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Unformatted text preview: 1(a) ”6
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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—D2] (6) Now, back to computing the transmission coefﬁcient. Setting D = 0 the
outgoing flux is jam 2 7‘“LqC'[2 / m while the incident ﬂux is jin = hkAl2 / m and
the transmission coefﬁcient measures how much ﬂux penetrates the barrier T=j°ut=gﬂ= E_V°£E
jin kA2 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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 Fall '08
 Vachon
 mechanics

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