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Sol5 - 1(a ”6 [email protected]flgdx QAZJMC‘Q‘Y'XOW Ray 6:300...

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Unformatted text preview: 1(a) ”6 jaw [@cxpflgdx : QAZJMC‘Q‘Y'XOW : Ray. 6:300! 00:142J 5:;- «‘1 27F dx + F 0(‘X‘i19X 37T_m€, dx : g[ 6'1““)41 ’06 L(“”L))( 0 7T ‘ . (cx HL) 0 OC—t'}; ~00] : 9: ___!~ + 1 J; ouik lek] : g 0:: 9% 4. 2 ‘0 WW) = JV d’W’BEfi) M \57‘; L N m kg???“ (:1 filth/12?? OWL M 490:) MS wSLQJLf . ~ k , TM: m {Mm m MM (Mm POWJUW MT wot/L WM “Cw/n, Tor SWAN og 5/ - k 2 6130:) 1‘: 1!, :33 I M WOLUQ, ¥mct¢olvx {'3 313141“ N - 1» 7? L, k: Obi/l k Wd’ PVVDID-GJQS (1% E 34rd PourJUdL milk my? > L WLME$A>dwm1mChW€4 K W1 kmyw w W W943 gr 0L Whnb IS‘ Wit/in Av E: PC. M50, U/L ACLU—(.2 W W1 SW 02.18 CMS‘MIO’W # & WILL/001d W Eflxw m I VVQ/ W :5 C. ; KL : 7%: 75‘ 2” 0‘00 2 C/ 6—— (Von! Mamba} pourt'bm‘ ii 4 wt d2 1%. W <9th \_°>. 3.1 Part a For E < Vo the solution is Aei’“ + Be‘i’“ if :1: < 0; “(3’) ‘ { C’e‘“ ifz > 0. (5) Where k and It are both real number defined by V 2mE k: h and / x/ZmWo — E) —————h . Notice that there’s no term De+m in the solution for a: > 0. This is because such a term would blow up as :r —> co and could not be normalizable. Thus we have set D = O. The matching conditions are [$2 A+B = C ik(A—B) ~50 so that it’s easy to solve for the relection coefficient 2 =1. B R—b 2 |k+il~t k—‘i/‘i This is intuitive because the particle doesn’t have the energy to make it over the barrier and tunneling is not an option because the barrier has infinity Width (there is no region with V(x) = 0 to the right of the barrier). 3.2 Part b The case E > V0 is a little different. Now the solution is ”(13) __ Aeik“ + Be‘il“c if :r < 0; — Ceiq‘c + De‘i‘“c if :r > 0. The matching conditions are still eaSy and give 1 q 1 ( q) A 2 (1 + E) C + 2 1 k D _l _2 1( 2) B _ 2(1 k)0+2 1+_k D. For a wave incident from the left we set D = 0 and get 2 k _ q 2 = (k + q) ' 3.3 Part c The point here is that it’s the flux (or probability current) which is the physically interesting quantity to transmit through the barrier. The flux "m defined by j(t,$) : i (1111*6—1/1 _ 61/) ) . «1/ R: ‘ B A Zmi 6x 8st The states we’re interested in are energy eigenstates which have the form 1/1(t,:c) = e‘iEt/r‘uCE) so that the probability density |1,[1(t,:c)|2 = |u(ac)[2 is independent of time. Th'm means that the conservation law 3p 33' _ a+a*° implies that j(t, ac) = constant. 4 For :1: < 0 we have flux his - _ _ 2 _ 2 M - m [IAI IBI ] and for :1: > 0 we have . fig 2 2 Jz>o = 7n— [ICI — IDI] which gives the relation 4 IA? — IBI2 = ;[|012—|D|2]- (6) Now, back to computing the transmission coefficient. Setting D = 0 the outgoing flux is jam 2 7‘“Lq|C'[2 / m while the incident flux is jin = hk|Al2 / m and the transmission coefficient measures how much flux penetrates the barrier T=j°ut=gfl= E_V°|£E jin k|A|2 E IA!2 as desired. For E < V0 it’s easy to see that the solution (5) on a: > 0 has flux j$>0 = jout = 0 so that T = 0 as you should expect from part a. 3.4 Part (1 For E > Vo we have 2 2 . q ICI IBI T = '—'— _. 2/ + R k W + W2 Using equation (6) with D = 0 gives T + R = 1 as desired. ...
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