each carries energy and “momentum”

• “Momentum” conserved modulo any G vector • Measured directly by inelastic diffraction - difference in in

and out energies is the quantized phonon energy• Neutrons, X-rays, …..

Phonons II - Thermal Properties(Kittel Ch. 5)

Hea

t Cap

acity

C

T

T3

Approachesclassical limit

3 N kB

Heat capacity• Heat capacity is the measure of how much energy it takes to

raise the temperature of a unit mass of an object a certain amount.

• Two heat capacities: constant volume, CV , and constant pressure, CP . For a gas CP > CV . For a solid CP ≈ CV .

• The contribution of the phonons (lattice vibrations) to the heat capacity of a crystal is called the lattice heat capacity.

• Classical result: C ≈ 3NNcell kB , where N is the number of atoms in a unit cell and Ncell is the number of cells in the crystal.

Lattice heat capacity

• The heat that goes into a solid to raise its temperature shows up as internal vibrational energy U (phonons).

• Determine U(T), then we can calculate CV = (∂U/∂T)V .

• In calculating U we need to consider the following: – (1) what is the average energy of each phonon; – (2) what is the average number of phonons existing at any T for

each type (mode) of vibration; – (3) how many different types (modes) of vibration are there.

Quantization of vibration waves• Vibrations in a crystal are independent waves, each labeled by k.• There are 3N wave types in a 3D crystal with N atoms per cell

• Each independent harmonic oscillator has quantized energies:En = (n + 1/2) hν = (n + 1/2) hω

• Quanta are called phonons, each phonon carries energy hω• These can be viewed as particles• For each independent oscillator (i.e., for each independent wave in a

crystal), there can be any integer number of phonons • Need to find out the average number of phonons and the average energy

associated with each independent oscillator (mode).

Thermal Properties - Key Points

• Fundamental law of a system in thermal equilibrium: If two states of the system have total energies E1 and E2, then the ratio of probabilities for finding the system in states 1 and 2 is P1 / P2 = exp ( - (E1 - E2) / kB T) where kB is the Boltzman constant

• Larger probability of smaller energy

• Applies to all systems - whether classical or quantum and whether the particles are bosons (like phonons) or fermions (like electrons)

Thermal Properties - Phonons• Phonons are examples of bosons. • There can be any number n phonons for each oscillator, i.e.,

the energy of each oscillator can be En = (n + ½ ) h ω, n = 0,1,2,. . .

The probability of finding an oscillator with n phonons (and not another value) Pn = exp ( - En / kB T) / ∑n’=0 exp ( - En’ / kB T) and the average phonon occupation is

<n> = ∑n =0 Pn n = ∑n =0 n exp ( - En / kB T) / ∑n’ =0 exp ( - En’ / kB T)

∞

∞

∞

• Using the formulas: 1/(1 - x) = ∑s=0 xs and x/(1 - x) 2 = ∑s=0 s xs

<n> = 1 / [ exp ( h ω / kB T) - 1 ] Planck Distribution

Average energy of an oscillator at temperature T:U = < (n + ½ ) h ω > = h ω (<n> + ½ )

= h ω (1 / [ exp ( h ω / kB T) - 1 ] + ½ )

• At high T, U → h ω / [ h ω / kB T ] → kB Twhich is the classical result

Planck Distribution∞ ∞

Total thermal energy of a crystal • The crystal is a sum of independent oscillators (in the

harmonic approximation). The independent oscillators are waves labeled by k and an index m = 1, ..., 3N. Therefore, the total energy of the crystal is:

U = U0 + ∑k,m h ωk,m (1 / [ exp ( h ωk,m / kB T) - 1 ] + ½ )

Fixed atoms

Added thermal energy Zero point energy

Question: How to do the sum over k ??

3 dimensionsN atoms per cell

Sum over vibration modes of a crystal • The sum over k and the index m = 1, ..., 3N can be thought

of as follows:One k point for each unit cellThe index m counts the atoms per unit cell N multiplied by the number of independent ways each atoms can move (3 in 3D).

• The entire crystal has 3 N Ncell degrees of freedom(i.e. number of ways the atoms can move) . This must not change when we transform to the independent oscillators.

1D, two boundary conditions • Demonstration that the sum over k is equivalent to one k

point for each unit cell• N atoms at separation a, us = u exp(ik (s a) - iωk,m t)• Fixed boundary conditions: u0= uN=0

– Standing waves only– Possible k values: k= π/Na , 2π/Na, .. nπ/Na, (N-1)π/Na– One k value per mobile atom, one k value per cell

• Periodic boundary conditions: us= uN+s– traveling waves, need kNa=+-2nπ– Possible k values: k= 0, +-2π/Na , +-4π/Na, .. 2nπ/Na– One k value per mobile atom, one k value per cell

Density of states • All we need is the number of states per unit energy, and we

can integrate over energy to find the thermal quantities

• Total energy• We know that there are Ncell possible k values• In a large crystal one can replace the sum over k with an

integral• Since ω and k are related by the dispersion relation we can

change variables

• Dm(ω)dω − number of modes (states) in frequency range ωtο ω+dω

∑∑= = −

=cellN

k

N

m Bm,k

m,k

)Tkexp(U

1

3

1 1ωω

h

h

∑∫ −=

m Bm )Tkexp(

)(DdU1

ω

ωωω

h

h

Relation between k and ω

• Total energy

• Dm(ω)dω − number of modes (states) in frequency range ωtο ω+dω for branch mdispersion relation: ωk = 2 ( C / M ) 1/2 | sin (ka/2) |

• Modes in interval (ω, k, E) to (ω+∆ω, k+ ∆k, E+ ∆E) ∆N= D(ω) ∆ω=Ν(k) ∆k=N(E) ∆E

∑∫ −=

m Bm )Tkexp(

)(DdU1

ω

ωωω

h

h

ωωω

ωω dv

)k(Ndddk)k(Nd)(D

g

==

Group velocity

Density of states in 1D • Dm(ω)dω − number of modes (states) in frequency range ω

tο ω+dω

• N(k) - number of modes per unit range of k• number of modes between -π/a <k< π/a = N = L/a (the

number of atoms)

• N(k)=N/(2π /a)

ωωω dv

)k(Nd)(Dg

=

ωπ

ωπ

ωω dv

Ldv

Nad)(Dgg

12

12

==

Possible wavevectors in 3D • Assume Ncell= n3 primitive cells, each a cube of side a• volume of the crystal V=(na)3 =L3

• vibrations:

