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Lecture VIII Band theory dr hab. Ewa Popko

Lecture VIII Band theory dr hab. Ewa Popko. The calculation of the allowed electron states in a solid is referred to as band theory or band structure theory.

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Lecture VIII Band theory dr hab. Ewa Popko

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  1. Lecture VIIIBand theorydr hab. Ewa Popko

  2. 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. Band Theory

  3. 0 V(r) E2 E1 E0 Increasing Binding Energy r Bound States in atoms Electrons in isolated atoms occupy discrete allowed energy levels E0, E1, E2 etc. . The potential energy of an electron a distance r from a positively charge nucleus of charge q is

  4. + + + + + Nuclear positions a Bound and “free” states in solids 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. 0 V(r) E2 E1 E0 V(r) Solid V(r) lower in solid (work function). Naive picture: lowest binding energy states can become free to move throughout crystal r 0

  5. E + + + + + position 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).

  6. 0 Influence of the lattice periodicity In the free electron model, the allowed energy states are where for periodic boundary conditions nx , ny and ny positive or negative integers. Periodic potential Exact form of potential is complicated Has property V(r+ R) = V(r) where R = m1a + m2b + m3c where m1, m2, m3 are integers and a ,b ,c are the primitive lattice vectors. E

  7. Wave moving to right Scattered waves moving to left a Waves in a periodic lattice Recall X-ray scattering in Solid State: nl = 2asina Consider a wave, wavelength l moving through a 1D lattice of period a. Strong backscattering for nl = 2a Backscattered waves constructively interfere. Wave has wavevector k = 2p/l. Scattering potential period a 1D lattice: Bragg condition is k = np/a (n – integer) 3D lattice: Scattering for k to k' occurs if k' = k + G where G = ha1 + ka2 + la3 h,k,l integer and a1 ,a2 ,a3 are the primitive reciprocal lattice vectors k' G k

  8. R for a crystal can be expressed in general as: R=n1a1+n2a2+n3a3 where a1, a2 and a3 are the primitive lattice vectors and n1,n2 and n3 are integers Corresponding to a1, a2 and a3 there arethree primitive reciprocal lattice vectors: b1, b2 and b3 defined in terms of a1, a2 and a3 by: Real and Reciprocal Lattice Spaces

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

  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+(2p/a) thenFk(x)=exp(ikx)u(x) and Fk’(x)=exp(ik’x)u(x)=exp(ikx)exp(i2px/a)u(x) =exp(ikx)u’(x) essentially have the same “wavelength”

  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(px/a) Sin(px/a) In a periodic lattice the allowed wavefunctions have the property where R is any real lattice vector. Cos2(px/a) Sin2(px/a)

  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 ψ(+) = Acos(px/a) gives so Solving Schrödinger’s equation with

  13. Gaps at the Brillouin zone boundaries At points A ψ(+)= 2 cos(px/a) andE=(hk)2/2me - V0/2 . At points B ψ(-) = 2isin(px/a)and E=(hk)2/2me + V0/2 .

  14. 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 + R1 + R2). This can be written Or Therefore a(R1 + R2) = (R1) + (R2) (R) is linear in R and can be written (R) = kxRx + kyRy + kzRz = k.R. where kx, ky and kz are the components of some wavevector k so (Bloch’s Theorem)  Bloch States

  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 ≈ eikx for a large fraction of the crystal volume.

  16. ψ(r) = exp[ik.r]u(r) Bloch Wavefunctions: allowed k-states 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 Therefore i.e. kx =2pnx/L nx 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. 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 ~p/a E=(hk)2/2me+ or - V0/2 Nearly Free Electrons Construct Bloch wavefunctions of electrons out of plane wave states.

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

  19. Consider a electron in the ground, 1s, state of a hydrogen atom The Hamiltonian is Solving Schrödinger’s equation : E = E1s = -13.6eV E1s F(r) + V(r) Molecular orbitals and bonding

  20. Consider the H2+ molecular ion in which one electron experiences the potential of two protons. The Hamiltonian is We approximate the electron wavefunctions as and e- r R p+ p+ Hydrogen Molecular Ion

  21. Solution: E = E1s – g(R) for E = E1s + g(R) for g(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) Bonding andanti-bonding states

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

  23. Simple cubic: nearest neighbour atoms at So E(k) =- a -2g(coskxa + coskya + coskza) Minimum E(k) =- a -6g for kx=ky=kz=0 Maximum E(k) =- a +6g for kx=ky=kz=+/-p/2 Bandwidth = Emav- Emin = 12g For k << p/a cos(kxx) ~ 1- (kxx)2/2 etc. E(k) ~ constant + (ak)2g/2 c.f. E = (hk)2/me • = 10 g = 1 E(k) -p/a p/a k [111] direction E(k) for a 3D lattice Behave like free electrons with “effective mass” h/a2g

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

  25. Bloch states Let k = ḱ + 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. • = 10 g = 1 E(k) -p/a p/a k [111] direction Independent Bloch states 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. The only independent values of k are those in the first Brillouin zone.

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

  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. Independent k-states in the first Brillouin zone, i.e. kx</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/Vin k-space. Volume of first BZ is (2/a)3 Total number of allowed k-states in a band is therefore The number of states in a band 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. 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 EF Metal due to overlapping bands Metals and insulators

  30. Part Filled Band Empty Band Partially Filled Band Part Filled Band Energy Gap Full Band Energy Gap Full Band EF EF INSULATOR METAL METAL or SEMICONDUCTOR or SEMI-METAL

  31. 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. Germanium Bands in 3D Germanium Figure removed to reduce file size

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