1549380323-Statistical Mechanics Theory and Molecular Simulation

470 The Feynman path integral

y(τ) to an integral over the coefficientscn. Using eqn. (12.4.27), let us first determine
the argument of the exponential. We first note that

y ̇(τ) =

∑∞

n=1

ωncncos(ωnτ). (12.4.29)

Thus, the terms in the action are:

∫β ̄h

0

dτ

m

2

y ̇^2 =

m

2

∑∞

n=1

∑∞

n′=1

cncn′ωnωn′

∫β ̄h

0

dτcos(ωnτ) cos(ωn′τ). (12.4.30)

Since the cosines are orthogonal on the intervalτ∈[0,β ̄h], the integral simplifies to

∫β ̄h

0

dτ

m

2

y ̇^2 =

m

2

∑∞

n=1

c^2 nωn^2

∫β ̄h

0

dτcos^2 (ωnτ)

=

m

2

∑∞

n=1

c^2 nωn^2

∫β ̄h

0

dτ

[

1

2

+

1

2

cos(2ωnτ)

]

=

mβ ̄h

4

∑∞

n=1

c^2 nωn^2. (12.4.31)

In a similar manner, we can show that

∫β ̄h

0

dτ

1

2

mω^2 y^2 =

mβ ̄h

4

ω^2

∑∞

n=1

c^2 n. (12.4.32)

Next, we need to change the integration measure fromDy(τ) to an integration over
the coefficientscn. This is rather subtle since we are transforming from a continuous
functional measure to a discrete one, and it is not immediately clear how the Jacobian
is computed. The simplest way to transform the measure is to assume that

Dy(τ) =g 0

∏∞

n=1

gndcn, (12.4.33)

where theg 0 andgnare constants, and then adjust these parameters so that the final
result yields the correct free particle limitω= 0. With this change of variables,I 0
becomes

I 0 =g 0

∏∞

n=1

∫∞

−∞

gndcnexp

[

−

mβ

4

(ω^2 +ω^2 n)c^2 n

]

=g 0

∏∞

n=1

gn

[

4 π

mβ(ω^2 +ω^2 n)

] 1 / 2

. (12.4.34)

From eqn. (12.4.34), we see that the free particle limit can be recovered by choosing