• periodic boundary conditions:

na= L• There is one allowed value of k in each volume

(2π/L)3=8π3/VV - volume of the crystal

kkjkikk zyx ++=

)]}Lz(k)Ly(k)Lx(k[iexp{)]zkykxk(iexp[ zyxzyx +++++=++

t) - iωrk(iexp u umk,s ⋅=

kx, ky , kz = 0, +-2π/L , +-4π/L, .. 2mπ/L

kzjyixr ++=

Density of states in 3D

• one allowed value of k per volume 8π3/V• D(ω)dω − number of modes (states) in frequency range ω tο

ω+dω∆N= D(ω) ∆ω=Ν(k) ∆k

In a cubic lattice the 1D dispersion relation holds - ω onlydepends on k - spherical symmetry

Ν(k) = V/8π3

• D(ω)=k2V/2π2 dk/dω

ωω

πωω dddkkkNdD 24)()( =

Heat Capacity• The internal energy is found by summing over all modes

• The heat capacity is found by differentiating U with respect to temperature, C = dU/dT

• Need to express D(ω) and dω as a function of x to do the integral

∑∫ −=

m Bm )Tkexp(

)(DdU1

ω

ωωωh

h

∑∫ −=

mmB ])x[exp(

)xexp(x)(DdkC 2

2

1 ωω Tk

xB

ωh=

Debye Approximation• Approximate crystal with an elastic continuum,• ω=vk (v= sound velocity)• In 3D cubic crystal D(ω)= k2V/2π2 dk/dω• D(ω)=Vω2/2π2v3

• In this approximation the maximum ω is not known, we need to determine it from the fact that there are Ncell modes in each branch

• ωD3 =6π2 v3 Ncell /V Ncell /V =1/Vcell

• maximum wavevector kD =ωD/v =(6π2 Ncell /V)1/3

cellN)(DdD

=∫=

ω

ω

ωω0

Debye temperature• Thermal energy ( 3 polarizations)

• Characteristic Debye temperature θ= hv/kB(6π2 Ncell /V)1/3

∫∫ −=

−=

DD xB

B )xexp(xdxTVk

)Tkexp()(DdU

0

3

332

44

0 1v23

1 3

hh

h

πωωωω

ω

TTkx

B

DD

θω==

h

∫ −⎟⎠⎞

⎜⎝⎛=

Dx

Bcell )xexp(xdxTTkNU

0

33

1 9

θ

∫ −⎟⎠⎞

⎜⎝⎛=

Dx

BcellV ))x(exp()xexp(xdxTkNC

02

43

1 9

θ

Ex. Debye temperature θ= hv/kB(6π2 Ncell /V)1/3

What material characteristics does the Debye temperaturedepend on?

Limits of heat capacity in the Debye appr.

∫ −⎟⎠⎞

⎜⎝⎛=

Dx

BcellV ))x(exp()xexp(xdxTkNC

02

43

1 9

θ

• For T>>θ, xD 0, exp(x)~1+xCV~3Ncell kB - equal to classical limit

• For T<<θ, xD ∞,

CV~T3

15

491

943

02

43 πθθ

⎟⎠⎞

⎜⎝⎛=

−⎟⎠⎞

⎜⎝⎛= ∫

∞TkN

))x(exp()xexp(xdxTkNC BcellBcellV

Debye Approximation

Cv has correct general behavior that must be found in allcrystals. For 3 dimensions

Hea

t Cap

acity

C

T

T3

Approachesclassical limit

3 Ncell kB

∫ −⎟⎠⎞

⎜⎝⎛=

Dx

BcellV ))x(exp()xexp(xdxTkNC

02

43

1 9

θ

Einstein Approximation• The Debye approximation is valid for acoustic modes• For optic modes one can assume a constant frequency-

Einstein approximation

0 2π/aπ/a

ωk

3 Acoustic modesEach has ω ~ k at small k

3 (N -1) Optic Modes

Einstein Approximation• Assume each oscillator has ω=ω0

• D(ω)=Ncellδ(ω−ω0)

• For T 0, CV~exp(-hω/kBT)• For T>>0, CV~3Ncell kB - equal to classical limit

13

0

0

−=

)Tkexp(NU

Bcell ω

ωh

h

20

0

2

0

13

))Tk(exp()Tkexp(

TkkNC

B

B

BBcellV −⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ωωω

h

hh

General expression for D(ω)• In 1D , vg=group velocity

• In 3D ∆N= D(ω) ∆ω=Ν(k) ∆k

• integral over the volume of the shell in k space bounded by the surfaces with ω and ω+dω

integral over ω=const surface

• vg=0 - Van Hove singularities

ωkgv ∆=

gv)k(N)(D =ω

∫=shell

kd)k(Nd)(D 3ωω

∫∫ ∫ == ⊥gshell v

ddSdkdSkd ωωω

3

( ) ∫=shell gv

dSV)(D ω

πω 32

• Plot the density of states as a function of ω for a Debyesolid, D(ω)=Vω2/2π2v3

• How does this plot change for an actual crystal structure?

( ) ∫=shell gv

dSV)(D ω

πω 32

Energy & Force due to Displacements

• The energy of the crystal changes if the atoms are displaced. • The change in energy can be written as a function of the positions of

all the atoms:E(R1, R2, R3, …)=E(R1

0 +∆R1, R20 +∆R2, ..)

• To lowest order in the displacements the energy is quadratic - Hooke’slaw - harmonic limit

E = E0 + (1/2) Σi j ∆Ri . Di j . ∆Rj + ….

(There are no linear terms if we expand about the equilibrium positions)

Consequences of anharmonicity• If we expand the energy beyond the harmonic order:

E = E0 + (1/2) Σi j ∆Ri . Di j . ∆Rj + (1/6) Σi jk Di jk . ∆Ri ∆Rj ∆Rk + . . .

• The problem is fundamentally changed:No longer exactly solvable

• Consequences:There is thermal expansion and other changes with temperatureThe heat capacity is not constant at high TPhonons decay in timeTwo phonons can interact to create a thirdPhonons can establish thermal equilibrium andconduct heat like a “gas” of particles

Difficult and Messy

Inelastic Scattering and Fourier Analysis

Inelastic diffraction occurs for kin - kout = G ± kphononωin - ωout = ± ωphonon or Εn - Εout = ± hωphonon

kin, ωinkout, ωout

kphonon ,, ωphonon

From Before

Scattering of Phonons - I

• The same idea applies to phonons. One phonon can scatter to create two.

