QMGreensite_merged

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

A (8.35)