# 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?

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