• Scattering can occur for kin phonon = kout phonon 1 + kout phonon 2 ± G

ωin phonon = ωout phonon 1 + ωout phonon 2

kin phonon , ωin phononkout phonon 1, ωout phonon 1

kout phonon 2 , ωout phonon 2AnharmonicInteraction

Scattering of Phonons - II• Two phonons can interact to create one.• This is called “up conversion (umklapp)”, which

can be done with intense phonon beams and occurs for kout phonon = kin phonon 1 + kin phonon 2 ± Gωout phonon = ωin phonon 1 + ωin phonon 2

kout phonon , ωout phononkin phonon 1 , ωin phonon 1

kin phonon 2 , ωin phonon 2

AnharmonicInteraction

Thermal expansion• The energy of a pair of atoms depends on the displacement

x from their equilibrium separations, U(x)=cx2-gx3-fx4

• The average displacement <x> is determined by the condition that the average force vanishes.

•• In the harmonic approximation F = cx. Therefore <x> = 0,

<F> = 0, and there is no thermal expansion

• Anharmonicity adds additional terms: F = cx - 1/2 g x2 - …, <F> = 0 ⇒ <x> = 1/2 (g/c) <x2>

• In general, this means thermal expansion.

Thermal expansion

Distance Between Atoms

Ene

rgie

s of C

ryst

al

Absolute Minimum

Thermal Expansion -Average distance increases

as vibration amplitude increases

Characterizing thermal dilation• Average displacement of atoms, <x>

• U(x) – potential energy of a pair of atoms.• Thermal expansion of a crystal or cell: δ =dV/V • For cubic crystals dV/V = 3 dx/x

From the theory of elastic media:• Potential energy of a unit cell due to dilation:

Ucell(T)=Vcell B δ2/2, where B is the bulk modulus

[ ]

[ ]Tk

cg

Tk/)x(Uexpdx

Tk/)x(Uexpxdxx B

B

B

243

≅

−

−=

∫

∫∞

∞−

∞

∞−

Transport of heat in an ordinary gas• Molecules move in all directions and scatter so that they

come to local thermal equilibrium in each region.• How can random motion cause heat flow in one direction? • On average, in hotter regions molecules have more kinetic

energy. A molecule that moves from a hotter region to a colder one brings energy above the local average. The opposite for a molecule moving from a colder to a hotter region. Either way, there is transport of energy from hotter to colder regions.

Heat Flow

coldhot

Phonons also act like a gas• A phonon is a particle - a quantum of vibration• It carries energy just like a molecule.• Phonons can come to equilibrium by scattering just like

molecules (phonon scattering is due to defects and to anharmonicity).

• What is different?Phonons can be created and destroyed. But we will see that we can treat them exactly like gas.

Heat Flow

coldhot

Thermal conductivity of phonons• Definition: j = heat flow (energy per unit area per unit

time ) = - K dT/dx; K – thermal conductivity• If a phonon moves from a region with local temperature T

to one with local temperature T - ∆T, it supplies excess energy c ∆T, where c = heat capacity per phonon. (Note ∆T can be positive or negative).

• Temperature difference between the ends of a free path :∆T = (dT/dx) vx τ, where τ = mean time between collisions

•• Then j = - n vx c vx τ dT/dx = - n c vx

2 τ dT/dx

Density Flux

Phonon Heat Transport - continued• This can be simplified in an isotropic case, since averaging

over directions gives ( vx2 ) average = (1/3) v2

• This leads to j = - (1/3) n c v2 τ dT/dx

• Finally we can define the mean free path:L = v τ and C = nc = total heat capacity

• Then j = - (1/3) C v L dT/dxand

K = (1/3) C v L = thermal conductivity(just like an ordinary gas!)

Phonon Heat Transport - continued• What determines mean free path L = v τ ? • At low temperature, the thermal phonons are sound waves

that have long mean free paths -L ~ sample size

• At high temperature, phonons scatter from other phonons. • ONLY Umklapp scattering limits the energy flow.

The density of other phonons is ~ T, so L ~ 1/T

• At intermediate temperature, phonon scatter from defects and other phonons.

Phonon Heat Transport - continued• Low T - K increases with T because density of phonons

increases with roughly constant v and L• High T - K decreases with T as Umklapp scattering

increases

Heat Flow

LowT

HighT

coldhot

Phonon Heat Transport - continued• Behavior in an excellent quality crystal:

Ther

mal

con

duct

ivity

K

T

Low T -- K increases as density

of phonons increases (v and L are ~ constant)

Approacheshigh T limit

1/T

1

10

100

1 10

K decreasesbecause Umklappscattering increases rapidly

Maximum controlled by defects

Umklapp Scattering• Scattering that changes total crystal momentum by a

reciprocal lattice vector.• kin phonon = kout phonon 1 + kout phonon 2 ± G

ωin phonon = ωout phonon 1 + ωout phonon 2

kin phonon , ωin phonon kout phonon 1 , ωout phonon 1

kout phonon 2 , ωout phonon 2AnharmonicInteraction

Unless G ≠ 0, the scattering does not change the total phonon momentum or energy. Therefore only Umklapp scattering limits the heat flow. It also leads to thermal equilibrium.

• Vibrations of atoms Harmonic approximationExact solution for waves in a crystal

Labeled by k and index m = 1, …, 3N• Quantization of vibrations

Phonons act like particlesCan be created or destroyed by inelastic scattering experiments

• Thermal propertiesFundamental law of probabilitiesPlanck distribution for phononsHeat Capacity C

Low T: C ~ T3 -- High T: C ~ constant Thermal conductivity KMaximum as function of T

Free Electron Fermi Gas(Kittel Ch. 6)

Role of Electrons in Solids• Electrons are responsible for binding of crystals --

they are the “glue” that hold the nuclei togetherTypes of binding (see next slide)

Van der Waals - electronic polarizabilityIonic - electron transferCovalent - electron bonds

• Electrons are responsible for important properties:Electrical conductivity in metals(But why are some solids insulators?)MagnetismOptical properties. . . .

Characteristic types of binding

Closed-Shell BindingVan der Waals

Metallic BindingCovalent Binding

Ionic Binding

Starting Point for Understanding Electrons in Solids

• Nature of a metal:Electrons can become “free of the nuclei” and move between nucleisince we observe electrical conductivity

• Electron GasSimplest possible modelfor a metal - electrons arecompletely “free of the nuclei” - nuclei are replacedby a smooth background --“Electrons in a box”

Electron Gas - History• Electron Gas model predates quantum mechanics

• Electrons Discovered in 1897

• Drude-Lorentz Model (1905)-Electrons - classical particlesfree to move in a box

• Model: All electrons contribute to conductivity. Works! Still used!

