>> Mathematical Expression of Uncertainty Principle:

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)

Click for more:
                       - The Heisenberg Uncertainty Relations(1): 

                       - Heisenberg recalls his Early thought on Uncertainty Principle(2):