ognik said:
Frustratingly I can't find the source I found this in, but this was in a section on wave packets and the uncertainty principle - and my apologies for using \nu instead of v ...
##\Delta x_0 \Delta p_0 \ge \frac{\hbar}{2} ##
## \therefore \Delta v_0 \ge \frac{\hbar}{2 m \Delta x_0 } ##
## \therefore \Delta x \ge \frac{\hbar}{2 m \Delta x_0 } t ##
So my question is just why ## \Delta v_0 t = \Delta x ## and not ## \Delta x_0 ##?
then its a 'riddle' in the sense that your uncertainty in x(0) and p(0) is related by your given equation -
if its at time say t=t(0) then as time advances -goes to t=t they wish to calculate the delta change in x i.e. equivalent to ( x- x(0))
its an infinitesimal change in x so its written in that manner- this has been effected by the uncertainty in v(0) times the time elasped , so if somebody has plotted a shape of the packet with time with t it may be seen as spread of the wave packet in x and naturally its momentum will be /may be more sharp-
but the above is a wild guess only and any question /problem should be framed in a context- if you look up any text on QM and see the discussion on time development of a packet -then perhaps you can get open the riddle yourself
we know how a gaussian packet spreads with time - as the graphs are given in textbooks
For instance, if an electron is originally localized in a region of atomic scale (
i.e., [PLAIN]http://farside.ph.utexas.edu/teaching/qmech/Quantum/img395.png) then the doubling time is only about [PLAIN]http://farside.ph.utexas.edu/teaching/qmech/Quantum/img396.png. Evidently, particle wave packets (for freely moving particles) spread very rapidly..
for details see <http://farside.ph.utexas.edu/teaching/qmech/Quantum/node26.html>