• But same model predicted that all electrons contribute to heat capacity. Disaster. Heat capacity is MUCH less than predicted.

Quantum Mechanics• 1911: Bohr Model for H • 1923: Wave Nature of Particles Proposed

Prince Louis de Broglie• 1924-26: Development of Quantum

Mechanics - Schrodinger equation• 1924: Bose-Einstein Statistics for

Identical Particles (phonons, ...)• 1925-26: Pauli Exclusion Principle,

Fermi-Dirac Statistics (electrons, ...)• 1925: Spin of the Electron (spin = 1/2)

G. E. Uhlenbeck and S. GoudsmitSchrodinger

Schrodinger Equation • Basic equation of Quantum Mechanics

[ - ( h/2m ) ∆2 + V( r ) ] Ψ ( r ) = E Ψ ( r )

wherem = mass of particleV( r ) = potential energy at point r∆2 = (d2/dx2 + d2/dy2 + d2/dz2)E = eigenvalue = energy of quantum stateΨ ( r ) = wavefunctionn ( r ) = | Ψ ( r ) |2 = probability density

Schrodinger Equation – 1D line • Suppose particles can move freely on a line with

position x, 0 < x < L

• Schrodinger Eq. In 1D with V = 0- ( h2/2m ) d2/dx2 Ψ (x) = E Ψ (x)

• Solution with Ψ (x) = 0 at x = 0,L Ψ (x) = 21/2 L-1/2 sin(kx) , k = n π/L, n = 1,2, ...

(Note similarity to vibration waves)

Factor chosen so ∫0L dx | Ψ (x) |2 = 1

• E (k) = ( h2/2m ) k 2

0 L

Boundary Condition

Electrons on a line • Solution with Ψ (x) = 0 at x = 0,L

Examples of waves - same picture as for lattice vibrations except that here Ψ (x) is a continuous waveinstead of representing atom displacements

0 L

Ψ

Electrons on a line• For electrons in a box, the energy is just the kinetic

energy which is quantized because the waves must fit into the box

E (k) = ( h2/2m ) k 2 , k = n π/L, n = 1,2, ...

E

k

Approaches continuum

as L becomes large

Schrodinger Equation – 1D line • E (k) = ( h2/2m ) k 2 , k = n π/L, n = 1,2, ...

• Lowest energy solutions with Ψ (x) = 0 at x = 0,L

Ψ (x)

x

Electrons in 3 dimensions-(h2/2m ) [d2/dx2 + d2/dy2 + d2/dz2 ] Ψ (x,y,z) = E Ψ (x,y,z)

Ψ (x) = 0 at x = 0,L; Ψ (y) = 0 at y = 0,L ; Ψ (z) = 0 at z = 0,LΨ = 23/2 L-3/2 sin(kxx) sin(kyy) sin(kzz) ,

kx = n π/L, n = 1,2, …, same for y,z

E (k) = ( h2/2m ) (kx2 + ky

2 + kz2 ) = ( h2/2m ) k2

E

k

Approaches continuum

as L becomes large

Electrons in 3 dimensions - continued• Just as for phonons it is convenient to define Ψ with

periodic boundary conditions• Ψ is a traveling plane wave:

Ψ = L-3/2 exp( i(kxx + kyy + kzz) , kx = ± n (2π/L), etc., n = 0,1,2,..

E (k) = ( h2/2m ) (kx2 + ky

2 + kz2 ) = ( h2/2m ) k2

E

k

Approaches continuum

as L becomes large

Density of states • Key point - exactly the same as for vibration waves • We need the number of states per unit energy to find

the total energy and the thermal properties of the electron gas.

• Difference: density of states is defined in terms of energy E, not angular frequency.

• D(E)dE - number of states in energy range E to E+dE• States in interval (k, E) to (k+ ∆k, E+ ∆E)

∆N= N(k) ∆k=N(E) ∆E dN/dE=(dN/dk)/(dE/dk)

Density of States in 3D• The values of kx ky kz are equally spaced: ∆kx = 2π/L ,.Thus the volume in k space per state is (2π/L)3

and the number of states N with |k| < k0 is N = (4π/3) k0

3 / (2π/L)3 = V/6π2 k03 L3=V

• The density of states per unit energy is D(E) = dN/dE = (dN/dk) (dk/dE)

E = ( h2/2m ) k2 , dE/dk = ( h2/m ) k⇒ D(E) = (V/2π2) k2 / (h2/m ) k = (V/2π2) k / (h2/m )

= (V/4π2) E1/2 (2m / h2)3/2

Kittel adds a factor of 2 for multiplicity of electrons in thesame state (spin): D(E) = (V/2π2) E1/2 (2m / h2)3/2

Electron orbitals• In 1D E (k) = ( h2/2m ) k 2 , k = n π/L, n = 1,2, …• In 3D E (k) = ( h2/2m ) (kx

2 + ky2 + kz

2 ) = ( h2/2m ) k2

kx, ky kz = ± n (2π/L), etc., n = 0,1,2,..• Thus E1D (k) = n2 ( h2/2m ) (π /L)2

• E3D (k) = (nx+ny+nz)2 ( h2/2m ) (2π /L)2

• To describe a system of Nelec electrons, we assign the electrons to orbitals of increasing energy, until all orbitals are filled.

• Order of filling: n=1, 2, .. nF

• nF - topmost filled energy level (Fermi level)

Electron orbitals• D(E)= (V/2π2) E1/2 (2m / h2)3/2

• Now we need to figure out how many electrons are on a given orbital (electron occupancy)

E

D(E) EF

FilledEmpty

What is special about electrons?

• Fermions - obey exclusion principle• Fermions have spin s = 1/2 - two electrons (spin up and

spin down) can occupy each state• Kinetic energy = ( p2/2m ) = ( h2/2m ) k2

• Thus if we know the number of electrons per unit volume Nelec/V, the lowest energy allowed state is for the lowest Nelec/2 states to be filled with 2 electrons each,and all the (infinite) number of other states to be empty.

