A Short Course on Quantum Mechanics and Methods of Quantization 41We now insert a resolution of the identity in momentum space (I=∫
dp|pj〉〈pj|)
in each factor of the integrand, to get:
〈xn|e−ıT
e−ıV
|xn− 1 〉=∫
dpn〈xn|e−ıT
|pn〉〈pn|e−ıV
|xn− 1 〉=∫
dpne−ıp^2 n
2 m〈xn|pn〉e−ıV(xn− 1 )
〈pn|xn− 1 〉=^1
2 π∫
dpne−ıp^2 n
2 me−ıV(xn− 1 )
eıpn(xn−xn− 1 )
,(1.197)where we have used the fact that:
〈xj|pn〉=√^1
2 πeıxjpn
. (1.198)Now we insert (1.197) in (1.196), to find:
〈x|e−ıtH
|x′〉= lim→ 0∫
dx 1 ...dxM− 1dp 1 ...dpm
(2π)M∏M
n=1[
exp{
ıpn(xn−xn− 1 )
−ıp(^2) n
2 m
−ıV(xn−^1 )
}]
. (1.199)
Thanks to the form of the Hamiltonian, we see that the integrals in thepn’s are
of Gaussian type and thus they can be easily performed, to finally achieve:
〈x|e−ıtH
|x′〉= lim→ 0∫
dx 1 ...dxM− 1( m
2 πı)M 2
×exp{
ı
∑M
n=1[
m
2(
xn−xn− 1) 2
−V(xn− 1 )]}
. (1.200)
Let us remark that we have made no approximations to get this formula, which
is mathematically sound as long as the problem of domains is taken into account,
and the limit is taken after all integrals have been calculated.
Despite the last comment, it is very tempting to bring the limit inside the
integrals. Noticing that:
xn−xn− 1
→x ̇(t′),V(xn− 1 )→V(x(t′)),∑M
n=1→
∫t0dt′, (1.201)one can write:
∑Mn=1[
m
2(
xn−xn− 1) 2
−V(xn− 1 )]
→
∫t0dt′[
mx ̇^2 (t′)
2 −V(x(t