The Uncertainty Principle 'Protects' Quantum Mechanics.
Because nobody could figure out a way to measure the position & the momentum of anything simultaneously - a screen an electron with any greater acurracy.
So Heisenberg proposed that It is impossible to measure the momentum & the position simultaneously with a greater acuuracy.
Don't get too excited: the uncertainty principle still stands, says Steinberg: “In the end, there's no way you can know [both quantum states] accurately at the same time.” But the experiment shows that the act of measurement isn't always what causes the uncertainty. “If there's already a lot of uncertainty in the system, then there doesn't need to be any noise from the measurement at all,” he says [Click].
If it is possible to measure, the Quantum Mechanics would Collapse.
Scientific Proof:
Let $ \Omega $ & $ \Lambda $ be two Hermitian operator, with a commutator:
[$ \Omega , \Lambda ] = i\Gamma $ ............................ (1)
You may readily verify that $ \Gamma$ is also Hermitian. Let us start with the uncertainty product in a normalized state $ \mid \Psi> $:
$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} = < \Psi \mid (\Omega -<\Omega > )^{2}\mid \Psi > < \Psi \mid (\Lambda -<\Lambda > )^{2}\mid \Psi > $ ............. (2)Where $<\Omega> = <\Psi|\Omega|\Psi>$ & $<\Lambda> = <\Psi|\Lambda|\Psi>$.Let us next define the pair:
$\left.\begin{matrix}
\hat{\Omega } = \Omega -<\Omega > & \\
\hat{\Lambda } = \Lambda -<\Lambda > &
\end{matrix}\right\}$ ....................................... (3)Which has the same commutator as $\Omega$ & $\Lambda$ (Verify this). In terms of $\hat{\Omega }$ and $\hat{\Lambda }$,$\left.\begin{matrix}
(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} = < \Psi \mid \hat{\Omega }^{2}\mid \Psi > < \Psi \mid \hat{\Lambda }^{2}\mid \Psi > & \\
= <\hat\Omega\Psi \mid \hat\Omega\Psi > <\hat\Lambda\Psi \mid \hat\Lambda\Psi > \end{matrix}\right\}$ ......................... (4)Since$\hat{\Omega ^{2}} = \hat{\Omega }\hat{\Omega } = \hat{\Omega }^{\dagger }\hat{\Omega }$ & $\hat{\Lambda}^{2} = \hat{\Lambda}^{\dagger}\hat{\Lambda}$ ............................. (5)If we apply the Schwartz inequality$| V_{1}|^{2}| V_{2}|^{2} \geq | <V_{1}| V_{2}>|^{2}$ ....................................... (6)(Where the equality sign holds only if $| V_{1}> = c| V_{2}>$, where $c$ is a constant) to the states $|\hat\Omega \Psi>$ and $|\hat\Lambda \Psi>$, we get from Eq. (4),$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq |<\hat\Omega \Psi|\hat\Lambda \Psi>|^{2}$ ................................. (7)Let us now use the fact that$<\hat\Omega \Psi|\hat\Lambda \Psi> = <\Psi|\hat{\Omega }^{\dagger }\hat{\Lambda}|\Psi> = <\Psi|\hat{\Omega }\hat{\Lambda}|\Psi>$ .................................. (8)to rewrite the above inequality as$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq |<\Psi|\hat{\Omega }\hat{\Lambda}|\Psi>|^{2}$ .................................... (9)Now, we know that the commutator has to enter the picture somewhere. This we arrange through the following identity:$\left.\begin{matrix}
\hat{\Omega}\hat{\Lambda} = \frac{\hat{\Omega}\hat{\Lambda} + \hat{\Lambda} \hat{\Omega}}{2} + \frac{\hat{\Omega}\hat{\Lambda} - \hat{\Lambda} \hat{\Omega}}{2} &\\
&\\ = \frac{1}{2}[\hat{\Omega},\hat{\Lambda}]_+ + \frac{1}{2}[\hat{\Omega},\hat{\Lambda}]
\end{matrix}\right\}$ .................................. (10)Where $[\hat{\Omega},\hat{\Lambda}]_+$ is called the $anticommutator$. Feeding Eq. (10) into the inequality (9), we get$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq |<\Psi|\frac{1}{2}[\hat{\Omega},\hat{\Lambda}]_+ + \frac{1}{2}[\hat{\Omega},\hat{\Lambda}]|\Psi>|^{2}$ .............................. (11)We next use the fact that:(1) Since $[ \Omega , \Lambda ] = i\Gamma $, where $\Gamma$ is $Hermitian$, the expectation value of the commutator is pure imaginary;(2) Since [$ \Omega , \Lambda ]_+$ is $Hermitian$, the expectation value of the anticommutator is real.Recalling that $|a + ib|^{2} = a^{2} + b^{2}$, we get$ \left.\begin{matrix}
(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq \frac{1}{4}|<\Psi|[\hat{\Omega},\hat{\Lambda }]_+|\Psi> + i<\Psi|\Gamma |\Psi>|^{2} \\
\geq \frac{1}{4}{<\Psi|[\hat{\Omega},\hat{\Lambda }]_+|\Psi>}^{2} + \frac{1}{4}<\Psi|\Gamma |\Psi>^{2}
\end{matrix}\right\}$ .................... (12)This is the general uncertainty relaton between any two Hermitian operator and is evidently state dependent. Consider now canonically conjugate operators, for which $\Gamma$ = $\hbar$. In this case$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq \frac{1}{4}{<\Psi|[\hat{\Omega},\hat{\Lambda }]_+|\Psi>}^{2} + \frac{\hbar^{2}}{4}$ ............................... (13)Since the first term is positive definite, we may assert that for any |\Psi>$(\Delta\Omega ) ^{2}(\Delta\Lambda ) ^{2} \geq \frac{\hbar^{2}}{4}$
or,.................... (14)
Which is the celebrated $ Uncertainty$ $Relation $. Let us note that the above inequality becomes an equality only if(1) $\hat{\Omega}|\Psi > = c\hat{\Lambda}|\Psi> $and
(2) $ <\Psi|[\hat{\Omega},\hat{\Lambda}]_+|\Psi> = 0$ ................. (15)