• The number of states with |k| < k0 is N = (V/6π2) k03

(from before)

Fermi momentum and energy

Fermi surface in 2D

Thus all states are filled up to the Fermi momentum kF and Fermi energy EF = ( h2/2m ) kF

2, given byNelec/2 = (V/6π2) kF

3

⇒kF = (3π2 Nelec/V )1/3 and EF = (h2/2m) (3π2 Nelec/V )2/3

Reciprocal space

Possible k valuesFilled states

Fermi Distribution • At finite temperature, electrons are not all in the lowest energy

states• Applying the fundamental law of statistics to this case (occupation

of any state and spin only can be 0 or 1) leads to the Fermi Distribution giving the probability that an orbital of energy E is occupied (Kittel appendix)

f(E) = 1/[exp((E-µ)/kBT) + 1]

E

D(E)

µf(E)

1

1/2

Chemical potential for electrons =

Fermi energy at T=0

kBT µ is temperature dependent

Ex. How does the Fermi distribution

f(E) = 1/[exp((E-µ)/kBT) + 1]

compare with the Planck distribution for phonons? n(E) = 1 / [ exp ( E / kB T) - 1 ]

Sketch them as a function of energy for differenttemperatures.

Typical values for electrons?• Here we count only valence electrons (see Kittel table)

• Element Nelec/atom EF TF = EF/kB

Li 1 4.7 eV 5.5 x104 K

Na 1 3.23eV 3.75 x104 K

Al 3 11.6 eV 13.5 x104 K

• For typical metals the Fermi energy temperature is much greater than ordinary temperatures – transition from f(E)=1 to f(E)=0 is sharp at room temperature

Heat Capacity for Electrons Just as for phonons the definition of heat capacity is C = dU/dTwhere U = total internal energy

• When heated from T=0 only electrons within an energy range kBT of the Fermi energy can be excited thermally• For T << TF = EF /kB roughly U ~ U0 + Nelec (T/ TF) kB T so that

C = dU/dT ~ Nelec kB (T/ TF)

E

D(E)µ

f(E)1

1/2

Chemical potential

for electrons

kBT

Heat Capacity for Electrons • More precisely, the change in energy when heated

from 0 to T is

∆U = ∫0∞ dE E D(E) f(E) - ∫0

EF dE E D(E)

• Using the fact that T << TF:C = dU/dT = ∫0

∞ dE (E - EF) D(E) (df(E)/dT) ≈ D(EF) ∫0

∞ dE (E - EF) (df(E)/dT)

• The integral can be done almost exactly (exact in the low T limit) to giveC = (π2/3) D(EF) kB

2 T (valid for any metal) → (π2/2) (Nelec/EF) kB

2 T (for the electron gas)(using D(EF) = 3 Nelec/2EF )

• Key result: C ~ T - agrees with experiment!

Heat capacity• Comparison of electrons in a metal with phonons

Hea

t Cap

acity

C

T

T3

Phonons approachclassical limitC ~ 3 Natom kB

Electrons have C ~ Nelec kB (T/TF)

Electrons dominateat low T in a metal

T

Phonons dominateat high T because of reduction factor (T/TF)

Heat capacity• Experimental results for metals

C/T = γ + A T2 + ….• Find the ratio γ / γfree, γfree = (π2/2) (Nelec/EF) kB

2 is the free electron gas result. Equivalently since EF ∝1/m, we can consider the ratio γ / γfree = mfree/mth*, where mth* is an thermal effective mass for electrons in the metal

Metal mth*/ mfreeLi 2.18Na 1.26K 1.25Al 1.48Cu 1.38

• mth* close to m(free) is the “good”, “simple metals” !

Electrical Conductivity & Ohm’s Law• Consider electrons in an external field E. They

experience a force F = -eE• Now F = dp/dt = h dk/dt , since p = h k• Thus in the presence of an electric field all the

electrons accelerate and the k points shift, i.e., the entire Fermi surface shifts E

Equilibrium - no field With applied field

Electrical Conductivity & Ohm’s Law• What limits the acceleration of the electrons? • Scattering increases as the electrons deviate more

from equilibrium• After field is applied a new equilibrium results as a

balance of acceleration by field and scatteringE

Equilibrium - no field With applied field

Electrical Conductivity and Resistivity• The conductivity σ is defined by j = σ E,

where j = current density• How to find σ?• From before F = dp/dt = m dv/dt = h dk/dt• Equilibrium is established when the rate that k

increases due to E equals the rate of decrease due to scattering, then dk/dt = 0

• If we define a scattering time τ and scattering rate1/τh ( dk/dt + k /τ ) = F= q E (q = charge)

• Now j = n q v (where n = density) so that j = n q (h k/m) = (n q2/m) τ E⇒ σ = (n q2/m) τ

• Resistance: ρ = 1/ σ ∝ m/(n q2 τ) Note: sign of charge

does not matter

Scattering mechanisms• Impurities - wrong atoms, missing atoms, extra atoms,

….

Proportional to concentration

• Lattice vibrations - atoms out of their ideal places

Proportional to mean square displacement

• (Really these conclusions depend upon ideas from the next section that there is no scattering in a perfect crystal.)

Electrical Resistivity• Resistivity ρ is due to scattering: Scattering rate

inversely proportional to scattering time τ

ρ ∝ scattering rate ∝ 1/τ

• Matthiesson’s rule - scattering rates add

ρ = ρvibration + ρimpurity ∝ 1/τvibration + 1/τimpurity

Temperature dependent∝ <u2>

Temperature independent- sample dependent

Electrical Resistivity• Consider relative resistance R(T)/R(T=300K)• Typical behavior (here for samples of potassium)

Rel

ativ

e re

sist

ance

TIncrease as T2

Inpurity scattering dominatesat low T in a metal

(Sample dependent)

Phonons dominate at high T because mean square

displacements <u2> ∝ TLeads to R ∝ T

(Sample independent)

0.01

0.05

Interpretation of Ohm’s lawElectrons act like a gas

• A electron is a particle - like a molecule.• Electrons come to equilibrium by scattering like

molecules (electron scattering is due to defects, phonons, and electron-electron scattering).

• Electrical conductivity occurs because the electrons are charged, and it shows the electrons move and equilibrate

• What is different from usual molecules?Electrons obey the exclusion principle. This limits the allowed scattering which means that electrons act like a weakly interacting gas.

