How Does Shankar Transition from Sums to Integrals in Feynman Path Integrals?

[Opus_723]

I am going through Shankar's treatment of Feynman Integrals right now, and I have one lingering doubt that I can't quite seem to work out.

I was pretty happy with the idea of discretizing time, then doing independent sums over x_i at each time. But Shankar simply says that we can consider the sums over x_i to be integrals. I don't quite follow this. Normally when you pass from a sum to an integral there has to be some infinitesimal factor in each term of your sum. Shankar just says that the sum over the phase factors becomes an integral in x_i over the phase factors, but I don't see where the "dx" comes from in order to let us do that. I have a suspicion that something deeper is going on, but I can't quite grasp it.

I've actually noticed this sort of thing in a couple of other places. For example, the completeness relation for operators with discrete spectra seems to pick up an infinitesimal "d_" that doesn't seem to have a counterpart in the completeness relation for operators with discrete spectra. This may be completely unrelated to the path integrals. But I get the feeling that I'm missing something either very obvious or very subtle, because this sort of thing keeps coming up in QM. Could anyone help me clear this up?

 

[hilbert2]

I've actually noticed this sort of thing in a couple of other places. For example, the completeness relation for operators with discrete spectra seems to pick up an infinitesimal "d_" that doesn't seem to have a counterpart in the completeness relation for operators with discrete spectra. This may be completely unrelated to the path integrals. But I get the feeling that I'm missing something either very obvious or very subtle, because this sort of thing keeps coming up in QM. Could anyone help me clear this up?

In short, this is because the orthonormality of basis vectors is like $\left<\psi_i \right.\left|\psi_j \right.\left.\right> = \delta_{ij}$ in the discrete case and $\left<x \right.\left|x' \right.\left.\right> = \delta (x-x' )$ in the continuum case (in the first case there's a Kronecker delta and in the second there's a Dirac delta).

 

[bhobba]

Actually defining path integrals properly is rather difficult and touches on some rather interesting, but advanced areas of applied math such as White Noise Theory:
http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1797-03.pdf

At the level below that ie at the level of the typical undergrad treat it like you would the Dirac Delta function - a normal function that for all practical purposes has that property - simply think of the path integral like a post I did about it ages ago:

You start out with <x'|x> then you insert a ton of ∫|xi><xi|dxi = 1 in the middle to get ∫...∫<x|x1><x1|...|xn><xn|x> dx1...dxn. Now <xi|xi+1> = ci e^iSi so rearranging you get ∫...∫c1...cn e^ i∑Si.

To get the path integral you are supposed to take the limit - but rigorously defining such is rather difficult. So just think of it as something where the difference between the xi is very small - but not actually zero. Sort of like you do in the early stages on calculus with how you view the integral as a sum with the Δx really small.

Thanks
Bill