# More on the Feynman Path Integral Formula in Brian Cox' Lecture and its Consequences

This is a continuation of this question about Brian Cox' lecture Night with the Stars.

I know the main steps to get from $K(q",q',T)=\sum_{paths}Ae^{iS(q",q',T)/h}$ to $\Delta t > \dfrac{m(\Delta x)^2}{h}$ as stated below, but can you expand? (just read below)

PART 1

The action function $S(q",q',T)$ is given by $S = \displaystyle\int dt\left( \dfrac{1}{2} m v^2 -U\right)$. For a classical path that goes uniformly from one point to the other you have $v = \dfrac{\Delta x}{\Delta t}$ and so you get $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$. What are the processes and steps taken to you get to $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$? (explained clearly please).

PART 2

$S/h$ appears as a complex phase term. To make it small we set $S/h < 1$, and we can then deduce that $\Delta t > \dfrac{m(\Delta x)^2}{h}$.

What are the processes and steps taken to then get to $\Delta t > \dfrac{m(\Delta x)^2}{h}$?

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I forgot to mention that in this the -U from the first equation is ignored and when I say how to get to the next step I mean in the sense that for example: "$x+5=8$ , $x=3$".......but rather "$x+5=8$ , minus 5 both sides $x=3$ –  Olly Price Mar 29 '12 at 18:58
So basically I need an explanation as to the mathematics done that got to the next step in the formulas –  Olly Price Mar 29 '12 at 19:02
I don't understand how you can take S/h<1 when you have assumed semi-classicality with v = delta x/delta t? –  user24703 May 20 '13 at 5:35

Part 1:

Let's say the velocity in the integral is constant in time, and the integral is from 0 to $\Delta t$. We now have a trivial integral of a constant. So (ignoring U)

$$S = \int_0^{\Delta t} \frac{1}{2}mv^2\ dt\\ =\left[\frac{1}{2}mv^2t\right]_{t=0}^{t=\Delta t}\\ =\frac{1}{2}mv^2\Delta t$$

So, substituting $v = \dfrac{\Delta x}{\Delta t}$, and ignoring any constants we have before them, except for mass, we get

$$S \propto mv^2 \Delta t\\ \propto m\left(\frac{\Delta x}{\Delta t}\right)^2 \Delta t\\ \propto m\frac{\Delta x^2}{\Delta t}$$

So $S \propto m \dfrac{\Delta x^2}{\Delta t}$ as required.

Part 2:

We have

$$S \propto m \dfrac{\Delta x^2}{\Delta t}$$, so then, saying the constant of proportionality is $k$, and that it is approximately 1 (we'll want this not to be huge in a minute), we get

$$S = k m \frac{\Delta x^2}{\Delta t}$$

Now, setting $S/h < 1$, we get

$$\frac{S}{h} = k m \frac{\Delta x^2}{h \Delta t} < 1$$

Setting $k = 1$, we can now say

$$m \frac{\Delta x^2}{h \Delta t} < 1 \Rightarrow m \frac{\Delta x^2}{h} < \Delta t$$ as required.

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when you're at the stage of $$\propto m\left(\frac{\Delta x}{\Delta t}\right)^2 \Delta t\\$$ what happened to $\Delta t$ at the end of the equation to get you to $\propto m\frac{\Delta x^2}{\Delta t}$? –  Olly Price Mar 29 '12 at 21:24
Also, at $\frac{S}{h} = k m \frac{\Delta x^2}{h \Delta t} < 1$, why can you just add Planck's Constant, h, into the equation? –  Olly Price Mar 29 '12 at 21:28
And how is your resultant $\frac{\Delta x^2}{h} < 1$ the same as my $\Delta t$ > $\frac{m(\Delta x)^2}{h}$ –  Olly Price Mar 29 '12 at 21:37
When answering the 3 questions, label the answer to the first comment '1', the second comment '2' and the third comment '3' please. Thanks –  Olly Price Mar 29 '12 at 21:41
Comment 1: $m(\dfrac{\Delta x}{\Delta t})^2 \Delta t = m \dfrac{\Delta x^2}{\Delta t^2} \Delta t$. The $\Delta t$ on the left cancels out the $\Delta t$ under the line. –  P O'Conbhui Mar 29 '12 at 22:36