Definition of the Path Integral

1. Jul 5, 2015

cpsinkule

Why is it that, in the definition of the path integral, we have the product of neighboring integrals of the form : ∫Φdx1.....dxn when the whole idea is based on adding the contribution of neighboring paths. I need some help understanding why it is of the form ∫Φdx1.....dxn and not the form ∫Φdx1+∫Φdx2+...∫Φdxn where Φ is eiS/ħ. I hope I'm making sense. In the formulation of the idea, we have discrete points in time where we sum over all values of x at that point, namely ∫Φdxi and we add each contribution in the discrete limit Σ∫Φdxi. Why is it that this turns into a chain of differentials dxi in the limit that the length of the time intervals goes to 0 and the number of intervals goes to infinity? Again, I hope this makes sense and that you can shed some light on this for me.

2. Jul 5, 2015

Chopin

The path integral is a multi-dimensional integral over all of the variables in the configuration space. For that reason you need to add the contribution from every combination of the variables' values, which means you have to multiply the differentials together.

This is exactly the same as a double integral over a two-dimensional function--you write the integral as $\int dx\:dy\:f(x,y)$. You couldn't write it as $\int dx\:f(x,y) + \int dy\:f(x,y)$, because $f(x,y)$ is a two-dimensional function, and neither of those terms could specify both variables for the function. Only by writing one integral with both differentials can you provide all of the values the function needs. The path integral is just this same idea, with the number of integration variables taken to infinity.

3. Jul 6, 2015

Staff: Mentor

Here is the detail.

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.

Focus in on ∑Si. Define Li = Si/Δti, Δti is the time between the xi along the jagged path they trace out. ∑ Si = ∑Li Δti. As Δti goes to zero the reasonable physical assumption is made that Li is well behaved and goes over to a continuum so you get ∫L dt.

Now Si depends on xi and Δxi. But for a path Δxi depends on the velocity vi = Δxi/Δti so its very reasonable to assume when it goes to the continuum L is a function of x and the velocity v.

Its a bit of fun working through the math with Taylor approximations seeing its quite a reasonable process.

In this way you see the origin of the Lagrangian. And by considering close paths we see most cancel and you are only left with the paths of stationary action.

Now as the path goes to a continuum ∫.....∫c1....cn is written as ∫Dx(t) and you get ∫Dx(t) e^( i ∫Ldt).

You can find more detail here:
http://hitoshi.berkeley.edu/221a/pathintegral.pdf

Thanks
Bill