Homework 4. ω 2, (1) where ɛ 0 is the vacuum permittivity. In the metal a solution of Laplace s equation 2 φ =0is. φ 0 (x, z) = A cos(kx) e kz, (3)

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1 Homework 4 1. Problem. Consider the metal-vacuum interface located at z =0,the metal filling the entire half-space z 0, vacuum filling (!?) the half-space z<0. The dielectric function in the metal in the long-wavelength limit is given by: ( ) ɛ M (ω) = ɛ 0 1 ω2 P ω 2, (1) where ɛ 0 is the vacuum permittivity. In the metal a solution of Laplace s equation 2 φ =0is φ M (x, z) = A cos(kx) e kz, (2) so that E Mz = ka cos(kx) e kz and E Mx = ka sin(kx) e kz are the z and x components of the electric field. a. Show that in the vacuuum (that is,for z<0) φ 0 (x, z) = A cos(kx) e kz, (3) satisfies the boundary condition that the tangential component of E be continuous at the metal-vacuum boundary; i. e.,find E 0x. b. Recalling that D M = ɛ M (ω)e M and that D 0 = ɛ 0 E 0,show that the boundary condition that the normal component of D be continuos at the boundary requires ɛ M (ω) = ɛ 0,so that,by Eq. (1),the frequency of the interface plasma oscillation, ω s,will be ω P / 2. Solution. The component of the electric field, E,parallel to the plane of the interface is: E (x, z) = φ 0(x, z) x = { Ak sin(kx) e kz (z 0) Ak sin(kx) e kz (z>0), (4) ECE618 Spring

2 while the component of the electric displacement field, D,normal to the plane of the interface is: D (x, z) = ɛ φ { 0(x, z) Akɛ = 0 cos(kx) e kz (z 0) z Akɛ M cos(kx) e kz. (5) (z>0) Therefore the continuity of E at z =0is obvious from Eq. (4),while from Eq. (5) the continuity of D implies ɛ 0 + ɛ M (ω) = 0. (6) Inserting Eq. (1) into this equation,we have ( ) ɛ 0 + ɛ 0 1 ω2 P ω 2 = 0, (7) so that ω = ω P / 2. This is the frequency of the surface plasmon. 2. Problem. Consider now the interface between an insulator (for z<0) and a semiconductor (z 0) whichat z =0behaves like a two-dimensional electron gas. The dielectric function of the insulator can be written as: ɛ i (ω) = ɛ i + (ɛ i0 ɛ i ) ω 2 TO ω 2 TO ω2. (8) The dielectric function of the 2D electron gas can be written as: ( ɛ 2D (Q, ω) = ɛ 1 ω2 P (Q) ) ω 2, (9) where Q is the 2D wavector on the plane of the interface,and ω 2 P (Q) =e2 n 2D Q/(ɛ m ) is the frequency of the two-dimensional plasma of density n 2D (in units of carriers per unit area) which depends on Q (unlike the bulk plasma). Following the same procedure of Problem 1 (that is: expressing the potential in the insulator ECE618 Spring

3 and semiconductor with equations (2) and (3) and imposing the continuity of E x and D z = ɛ(ω)e z across the interface) find the dispersion of the coupled phonon-plasmon modes at the interface as a function of Q. Note that you should get something resembling the results of pages of the Notes (but now Q plays the role of the density n in the Notes). Solution. We proceed as in Problem 1,obtaing from Eq. (5): ɛ i (ω) +ɛ 2D (ω, Q) = 0. (10) Expressing ɛ i (ω) with Eq. (8) and ɛ 2D (ω, Q) via Eq. (9) we find with some algebra: (ɛ i + ɛ )ω 4 {ɛ [ω 2 TO + ω2 P (Q)] + ɛ i0ω 2 TO }ω2 ɛ ω 2 P (Q)ω2 TO = 0. (11) The two roots are: 1 { ω ± = ɛ [ωto 2 2(ɛ + ɛ i ) + ω2 P (Q)] + ɛ i0ωto 2 ± ±{[ɛ [ωto 2 + ω2 P (Q)] + ɛ i0ωto 2 ]2 4(ɛ + ɛ i )ɛ ωto 2 ω2 P (Q)}1/2} (12) Note the following behavior at small and large Q. ForQ 0, ω P (Q) 0 and from Eq. (12) we have ω 2 ± { ɛ +ɛ i0 ɛ +ɛ i ω 2 TO (+) ω 2 P (Q)/(1 + ɛ i0 ɛ ) ( ). (13) The first solution is just the frequency of the surface optical phonons (compare this expression with the Lyddane-Sachs-Teller expression,eq. (195),we have derived on page 169 of the Notes). The second solution is just the surface plasma frequency ω P (Q)/ 1+ɛ i0 /ɛ. This expression reduces to ω P / 2 (the result of Problem 1) when the insulator is the vacuum and the semiconductor is replaced by a metal. In the opposite limit Q,and so ω P (Q),we have: { ω± 2 ωp 2 (Q)/(1 + ɛ i ɛ ) (+). (14) ( ) ω 2 TO ECE618 Spring

4 The first solution is,once more,the surface plasma frequency,like the second solution in Eq. (13). However, since we are at high frequencies, ɛ i0 is repleced here by ɛ i. The second solution is just the uncreened optical phonon. The following figure shows the dispersion of the two modes, ω ±,as well as their scattering strength (analogous to Eq. (54),page 224) and the phonon-content (analogous to Eq. (51),page 223 of the Notes,Part IV. ω (mev) Si SiO 2 n s =10 12 cm 2 coupled ω SO ω TO ω LO ω ps SCATTERING STRENGTH Λ bare scattering strength PHONON CONTENT Φ IN PLANE WAVEVECTOR Q (10 8 cm 1 ) IN PLANE WAVEVECTOR Q (10 8 cm 1 ) IN PLANE WAVEVECTOR Q (10 8 cm 1 ) 3. Problem. Plot the non-polar scattering rate with acoustic and optical phonons given at page 214 of the Notes as a function of electron energy in the range 0 <E 1 ev. Use the following values (roughly related to Si): (D t K) op = 10 9 ev/cm, ac =10eV, m =0.32m 0 (where m 0 isthefreeelectronmass), ρ =2.33gr/cm 3, hω op = ev, c s = cm/sec. Solution. See figure and comments below. 4. Problem. a. DothesameforthepolarFröhlich scattering,eq. (44) of the Notes,Part 4. Use the following parameters (approximately correct for GaAs): m = m 0, ECE618 Spring

5 hω LO = ev, ɛ 0 = ɛ vac, ɛ 0 = 12.9 ɛ vac. b. Compare the result with the result of Problem 3. Discuss how the different energy dependence will affect electron transport. Solution. See the figure. Note how the polar scattering rate for GaAs is smaller then the total nonpolar scattering rate for Si. This suggests a higher mobility for GaAs. ECE618 Spring

6 10 14 nonpolar acoustic SCATTERING RATE (1/s) nonpolar optical polar ELECTRON ENERGY (ev) ECE618 Spring