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132 CHAPTER8. RECTANGULARPOTENTIALS

Thisproves that φ(x)is alsoa solution. Thecrucial factusedin(8.46) wasthe
symmetryofV(x)aroundx= 0

V(x)=V(−x) (8.47)

Withoutthissymmetry,φ(x)=φ(−x)wouldnotbeasolutionoftheSchrodinger
equation.
IfEn isdegenerate,thenφ(x)andφ(x)couldbetwodifferentstates. Onthe
otherhand,ifEnisnon-degenerate,thenthetwofunctionsmustbeproportionalto
oneanother:
φ(x)=cφ(x) (8.48)

wherecisareal-valuedconstant,sinceφandφarebothreal-valued. Finally,the
normalizationconditionrequires
∫
dxφ^2 =c^2

∫

dxφ^2 = 1 (8.49)

whichmeans, sinceφisnormalized,thatc=±1. Weconcludethat ifEn isnon-
degeneratetherearetwopossibilities:

1. ”EvenParity”

φ(−x)=+φ(x) (8.50)

2. ”OddParity”

φ(−x)=−φ(x) (8.51)
Assumingthattheenergyeigenvaluesarenon-degenerate, thealgebrasimplifies
alot.InregionII,theSchrodingerequationisoftheform

d^2 φII

dx^2

=−

2 m

̄h^2

(V 0 −E)φII (8.52)

withsolutions
φII=c 1 ei

√

2 m(V 0 −E)x/ ̄h+c

2 e

−i

√

2 m(V 0 −E)x/ ̄h (8.53)

However,sinceφIImustbearealfunction,weexpandtheexponentialsinacombi-
nationofsinandcosine

φII=CEcos

[√

2 m(V 0 −E)x/ ̄h

]

+COsin

[√

2 m(V 0 −E)x/ ̄h

]

(8.54)

Wehaveseenthatthenon-degeneratesolutionshavetobeeitherevenoroddparity.
Forevenparity, CO =0,while foroddparityCE =0. Wedealwiththesecases
separately.