Homework 1. 0 for z L. Ae ikz + Be ikz for z 0 Ce κz + De κz for 0 < z < L Fe ikz

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Regina Wilkinson
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1 Homework. Problem. Consider the tunneling problem with the potential barrier: for z (z) = > for < z < L for z L. Write the wavefunction as: ψ(z) = Ae ikz + Be ikz for z Ce κz + De κz for < z < L Fe ikz for z L, with k =(me) / / h, κ =[m( E)] / / h. a. Write the system of four equations expressing the continuity of the wavefunction and its derivative at z =and at z = L. b. Find the transmission coefficient, T = F / A. There s no need to solve the full system. Be creative. Hint: Multiply the equation expressing continuity of ψ at z =by ik and add and substract it from the equation expressing continuity of the derivatives at z =. Do a similar thing with the other two equations (by multiplying one by κ). Now it should be relatively easy to solve for F in terms of A alone. This gives you T. Solution. a. The equations expressing continuity of the wavefunction and its derivatives at z =and z = L are: ψ( ) = ψ( + ) A + B = C + D, () dψ( ) = dψ(+ ) ika ikb = κc κd, () dz dz ψ(l ) = ψ(l + ) Ce κl + De κl = Fe ikl, (3) ECE68 Spring

2 dψ(l ) = dψ(l+ ) κce κl κde κl = ikf e ikl. (4) dz dz b. Multiply Eq. () by ik,add and substract it from Eq. (): ika = (ik + κ)c +(ik κ)d, (5) ikb = (ik κ)c +(ik + κ)d. (6) Multiply Eq. (3) by κ,add and substract it from Eq. (4): κce κl = (ik + κ)fe ikl, (7) κde κl = (ik κ)fe ikl. (8) From Eqns. (7) and (8) we can express C and D in terms of F. Inserting these expressions into Eq. (5) we can finally express F in terms of A: so that the transmission coeficient is: ika = eikl F κ [(κ k )sinh(κl) + iκk cosh(κl)], (9) T = F A = 4κ k (κ k ) sinh (κl) + 4κ k cosh (κl). (). Problem. From the Schrödinger equation derive the continuity equation: ρ t + S =, where ρ = Ψ is the probability density and S = i h m [Ψ Ψ Ψ Ψ] is the probability density current. ECE68 Spring

3 Solution. By the definition of the probability density ρ: ρ t = t [ Ψ Ψ ] = Ψ t Ψ + Ψ Ψ t. Since Ψ/ t =/(i h)hψ and Ψ / t = /(i h)hψ, ρ t = i h (HΨ )Ψ + i h Ψ HΨ = [ h i h m Ψ Ψ ΨΨ h m Ψ Ψ + Ψ Ψ ]. But the second and fourth terms inside the square bracket cancel,so that: ρ t = i h [ Ψ Ψ Ψ ] Ψ m On the other hand,using the definition of the probability-flux S: S = i h m [Ψ Ψ Ψ Ψ ] = = i h m i h m. () [ Ψ Ψ + Ψ Ψ Ψ Ψ Ψ Ψ [ Ψ Ψ Ψ Ψ ]. () From Eqns. () and () we have ρ/ t = S,which is the desired result. 3. Problem. The Wentzel-Kramers-Brillouin (WKB) approximation to solve the Schrödinger equation consists in writing the solution of the time-independent problem: ] as h d ψ(x) m dx + (x)ψ(x) = Eψ(x), ψ(x) k / exp { i x k(x ) dx }, ECE68 Spring 3

4 where k(x) = {m[e (x)]} / / h. This is a good approximation if the potential (x) varies slowly (that is,it does not change much compared to the electron energy E when x varies over several de Broglie wavelengths). If E (x) <,the WKB wavefunction becomes ψ(x) k / exp { x } κ(x ) dx where now κ(x) = {m[ (x) E]} / / h. Let s now ignore the factor k / (which simply ensures continuity of probability current). Consider now the previous tunneling problem (problem ) and identify the WKB approximation to the transmission coefficient as: { } L T WKB = ψ(l) = exp κ(x ) dx. (3) Compare T WKB with the exact transmission coefficient T of the previous tunneling problem (problem ) in the limit in which κl >>. Solution. From Eq. () above,for κ>>lwe have sinh(κl) cosh(κl) e κl /,so, T 6κ k e κl (κ + k ). (4) On the other hand,from the WKB expression,eq. (3),we have trivially (since inside the barrier ( <x L) κ(x) =[m(e )] / / h is a constant independent of x): T WKB e κl. (5) Except for prefactors which we have ignored in the WKB expression,eq. (5) agrees with the exponential behavior of Eq. (4). ECE68 Spring 4

5 4. Problem. Calculate the matrix element between two wavefunctions of the form ψ(k, r) = /eik r and ψ(k, r) = /eik r, and the pertubation potentials of the form: a. H e iq r b. H δ(r) c. H r d. H e r /r Polar coordinates are useful in c and d. Solution. In each case we must evaluate the integral /e ik r H(r) /eik r dr = e i(k k ) r H(r) dr. So: a. For H(r) = e iq r : e i(k k +q) r dr = δ(k k + q), thanks to the definition of the δ function. b. For H(r) = δ(r): e i(k k ) r δ(r) dr =. c. For H(r) = r,going to polar coordinates and assuming a spherical volume 4πR 3 /3= : e i(k k ) r r dr = π dφ π dθ sin θ drr i kr cos θ e r = ECE68 Spring 5

