QuantumPhysics.dvi

for the operatorM(t), defined by

M(t) =−m

d^2

dt^2

−V′′(q 0 (t)) (14.64)

and subject to the boundary conditionsy(ta) =y(tb) = 0. We will now show that, if the complete
spectrum of the self-adjoint operatorM(t) (subject to the above boundary conditions) is known,
then this functional integral of Gaussian fluctuations can be computed. Let the eigenfunctions
φn(t) satisfy

M(t)φn(t) =λnφn(t) (14.65)

subject toφn(ta) =φn(tb) = 0. SinceM(t) is self-adjoint, the eigenvaluesλnare real and the
eigenfunctionsφn(t) span an orthonormal basis of the function space fory(t). SinceM(t) is actually
real, we may choose a basis of real functionsφn(t), obeying
∫tb

ta

dtφn(t)φn′(t) = δn,n′

∑

n

φn(t)φn(t′) = δ(t−t′) ta≤t,t′≤tb (14.66)

The completeness ofφn(t) allows us to expand

y(t) =

∑

n

cnφn(t) (14.67)

for real coefficientscn. Since the metric on function space and on thecnare related by
∫tb

ta

dtδy(t)^2 =

∑

n

(δcn)^2 (14.68)

the change of variablesy(t)→{cn}has unit Jacobian, and we have

Dy=

∏

n

(dcn) (14.69)

It is now also straightforward to evaluate
∫tb

ta

dty(t)M(t)y(t) =

∑

n

λn(cn)^2 (14.70)

so that the entire functional integral over Gaussian fluctuations becomes,
∫
Dyexp

{

i

2

∫tb

ta

dty(t)M(t)y(t)

}

=

∏

n

(∫

dcnexp

{

i

2

λn(cn)^2

})

=

∏

n

λ

−^12

n (14.71)

Relating the product of the eigenvalues to the determinant,one often writes,

∫

Dyexp

{

i

2

∫tb

ta

dty(t)M(t)y(t)

}

= Det

(

−m

d^2

dt^2

−V′′(q 0 (t))

)−^12

(14.72)