QuantumPhysics.dvi

of paths around the classical one. As before, we assume that we have a classical solutionq 0 (t)
connecting initial and final positions. If ̄hwere zero, that is all there would be. So, it makes sense
to think of the quantum fluctuations aroundq 0 (t) as small whenS/ ̄h≫π. The correct expansion
order is given by

q(t) =q 0 (t) +

√

̄hy(t) y(ta) =y(tb) = 0 (14.52)

The contribution from the Gaussian fluctuations is then obtained by expandingS[q 0 +

√

̄hy] in
powers of ̄h. By the very fact thatq 0 (t) is a classical solution, we know that the term of order

√

̄h
in this expansion must vanish. Hence, we have

S[q 0 +

√

̄hy] = S[q 0 ] + ̄hS 2 [y;q 0 ] +O( ̄h^3 /^2 )

S 2 [y;q 0 ] =

∫tb

ta

dt

(

1

2

Lq ̇q ̇y ̇^2 +Lqq ̇yy ̇ +

1

2

Lqqy^2

)

(14.53)

where the coefficients are given by

Lq ̇q ̇=

∂^2 L

∂q ̇^2

∣∣

∣∣

q=q 0

Lqq ̇ =

∂^2 L

∂q∂q ̇

∣∣

∣∣

q=q 0

Lqq=

∂^2 L

∂q^2

∣∣

∣∣

q=q 0

(14.54)

Notice that, for a general LagrangianL, all three coefficientsLq ̇q ̇,Lqq ̇,Lqqwill be non-vanishing
and may beτ-dependent. In the case of the standard Lagrangian,

L=

1

2

mq ̇^2 −V(q) (14.55)

the problem is much simplified though and we have

Lq ̇q ̇=m Lqq ̇ = 0 Lqq=−V′′(q 0 (t)) (14.56)

The Gaussian action is then simply,

S 2 [y;q 0 ] =

∫tb

ta

dt

(

1

2

my ̇^2 −

1

2

V′′(q 0 )y^2

)

=

1

2

∫tb

ta

dt y(t)

(

−m

d^2

dt^2

−V′′(q 0 (t))

)

y(t) (14.57)

The energy levels and wave-functions of the harmonic oscillator could be recovered in this way (see
homework problem).

14.9 Gaussian integrals

We shall compute the Gaussian integrals for real and complexvariables, and show that
∫
dNXexp

{

−πτXtMX

}

=

1

(det(τM))

1

2

∫

dNZdNZ ̄exp

{

− 2 πτZ†HZ

}

=

1

det(τH)

(14.58)