QMGreensite_merged

372 CHAPTER24. THEFEYNMANPATHINTEGRAL

(recallthatx=xN, y=x 0 ),andlet

a=

m

2 i! ̄h

(24.29)

Then

GT(x,y)

= lim

N→∞

B−N

∫

dzN− 1 dzN− 2 ...dz 1 exp

[

−a(x−xN− 1 )^2 −a

N∑− 1

k=1

z^2 k

]

= lim

N→∞

B−N

∫

dzN− 1 dzN− 2 ...dz 1 e−a

∑N− 1

k=1zk^2 exp

[

−a(x−y−

N∑− 1

k=1

zk)^2

]

= lim

N→∞

B−N

∫

dzN− 1 dzN− 2 ...dz 1 e−a

∑N− 1

k=1zk^2

∫

due−a(x−y−u)

2

δ

[

u−

N∑− 1

k=1

zk

]

= lim

N→∞

B−N

∫

dzN− 1 dzN− 2 ...dz 1 e−a

∑N− 1

k=1zk^2

∫

due−a(x−y−u)

2

∫ dq

2 π

eiq(u−

∑N− 1

k=1zk)

= lim

N→∞

B−N

∫ dq

2 π

∫

due−a(x−u−y)

2

eiqu

N∏− 1

n=1

∫

dzke−az

k^2 +iqzk

(24.30)

Atthispoint,weagainusethegaussianintegralformulato doeachoftheN− 1
integralsoverthe{zk}.Thisgivesus

GT(x,y) = lim

N→∞

B−N

∫ dq

2 π

∫

due−a(x−u−y)

2

eiqu

[√π

a

e−q

(^2) / 4 a]N−^1

=

(

π

a

)(N−1)/ 2

lim

N→∞

B−N

∫ dq

2 π

∫

due−a(x−u−y)

2

eique−(N−1)q

(^2) / 4 a
(24.31)
Finally,inrapidsuccession,wedothegaussianintegralsoverqandthenu,
GT(x,y) = lim
N→∞

B−N

(π

a

)(N−1)/ (^21)
2 π
√
π
(N−1)/ 4 a
∫
due−a(x−u−y)
2
exp[−
u^2
4(N−1)/ 4 a

]

= lim

N→∞

B−N

(π

a

)(N−1)/ (^21)
2 π
√
π
(N−1)/ 4 a
∫
due−a(x−u−y)
2
e−a(x−y)
(^2) /(N−1)
= lim
N→∞

B−N

(

π

a

)(N−1)/ 2

1

2 π

√

4 aπ

(N−1)

√

π

a

e−a(x−y)

(^2) /(N−1)
(24.32)
InsertingtheexpressionforB,weget
GT(x,y) = lim
N→∞
[ m
2 πi ̄h!
]N/ 2 [ 2 πi! ̄h
m
]N/ 2 √
a
π(N−1)
e−a(x−y)
(^2) /(N−1)
= lim
N→∞
√
m
2 πih ̄!(N−1)
exp
[
−
m(x−y)^2
2 i ̄h!(N−1)
]
(24.33)