Why the uncertainty principle can be used for estimation?

It is usually said/done in textbooks and classes that if $\Delta x$ is known then $\Delta p_x$ can be estimated using the uncertainty principle as $\Delta p_x \sim \hbar/\Delta x$.

But the uncertainty principle does not say that, it says $\Delta p_x\ge\hbar/\Delta x$. That means we can only set a lower bound on $\Delta p_x$, i.e. $\Delta p_x$ could be anything between $\hbar/\Delta x$ and $\infty$

Why then the lower bound is chosen for estimation?

Are there certain situations where the products in uncertainties is of the same order as $\hbar$?

It assumes a sort of democracy between $x$ and $p$, and is obviously not valid everywhere. The connection between classical and quantum mechanics happens through coherent states, which are states which minimize the product $\Delta p \Delta x$, and in a sense, behave most classically (and have well-defined classical limits as $\hbar \to 0$).
• I did not ask for rigorous. It is only an estimation. My problem is the estimation is based on an inequality. So I can estimate something to be n or one million times bigger than n. Sorry I am not sure that I understand your answer. I do not know what coherent states are. Are you saying that in certain situations the uncertainty product can be of the same order as $\hbar$? – Revo Nov 28 '13 at 20:31
• Yes. The ground state of the harmonic oscillator is an example of a coherent state, and satisfies $\Delta x \Delta p = \frac{\hbar}{2}$. – lionelbrits Nov 28 '13 at 20:33