qm theory of solids: part - carleton universitytjs/4700_pdf/lecphys5.pdf · qm theory of solids:...

QM theory of Solids: Part I

Band structures, bonding, Fermions

Fig 4.1

From Principles of Electronic Materials and Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)

Introduction

•  We can now develop a QM theory of solids. •  We deal with crystals (otherwise too difficult) •  Band structure with electron “filling”. •  The band structure is determined by the lattice and atomic structure. •  Primarily interested in two top most bands (valance and conduction) •  Wish to understand the effect of T, applied voltage and incident

light. •  Three types of materials:

–  Metals – overlapping bands –  Semiconductors – small band gap ~ 1ev –  Insulators – large band gap > 2ev

•  Start with quantum bonding of atoms (in simplistic manner).

Fig 4.1


Fig 4.1


Bring together two H atoms and we have an interaction which produces two possible solutions with different energies. The two electrons pair their spins and occupy the bonding orbital. Using a linear combination of the single atom we can get an approximate solution. A exact solution would require solving Sch. Eq.

Quantum Mechanics and Bonding


Linear Combination of Atomic Orbitals

Two identical atomic orbitals ψ1s on atoms A and B can be combined linearly in two different ways to generate two separate molecular

orbitals ψσ and ψσ*

ψσ and ψσ* generated from a linear combination of atomic

orbitals (LCAO)

)()( 11 BsAs rr ψψψσ +=

)()( 11* BsAs rr ψψψσ

−=

This is a very approximate model. We will see a variety of approaches when we look at material modeling. Math becomes tough.

Fig 4.2


(a)  Electron probability distributions for bonding and antibonding orbitals, ψσ and (b) ψσ*. (b) Lines representing contours of constant probability (darker lines represent greater relative probability).

Lower Energy level – due to electron density between nuclei. Higher Energy Level

Fig 4.3


(a)  Energy of and * vs. the interatomic separation R. (b)  Schematic diagram showing the changes in the electron energy as two isolated H atoms, far left and far right, come together to form a hydrogen molecule.

Splitting of levels as be form H2 Minimum energy when both electron (spin up/down) occupy the lower state. This is the covalent bond from a QM perspective.

Fig 4.5


Two He atoms have four electrons. When He atoms come together, two of the electrons enter the Eσ level and two the Eσ* level, so the overall energy is greater than two isolated He atoms.

Situation is similar for He, but we have 4 electron so occupy both levels

Fig 4.7


(a) Three molecular orbitals from three Ψ1s atomic orbitals overlapping in three different ways. (b) The energies of the three molecular orbitals, labeled a, b, and c, in a system with three H atoms.

Add another atom – get another energy level

Fig 4.8


The formation of 2s energy band from the 2s orbitals when N Li atoms come together to form the Li solid. There are N 2s electrons, but 2N states in the band. The 2s band is therefore only half full. The atomic 1s orbital is close to the Li nucleus and remains undisturbed in the solid. Thus, each Li atom has a closed K shell (full 1s orbital).

Many atoms – many levels

Fig 4.9


As Li atoms are brought together from infinity, the atomic orbitals overlap and give rise to bands. Outer orbitals overlap first. The 3s orbitals give rise to the 3s band, 2p orbitals to the 2p band, and so on. The various bands overlap to produce a single band in which the energy is nearly continuous.

Enough atoms and we have energy bands. Just as we did for solving Ψ for a crystal using Bloch solutions. Note: the number of states is the number of atoms!

Fig 4.10


In a metal, the various energy bands overlap to give a single energy band that is only partially full of electrons. There are states with energies up to the vacuum level, where the electron is free.

Bands can overlap or can be separated by “band-gaps”

Fig 4.10


Bands and electron transport

• We saw the presence of a periodic electron structure produces bands when analyzed using Bloch functions and Sch.Eq. • Intuitively we can also look at it a scattering problem with the electrons undergoing Bragg diffraction.

Fig 4.50


An electron wave propagation through a linear lattice. For certain k values, the reflected waves at successive atomic planes reinforce each other, giving rise to a reflected wave traveling in the backward direction. The electron cannot then propagate through the crystal.

If the wavelength of the electron is 2a = nλ then we have destructive interference and no transmission. For this case k = nπ/a exactly the 1D edges for the Brillioun zone. Bragg Diffraction condition

Bragg Diffraction

Fig 4.51


Forward and backward waves in the crystal with k = +/- π/a give rise to two possible standing Waves Ψc and Ψs. Their probability density distributions | Ψc |2 and | Ψs |2 have maxima either at the ions or between the ions, respectively.

We can also imagine two configurations for a electron with k = π/a at the edge of the zone. These have two different energies (as with the symmetric and anti-symmetric solutions for H2) producing an energy gap.

