# Feynman Path Integral Formula in Brian Cox' “A Night with the Stars” Lecture

The Youtube link keeps breaking, so here is a search on Youtube for Brian Cox' A Night with the Stars lecture. Pause the video on 40.32minutes.

What you see he said is called Feynman's Path Integral.

$K(q",q',T)=\sum_{paths}Ae^{iS(q",q',T)/h}$

Am I right in thinking this adding all the different paths a particle can take predicting the probability of it landing in a certain position?

Once that question is answered how in the world did he simplify it to:

$t > \dfrac{x \Delta{x} m}{ h}$--- (42.39minutes)

This second equation he got from simplifying was what is used to "predict how long it would take" for his "diamond to jump out of the box", how fascinating.

So: what is all this in Feynman's Path Integral? and how did he simplify it to get the other equation? and if you know, what is this second equation called?

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Well, first of all the "action" function $S(q",q',T)$ that appears in his formula is given by $S = \int dt\left( \dfrac{1}{2} m v^2 -U\right)$, where the integral is taken over the path in question. $U$ is a potential term which we will ignore. For the "classical" path which goes uniformly from one point to the other, we have $v = \dfrac{\Delta x}{\Delta t}$ and so you get $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$.
Now, since $S/h$ appears as a complex phase term, we want it to be small so that neighboring paths do not destructively interfere (cancel each other out). So if we set $S/h < 1$, we get $\Delta t > \dfrac{m(\Delta x)^2}{h}$. Now in his formula it wasn't really clear what the difference was between $x$ and $\Delta x$, he used similar values for each, and in any event this is just a very crude order of magnitude type estimation.
Thank you, that was well described and I'm pretty near to understanding now. I think that x = displacement and $\Delta x$ = uncertainty in position (I THINK) – ODP Mar 14 '12 at 20:57
i just noticed the final formula you got was $\Delta t > \dfrac{m(\Delta x)^2}{h}$ but this is not the same as what I stated in my question, or is it? – ODP Mar 24 '12 at 17:46
is '$S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$' all one equation or did you mean that $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2$ is the same as $\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$ ??? I'm guessing it's all one equation but just to be sure – ODP Mar 29 '12 at 16:58