QMGreensite_merged

368 CHAPTER24. THEFEYNMANPATHINTEGRAL

Toseehowasumoverpathscomesin,letssupposethatweknewthepropagator,
foranyHamiltonian,whenthetimedifferenceT=!istiny,i.e.

ψ(x,t+!)=

∫

dyG!(x,y)ψ(y,t) (24.5)

Inthat case, we could getthe propagatorforlarge timedifferences by usingthe
propagatorsuccessively,i.e.

ψ(x,t+ 2 !) =

∫

dydx 1 G!(x,x 1 )G!(x 1 ,y)ψ(y,t)

ψ(x,t+ 3 !) =

∫

dydx 2 dx 1 G!(x,x 2 )G!(x 2 ,x 1 )G!(x 1 ,y)ψ(y,t) (24.6)

andbyinduction

ψ(x,t+N!)=

∫

dy

∫ (N∏− 1

i=1

dxn

)

G!(x,xN− 1 )G!(xN− 1 ,xN− 2 )...G!(x 1 ,y)ψ(y,t)

(24.7)
Sothepropagator,forthetimedifferenceT=N!,hastheform

GT(x,y)=

∫ (N∏− 1

i=1

dxn

)

G!(x,xN− 1 )G!(xN− 1 ,xN− 2 )...G!(x 1 ,y) (24.8)

Nowasetofpoints:
y attime t
x 1 attime t+!
x 2 attime t+ 2!
... ........ ....
xN− 1 attime t+(N−1)!
x attime t+T

(24.9)

connectedbystraightlines,asshowninFig.24.1,isapathbetweenpoint(y,t)and
point(x,t+T).Theintegralineq. (24.8)isthereforeasumoverallpathsbetween
thesetwopoints, whichconsist ofN straight-linesegmentsof equaltime-duration
!=T/N. Clearly,asN→∞,anycontinuouspathbetween(y,t)and(x,t+T)can
beapproximatedtoarbitraryprecision.
Itsclear,then,thatapropagatorcanberepresentedbyasumoverpaths. The
problemistofindanexpressionforG!(x,y),whichdoesn’tinvolvefirstsolvingthe
eigenvalueequation(24.2)andpluggingtheresultinto(24.3).Fortunately,if!isvery
small,wecandeduceG!(x,y)fromthetime-dependentSchrodingerequation. First,
approximatethetime-derivativewithafinitedifference

i ̄h

ψ(x,t+!)−ψ(x,t)

!

≈Hψ(x,t) (24.10)