Fig 4.52


The energy of the electron as a function of its wavevectore k inside a one-dimensional crystal. There are discontinuities in the energy at k = +/-nπ/a, where the waves suffer Bragg reflections in the crystal. For example, there can be no energy value for the electron between Ec and Es. therefore, Es-Ec is an energy gap at k = +/- π/a. Away from the critical k values, the E-k vector is like that of a free electron, with E increasing with k as . In a solid, these energies fall within an energy band.

emkE 2/2!=

At the edge of the zones we have diffraction and an energy gap. Away from the zone edges we have no diffraction and basically free electrons.

Fig 4.53


Diffraction of the electron in a two dimensional cubic crystal. Diffraction occurs whenever k has a component satisfying k1 = ±nπ/a, k2 = ±nπ/a or k3 = ±21/2nπ/a. In general terms, when ksinθ = nπ/d.

In a 2D or 3D crystal the Bragg Diffraction condition is function of the angle of incidence. For a given direction in the electron will scatter off the atom planes. This gives rise to 2D/3D Brillioun zones in k-space. d

nk πθ =sin

Fig 4.15


The electronic structure of Si

Si has an unfulfilled outer shell of electrons. With 4 valence electrons.

Single Silicon Atom electronic structure

Silicon band structure and bonding.

Fig 4.16


(a) Si is in Group IV in the Periodic Table. An isolated Si atom has two electrons in the 3s and two electrons in the 3p orbitals. (b) When Si is about to bond, the one 3s orbital and the three 3p orbitals become perturbed and mixed to form four hybridized orbitals, ψhyb, called sp3 orbitals, which are directed toward the corners of a tetrahedron. The ψhyb orbital has a large major lobe and a small back lobe. Each ψhyb orbital takes one of the four valence electrons.

When Si is solidified bonding is complicated but 4 strong hybridized bonds form with neighboring atoms in a tetrahedral lattice.

Bond Hybridization

Fig 4.17


(a) Formation of energy bands in the Si crystal first involves hybridization of 3s and 3p orbitals to four identical ψhyb orbitals which make 109.5o with each other as shown in (b). (c) ψhyb orbitals on two neighboring Si atoms can overlap to form ψB or ψA. The first is a bonding orbital (full) and the second is an anti-bonding orbital (empty). In the crystal yB overlap to give the valence band (full) and ψA overlap to give the conduction band (empty).

As with H and He we get symmetric and anti-symmetric solutions to Sch. Eq. Creates a tetrahedral crystal And produces two bands (Conduction and Valance) each with N electronic states.

Fig 4.18


Energy band diagram of a semiconductor. CB is the conduction band and VB is the valence band. AT 0 K, the VB is full with all the valence electrons.

At T=0 the valance band is full (Si has an even number of electrons and each band accepts 2N electrons. For T > 0 a few electron have enough KE to get excited over the band gap and become conduction (free) electrons.

Simplified Band model

Fig 4.55


(a) For the electron in a metal, there is no apparent energy gap because the second BZ (Brillouin zone) along [10] overlaps the first BZ along [11]. Bands overlap the energy gaps. Thus, the electron can always find any energy by changing its direction. (b) For the electron in a semicondcuctor, there is an energy gap arising from the overlap of the energy gaps along the [10] and [11] directions. The electron can never have an energy within this energy gap Eg.

Metal, Semiconductor and Insulator Band Structures

Fig 4.11


Typical electron energy band diagram for a metal. All the valence electrons are in an energy band, which they only partially fill. The top of the band is the vacuum level, where the electron is free from the solid (PE = 0).

In a metal we have a half filled band(s). Define the Fermi level as the energy level at which for T= 0 (no kinetic energy) all the lower states are filled. Define work function as energy needed to remove an electron from the Fermi energy.

Filling the Bands with Electrons


Work function Φ

The energy required to excite an electron from the Fermi level to the vacuum level, that is, to liberate the electron from the metal, is called the work function Φ of the metal.

The electrons in the energy band of a metal are loosely bound valence electrons, which become free in the crystal and thereby form a kind of electron gas within the crystal. It is this electron gas that holds the metal ions together in the crystal structure and constitutes the metallic bond.

Electron gas in a metal

Effect of a field

•  An applied electric field produces a force on the electron (a term in SE), but assume a small field and perturbation of the zero field case.

•  This changes the momentum of the electron and it’s energy. Will only be able to do this if a “space” or state is available in the band

•  As the field represents a force (a potential energy) the field changes the zero point energy of the electrons.

•  This means that the band structure moves up or down. •  When a simplified band structure is plotted versus x this is shown as

“band bending”.


