Electronic structure of solids - :: DST Unit of · PDF fileElectronic structure of solids ......

2007 T. Pradeep CY306 Lectures

Eigenvalue equation:

Áf(x) = af(x)

KNOWN: Á is an operator.

UNKNOWNS:

f(x) is a function (and a vector), an ‘eigenfunction’ of Á;

a is a number (scalar), the ‘eigenvalue’.

Electronic structure of solids

Ackowledgement: These slides have used materials from the following link:http://www.tyndall.ie/research/computational-modelling-group/cm4107/


Problem….


PE operator V = ? to describe system, boundary conditions

KE operator T = -( ħ2/2m) d2/dx2,where m is mass and ħ = h/2π

Hamiltonian H = T+V

Time-independent Schrödinger equation, HΨ =E Ψ

Suggest a general solution Ψ(x)

Solve for E

Apply boundary conditions , get quantum number -- n

Specific set of solutions Ψn(x), En

How

do we approach the problem

?


eix = exp(ix) = cos(x) + i sin(x) ... wavelength = 2πphase difference = π/2 between sin and cos functions

For other functionseikx = exp(ikx) = cos(kx) + i sin(kx) ... Wavelength = 2π/k


Wavefunction for a free electron

Potential energy, V(x) = 0 everywhereKinetic energy, T= -(ħ2/2m)d2/dx2

H = T+V

Schrödinger equation, HΨ=EΨ


Solutions:Ψk(x)=[1/√a].exp(ikx) = [/√a][cos(kx)+isin(kx)] are plane waves

k is the quantum number; any value of k is possible for the free electron.

Ek= ħ2k2/2m is kinetic, also the total energy, as V = 0.

Monochromatic: wavelength = 2π/kPrecisely defined momentum, px = ħk and Δpx = 0

py=pz=0, Δpy= Δpz=0

Totally delocalised position: Ψ(x)Ψ*(x) = 1/a = constant over x

ie. Δx = ∞

But Heisenberg uncertainty principle demands, Δpx . Δx ≥ ħ/2

Real beams of free electrons are somewhat localised (Δx ≠∞)It can be represented as a wave packet, with a distribution of momenta (Δpx >0).


Question:

Is free electron a good approximation for the electrons in a crystalline solid?


Hückel Theory

Aim: A model for π systems such as CnHm

1. Hückel theory is wavefunction-based.MO’s are formed by LCAO

Use the φ=C:2pz as the atomic orbital ‘basis’n φ’s are used LCAO to form n MOs ψj labeled with j=0..(n-1):

Ψj = N.Σm cjmφmsum runs over m = 1 to n, cjm are coefficients to be determined and Nis a normalization constant.


2. Hückel theory is semi-empirical.

We evaluate the electronic energy of a system in terms of two integrals.

Hii = ∫ φiHφidτHij = ∫ φiHφjdτ

The values of these integrals are used by fitting. Use experimental data or use higher levels of theory ( ”semi-empirical” method).

For φ = C:2pz we use the AO energy: α= E(C:2pz) = Hii = coulomb integral = ∫ φ1Hφ1dτ = 1050 kJ/mol > 0

Interaction between two AOs:β = Hij = exchange or overlap integral = ∫ φ1Hφ2dτ = E(pz-pz overlap) < 0

Secular determinant


Φ = Σn = 1…N cnfn

H is known

Apply variational principle to get energy

Φ = c1f1 + c2f2

∫ ΦH Φdτ = ∫(c1f1 + c2f2) H (c1f1 + c2f2) dτ= c1

2H11 + c1c2H12 + c1c2H21 + c22H22

Hij = Hji

∫ Φ*Φdτ = c12S11 + c1c2S12 + c1c2S21 + c2

2S22

Sij = Sji

E(c1,c2) = c12H11 + c1c2H12 + c1c2H21 + c2

2H22

c12S11 + c1c2S12 + c1c2S21 + c2

2S22


Rewrite the eq. Differentiate w.r.t c1 and c2 ∂E/ ∂c1 = 0

c1(H11-ES11) + c2 (H12 – ES12) = 0

Secular determinant.

H11-ES11 H12 – ES12= 0

H21-ES21 H22 – ES22


C2H4

C: 2s+2px+2py = sp2 forms σ skeleton for C-C and C-H

C: 2pz form C-C π

2 ‘free’ electrons occupy the π system.

In phase, lower energy than α

Out of phase, higher energy than α

Bonding

Antibonding


We have two atoms, two AOs (m) labeled 1 and 2Two MOs (j) labeled 0,1.

