Re: What's the correct form of Heisenberg's uncertaint

Hallo, Flavio!

To make a long story short, the answer to your question is that they both 
are correct, the cause for difference beeing different normalisation 
convention or something like that.

Now, to extend this long story made short, let me just briefly tell you 
(what you may know already) where the root of the Heisenberg relation lies.
If you are familiar with the fourier analysis (if not, you might want to 
check on it in any book on mathematical physics - I suggest 'Mathematical 
methods for physicists' by G. Arfken and H. Weber, Academic press, San 
Diego 1995 - it's really neat and easily comprehensible method), then you 
will know that every square-integrable function y(x) (you can think of it 
as a spatial distribution of something - charge, probability density or 
whatewer - with x a spatial coordinate) can be decomposed into the 
superposition of continous spectrum of sinusoidal and cosinusoidal 
functions (of the same spatial coordinate x), the magnitude of each of them 
beeing uniquely determined by the original function. 

Now, starting from y(x), its fourier transform gives you the g(omega), 
where omega is the angular frequency. g(omega) is also called a 'fourier 
spectrum' of y(x). One property of the spectrum is that if the original 
function is localised in x space (i.e. your distribution y(x) is different 
from zero only within some Dx around some point x0), its fourier spectrum 
will also be localised in omega space (that is, it will be different from 
zero only within some Domega around omega0). Interesting, but simple to 
grasp property of the Dx and Domega is that Dx*Domega is a certain nonzero 
number, whose detailed value depends on the choice of the form of fourier 
transform and the shape of y(x) (the favourite among the physicists beeing 
the gaussian curve, as it is a very nice analytical localised curve with 
well defined width). What this means is that if you make a fourier 
transform of some other (spatial distribution) function, let's name it z(x)
which is much more (smaller Dx) or much less (greater Dx) localised than 
the y(x), you will get the fourier spectrum which is respectively much 
broader (greater Domega) or much sharper (smaller Domega) then the one you 
got from the y(x).

Now Heisenberg relations are just a byproduct of dealing with 
spatial distribution functions (wavefunctions) and their fourier transform 
(momentum). Its particular form is less significant (if indeed it is 
significant at all) than the fact that the right hand side is nonzero!

hope I'w been helpfull

Duje