Classical limit of the path integral

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In Feynman's quantum mechanics and path integrals, small changes in the action S lead to significant phase changes, causing contributions from various paths to oscillate rapidly and cancel each other out. The discussion highlights that in the classical limit, where S is much larger than the reduced Planck's constant h, only paths close to the path of least action significantly contribute to the overall sum. The cancellation occurs because paths with large action values result in rapid oscillations between positive and negative contributions. The threshold for S to be considered "classical" varies by context, but it is generally much larger than h. For example, the action for a single 8 carbon polymer is significantly larger than h, allowing its motion to be treated classically.
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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 Feynman 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"?
 
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The claim that the paths "cancel each other out" is true, but a more detailed explanation must be given. Essentially, what happens is that for large S (on the classical scale) the phase difference between paths is very large, so the contributions of each path oscillate rapidly between positive and negative values. The net result is that these contributions cancel each other out, and only those paths with relatively small phase differences will contribute to the sum. This is why only paths close to the least action path are important in the classical limit.The size of S required for this effect to kick in depends on the particular problem, but generally speaking it should be much larger than h. For example, for a single 8 carbon polymer, the action S will be on the order of 10^-26 joules, which is much larger than h (10^-34 joules). So in this case, we can safely consider the motion to be "classic".
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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