QMGreensite_merged

369

sothat

ψ(x,t+!) =

(

1 +

!

i ̄h

H

)

ψ(x,t)+O(!^2 )

=

(

1 −i

!

̄h

V+

i! ̄h

2 m

d^2

dx^2

+O(!^2 )

)

ψ(x,t)

= e−i!V(x)/ ̄h

(

ψ(x,t)+

i! ̄h

2 m

ψ′′(x,t)

)

+O(!^2 ) (24.11)

Nowletswritey=x+η,andexpandψ(y,t)ineq. (24.5)inaTaylorseries

ψ(x,t+!) =

∫

dyG!(x,y)ψ(y,t)

=

∫

dηG!(x,x+η)

[

ψ(x)+ψ′(x)η+

1

2

ψ′′(x)η^2 +...

]

(24.12)

Comparing(24.11)and(24.12)weseethat,tofirstorderin!,thepropagatorG!(x,x+
η)mustsatisfy

∫

dηG!(x,x+η) = e−i!V(x)/ ̄h

∫

dηG!(x,x+η)η = 0

∫

dηG!(x,x+η)η^2 =

i! ̄h

m

(24.13)

Itsalsoclearfrom(24.5)that,inthe!→ 0 limit, thepropagatormustbecomea
deltafunction,i.e.
lim
!→ 0
G!(x,x+η)=δ(η) (24.14)

Therefore, G!(x,x+η)isadelta sequence. We haveseen two examples ofdelta
sequencesinLecture4;oneinvolvedtheintegralofeikxoverafiniterange,theother
wasagaussian.Thegaussiandelta-sequence,togetherwiththefirstoftheconditions
ineq.(24.13),motivatestrying,asanansatz

G!(x,x+η)=e−i!V(x)/ ̄h

√

A

π

e−Aη

2

(24.15)

whichwillbeadelta-sequenceifA→∞as!→0.Itiseasytocheckthatthisansatz
satisfiesthefirstandsecondconditionsin(24.13).TheconstantAisdeterminedby
thethirdcondition

i! ̄h

m

=

√

A

π

∫

dηη^2 e−Aη

2

=

1

2 A

(24.16)