Electronic structure of solids

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1 Electronic structure of solids 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. Ackowledgement: These slides have used materials from the following link:

2 Problem.

3 How do we approach the problem? PE operator V =? to describe system, boundary conditions KE operator T = -( ħ 2 /2m) d 2 /dx 2,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 Ψ n (x), E n

4 e ix = exp(ix) = cos(x) + i sin(x)... wavelength = 2π phase difference = π/2 between sin and cos functions For other functions e ikx = exp(ikx) = cos(kx) + i sin(kx)... Wavelength = 2π/k

5 Wavefunction for a free electron Potential energy, V(x) = 0 everywhere Kinetic energy, T= -(ħ 2 /2m)d 2 /dx 2 H = T+V Schrödinger equation, HΨ=EΨ

6 Solutions: Ψ k (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. E k = ħ 2 k 2 /2m is kinetic, also the total energy, as V = 0. Monochromatic: wavelength = 2π/k Precisely defined momentum, p x = ħk and Δp x = 0 p y =p z =0, Δp y = Δp z =0 Totally delocalised position: Ψ(x)Ψ*(x) = 1/a = constant over x ie. Δx = But Heisenberg uncertainty principle demands, Δp x. 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 (Δp x >0).

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

8 Hückel Theory Aim: A model for π systems such as C n H m 1. Hückel theory is wavefunction-based. MO s are formed by LCAO Use the φ=c:2p z as the atomic orbital basis n φ s are used LCAO to form n MOs ψ j labeled with j=0..(n-1): Ψ j = N.Σ m c jm φ m sum runs over m = 1 to n, c jm are coefficients to be determined and N is a normalization constant.

9 2. Hückel theory is semi-empirical. We evaluate the electronic energy of a system in terms of two integrals. H ii = φ i Hφ i dτ H ij = φ i Hφ j dτ The values of these integrals are used by fitting. Use experimental data or use higher levels of theory ( semi-empirical method). For φ = C:2p z we use the AO energy: α= E(C:2p z ) = H ii = coulomb integral = φ 1 Hφ 1 dτ = 1050 kj/mol > 0 Interaction between two AOs: β = H ij = exchange or overlap integral = φ 1 Hφ 2 dτ = E(p z -p z overlap) < 0 Secular determinant

10 Φ = Σ n = 1 N c n f n H is known Apply variational principle to get energy Φ = c 1 f 1 + c 2 f 2 ΦH Φdτ = (c 1 f 1 + c 2 f 2 ) H (c 1 f 1 + c 2 f 2 )dτ = c 12 H 11 + c 1 c 2 H 12 + c 1 c 2 H 21 + c 22 H 22 H ij = H ji Φ*Φdτ = c 12 S 11 + c 1 c 2 S 12 + c 1 c 2 S 21 + c 22 S 22 S ij = S ji E(c 1,c 2 ) = c 12 H 11 + c 1 c 2 H 12 + c 1 c 2 H 21 + c 22 H 22 c 12 S 11 + c 1 c 2 S 12 + c 1 c 2 S 21 + c 22 S 22

11 Rewrite the eq. Differentiate w.r.t c1 and c2 E/ c 1 = 0 c 1 (H 11 -ES 11 ) + c 2 (H 12 ES 12 ) = 0 Secular determinant. H 11 -ES 11 H 12 ES 12 = 0 H 21 -ES 21 H 22 ES 22

12 C 2 H 4 C: 2s+2p x +2p y = sp 2 forms σ skeleton for C-C and C-H C: 2p z form C-C π 2 free electrons occupy the π system. Antibonding Out of phase, higher energy than α Bonding In phase, lower energy than α

