1. 1 Lecture VIII Band theory dr hab. Ewa Popko

2. 2 Band Theory The calculation of the allowed electron states in a solid is referred to as band theory or band structure theory. To obtain the full band structure, we need to solve Schrödinger’s equation for the full lattice potential. This cannot be done exactly and various approximation schemes are used. We will introduce two very different models, the nearly free electron and tight binding models. We will continue to treat the electrons as independent, i.e. neglect the electron-electron interaction.

3. 3 Bound States in atoms Electrons in isolated atoms occupy discrete allowed energy levels E 0, E 1, E 2 etc.. The potential energy of an electron a distance r from a positively charge nucleus of charge q is V(r) E2E1E0E2E1E0 r 0 Increasing Binding Energy

4. 4 Bound and “free” states in solids V(r) E2E1E0E2E1E0 The 1D potential energy of an electron due to an array of nuclei of charge q separated by a distance a is Where n = 0, +/-1, +/-2 etc. This is shown as the black line in the figure. r 0 0 + ++++ a Nuclear positions V(r) lower in solid (work function). Naive picture: lowest binding energy states can become free to move throughout crystal V(r) Solid

5. 5 Energy Levels and Bands Isolated atoms have precise allowed energy levels. In the presence of the periodic lattice potential bands of allowed states are separated by energy gaps for which there are no allowed energy states. The allowed states in conductors can be constructed from combinations of free electron states (the nearly free electron model) or from linear combinations of the states of the isolated atoms (the tight binding model). + E +++ + position

6. 6 Influence of the lattice periodicity In the free electron model, the allowed energy states are where for periodic boundary conditions n x, n y and n y positive or negative integers. 0 E Periodic potential Exact form of potential is complicated Has property V(r+ R) = V(r) where R = m 1 a + m 2 b + m 3 c where m 1, m 2, m 3 are integers and a,b,c are the primitive lattice vectors.

7. 7 Waves in a periodic lattice Recall X-ray scattering in Solid State: n  = 2asin  Consider a wave, wavelength moving through a 1D lattice of period a. Strong backscattering for n  = 2a Backscattered waves constructively interfere. Wave has wavevector k = 2  a Wave moving to right Scattered waves moving to left Scattering potential period a 1D lattice: Bragg condition is k = n  /a (n – integer) 3D lattice: Scattering for k to k' occurs if k' = k + G where G = ha 1 + ka 2 + la 3 h,k,l integer and a 1,a 2,a 3 are the primitive reciprocal lattice vectors k k'k' G

8. 8 Real and Reciprocal Lattice Spaces R for a crystal can be expressed in general as: R=n 1 a 1 +n 2 a 2 +n 3 a 3 where a 1, a 2 and a 3 are the primitive lattice vectors and n 1,n 2 and n 3 are integers Corresponding to a 1, a 2 and a 3 there are three primitive reciprocal lattice vectors: b 1, b 2 and b 3 defined in terms of a 1, a 2 and a 3 by:

9. 9 Bragg scattering & energy gaps 1D potential period a. Reciprocal lattice vectors G = 2n  /a A free electron of in a state exp( i  x/a), ( rightward moving wave) will be Bragg reflected since k =  /a and a left moving wave exp( -i  x/a) will also exist. In the nearly free electron model allowed un-normalised states for k =  /a are  (+) = exp(i  x/a) + exp( - i  x/a) = 2 cos(  x/a) ψ(-) = exp(i  x/a) - exp( - i  x/a) = 2i sin(  x/a) + E + ++ + position a N.B. Have two allowed states for same k which have different energies

10. 10 Reciprocal lattice Use of reciprocal lattice space:  Wave vectors k for Bloch waves lie in the reciprocal lattice space.  Translation symmetry=> a Bloch wave can be characterized by two wavevectors (or wavelengths) provided they differ by a reciprocal lattice vector!  Example in 1D: Suppose k’=k+(2  /a) then  k (x)=exp(ikx)u(x) and  k’ (x)=exp(ik’x)u(x)=exp(ikx)exp(i2  x/a)u(x) =exp(ikx)u’(x) essentially have the same “wavelength”

11. 11 Cosine solution lower energy than sine solution Cosine solution ψ(+) has maximum electron probability density at minima in potential. Sine solution ψ(-) has maximum electron probability density at maxima in potential. Cos(  x/a) Sin(  x/a) Cos 2 (  x/a) Sin 2 (  x/a) In a periodic lattice the allowed wavefunctions have the property where R is any real lattice vector.

12. 12 Magnitude of the energy gap Let the lattice potential be approximated by Let the length of the crystal in the x-direction to be L. Note that L/a is the number of unit cells and is therefore an integer. Normalising the wavefunction  (+) = Acos(  x/a) gives so Solving Schrödinger’s equation with

13. 13 Gaps at the Brillouin zone boundaries At points A ψ (+) = 2 cos(  x/a) and E=(  k) 2 /2m e - V 0 /2. At points B ψ (-) = 2isin(  x/a) and E=(  k) 2 /2m e + V 0 /2.

14. 14 Bloch States In a periodic lattice the allowed wavefunctions have the property where R is any real lattice vector. Therefore where the function  (R) is real, independent of r, and dimensionless. Now consider ψ (r + R 1 + R 2 ). This can be written Or Therefore  (R 1 + R 2 ) =  (R 1 ) +  (R 2 )  (R) is linear in R and can be written  (R) = k x R x + k y R y + k z R z = k.R. where k x, k y and k z are the components of some wavevector k so (Bloch’s Theorem)

15. 15 (Bloch’s Theorem) For any k one can write the general form of any wavefunction as where u(r) has the periodicity ( translational symmetry) of the lattice. This is an alternative statement of Bloch’s theorem. Alternative form of Bloch’s Theorem Real part of a Bloch function. ψ ≈ e ikx for a large fraction of the crystal volume.

