# Classical limit of the path integral

## Main Question or Discussion Point

In feynman's quantum mechanics and path integrals,

he makes the following claim:

"Now if we move the path by a small amount dx, small on the classical scale, the change in S (the action), is likewise small on the classical scale, but not when measured in the tiny unit of reduced planck's constant h. These small changes in path will, generally, make enormous changes in phase, and our cosine or sine will oscillate exceedingly rapidly between plus and minus values. The total contribution will then add to zero.

I have difficulty mathematically showing this.

The contribution of the action along a path j, $$\phi_j[x(t)]=e^{(i/h)S_i[x(t)]}$$ , k constant and $$S_j[x(t)]$$ is the action along the path j

$$K(b,a) = \sum_{over all paths}\phi[x(t)] = \phi_1[x(t)]+\phi_2[x(t)]+...+\phi_N[x(t)] = e^{(i/h)S_1[x(t)]}+e^{(i/h)S_2[x(t)]}+ ... +e^{(i/h)S_N[x(t)]}$$

Using Euler's identity,

$$K(b,a) = cos((i/h)S_1[x(t)]) + i*sin((i/h)S_1[x(t)]) + cos((i/h)S_2[x(t)]) + i*sin((i/h)S_2[x(t)])+... + cos((i/h)S_N[x(t)]) + i*sin((i/h)S_N[x(t)])$$

In the classical limit where $$S_i >> h$$ :

Let's assume that the action $$S_1$$ has the least action, and all other paths differ from $$S_1$$ by $$ds$$

Then $$K(b,a) = cos((i/h)S_1[x(t)]) + i*sin((i/h)S_1[x(t)])$$

since Feynmann claims that:

$$cos((i/h)S_2[x(t)]) + i*sin((i/h)S_2[x(t)]) ... + cos((i/h)S_N[x(t)]) + i*sin((i/h)S_N[x(t)] = 0$$ as they "cancel each other out"

But what I don't understand is that, because no matter how large S_N becomes in the classical limit (ie for objects of very large mass), $$cos(S_N) \leq 1$$, so how exactly do they cancel out?

So all paths that are far away from the path of least action is NOT important so long as the action S >> h? What exactly is >>, ie is 10^2 bigger, 10^3 bigger? How about the action of the rotational and translational movement for single 8 carbon polymer?

Also, at molecular mass and time scale do we start considering it to be "classic"?