Examples 149I 2 =
(π
2 α) 3 / 2
e−α(r(^2) +2r (^23) )/ 2
eα(r+2r^3 )
(^2) / 6
∫
all spacedr 2 e−^3 α[r^2 −(r+2r^3 )](^2) / 2
=
(π
2 α) 3 / 2 ( 2 π
3 α) 3 / 2
e−α(r−r^3 )(^2) / 3
=
(
π^2
3 α^2) 3 / 2
e−α(r−r^3 )(^2) / 3
. (4.5.30)
From the calculation ofI 1 andI 2 , a pattern can be discerned from which the result of
performing allNintegrations can be predicted. Specifically, after performingn < N
integrations, we find
In=(
πn
(n+ 1)αn) 3 / 2
e−α(r−rn+1)(^2) /(n+1)
. (4.5.31)
Thus, settingn=N, we obtain
IN=
(
πN
(N+ 1)αN) 3 / 2
e−α(r−rN+1)(^2) /(N+1)
. (4.5.32)
IdentifyingrN+1=r′ and attaching the prefactor (2πm/βh^2 )^3 N/^2 , we obtain the
partition function for fixedrandr′as
Q(N,T,r,r′) =(
2 π
βhω) 3 N
1
(N+ 1)^3 /^2
e−βmω(^2) (r−r′) (^2) /(N+1)
. (4.5.33)
The volume dependence has dropped out because the integrationswere extended over
all space. Eqn. (4.5.33) can be regarded as a probability distributionfunction for the
distance|r−r′|^2 between the endpoints of the polymer. Note that this distribution is
Gaussian in the end-to-end distance|r−r′|.
If we now allow the endpoints to move, then the full partition function can be
calculated by introducing the momentap 0 andpN+1of the endpoints and performing
the integration
Q(N,V,T) =
1
h^6(
2 π
βhω) 3 N
1
(N+ 1)^3 /^2
∫
dp 0 dpN+1e−β(p(^20) +p (^2) N+1)/ 2 m
×
∫
dr 0 drN+1e−βmω(^2) (r 0 −rN+1) (^2) /(N+1)
. (4.5.34)
Here, the extra factor of 1/h^6 has been introduced along with the kinetic energy of
the endpoints. Performing the momentum integrations gives