13 We have two atoms, two AOs (m) labeled 1 and 2 Two MOs (j) labeled 0,1. ψ j =(1/ 2)Σ m exp[iπj(m-1)].φ m E j =α+2βcos(πj) Consider j = 0, in-phase ψ 0 =(1/ 2)(exp[0].φ 1 +exp[0].φ 2 ) =(1/ 2) (cos[0].φ 1 +isin[0].φ 1 +cos[0].φ 2 +isin[0].φ 2 ) =(1/ 2) ([1].φ 1 +i[0].φ 1 +[1].φ 2 +i[0].φ 2 ) =(1/ 2) (φ 1 +φ 2 ) E 0 =α+2βcos(πj) = α+2β and Consider j=1, out of phase ψ 1 =(1/ 2) (exp[0].φ 1 +exp[iπ].φ 2 ) =(1/ 2) (cos[0].φ 1 +isin[0].φ 1 +cos[π].φ 2 +isin[π].φ 2 ) =(1/ 2) ([1].φ 1 +i[0].φ 1 +[-1].φ 2 +i[0].φ 2 ) =(1/ 2) (φ 1 - φ 2 ) E 1 =α+2βcos(π) = α-2β

14 Cyclic polyenes For a general π system of cyclic C n H n There are n atomic orbitals φ=c:2p z n MO ψ j labeled with j=0..(n-1) formed by LCAO: ψ j =NΣ m c jm φ m ψ j =(1/ n) Σ m=1..n exp[i(m-1).phase-angle].φ m E j = α+ 2βcos(phase-angle) where phase-angle=2πj/n

15 j=3, ψ 3, phase = π ψ j =(1/ 6) Σ m=1..6 exp[i.j(m-1).π/3].φ m, j=0..5 ψ3= 1/ 6) Σ m=1..6 exp[iπ(m-1)].φ m = 1/ 6) Σ m=1..6 (-1) m-1.φ m =(φ 1 -φ 2 +φ 3 -φ 4 +φ 5 -φ 6 )/ 6 Phase changes as 0, π/3, 2π/3, 5π/3 ψ j =(1/ 6)Σ m=1..6 exp[ij(m-1).π/3].φ m, j = 0..5 e.g. ψ 0 =(1/ 6)Σ m=1..6 exp[i0(m-1)].φ m =(1/ 6)Σ m=1..6 (+1) m-1.φ m =(φ 1 +φ 2 +φ 3 +φ 4 +φ 5 +φ 6 )/ 6) phase = 0

17 For benzene: E j = α+ 2βcos(phase-angle) phase-angle=2πj/n n = 6, j = 0..5

18 Energy as a function of phase angle Dispersion diagram

19 Density of states

20 Total MO ψ is formed by combining atomic wavefunctions φ. phase factor exp(iθ). Because of periodic/cyclic boundary conditions, ψ can 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.

21 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 ψ k are obtained by combining the wavefunction for a single cell with the phase factor k for the interactions between cells. Combining energy for single cell energy (~Hückel α) with interaction between cells (β and phase factor k) gives a band of energies E(k).

22 Block Theorem Solution of the Schrodinger equation for a periodic potential will be of the type, Ψ k (r) = u k (r) exp (ik.r) Where u k (r) has the period of the crystal lattice. exp(ik.r) introduces correct phase, where k.r =k x x+k y y+k z z Solutions continuous at cell boundaries will be chosen. Ψ n,k (x) and E n (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.

23 a) from this origin, lay out the normal to every family of parallel planes in the direct lattice b) set the length of each normal equal to 2p times the reciprocal of the interplanar spacing for its particular set of planes c) place a point at the end of each normal.

24 Energy bands in one dimension

25 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 points 2. At the midpoint of these lines draw new lines, normal to these lines 3. Smallest volume occupied this way is the Wigner-Seitz primitive cell. All available space of the crystal can be filled with this. Brillouin zone

26 Crystal and reciprocal lattice in 1D a O A k = -π/a k = π/a Brillouin zone k

27 First Brillouin zone for fcc

28 Dispersion Each solution u n (x) leads to a band of energies E n (k). There exist forbidden regions of E where boundary conditions can not be fulfilled at any k; these are band gaps. E n (k) is called the dispersion of the band.

29 Dispersion diagram of p z orbitals for infinite atoms arranged on x or y direction

30 s orbitals

31 p x orbitals

32 An arbitrary system

33 Crystal potential splits E k

34 origin Symmetry points in the Brillouin zone

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

Introduction to Condensed Matter Physics

Introduction to Condensed Matter Physics The Reciprocal Lattice M.P. Vaughan Overview Overview of the reciprocal lattice Periodic functions Reciprocal lattice vectors Bloch functions k-space Dispersion