Question

Answers can be more than one:

VII. (12pts) Consider the following potential energy: region 1: U(X) = U. x < 0 region 2: U(X) = 0 0<x</ Uo region 3: U(x) = U. x>L ТЕ where U. >0. We want to consider a particle with energy E such that 0 < E<Uo. There are two possible forms for the wave function that might be used to represent the particle: (x) = 4 sin kyx+ B, coskx v(x) = 4e** + Bje tik where i = 1,2,3 indicates the three regions. (a) Write down wave functions that describe the behavior of the particle in region 1, region 2, and region 3. Use appropriate subscripts to label all parameters. If any of the coefficients Aior Bi are zero, identify those coefficients and explain why they are equal to zero. Write down the expression of Kį as well. (b) Sketch the probability distributions you would expect for the ground state and the first excited state. (c) Use the continuity conditions at x = 0 to show how the coefficients of the wave function in region 2 are related to the coefficients of the wave function in region 1.

Answer 1

Given ( Consider) three region U(X)= U. Kro Veu) 0 o KALL Region 2 : Region 2 : Region 3 : 1 TE (a): do KYL where U, YO We want to consider parhile lui 14 energy such that OCEKU and there are two possible forms for the wave function that might be to reposent the pastele y (x) Ai sinkin t B. Coskix y (a) kin iz 1,2,A for regions used AU e & Bi e (a) Now by the schrondinger wave equation d'y (t-1) Ý - am co 7 du ² h2 For region - 1 (20) ETU D = a on dey dre CS Scanned with h² im (0-€) 4 = 0 CamScanne (0-4) 420 : (0.5-6)

Now solution become for these roots are ta tu da - 2x + Ве 4 = Ae because we know that when two rools mi, ma then solution of second order differential equation A emint be hish M₂ x become so How using just boundary concution then Por region First (1) - ax Ae B-o YA for region thind (11) Be Aco Scanned will Cam Scanner

Гал 4e2107 - 11 lor121) so equation become 0² + 2ME 4:0 h (0'+k%) 4 =0 {k k² ame h² so 0²/k² = co = D = tik are You (n)= We take so its rools Complex so wavefumehion beome e sin (Ku) 7 0 Cos (ku) take cofficient A.B.C.D in warefumtion YE YI & Y 10 Pr (in this B=0 , because expl-00) = 6 0] ( wavefrometion vamich af x= 15) Ya c son/kn) to cos (ken) an A e -dn x = -6 You Be (AEO, because wavefumchon vanish at infinite when we take a then Lero A will be CS Scanned with CamScanner

(b) Pou ground share (x) probability distribution Function for ground stafe Vo vo 19,011? E, - ra 20 x=L for first exuted state V (2) vo 2 rn 14.5-2 ty Scanned-with CamScanner X70 x=L

x=0 we (C) Mow use the continuity condenon at and check How the cofficient are related to legion region 1 -2 SO t ro, o continuity condition 471012 Yn 10) V 10) = 4010) Yo = A ex Pq (0) = A / =P Ya ac sin(km)fo coglken) 4210) Now by Conchition o Lolo) Yelo)= TA A=D So cofficient A&D will be equal telo) = AX . = Adean yam O) = CK ck Losken - Dk sinka Yo Mow by condition Aa= ck © show the relation b/w cofficients SO . vitº er
If any doubt in solution then you can do a comment and I will help in solve your doubt definitely.

If you like it then also give thumbs up.

Thanks