Fig 4.12


(a) Energy band diagram of a metal. (b) In the absence of a field, there are as many electrons moving right as there are moving left. The motions of two electrons at each energy cancel each other as for a and b. (c) In the presence of a field in the -x direction, the electron a accelerates and gains energy to a’ where it is scattered to an empty state near EFO but moving in the -x direction. The average of all momenta values is along the +x direction and results in a net electrical current.

Applied E field increase the momentum (p) for all electrons. Sliding them to the right. Scattering stops them sliding all the way to infinity. Classical model is useful.

Fig 4.13


Conduction in a metal is due to the drift of electrons around the Fermi level. When a voltage is applied, the energy band is bent to be lower at the positive terminal so that the electron’s potential energy decreases as it moves toward the positive terminal.

Band structure as function of position with an applied potential. Potential raises the band structure (potential energy). Electron move to the right gaining KE losing PE then undergoes a scattering and drops energy by passing heat to the lattice.

Fig 4.19


As we saw in QM lecture the effective mass of the electron is determined by the E(k) relationship

GaAs Bandstructure

Effective Mass

Fig 4.13


Semiconductor/Metal Carrier Densities • Metals have effectively 1⁄2 filled bands • Semiconductors have a small band gap < 2ev • This leads to excitation of conduction electrons and production of holes. • We are interested in calculating the number of electrons (only this for metals) and holes in the bands. • To do this we need two things:

• The density of states in a band (the number of states/dE) -- g(E) • The probability of filling a state at a given temperature -- f(E)

• The electron density in E is then n(E) = g(E)f(E) • To obtain n integrate over the band! That’s simple!

Fig 4.19


Density of states.

• We saw that each atom contributes one electronic state to each band. • However, these states are not distributed uniformly over the band. • We need a function that prescribes the distribution over the band • This function provides the number of states/dE – and is usually denoted by g(E). The density of states function. • This function is determined by the geometric order of the structure.

• 0D – quantum dot (Quantum well) • 1D – quantum line (1D electron gas) • 2D – quantum plane (2D electron gas) • 3D – quantum solid (3D bulk crystal).


(a) In the solid there are N atoms and N extended electron wavefunctions from ψ1 all the way

To ψN. There are many wavefunctions, states, that have energies that fall in the central regions

Of the energy band. (b) The distribution of states in the energy band; darker regions have a

higher number of states. (c) Schematic representation of the density of states g(E) versus energy E.

One ψ state associated with each atom. Use a superposition of single atom states. Many closely related states (similar energies) in the band middle. g(E) has to go to zero at band edges

Fig 4.21


Each state, or electron wavefunctions in the crystal, can be represented by a box at n1, n2.

To calculate a 2D density of states. Assume parabolic E(k) Crystal of size L2 and a 2D Quantum well which means two quantum numbers n1 and n2 Each state has an energy of And we have constant energy surfaces given by circles. We can derive a formula for the number of states in ring of thickness dE and that is the density of states.

E =h2

8meL2(n2

1 + n22)

Fig 4.22


In three dimensions, the volume defined by a sphere of radius n' and the positive axes n1, n2, and n3, contains all the possible combinations of positive n1, n2, and n3 values that satisfy 22

322

21 nnnn ʹ′≤++

Likewise to calculate a 3D density of states. Assume parabolic E(k) Crystal of size L3 and a 3D Quantum well which means two quantum numbers n1, n2 and n3 Each state has an energy of And we have constant energy surfaces given by spheres. We can derive a formula for the number of states in shell of thickness dE and that is the density of states.

E =h2

8meL2(n2

1 + n22 + n2

3)


Density of States – 3D solid

g(E) = Density of states g(E) dE is the number of states (i.e., wavefunctions) in

the energy interval E to (E + dE) per unit volume of the sample.

( ) 2/12/3

22/128)( EhmE e ⎟

⎠

⎞⎜⎝

⎛= πg

Other systems.

•  Blue 3D (solid) •  Red 2D (plane) •  Green 1D (wire) •  What’s 0D (dot)?


Fig 4.24


• Classically we know an electron gas should follow a Boltzmann distribution

• But this distribution is for non-interacting particles and electrons strongly interact as they are fermions.

• The Pauli exclusion principle says only two electrons can occupy a state (spin up/down)

• The Boltzmann distribution only works when there are many more states then particles or the particle is not a Fermion.

• This is not true for electrons at the bottom of a band (where they all are!)

• We have to use Fermi-Dirac statistics which work for fermions.

• The classical expression is still useful at the top of the band.