16. 16 Bloch Wavefunctions: allowed k-states ψ (r) = exp[ik.r]u(r) Periodic boundary conditions. For a cube of side L we require ψ (x + L) = ψ (x) etc.. So but u(x+L) = u(x) because it has the periodicity of the lattice therefore Thereforei.e. k x = 2  n x /Ln x integer. Same allowed k-vectors for Bloch states as free electron states. Bloch states are not momentum eigenstates i.e. The allowed states can be labelled by a wavevectors k. Band structure calculations give E(k) which determines the dynamical behaviour.

17. 17 Nearly Free Electrons Need to solve the Schrödinger equation. Consider 1D write the potential as a Fourier sum where G = 2  n/a and n are positive and negative integers. Write a general Bloch function where g = 2  m/a and m are positive and negative integers. Note the periodic function is also written as a Fourier sum Must restrict g to a small number of values to obtain a solution. For n= + 1 and –1 and m=0 and 1, and k ~  / a E=(  k) 2 /2m e + or - V 0 /2 Construct Bloch wavefunctions of electrons out of plane wave states.

18. 18 Tight Binding Approximation NFE Model: construct wavefunction as a sum over plane waves. Tight Binding Model: construct wavefunction as a linear combination of atomic orbitals of the atoms comprising the crystal. Where  (r)  is a wavefunction of the isolated atom r j are the positions of the atom in the crystal.

19. 19 Molecular orbitals and bonding Consider a electron in the ground, 1s, state of a hydrogen atom The Hamiltonian is Solving Schrödinger’s equation : E = E 1s = -13.6eV + E 1s V(r)  (r)

20. 20 Hydrogen Molecular Ion Consider the H 2 + molecular ion in which one electron experiences the potential of two protons. The Hamiltonian is We approximate the electron wavefunctions as and p+ e- R r

21. 21 Bonding and anti-bonding states Solution: E = E 1s –  (R) for E = E 1s +  (R) for  (R) - a positive function Two atoms: original 1s state leads to two allowed electron states in molecule. Find for N atoms in a solid have N allowed energy states V(r)

22. 22 The tight binding approximation for s states + ++++ a Nuclear positions Solution leads to the E(k) dependence!! 1D:

23. 23 E(k) for a 3D lattice Simple cubic: nearest neighbour atoms at SoE(k) =  2  (cosk x a + cosk y a + cosk z a) Minimum E(k) =  6  for k x =k y =k z =0 Maximum E(k) =  6  for k x =k y =k z =+/-  /2 Bandwidth = E mav - E min = 12  For k <<  a cos(k x x) ~ 1- (k x x) 2 /2 etc. E(k) ~ constant + (ak) 2  /2 c.f. E = (  k ) 2 /m e k [111] direction  /a  /a   E(k) Behave like free electrons with “effective mass”  /a 2 

24. 24 Each atomic orbital leads to a band of allowed states in the solid Band of allowed states Gap: no allowed states

25. 25 Independent Bloch states Bloch states Let k = k + G where k is in the first Brillouin zone and G is a reciprocal lattice vector. But G.R = 2  n, n-integer. Definition of the reciprocal lattice. So k is exactly equivalent to k. k [111] direction  /a  /a   E(k) The only independent values of k are those in the first Brillouin zone. Solution of the tight binding model is periodic in k. Apparently have an infinite number of k-states for each allowed energy state. In fact the different k-states all equivalent.

26. 26 Reduced Brillouin zone scheme The only independent values of k are those in the first Brillouin zone. Results of tight binding calculation Results of nearly free electron calculation Discard for |k| >  /a Displace into 1 st B. Z. Reduced Brillouin zone scheme -2  /a 2  /a

27. 27 Extended, reduced and periodic Brillouin zone schemes Periodic Zone Reduced Zone Extended Zone All allowed states correspond to k-vectors in the first Brillouin Zone. Can draw E(k) in 3 different ways

28. 28 The number of states in a band Independent k-states in the first Brillouin zone, i.e.  k x  <  /a etc. Finite crystal: only discrete k-states allowed Monatomic simple cubic crystal, lattice constant a, and volume V. One allowed k state per volume (2  ) 3 /V in k-space. Volume of first BZ is (2  /a) 3 Total number of allowed k-states in a band is therefore Precisely N allowed k-states i.e. 2N electron states (Pauli) per band This result is true for any lattice: each primitive unit cell contributes exactly one k-state to each band.

29. 29 Metals and insulators In full band containing 2N electrons all states within the first B. Z. are occupied. The sum of all the k-vectors in the band = 0. A partially filled band can carry current, a filled band cannot Insulators have an even integer number of electrons per primitive unit cell. With an even number of electrons per unit cell can still have metallic behaviour due to band overlap. Overlap in energy need not occur in the same k direction EFEF Metal due to overlapping bands

30. 30 Full Band Empty Band Energy Gap Full Band Partially Filled Band Energy Gap Part Filled Band EFEF INSULATORMETAL METAL or SEMICONDUCTORor SEMI-METAL EFEF

31. 31 Bands in 3D In 3D the band structure is much more complicated than in 1D because crystals do not have spherical symmetry. The form of E(k) is dependent upon the direction as well as the magnitude of k. Figure removed to reduce file size Germanium