Hall Effect I• Electrons moving in an electric and a perpendicular

magnetic field• Now we must carefully specify the vector force

F = q( E + (1/c) v x B ) (note: c → 1 for SI units)(q = -e for electrons)

E

B

vFE

FB

Vector directions shown for positive q

Hall Effect II• Relevant situation: current j = σ E = nqv flowing along

a long sample due to the field E• But NO current flowing in the perpendicular direction• This means there must be a Hall field EHall in the

perpendicular direction so the net force F⊥ = 0F⊥ = q( EHall + (1/c) v x B ) = 0

E

vF⊥

j

j

EHall

B

x

zy

Hall Effect III• Since

F⊥ = q( EHall + (1/c) v x B ) = 0 and v = j/nq

then defining v = (v)x, EHall = (EHall )y, B = (B )z, EHall = - (1/c) (j/nq) (- B )

and the Hall coefficient isRHall = EHall / j B = 1/(nqc) or RHall = 1/(nq) in SI

E

vF⊥ j

EHall

B

Sign from cross product

Hall Effect IV• Finally, define the Hall resistance as

ρHall = RHall B = EHall / j

which has the same units as ordinary resistivity• RHall = EHall / j B = 1/(nq)• Note: RHall determines sign of charge q

Since magnitude of charge is known RHall also determines density n

• The sign of charge in several metals (Mg, Al) is positive

Each of these quantities can be measured directly

Electrons act like gas - heat transport• A electron is a particle that carries energy - just like a

molecule.• Electrical conductivity shows the electrons move,

scatter, and equilibrate• What is different from usual molecules?

Electrons obey the exclusion principle. This limits scattering and helps them act like weakly interacting gas.

Heat Flow

coldhot

Heat Transport due to Electrons• Definition (just as for phonons):

jthermal = heat flow (energy per unit area per unit time ) = - K dT/dx

• If an electron moves from a region with local temperature T to one with local temperature T - ∆T, it supplies excess energy c ∆T, where c = heat capacity per electron. (Note ∆T can be positive or negative).

• On average :∆T = (dT/dx) vx τ, where τ = mean time between collisions

• Then jthermal = - n vx c vx τ dT/dx = - n c vx2 τ dT/dx

DensityFlux

Electron Heat Transport - continued• Just as for phonons:

Averaging over directions gives ( vx2 ) average = (1/3) v2

and j = - (1/3) n c v2 τ dT/dx

• Finally we can define the mean free path L = v τand C = nc = total heat capacity,Then

j = - (1/3) C v L dT/dxandK = (1/3) C v L = (1/3) C v2 τ = thermal conductivity

(just like an ordinary gas!)

Electron Heat Transport - continued• What is the appropriate v? • The velocity at the Fermi surface = vF

• What is the appropriate τ ? • Same as for conductivity (almost).

• Results using our previous expressions for C:

K = (π2/3) (n/m) τ kB2 T

• Relation of K and σ -- From our expressions:K / σ = (π2/3) (kB/e)2 T

• This justifies the Weidemann-Franz Law thatK / σ ∝ T

Electron Heat Transport - continued• K ∝ σ T• Recall σ → constant as T → 0, σ → 1/T as T → large

Ther

mal

con

duct

ivity

KW

/cm

K

T

Low T -- K increases as heat

capacity increases (v and L are ~ constant)

Approacheshigh T limit- K constant

0

50

0 100

Electron Heat Transport - continued

• Comparison to Phonons

Electrons dominate in good metal crystals

Comparable in poor metals like alloys

Phonons dominate in non-metals

Summary• Electrical Conductivity - Ohm’s Law

σ = (n q2/m) τ ρ = 1/σ• Hall Effect

ρHall = RHall B = EHall / jρ and ρHall determine n and the charge of the carriers

• Thermal ConductivityK = (π2/3) (n/m) τ kB

2 TWeidemann-Franz Law:K / σ = (π2/3) (kB/e)2 T

• Metallic Binding Kinetic repulsionCoulomb attraction to nuclei (not included in gas model - must be added)

Energy Bands for Electronsin Crystals (Kittel Ch. 7)

EnergyGap

kπ/a−π/a 0

Ener

gy

• Recall nature of free electron gasFree electrons in box of size L x L x L

(artificial but useful) Solved Schrodinger EquationStates classified by k with E(k) = (h2/2m) k2

Periodic boundary conditions convenient: Leads to kx = 2nπ/L, etc.

Pauli Exclusion Principle, Fermi StatisticsSimplest model for metals

• Why are some materials insulators, some metals?•

• Recall nature of free electron gasFree electrons in box of size L x L x L

(artificial but useful) Solved Schrodinger EquationStates classified by k with E(k) = (h2/2m) k2

Periodic boundary conditions convenient: Leads to kx = 2nπ/L, etc.

Pauli Exclusion Principle, Fermi StatisticsSimplest model for metals

• Why are some materials insulators, some metals?• First step - NEARLY free electrons in a crystal

Simple picture of how Bragg diffraction leads to standing waves at the Brillouin Zone Boundary and toenergy gaps

Understanding Electrons in Crystals

• Electron GasSimplest possible modelfor a metal - electrons arecompletely “free of the nuclei” - nuclei are replacedby a smooth background --“Electrons in a box”

• Real Crystal -Potential variation with the periodicity of the crystal

Attractive (negative) potential around each nucleus

Schrodinger Equation • Basic equation of Quantum Mechanics

[ - ( h2/2m ) ∆2 + V(r ) ] Ψ (r ) = E Ψ (r )

wherem = mass of particleV(r ) = potential energy at point r ∆2 = (d2/dx2 + d2/dy2 + d2/dz2)E = eigenvalue = energy of quantum stateΨ (r ) = wavefunctionn (r ) = | Ψ (r ) |2 = probability density

• Key Point for electrons in a crystal: The potentialV(r ) has the periodicity of the crystal

Schrodinger Equation • How can we solve the Schrodinger Eq.

[ - ( h2/2m ) ∆2 + V( r ) ] Ψ ( r ) = E Ψ (r )

where V( r ) has the periodicity of the crystal?

• Difficult problem - This is the basis of current research in the theory of electrons in crystals

• We will consider simple cases as an introductionNearly Free ElectronsKronig-Penney Model

Next Step for Understanding Electrons in Crystals

• Simplest extension of theElectron Gas model

• Nearly Free electron Gas -Very small potential variationwith the periodicity of the crystal

• We will first consider electrons in one dimension

Very weak potentials with crystal periodicity

Consider 1 dimensional example• If the electrons can move freely on a line from 0 to L

(with no potential),

we have seen before that :• Schrodinger Eq. In 1D with V = 0

- ( h2/2m ) d2/dx2 Ψ (x) = E Ψ (x) • Solution with Ψ (x) = 0 at x = 0,L

Ψ (x) = 21/2 L-1/2 sin(kx) , k = n π/L, n = 1,2, ...

or Ψ (x) = L-1/2 exp( ikx), k = ± n (2π/L), n = 0,1,..

