Homework 2: Solutions

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Meryl Nelson
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1 Homework : Solutions. Problem. We have approximated the conduction band of many semiconductors with an isotropic, parabolic expression of the form: E(k) h k m. () However,this approximation fails at some large k. A first correction to this expression consists in adding also terms of order k 4,so that the dispersion is usually written as: E(k) h k ( m α h k ) m, () where α (of dimension of inverse energy) is called the nonparabolicity parameter. a. Using Eq. () derive the density of states of the conducton band only to first-order in the parameter α. b. Also to first-order in α,find the group velocity E(k)/ h as a function of k. (Note that,like the energy, the group velocity depends only on the magnitude k of k and not on its direction.) Solution. Let s first derive a few useful relations: From Eq. (),to first order in α: so that From Eq. (3),differentiating: k [m E( + αe)] / h k E ( + αe), (3) m h (m E) / h ( + α E). (4) kdk m ( + αe) de. (5) h ECE69 Spring

2 a. Then: ρ(e) = (m ) / m π h 3 (π) 3 dk δ[e(k) E] = π dkk δ[e(k) E] de ( + αe ) ( + α E ) δ(e E) / m 3/ π h 3 E / ( + 5 αe ) (6). b. From Eq. (5),to first-order in α: v(k) = h de dk hk ( αe). (7) m This is the usual parabolic velocity hk/m but corrected by the factor ( αe). InSi,α =.5eV,so the velocity is reduced with respect to its parabolic expression.. Problem. For a parabolic conduction band with effective mass m,derive E F as a function of electron density n at zero temperature. Hint: At T = the electrons in k-space fill a sphere of radius k F,the Fermi wavevector. Thus,the electron density will be the volume of this sphere divided by the density of states in k-space. Solution. The electron density is,by definition: n = V k f FD (k) = (π) 3 dk f FD (k). (8) But at T =f FD vanishes for E(k) >E F,that is for k>k F,where k F =(m E F ) / / h. Also, F FD (k) =for k k F.So dk f FD (k) = 4 3 πk3 F, (9) ECE69 Spring

3 (the volume of a sphere of radius k F in k-space) and so n = 4 (π) 3 3 πk3 F = k3 F 3π = (m ) 3/ 3π h 3 E /3 F, () E F = 3/3 π 4/3 h m n /3. () So the Fermi level is always above the bottom of the conduction band at T =. 3. Problem. In a band diagram,plot the position of the Fermi level for Si at 3 K as a function of donor concentration, N D. The plot may be qualitative,but it should be qualitatively correct. Try to be as accurate as possible. Solution. We have discussed in class the behavior in the non-degenerate limit. The lecture notes (page 4) show nvs.e F,so that the requested plot can be obtained from that figure by turning it on its side,so to speak Problem. Prove in detail that the screend Coulomb potential has the form (in MKS units) V (r) = e e βr 4πɛ s r. Use Fourier transforms and the fact that the Fourier components of the screened Coulomb potential are: V (q) = e ɛ s q + β. Solution. We must take the inverse Fourier transform of the potential energy: V (r) = (π) 3 dq V (q) e iq r = e (π) 3 ɛ s dq eiq r q + β. () ECE69 Spring 3

4 Going to polar coordinates with the z axis aligned with r,integrating trivially over the azimuthal angle φ and noticing as usual that q r = qr cos θ,where θ is the polar angle,noticing also that V depends only on the magnitude of r: V (r) = e π ɛ s r e (π) ɛ s since,from any book on tables of integrals: dq q q + β π dθ sin θe iqr cos θ = dq q sin(qr) q + β = e 4πɛ s r e βr, (3) x sin(ax) x + β dx = π e βa. 5. Problem. Looking at the figure on the next page,consider a system of electrons uniformly distributed over a square area with a uniform background of positive charge compensating the negative electron charge,so that at (mechanical) equilibrium the system is charge-neutral. Let n be the density of the sheet of electron charge,let m be the mass of each electron,and let ɛ s be the dielectric constant of the medium surrounding the system. Now shift the electrons assumed to be all rigidly connected among themselves,so as to form a rigid sheet of negative charge by a displacement u to the left. This will leave a rectangle of positive charge exposed at the right hand side of the square. This positive charge will exhert a force which will attempt to call back the sheet of electrons to their equilibrium position. Assuming that the square extends to infinity up and down along the vertical direction (so that edge effects should be neglected and the expression for the electrostatic force is simply that between a point-charge and an infinite sheet of charge): a. Write the equation of motion for the displacement of the sheet of electrons and show that it will lead to harmonic oscillations. b. Show that the frequency of these oscillations is the plasma frequency (in MKS units) ω p = [e n/(m ɛ s )] /. ECE69 Spring 4

