Question

quantum mechanics: uncertainty principle

Answer 1

\[\Delta x \Delta p_x \ge \frac{ h }{ 2\pi }\]

means that uncertainty is always greater than or equal to h/2pi
but for deriving some relation we take it as approx. equal to h/2pi
for example: mass of meson
here \[\Delta x = r (size of nucleus)\]
p= mc
and hence\[m \approx \frac{ h }{ 2 \pi rc }\]
but my question is that uncertainty principle say that it must be equal to or greater than, it just give minimum possible value but it doesn't rule out the ">" case.
but above derivation agree with experimental results also. so, what is the basic logic that in these derivation, we are taking the uncertainty inequality as an equation.

Answer 2

i really didn't get your doubt :O

Answer 3

\[\Delta E \Delta t \ge h/ 4\pi\]
in alpha emission energy of alpha particle is less in nucleus to overcome the potential barrier. let the it need ΔE more. so from uncertainty principle it can have uncertainty in ΔE for the time interval Δt=h/4πΔE. if it is able to cross the potential wall in this time interval it can come out of the nucleus.
back to my question: why Δt=h/4πΔE,
why not Δt>h/4πΔE
since it is an inequality, i know it's wierd because in this case alpha particle will always be able to come out of the nucleus irrespective of its potential wall thickness.