The Energy Distribution


Boltzmann Classical Statistics

Boltzmann Statistics for two energy levels

⎟⎠

⎞⎜⎝

⎛−=kTEAEP exp)(

⎟⎠

⎞⎜⎝

⎛ −−=

kTEE

NN 12

1

2 exp

Boltzmann probability function

Boltzmann statistics describes systems of non-interacting (classical) particles under thermal equilibrium. Does not apply to electrons as they are Fermions subject to the Pauli exclusion principle. Exception is when there are many states and few electrons.

Fig 4.25


The Boltzmann energy distribution describes the statistics of particles, such as electrons, when there are many more available states than the number of particles. At T = 0 all particles to to E = 0

Fig 4.24


Two electrons with initial wavefunctions ψ1 and ψ2 at E1 and E2 interact and end up different energies E3 and E4. Their corresponding wavefunctions are ψ3 and ψ4.

Fermions - electrons

Electrons can only move to unoccupied states. So E4 and E3 must be empty. This changes the distribution function.

Fig 4.26


The fermi-Dirac f(E) describes the statistics of electrons in a solid. The electrons interact with each other and the environment, obeying the Pauli exclusion principle.

The distribution is radically different At T = 0 the distribution is 1.0 from 0 to the Fermi energy EF. At T > 0 we get smearing of the function.

Fermi-Dirac Function


Fermi-Dirac Statistics

where EF is a constant called the Fermi energy.

f(E) = the probability of finding an electron in a state with energy E is given

⎟⎠

⎞⎜⎝

⎛ −+

=

kTEE

EfFexp1

1)(

The Fermi-Dirac function

Fermi-Dirac Implications

•  We can now calculate the number of electrons that contribute to conduction.

•  This is the main “correction” that QM makes to the classical theory of conduction in a metal.

•  EF is the chemical potential of the electrons in the material. •  In general two materials will have different EF

•  When placed in contact electrons will flow from one material to the other until the Fermi levels align and thermal equilibrium is established.

•  The Fermi energy EF characterizes the material.



(a) Above 0K, due to thermal excitation, some of the electrons are at energies above EF.

(b) The density of states, g(E) versus E in the band. (c) The probability of occupancy of a state at an energy

E is f (E). (d) The product of g(E) f (E) is the number of electrons

per unit energy per unit volume, or the electron concentration per unit energy. The area

under the curve on the energy axis is the concentration of electrons in the band.

n =

Z 1

0f(e)g(e)dE

n =8⇡21/2m3/2

e

h3

Z 1

0

E1/2dE

1 + eE�EF

kT

Calculating n


Fermi Energy

Fermi energy at T (K)

3/22 38

⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

πn

mhEe

FO

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

22

121)(

FOFOF E

kTETE π

Fermi energy at T = 0 K

n is the concentration of conduction electrons (free carrier concentration)

We can express EF|T=0 in terms of n

And also obtain an expression for EF(T)

Metal Conductivity •  QM gives an Ohm’s Law

like conductivity. Assuming a scattering process.

•  But only the electrons at EF (ie. vF) contribute to the drift current.

•  The σ is proportional to g(EF) which is the density of states at the Fermi Energy.

•  Very important in determining the conductivity of a metal.


�QM =1

3e2vF ⌧g(EF )

�classical = e2n⌧

me

( ) 2/12/3

22/128)( EhmE e ⎟

⎠

⎞⎜⎝

⎛= πg


Average energy per electron

Average energy per electron at T (K)

FOEE53)0(av =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

22

av 1251

53)(

FOFO E

kTETE π

Average energy per electron at 0 K EF is related to the average energy of the electrons. At thermal equilibrium EF will be a constant value.

Fig 4.28


(a)  Electrons are more energetic in Mo, so they tunnel to the surface of Pt. (b)  Equilibrium is reached when the Fermi levels are lined up.

When two metals are brought together, there is a contact potential V.

At an interface of two metals a exchange of electrons will occur to bring the Fermi Energies into alignment at thermal equilibrium. Surface charge and E field is present at the interface. This produces a contact potential drop.

Fig 4.29


There is no current when a closed circuit is formed by two different metals, even though there is a contact potential at each contact. The contact potentials oppose each other.

In equilibrium a closed system must have I = 0. So the potential drops must cancel out. A simple ring of two metals show how the potentials will cancel.


Fermi Energy Significance

For a given metal the Fermi energy represents the free energy per electron called the electrochemical potential. The Fermi energy is a measure of the potential of an electron to do electrical work (e×V) or non-mechanical work, through chemical or physical processes.

Summary

•  This lecture primarily dealt with metals we will deal with semiconductors in detail later.

•  The QM model of the metal provides insight into: – Band structure – Fermi energy – Free electrons – Effective mass – Density of states – Number of electrons contributing to

conduction


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