5 Solution. a. The positive charge density (at the right in the figure) is enu. The force acting on any electron in the negative sheet of charge will be independent of the distance from the positive charge and will be directed opposite to u,that is,f = e nu/ɛ s. Thus,the equation of motion for the displacement u will be: m d u(t) dt = e nu(t) ɛ s. (4) This is the equation of a harmonic oscillator: The sheet of electrons will move to the left (or in whatever direction it was intially pushed) until the recalling force will grow so large as to stop it and invert its motion. The force will disppear when u will reach its zero value since there will be no exposed charges but the inertia will drive the sheet to the right and the cycle will repeat itself. b. Equation (4) can be written as: where ω p =[e n/(m ɛ s )] /. eigenfrequency of the electron gas. d u(t) dt + ωp u(t) =, (5) This is the frequency of the oscillations which is,in a sense,the proper 6. Problem. Show in detail the steps required to derive the expression for /τ op given at page 7 of the notes, using the matrix element in the same page. As usual,polar coordinates in k-space are handy. Solution. By the Fermi Golden Rule: τ op (k) = π h k < k H op k > δ[e(k ) E(k) ± hω op ], (6) where the upper sign holds for emission and the lower sign for absorption of optical phonons. Using the ECE69 Spring 5

6 expression for the matrix element given at page 7 of the lecture notes, τ op (k) = π h h(d t K) op ρω op ( n op + ± ) V k δ[e(k ) E(k) ± hω op ], (7) where,again,the upper signs hold for emission and the lower signs for absorption of optical phonons. The term (/V ) k δ(e E ± hω op ) is just one half the density of states at energy E hω op. Thus: τ op (k) = π(d tk) ρω op op ( n op + ± ) ρ[e(k) hω op ]. (8) Using the expression for the density of states,we finally get: op m 3/ τ op (k) = (D tk) / π h 3 ρω op ( n op + ) [E(k) hω op ] /. (9) ECE69 Spring 6

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8 7. Problem. Looking at the figure above,consider a linear chain of ions connected by springs with elastic constant α. The ions have alternating masses, M and m,as shown,with M>m. The equibrium distance between two adjacent atoms is a. a. Write the equations of motion. Hint: Write two equations,one for the displacement u of the atoms of mass M and another one for the displacement u of the atoms of mass m. b. Find the frequencies of the possible modes,assuming that the displacement of each atom is of the form e i(jqa ωt),where j is the index labeling the atom position,and q the wavevector of the mode. (Note the use of a instead of a since the lattice constant for this D lattice is indeed a.) c. Sketch as accurately as you can ωvs.qfor the possible modes and discuss their physical meaning. Solution. Denoting by u,j the displacement of the ion of mass M at location ja and by u,j the displacement of the ion of mass m at location (j +)a,than the equations of motion are: M d u,j dt = α(u,j u,j ) + α(u,j u,j ). () ECE69 Spring 8

9 Let s define: Then the equations of motion become: or: m d u,j dt = α(u,j+ u,j ) + α(u,j u,j ) () u,j = ɛ e i(jqa ωt), () u,j = ɛ e i(jqa ωt). (3) M ω ɛ = αɛ αɛ + αɛ e iqa αɛ, (4) m ω ɛ = αɛ e iqa αɛ + αɛ αɛ, (5) (M ω α)ɛ + α( + e iqa )ɛ =, (6) α( + e iqa )ɛ + (Mω α)ɛ =. (7) This linear homogeneous system admits nontrivial solutions only if the determinant of the coefficients vanish,that is: (M ω α)(m ω α) α ( + e iqa )( + e iqa ) =. (8) The two solutions are: ω ± = α mm { (m + M) ± [ (m + M) 4mM sin ] } / (qa). (9) The choice of the plus sign leads to an optical branch: { ( ) ω+ α m + M for q α m for q a π. ECE69 Spring 9

10 The choice of the minus sign yields an acoustic mode: ω { for q α M for q π a. The optical solution corresponds to a mode in which the two different ions oscillate out of phase,the acoustic solution to a mode in which the different ions oscillate in phase. ECE69 Spring

Electron-phonon scattering (Finish Lundstrom Chapter 2)

Electron-phonon scattering (Finish Lundstrom Chapter ) Deformation potentials The mechanism of electron-phonon coupling is treated as a perturbation of the band energies due to the lattice vibration. Equilibrium