By Charles Kittel
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Extra resources for Introduction to solid state physics: instructor's manual
To this we add the particular solution x = ⎣ ⎦ –e/mω and find A and φ to satisfy the initial conditions x(0) = 0 and x ( 0 ) = 0. 11. The Laplacian ∇ 2 ϕ = 0, whence d 2f − K 2f = 0 . 2 dz This has solutions f = AeKz for z < 0 f = Ae − K ( z −d ) for z > d f = B cosh K ( z − d 2 ) for 0 < z < d . This solution assures that ϕ will be continuous across the boundaries if B = A/cosh(Kd/2). To arrange that the normal component of D is continuous, we need ε(ω) ∂ϕ/∂z continuous, or ε(ω) = – tanh(Kd/2).
The proposed solution is seen directly to dx 2 λ 2 ⎛ 1 ⎞ satisfy this and to satisfy the boundary conditions B ⎜ ± δ ⎟ = Ba . (b) For δ < < λL, ⎝ 2 ⎠ 1a. 2 x 1⎛ x ⎞ cosh = 1 + ⎜ ⎟ + … 2 ⎝ λL ⎠ λ 1⎛ δ ⎞ δ cosh = 1+ ⎜ ⎟ +… 2λ 2 ⎝ 2λ ⎠ 2 ( )( ) therefore B ( x ) = Ba − Ba 1 8λ 2 δ 2 − 4x 2 . 2a. From (4), dFS = −MdB a at T = 0. From Problem 1b, M(x) = − ( ) ( ) 1 1 ⋅ 2 Ba ⋅ δ2 − 4x 2 , 4π 8λ whence FS ( x,Ba ) − FS ( 0 ) = 1 2 δ 2 − 4x 2 Ba . 2 64πλ b. The average involves ∫ ( 1 2δ 0 1 3 4 δ3 δ − ⋅ δ − 4x dx 2 3 8 = 2 δ2 , = 1 1 3 δ δ 2 2 2 2 ) whence 1 2⎛ δ⎞ ∆F = Ba ⎜ ⎟ , for δ << λ.
B) The angle between K and b1is 30o ; A right triangle is formed in the first BZ with two sides of length K and b1/2. Now b1 = 43πa , so: K = (b1/2)/cos(30o)= 4π/3a . (c) Quantization of k along x: kx(na)=2πj= kx =2π j/na. Assume n = 3i, where i is an integer. Then: kx = K(j/2i). For j = 2i, kx=K. Then ∆K = k y ˆj and there is a massless subband. (d) For n = 10, kx =2π j/10a =K(3j/20). The closest k comes to K is for j = 7, where ∆kx = K/20. 8 eV. The next closest is for j = 6, where ∆kx = K/10, twice the previous one.