• E (k) = ( h2/2m ) k 2

0 L

Periodic Boundary Condition

Fixed Boundary Condition

Electrons on a line• For electrons in a box, the energy is just the kinetic

energy E (k) = ( h2/2m ) k 2

• Values of k fixed by the box, k = ± n (2π/L), n = 0,1,..• Crystal: L = Ncell a

• The maximum (Fermi) wavevector is determined by the number of free electrons

• Nelec/2=2nF+1, thus nF~ Nelec/4• k = ± n (2π/Ncella), n=0,1,.. Nelec/4• define number of electrons per cell N• kF= N/2 (π/a)

Electrons on a line with potential V(x)• What happens if there is a potential V(x) that has the

periodicity a of the crystal?• An electron wave with wavevector k can suffer Bragg

diffraction to k ± G, with G any reciprocal lattice vector

E

kπ/a−π/a 0

G

Bragg Diffractionoccurs at

BZ boundary

State with k = π/adiffracts to k = - π/a

and vice versa

Electrons on a line with potential V(x)• Result:

Standing wave at zone boundaryEnergy gap where there are no waves that can travel in crystal

EnergyGap

kπ/a−π/a 0

Ener

gy

Energy Bands -Allowed energies for

electrons in the crystal(more later)

Interpretation of Standing waves at Brillouin Zone boundary

• Bragg scattering at k = π/a leads to two possible combinations of the right and left going waves:

Ψ(+)=(2L)-1/2[exp(iπx/a)+exp(-iπx/a)] = 21/2 L-1/2cos(πx/a)Ψ(-)=(2L)-1/2 [exp(iπx/a)-exp(-iπx/a)] = 21/2 i L-1/2 sin(πx/a)

with density n(+) = 2/L cos2(πx/a); n(-) = 2/L sin2(πx/a)

aAtoms

n(+) - high density at atoms n(-) - low density at atoms

Energy difference between solutions n(+) = 2/L cos2(πx/a); n(-) = 2/L sin2(πx/a)for n(+) the electrons are piled up on the positive ions,the magnitude of the negative potential energy ishigher, so the energy is lower

aAtoms - attractive

(negative) potential

n(+) - high density at atomslow energy

n(-) - low density at atomshigh energy

Electrons on a line with potential V(x)= Vcos(2πx/a)

Energy gap -- energies at which no waves can travelthrough crystal

EnergyGap

kπ/a−π/a 0

Ener

gy

n(+) - high density at atomslow energy

n(-) - low density at atomshigh energy

( ) V)(n)(n)x(dxVEL

g =−−+= ∫0

Understanding Electrons in Crystals• Real Crystal -

Potential variation with the periodicity of the crystal

• Potential leads to:Electron bands - E(k) different from freeelectron bandsBand Gaps

Ex. In a 1D crystal kF= N/2 (π/a), where N= # of electrons per cell. How many electron bands are expected for N=1,2,3,…?

Attractive (negative) potential around each nucleus

Representing V as a periodic function • We have seen (Kittel Ch 2) that any periodic function

can be written as Fourier seriesf(r) = ΣG fG exp( i G . r)

where the G ‘s are reciprocal lattice vectorsG(m1,m2,…) = m1 b1 + m2 b2 + m3 b3

• Check: A periodic function satisfies f(r) = f(r + T) where T is any translationT(n1,n2,…) = n1 a1 + n2 a2 + n3 a3the n’s are integers

• Thus V(r) = ΣG VG exp( i G . r)• And V(r) = real ⇒ VG = V*-G or if the crystal is

symmetric VG = V-G

Schrodinger Equation - Again • In a periodic crystal

[ - ( h2/2m ) ∆2 + ΣG VG exp( i G . r) ] Ψ ( r ) = E Ψ ( r )

• Now assume Ψ ( r ) = Σk ck exp( i k . r)

• Note we do NOT assume Ψ has the periodicity of the lattice! It is a superposition of waves!

• What is k? Just as before for electrons in a box, we assume Ψ ( r ) is periodic in a large box (L x L x L) which leads to

kx = ± nx (2π/L), n = 0,1,.. | k |= n (2π/L)

Schrodinger Equation - Continued • Then the Schrodinger Eq. becomes:

Σk ck λ k exp( i k . r) + Σk ck ΣG VG exp( i (k + G). r) ] = E Σk ck exp( i k . r)

where λ k = ( h/2m ) | k |2

• Introduce k’ = k+G then relabel k’ as kΣk { [λ k - E ] ck + ΣG VG ck-G } exp( i k . r) = 0

• Equating terms with the same r dependence on the two sides on the equation, we find the “Central Equation”

[λ k - E ] ck + ΣG VG ck-G = 0

“Central Equation” for electron bands • What is the interpretation of the equation:

[λ k - E ] ck + ΣG VG ck-G = 0

• If VG = 0 (no potential - free electrons) then each k is independent and each wavefunction is

Ψk ( r ) = ck exp( i k . r) ; E = λ k = ( h/2m ) | k |2

• If VG ≠ 0, then each k is mixed with k - G where G is any reciprocal lattice vector -- the solution is

Ψk ( r ) = ΣG ck-G exp( i (k - G). r)

Yet to be determined

Bloch Theorem • One of the most important equations of the course!• In a general crystal, the wave function for an electron

has the form:Ψk ( r ) = ΣG ck-G exp( i (k - G). r)

which can be written

Ψk ( r ) = exp( i k . r) uk ( r )

where uk ( r ) is the periodic functionwith the periodicity of the crystal lattice

uk ( r ) = ΣG ck-G exp( - i G . r)

Kronig-Penney model

square well potential

- ( h2/2m ) d2/dx2 Ψ (x)+ U(x)Ψ (x) = E Ψ (x)Combination of traveling waves where U=0, exponential penetration into the U=U0 region

The solution must satisfy the Bloch theorem

Ψk ( x) = exp (ik(a+b)) Ψk ( x-a-b)

⎩⎨⎧

++<<++++<<+

=)ba)(n(xnb)a(nU

nba)n(x)ba(n)x(U

11for 1for 0

0

⎩⎨⎧

+<<−+<<−+

=baxa)Qxexp(D)Qxexp(C

ax)iKxexp(B)iKxexp(A)x(

for 0for

ψ

Periodicity of the lattice, k is like a reciprocal vector

Kronig-Penney model

Ψ ( a) = exp (ik(a+b)) Ψ (-b)

boundary conditions: Ψ (x), dΨ (x)/dx continuousx=0: A+B = C+D, iK(A-B) = Q(C-D)x=a: A exp(iKa) +B exp(-iKa) = [C exp(-Qb) +D exp(Qb)]exp (ik(a+b))iK[A exp(iKa) -B exp(-iKa)] = Q [C exp(-Qb) -D exp(Qb)]exp (ik(a+b))

Solution if determinant vanishes – relationship between a, b, Q, K, k

⎩⎨⎧

+<<−+<<−+

=baxa)Qxexp(D)Qxexp(C

ax)iKxexp(B)iKxexp(A)x(

for 0for

ψ

Kronig-Penney model - limiting case

Assume b=0, U0~∞, Q2ba/2=P(P/Ka) sin Ka + cos Ka = cos ka

Solution only when l.h.s.<1energy gaps at k=0, +- π /a, ..