ψj=(1/√2)Σmexp[iπj(m-1)].φmEj=α+2βcos(πj)

Consider j = 0, in-phaseψ0=(1/√2)(exp[0].φ1+exp[0].φ2)=(1/√2) (cos[0].φ1+isin[0].φ1+cos[0].φ2+isin[0].φ2)=(1/√2) ([1].φ1+i[0].φ1+[1].φ2+i[0].φ2)=(1/√2) (φ1+φ2)

E0=α+2βcos(πj)= α+2β and

Consider j=1, out of phaseψ1=(1/√2) (exp[0].φ1+exp[iπ].φ2)=(1/√2) (cos[0].φ1+isin[0].φ1+cos[π].φ2+isin[π].φ2)=(1/√2) ([1].φ1+i[0].φ1+[-1].φ2+i[0].φ2)=(1/√2) (φ1 - φ2)

E1=α+2βcos(π) = α-2β


Cyclic polyenesFor a general π system of cyclic CnHnThere are n atomic orbitals φ=C:2pz

n MO ψj labeled with j=0..(n-1) formed by LCAO:

ψj=NΣm cjmφmψj=(1/√n) Σm=1..n exp[i(m-1).phase-angle].φm

Ej= α+ 2βcos(phase-angle)

where phase-angle=2πj/n


ψj=(1/√6)Σm=1..6 exp[ij(m-1).π/3].φm, j = 0..5e.g. ψ0=(1/√6)Σm=1..6 exp[i0(m-1)].φm=(1/√6)Σm=1..6 (+1)m-1.φm=(φ1+φ2+φ3+φ4+φ5+φ6)/ √6) phase = 0

j=3, ψ3, phase = πψj=(1/√6) Σm=1..6 exp[i.j(m-1).π/3].φm, j=0..5ψ3= 1/√6) Σm=1..6 exp[iπ(m-1)].φm= 1/√6) Σm=1..6 (-1)m-1.φm=(φ1-φ2+φ3-φ4+φ5-φ6 )/√6

Phase changes as 0, π/3, 2π/3,… 5π/3



For benzene:

Ej= α+ 2βcos(phase-angle)phase-angle=2πj/n

n = 6, j = 0..5


Energy as a function of phase angle

Dispersion diagram


Density of states


Total MO ψ is formed by combining

atomic wavefunctions φ . phase factor exp(iθ).

Because of periodic/cyclic boundary conditions, ψcan be labeled according to phase.

MO energies depend on phase. Because phase is limited (0..2π), the energies are bounded: in the Hückel case, α+2β ≤ E ≤ α-2β.

α is the contribution to the energy from an individual unit φ.

β is due to (i) geometrical arrangement in space; (ii) closeness in energy; (iii) shape, parity of φ.


Why are cyclic polyenes relevant for crystalline solids?

Periodicity = cyclic symmetry or translational symmetry.

Consider each C atom as a 1-D unit cell of length a.

Define ‘crystal momentum’ k = phase/a.

Wavefunctions ψk are obtained by combining the wavefunction for asingle cell with the phase factor k for the interactions between cells.

Combining energy for single cell energy (~Hückel α) with interaction betweencells (β and phase factor k) gives a ‘band’ of energies E(k).


Block Theorem

Solution of the Schrodinger equation for a periodic potential will be of the type,Ψk(r) = uk(r) exp (ik.r)

Where uk(r) has the period of the crystal lattice.

exp(ik.r) introduces correct phase, where k.r = kxx+kyy+kzz

Solutions continuous at cell boundaries will be chosen.

Ψn,k(x) and En(k) in the range of -π/a ≤ k ≤ π/a (first Billouin zone) will be of significance

The energies spanned for each n will form a band.


a) from this origin, lay out the normal to every family of parallel planes in the direct latticeb) set the length of each normal equal to 2p times the reciprocal of the interplanarspacing for its particular set of planesc) place a point at the end of each normal.


Energy bands in one dimension


Brillouin zone

A Brillouin zone is a Wigner-Seitz cell in the reciprocal lattice.

Wigner-Seitz cell

1. Draw lines to connect a lattice point to all the nearby points2. At the midpoint of these lines draw new lines, normal to these lines3. Smallest volume occupied this way is the Wigner-Seitz primitive cell. All

available space of the crystal can be filled with this.

Brillouin zone


Crystal and reciprocal lattice in 1D

a

A

k

O

k = -π/a k = π/aBrillouin zone


First Brillouin zone for fcc


Dispersion

Each solution un(x) leads to a band of energies En(k). There existforbidden regions of E where boundary conditions can not be fulfilled at any k; these are ‘band gaps’.

En(k) is called the dispersion of the band.


Dispersion diagram of pz orbitals for infinite atoms arranged on x or y direction


s orbitals


px orbitals


An arbitrary system


Crystal potential splits Ek


Symmetry points in the Brillouin zoneorigin


References:

C. Kittel, Introduction to Solid State Physics, Wiley Eastern Ltd., New Delhi, 1993.D.A. McQuarrie and J. D. Simon, Physical Chemistry A Molecular Approach, Viva Books Pvt. Ltd. New Delhi, 1998.

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