Question

Consider a particle encountering a barrier with potential U = U.>0 between x = -a and x = a with incoming energy E > U. a) Write the symbolic wave functions before and after passing through the barrier (i.e., for x<-a and x>a; regions I and III). U1 b) Write down the Schrodinger equation for the wave function in the middle (region II) where the potential is non-zero i.e., where -a<x<a; region II). c) What solution would you try for the differential equation in region II? (No need for detail calculations.) d) Calculate the flux of the outgoing wave in region III. (No need to calculate the normalization or any constants.) e) Now go ahead and actually calculate the solution for parts a and b, to find wave function in all three regions. f) Complete the calculation of the flux in part d. g) If we wanted to have bound solutions, what should be the condition on E?

Answer 1

CA when Exo it comesponds to scattoung phenomena here - x <-9 & nya In Region o UG): V-acaca Time - Independent Schrodinger êque is dy + VW)y of 4 In region & 42-2m£ ua) 20 so, dip 206 - Way Agelky - A e in of top In Region @ an Dy = p = 842 where, a= 2m (E-Uo). =>%2C) = 4, 2'94. A eine @ In Region , - 32 d 2 E4, [42 = Azelka as there is 1- A mand no incidence from the night side Now, to call the transmission flux we're to solve the coeffeciente In the boundary, &. (-a) = P2 (-a). a) A, elka, Alelkon - Aega A era a I e vita) = (-a) x ( A, et ka A, elkaj 2%q (Zeira Belang A eta A, reika a vt (Azelga Helaas 6 Adding a 4 we get 2A, eika - zera (164) & A (1-941 A = 44, e "(k-a) a (149/4) the carsa - 6 Similarly 9-6 we get Al 24 A₂ (1-1) e Carla ka. (1) (ark) a 6 1999

1 Similarly for Boundary conditions 42 (9) = 4(a) e a Aean zelka 4₂ (a) = 43(a). iq ( eiga A elaas aikAzelka : + @ we get, Az elga_A_eaa 44A elka @ Ax = Aze' Sealer kan. - ( & ® . we get;, A / A ze 1-16) - Now lesing C d 6. from 6 we get from A, as, A = het alaka). 12 Az el(ka) (14 %). (k-qa + (9+k)a '%. 2ikat NO = %likafaziza..to be (1221) 4, Pichaela 29.01 te nika = Aggelika [4 cosaga - ha Now, the transmission flex. cie ab eisin 29a7 @. the transmission coneff : U cos2q,a - 1 - 4 cos 2qa el sin ga KO ✓ 22 La a opet st-47 si 2n + REFJAPAM - Vi ve sobe Am @ 20 Shartest If we wanted to have bound solutions then Calculation from scattering state o>ED- to obtain bound state.

I replacing re by ik for bound state as I iz ove but asal In region 6 - 8 - Volle & E42 → = -2 CE + U)®z Then we will find in 2 where, and që - 2,59 (E+6) 4, = A, e-kx nogion 4, = he & Aieka here 220 E1, 1 by replacing A must be =0 &a ik G inters it will diverge at y=- Now, bom equr ③ we get, - Az eeka [ acosaga - - ta 1.28in 2qa] a cremains real but k ik. so, A=0 therefore, cotaqa = 92 77 & tan 2 tanaqa 2 2 2 2 = Y. het, a= tanga so tan 29a 2 tangan putting this we get, y= 22 - ² + 2xy 0 -- 2 2 Juruye = -1 t/ity tangazak 2-12 a={-14[1* uerita 32 S_1+ (8 + k3y 1 g?-?.. 1 97 -12 5. Zqke I (taking the tanga a 2 + ² + a² + ² a . – VE (taking are tanga tap (Even, synonatric case) colqa – – / Codd care) sqm) 29E sign ) care