8.3. THESTEPPOTENTIAL 129
whilethereflectioncoefficientis
R =
v 1 |B|^2
v 1 |A|^2
=
(p 2 −p 1 )^2
(p 2 +p 1 )^2
(8.29)
Noticethat
R+T= 1 (8.30)
whichhastheinterpretationthat
no.ofparticlesreflected/sec + no.ofparticlestransmitted/sec
= no.ofparticlesincident/sec (8.31)
andisaconsequenceoftheconservationofprobabilitybytheSchrodingerequation.
ItshouldalsobenotedthatR+=0,ingeneral.Thisisinsharpcontrasttoclassical
physics,wherenoparticlesofenergyE>V wouldbereflected.
Question:Whyis(8.31)aconsequenceofconservationofprobability?
Energies E<V
Inthiscase,thesolutions(8.21)inregionIand(8.23)inregionIIarethesame
asbefore,however
p 2 =
√
2 m(E−V)=iq 2 (8.32)
isanimaginarynumber.Thesolutionφ 2 intheclassicallyforbiddenregionisthen
φ 2 (x)=Ce−q^2 x/ ̄h+Deq^2 x/ ̄h where q 2 =
√
2 m(V−E) (8.33)
Likeanysolutionintheclassicallyforbiddenregion,φ 2 is,ingeneral,acombination
ofexponentiallygrowingandexponentiallydecayingfunctions.SolutionswithD+= 0
arenon-normalizable,however,anddonotcorrespondtophysicalstates. Soweneed
onlyconsidersolutionswithD=0.
Onceagainapplyingcontinuityof thewavefunction andits firstderivative, we
have
φ 1 (0)=φ 2 (0) =⇒ A+B=C
φ′ 1 (0)=φ′ 2 (0) =⇒ ip 1 (A−B)=−q 2 C (8.34)
SolvingforBandC,
B =
p 1 −iq 2
p 1 +iq 2
A
C =
2 p 1
p 1 +iq 2