Heisenberg Uncertainty Principle scientific proof Heisenberg's uncertainty principle states that:  
$$\sigma(x)\sigma(  p_x )\ge \frac {\hbar}{2}.$$
What is the scientific proof of this principle?
Operators Uncertainty
 A: A wide variety of experiments, of which the Double Slit experiment is the most dramatic, can be used to establish that matter is best represented as a wave on microscopic scales.  Once you represent matter as a wave, then it is natural to associate its position with the spread of the wave, and its momentum with the wavelength of the wave.  Once you do this, however, it should be clear that there is a tradeoff between a well defined ''location'' of the wave, and a well defined ``wavelength'' of the wave.  Therefore, one cannot simultaneously precisely define a particle's position and momentum.  Extra precision in one must come with a lost in precision in the other.
A: Not sure what you mean by scientific proof. A hypotheses can be validated by scientific method. Its not proof as in mathematics. Because physics does not deal with abstract mathematical ideas which can be proved by following some predefined axioms and rules. Which has nothing to do with observation.
If uncertainty principle is taken as truth then physical phenomenon regarding subatomic particles can be explained and to some extent predicated by  quantum mechanical framework. 
A: The uncertainty principle, in the variance formulation, states that in any quantum state $|\rangle$, the quantity
$$\langle (p-<p>)^2 \rangle \langle (x-\langle x\rangle)^2\rangle \ge {\hbar^2 \over 4} $$
To understand why shifting p and x by their expected value and squaring gives the squared uncertainty, see this answer.
The proof is by noting the following 
$$ |\langle \psi | \eta \rangle| \le \sqrt{ ||\psi||^2 ||\eta||^2}$$
This is the statement that the dot-product of two vectors is less than the product of their lengths. It is called the "Cauchy Schwartz inequality". For the special case above, defining the operators $P= p-\langle p\rangle$ and $Q=x-\langle x\rangle$ (and squaring both sides), 
$$ ( \langle P Q \rangle )^2 \le \langle PP\rangle\langle QQ\rangle $$
Where to see that the above is an instance of Cauchy Schwarz, take:
$$ |\psi\rangle = P|\rangle$$
$$ |\eta\rangle = Q|\rangle$$
While the product PQ can be decomposed into a real and imaginary part
$$ PQ = {1\over 2} (PQ+QP) + {1\over 2} (PQ-QP) $$
The first part is imaginary, because if you take the Hermitian conjugate, it changes sign. The second part is real (this is ultimately because P and Q are real, i.e. Hermitian). The expected value of PQ squared is the square of the imaginary and real parts separately
$$ (\langle P Q \rangle)^2 = {1\over 4} (\langle [P,Q]\rangle)^2 + {1\over 4}(\langle PQ+QP)\rangle)^2 $$
Since both square things are positive, this means that the left hand side is bigger than one quarter the square of the commutator. The commutator is unchanged by the shifting,
$$ [P,Q] = [p,x] = \hbar $$
So that
$$ \langle P^2 \rangle \langle Q^2\rangle \ge (\langle PQ \rangle)^2 \ge {1\over 4} (\langle [P,Q] \rangle)^2 = {\hbar^2 \over 4} $$
The proof is usually given in one line, as directly above, where the Cauchy Schwarz step (first inequality), the imaginary/real part decomposition (second inequality) and the shifted canonical commutation relations (last equality) are assumed internalized by the reader.
This proof appears on Wikipedia, it is used in all QM books, but perhaps this explanation is clearer.
A: For an observable $A$, write $\langle A\rangle$ for the expected value of $A$ and $\Delta A$ for its standard deviation (so that $\langle A\rangle$ and $\Delta A$ both depend on the current state $\phi$).   If $\langle A\rangle=0$ then $\Delta A=\langle A^2\rangle$.
Now given two observables $A$ and $B$, adjust so $\langle A\rangle=\langle B\rangle=0$.  Let $\phi$ be the current state and $x$ an arbitrary real number.
Then
$$\eqalign{
0\phantom{3}\le\phantom{3}&\Big\langle (A+ixB)\phi,(A+ixB)\phi\Big\rangle \cr
&= \overline{\phi} A^2 \phi + x^2 \overline{\phi} B^2 \phi -ix\overline{\phi} BA \phi
+ix \overline{\phi} AB\phi \cr
&= \langle A^2\rangle  +x \langle i[A,B]\rangle  + x^2 \langle B^2\rangle \cr}$$
This holds for all real $x$ so the quadratic in the last line has either no
real zeros or one double zero; either way, the discriminant is non-positive:
$$\langle i[A,B]\rangle ^2 - 4\langle A^2\rangle \cdot \langle B^2\rangle \phantom{3} \le\phantom{3}0$$
as needed.
