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SOLID STATE PHYSICS HW#7 Question 1. Square Lattice, free electron energies(Kittel 7.1). (a) The free electron kinetic energy is given by E= ~2 k 2 . 2m (1) At the corner of the first zone ( πa , πa ), |~k|2 for the two dimensional wave vector ~k = kx i + ky j is equal to to π2 a2 2π 2 , a2 which leads to E = leading to E = ~2 π 2 . 2ma2 ~2 π 2 . ma2 At midpoint of a side face of the zone ( πa , 0), |~k|2 is equal The ratio of former to latter is 2. (b) In 3 dimensions, |~k|2 for a wave vector ~k = kx i + ky j + kz k is equal to of the first zone ( πa , πa , πa ). Then the free electron energy is E = corresponding to an energy E = ~2 π 2 . 2ma2 3~2 π 2 2ma2 3π 2 a2 at the corner . At ( πa , 0, 0), |~k|2 is π2 , a2 The ratio is 3. In d dimensions, the ratio is equal to d. (c) For divalent metals (metals with even number of valence electrons per primitive cell) a band overlap occurs: Instead of one filled band giving an insulator, we can have two partially filled bands giving a metal. The kinetic energy difference between the boundary and the middle of the 1st BZ causes such an overlap which leads to partially filled orbitals, leaving room for electron hopping, namely resulting in a finite conductivity. Question 2. Kronig-Penney Model (Kittel 7.3). (a) For a square well potential (U = 0 for 0 < x < a, and U = U0 for −b < x < 0), consider two solutions of the Schrödinger equation: ψI = AeiKx + Be−iKx for U = 0 , ψII = CeQx + De−Qx for U = U0 , (2) For a square well potential, there are 4 boundary conditions. 2 of them are from continuity of wave functions and their derivatives: ψI (a) = ψII (a) , dψI dψII |x=a = |x=a , dx dx (3) The other 2 boundary conditions are derived from Bloch theorem, that is, from periodicity of the wavefunctions: ~ ψk (~r) = uk (~r)eik·~r where uk (~r) = uk (~r + T~ ) . T~ : lattice translation vector . (4) These 4 boundary conditions result in 4 equations, namely: A+B = C +D, (5) ik(A − B) = Q(C − D) , (6) AeiKa + Be−iKa = (Ce−Qb + DeQb )eik(a+b) , (7) iK(AeiKa − Be−iKa ) = Q(Ce−Qb − DeQb )eik(a+b) . (8) The determinant of matrix of the coefficients A, B, C, D is then equal to · 2 ¸ Q − K2 sinh Qb sin Ka + cosh Qb cos Ka = cos k(a + b) , 2QK (9) which should vanish in order to obtain finite coefficients. Consider the dirac delta potential limit of the determinant equation above. The limit in order to obtain dirac delta function is b → 0, U0 → ∞, Q >> K, and Qb << 1. Under these conditions, the equation reduces to µ P Ka ¶ sin Ka + cos Ka = cos ka , where P ≡ Q2 ab . 2 (10) Setting k = 0 as in the question and using Ka << 1 for finding the energy of the lowest energy band, one should obtain P K 2 a2 Ka + 1 − ' 1. Ka 2 Then the energy for the lowest energy band is E = ~2 K 2 2m = (11) ~2 P . ma2 (b) For k = πa , eq. (10) reduces to µ ¶ P sin Ka + cos Ka = −1 . Ka (12) In the vicinity of the zone boundary, in order to find the band gap, taking Ka = π + δ, where δ << 1 and looking for the solutions of δ is convenient. Eq. (12) is rewritten as µ ¶ P δ2 (−δ) − 1 + = −1 . π 2 There are two solutions for δ: δ1 = 0 and δ2 = 2P π . (13) Here, δ2 causes a finite energy gap. The total energy is E= ~2 π 2 ~2 πδ ~2 (π + δ)2 = + . 2ma2 2ma2 ma2 (14) The second term corresponds to the energy gap at k = πa . Question 3. Square Lattice (Kittel 7.6). The potential energy is given by U (x, y) = −4U cos 2πx 2πy cos . a a (15) In terms of Fourier components, the potential energy is equal to U (x, y) = XX G 0 UG,G0 eiGx eiG y = 4 G0 X cos Gx 0 1 UG,G0 cos G0 y . (16) G0 >0 G>0 The coefficient UG,G0 is found as follows: Z 1 Z UG,G0 = dx X dyU (x, y) cos Gx cos G0 y . (17) 0 Here the reciprocal lattice vectors G = G0 are equal to 2π a , since the energy gap we are looking for is equivalent to the energy difference of two ends of the Brillouin zone, namely ( πa , πa ) and (− πa , − πa ). Actually, the coefficient UG,G0 gives us the potential energy difference between two lattice layers corresponding to ( πa , πa ) and (− πa , − πa ) in k-space, respectively. UG,G0 is found using the following orthogonality condition, Z L 2nπx 2mπx L dx cos cos = δn,m , a a 2 0 (18) and is equal to UG,G0 = −U . In 2 dimensions, the central equation is (λk − ²)C(kx , ky ) + X C(kx − G, ky − G0 ) = 0 . (19) G,G0 Here, C(kx , ky ) are the coefficients of the Fourier components of the wave function (see Kittel), and λk ≡ ~2 k2 2m . Remember, the central equation is the characteristic equation (the determinant equation which is used for finding eigenvalues) of the Hamiltonian matrix in k-space. ² corresponds to energy eigenvalues. At the boundary of 1st Brillouin zone, k = ± πa = ± 12 G. Then the two equations derived from the central equation are as follows: 1 1 (λ − ²)C( G) − U C(− G) = 0 , 2 2 1 1 (λ − ²)C(− G) − U C( G) = 0 . 2 2 (20) The determinant of the 2 × 2 coefficient matrix above gives the two energies, λ1 = ² + U and λ1 = ² − U . Ultimately, the energy gap is 2U at the corner point of the Brillouin zone.