6 = π dr π d(cos θ) e i kr cos θ, where θ is the polar angle and we have assumed (without loss of generality) that the (polar) z-axis is along the direction of k k and k = k k. With a change of integration variable µ =cosθ,this can be written as: π i krµ dr ei krµ µ= µ= = π k dr sin( kr) r 4π π k in the limit R,since sin(x)/x = π/ (see any text of tables of integrals: The integral can be calculated rigorously by contour integrals in the complex plane). d. For H(r) = e r /r,going again to polar coordinates: e i(k k ) r e r /r dr = π dφ π π dθ sin θ drr e i kr cos θ e r/r = = π drr e r/r d(cos θ) e i kr cos θ. As before,let s set µ =cosθ: π R drr e r/r e i krµ µ= = π R drr sin( kr) e r/r. i krµ k µ= The last integral can be solved by replacing sin( kr) with e i kr and retaining only the imaginary part of the result. The integral is of the form: drre αr, ECE68 Spring 6

7 where α = /r + i k. Integrating by parts: drre αr = r α eαr R in the limit R. Therefore: π k drr sin( kr) e r/r = 4π [ k Im dr α eαr = r α eαr R + eαr α drre r(i k /r ) ] R = 4π = α, r ( /r + k ). 5. Problem. Extend the procedure followed on page 8 of the Notes to find the perturbed eigenvalues E n () to second order in α. Solution. Looking at page 8 of the Notes,if we equate terms of order α at the left- and right-hand side of the second equation,we have: H φ () n > + H φ() n > = E n φ () n > + E() n φ() n As done in the Notes,let s expand φ () n > and φ () n > over the (complete) basis ψ m >: > + E() n ψ n >. (6) φ () n > = m a () nm ψ m >, (7) and φ () n > = m a () nm ψ m >. (8) Insert now these expansions into Eq. (6): H m a () nm ψ m > + H m a () nm ψ m > = E n m a () nm ψ m > + E n () m a () nm ψ m > + E n () ψ n (9) ECE68 Spring 7

8 Let s now take the inner product of this equation with <ψ n : c () nn E n + m a () nm H nm = c () nn E n + a () nn E() n + E () n. () The first term on the lhs is equal to the first term on the rhs. Using now the fact that a () nm = H mn /(E n E m ) and E n () = H nn,we have: E () n = m n H mn H nm E n E m = m n H mn 6. Problem. Show in detail the equivalence between the two formulations of Bloch theorem: E n E m. () ψ(k, r + R l ) = e ik R l ψ(k, r), () and: where u k (r) is periodic: Solution. From Eq. (3): ψ(k, r) = e ik r u k (r), (3) u k (r + R l ) = u k (r). ψ(k, r + R l ) = e ik (r+r l ) u k (r + R l ). (4) By the periodicity of u, u k (r + R l )=u k (r),and from the previous equation: which is Eq. (). ψ(k, r + R l ) = e ik R le ik r u k (r) = e ik R l ψ(k, r), (5) ECE68 Spring 8

9 7. Problem. Find the reciprocal lattice vectors of the fcc lattice. As fundamental translation vectors use a = a (ˆx + ŷ), b = a (ŷ + ẑ), c = a (ẑ + ˆx). b. Find the volume of the primitive cell and the density of valence electrons in cm 3 for Si (a =.543 nm). c. Find the volume of the BZ. d. Using a computer program,list the first 5-to- G-vectors in order of increasing magnitude. Solution. a. Using the definition of reciprocal lattice vectors A, B,and C (see notes,page 36): A = π a (ˆx + ŷ ẑ), B = π a ( ˆx + ŷ + ẑ), C = π a (ˆx ŷ + ẑ). b. The volume of the primitive cell will be: a b c = a3 4. Since we have Si atoms in the primitive cell,each with 4 valence electrons,the electron density will be 3a 3 3 cm 3. c. The volume of the BZ will be: ( ) π 3 A B C = cm 3 a d. The G-vectors are of the form: G = l A + l B + l 3 C, where A, B,and C are the fundamental translation vectors in reciprocal space given above and l, l,and l 3 are integers. Therefore the first G-vectors are (in units of π/a): ECE68 Spring 9

10 the 8 vectors: the 6 vectors: the vectors: the 4 vectors: the 8 vectors:...etc. (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) of length 3; (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) of length 4; (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) of length 8; (3,, ) (3,, ) (3,, ) (3,, ) ( 3,, ) ( 3,, ) ( 3,, ) ( 3,, ) (, 3, ) (, 3, ) (, 3, ) (, 3,, ) (, 3, ) (, 3, ) (, 3, ) (, 3,, ) (,, 3) (,, 3) (,, 3) (,, 3) (,, 3) (,, 3) (,, 3) (,, 3) of length ; (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) of length ; ECE68 Spring

11 On the next pages I list the FORTRAN program used to generate G-vectors up to a magnitude of 59 and the output list. ECE68 Spring

Analogous comments can be made for the regions where E < V, wherein the solution to the Schrödinger equation for constant V is

Analogous comments can be made for the regions where E < V, wherein the solution to the Schrödinger equation for constant V is 8. WKB Approximation The WKB approximation, named after Wentzel, Kramers, and Brillouin, is a method for obtaining an approximate solution to a time-independent one-dimensional differential equation, in