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Unformatted text preview: Homework Solutions # 6 (Liboﬀ Ch. 7)
7.37 Use the 1D form, ∂ρ/∂t + ∂J/∂x = 0, and go through the derivation
in equations 7.102 to 7.105. You will get ∂ψ ∗ ψ ih ¯ ∂ 2ψ ∂ 2ψ∗ − ψ∗ 2 − ψ ∂t 2m ∂x ∂x2 i + ψ ∗ ψ (V − V ∗ ) = 0 h ¯ Since V = V ∗ , the last term does not vanish, and the continuity equation doesn’t work. The deeper reason it does not now work is not really given in Liboﬀ, so it can be hard to ﬁgure out. It is that the density ρ and current density J must depend on ψ and ψ ∗ only, and not on other functions such as V . More formally, J must be a functional, J [ψ, ψ ∗ ] only, and if the potential term does not vanish, it ends up depending on V . 7.40 The incoming wave is ψinc = Aeik1 x , where the charge density is
|A|2 = 1015 e/m, and h2 k1 /2me = 100 eV. Use the electron mass me = 0.511 ¯2 2 MeV/c , and the current becomes Jinc = hk1 2 ¯ |A| = 5.94 × 1021 e/s = 950 A me Use table 7.2 for the transmission coeﬃcient, with V /E = 1/2, T= So Jtrans = 0.97Jinc = 5.77 × 1021 e/s = 923 A Jreﬂ = 0.03Jinc = 1.78 × 1020 e/s = 27 A Sending a beam of electrons through a slit into a large capacitor would work. 4 1 − V /E [1 + 1 − V /E ]2 = 0.97 7.45 Resonant maxima occur for E > V . Using table 7.2, T must be a maximum, which happens when sin(2k2 a) = 0, or 2k2 a = nπ . The third maximum is at n = 3. So we solve for V in 2¯a 2me (E − V ) = 3π , with h 2a = 4.5 × 10−10 m and E = 100 eV. We get V = 83.4 eV.
Phys 580 HW# 6 Solutions 1 7.57 The wavefunction comes in three parts,
ψI = Aeik1 x + Be−ik1 x ψII = Ceκx + De−κx ψIII = F ek3 x h2 k1 ¯2 h2 κ2 ¯ =E =V −E 2m 2m The boundary conditions at x = ±a are where h2 k3 ¯2 = E + αV 2m Ae−ik1 a + Beik1 a = Ce−κa + Deκa ik1 (Ae−ik1 a − Beik1 a ) = κ(Ce−κa − Deκa ) Ceκa + De−κa = F ek3 a κ(Ceκa − De−κa ) = ik3 F ek3 a We solve for T , and after a lot of algebra, T= k3 F k1 A
2 = 4k1 k3 κ2 2 2 κ2 (k1 + k3 )2 + (k3 + κ2 )(k1 + κ2 ) sinh(2κa) Phys 580 HW# 6 Solutions 2 ...
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