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We all know that the Heisenberg uncertainity principle implies

$\Delta x\, \Delta p\geq\frac{\hbar}{2}.$

But is there an ideal condition where we can measure $\Delta x$ to a particular precision and also simultaneously measure $\Delta p$ such that we can predict the position and momentum of the particle (kind of like a balance in the weighing machine) , if not with arbitary precision ,atleast to some certainity ?

Consider a wave whose displacement and momentum can be altered to reach such an ideal condition that we can measure the displacement to some least accuracy but also predict where the wave is going.

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Yes, as you can see from the inequality form of the uncertainty principle it sets a lower bound on uncertainty. The minimum uncertainty is obtained for the equality, $$\Delta x \Delta p =\frac{\hbar}{2}$$ The types of states for which this happens and for which the uncertainty in both position and momentum are equal are sometimes called coherent states (in the study of harmonic oscillators). Specifically for the harmonic oscillator in the coordinate representation they are Gaussian wave packets, $$|\psi(x,t)|^2=\sqrt{\frac{m\omega}{\pi\hbar} } e^{-\frac{m\omega}{\hbar}\left(x- \langle \hat{x}(t) \rangle \right)^2 },$$ where $m$ is mass and $\omega$ is the frequency of the oscillator. Coherent states are important in other contexts, for more see: http://en.wikipedia.org/wiki/Coherent_states

But with that being said the sooner you get used to a fundamental uncertainty in nature the better. It is something we can minimize but it will always be there.

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