E (k) = ( h2/2m ) K 2

Since K depends on P E is plotted vs. kaE is not ~k2

E

kaπ 3π2π

bands

Bloch Theorem - II • The general form is

Ψkn ( r ) = exp( i k . r) uk

n ( r )

where ukn ( r ) is a periodic function. Here n labels

different bands

• Key Points:1) Each state is labeled by a wave vector k2) k can be restricted to the first Brillouin ZoneThis may be seen since Ψk+G’ ( r ) = exp( i (k + G’). r) u k+G’ ( r ) = exp( i k . r) u’k( r )where u’k ( r ) = exp( i G. r) u k+G’ ( r ) is just another periodic function

Bloch Theorem - III • Thus a wavefunction in a crystal can always be written

Ψkn ( r ) = exp( i k . r) uk

n ( r )

where: ukn ( r ) is a periodic function

n labels different bands k is restricted to the first Brillouin Zone

• In the limit of a large system k becomes continuousn is discrete index: n = 1,2,3, ….

The total number of k values • We can use the idea of periodic boundary conditions

on a box of size L x L x L - same as for phonons, electrons in a box,...

• Volume per k point = (2π/L)3

• Total number of k points in Brillouin zone Nk-point = VBZ /(2π /L)3 = (2π/a)3(L/2π)3 = (L/a)3 = Ncell

Each primitive cell contributes exactly one independentvalue of k to each energy band.Taking the two spin orientations into account, there are2Ncell independent orbitals in each energy band.

Solving the “Central Equation”• Simple cases where we can solve

[λ k - E ] ck + ΣG VG ck-G = 0

• If VG is weak, then we can solve the nearly free electron problem (and find the solution we saw earlier in the chapter).

• For k near BZ boundary, the wave exp( i k . r) is mixed strongly with exp( i (k - G). r), where G is the single vector that leads to | k | ~ | k - G |

• Let V = VG = V-G for that G

• Leads to two coupled equations

[λ k - E ] ck + V ck-G = 0

[λ k-G - E ] ck-G + V ck = 0

• or [λ k - E ] V

V [λ k-G - E ]• Solution

E = (1/2) (λ k + λ k-G ) +- [(1/4) (λ k - λ k-G )2 + V 2] 1/2

andck-G = [( -λ k + E)/V ] ck

Solving the Central Equation

= 0

E = (1/2) (λ k + λ k-G ) +- [(1/4) (λ k - λ k-G )2 + V 2] 1/2

ck-G = [( -λ k + E)/V ] ck

BZ boundary: k= π/a i, k-G= -π/a i (unit vector)λ k= λ k-G = λ, Ε−= λ−V, Ε+= λ+Vck-G = +- ck

Ψ (+) = ck [exp( i πx/a) + exp(-i πx/a)] Ψ (-) = ck [exp( i πx/a) - exp(-i πx/a)]

Solutions in 1D

Nearly Free Electrons on a line• Bands changed greatly only at zone boundary

Energy gap -- energies at which no waves can travel through crystal

EnergyGap

kπ/a−π/a 0

Ener

gy

Far from BZ boundarywavefunctions and energies

approach free electron values

Energies correspondingto next BZs are translatedinto first BZ

How to apply this idea in general• First find free electron bands plotted in BZ

• The energy is ALWAYS E (K) = ( h2/2m ) K 2but now we “reduce” K to first BZ, i.e., we find G such that K = k + G , and k is in the first BZ

• G(m1,m2,…) = m1 b1 + m2 b2 + m3 b3

• bi·aj= 2πδij

• Then add effects of potential – energy gaps

( ) ( )222222

22E(k) )Gk()Gk()Gk(

mhGk

mh

zzyyxx +++++=+=

Free Electrons, 1D, no gaps

E

kπ/a−π/a 0

G= 2π/a

K2

(k + 2π/a)2

2π/a

(k + 2π/a)2

−2π/a−3π/a

(k + 4π/a)2

3π/a

Free Electrons, 3D, simple cubic, K=(Kx,0,0)

E

kxπ/a−π/a 0

Gx

K2=Kx2

(kx + 2π/a)2

2π/a

(kx+2π/a)2+(2π/a)2

−2π/a

(kx )2 + (2π/a)2

Gy or Gz

(kx )2 + 2(2π/a)2

Nearly Free Electrons, 3D, schematic

E

kxπ/a−π/a 0

(kx + 2π/a)2

(kx )2 + (2π/a)2

Understanding Electrons in Crystals• Real Crystal -

Potential variation with the periodicity of the crystal

The nearly free electron cases show the general form of bands:

Continuous bands of allowed statesGaps where the are no states for the particular

k points

Attractive (negative) potential around each nucleus

Qualitative Picture of Electron EnergyBands and Gaps in Solids

Distance Between Atoms

Allo

wed

Ene

rgie

s for

Ele

ctro

ns

Forbidden Gapin Energies forValence Electrons

Atomic-like Core States

Metals vs Insulators • A band holds two electrons per each cell • Therefore a crystal with an odd number of electrons

per cell MUST* be a metal!Partially filled bands lead to Fermi energy and

“Fermi surface” in k spaceConductivity because states can change and

scatter when electric field is applied

• A crystal with an even number of electrons per cell MAY be an insulator!

Electrons “frozen”Gap in energy for any excitations of electrons

Metals vs Insulators • In 1d an even number of electrons per cell always

leads to an insulator!• In higher d, it depends on size of gaps

E

kx π/a−π/a 0 0 |k|

Different direction of k

Fermi Energy

Semi-metalE

Summary• Solving the “Central Equation” in Fourier space

Bloch Theorem Bloch states for electrons in crystals

• Nearly Free ElectronsGeneral Rules First Free electron bandsThen add effects of small potential

• Energy Bands and Band Gaps -- basis for understanding metals vs. insulators

• Read Kittel Ch 7