$\hat A$ is an operator. The uncertainty on $\hat{A}$, $\Delta A$ is defined by:
$$\Delta A:=\sqrt{\langle\hat A^2\rangle - \langle\hat A\rangle^2}.$$
What is difference between $\langle\hat A^2\rangle$ and $\langle\hat A\rangle^2$ that leads to Uncertainty Relation between two Operators?
More details: $$ \langle\hat A^2 \rangle=\langle\psi|\hat A^2|\psi \rangle.$$ What is the name of difference between absolute value of these two complex conjugates?
Although Qmechanics's answer is formally complete and correct, there is a more intuitive formulation of this identity that makes it self evident. Consider the operator B which is A minus its expectation value in some state.
$$B = A - \langle A\rangle $$
Then the expectation value of B is zero in the same state (obviously--- it has been shifted to make it so). The expected value of $B^2$ can be nonzero--- it is a measure of the spread in B in state $\psi$. It is positive, as you can see by the definition of matrix multiplication (or by "inserting the identity in a basis")
$$ \langle B^2 \rangle = \sum_i \langle |B|i\rangle\langle i|B\rangle $$
The last thing on the right is the sum of positive quatities of the form $c^*c$. If you now reexpress the expectation value of $B^2$ in terms of A,
$$ \langle B^2 \rangle = \langle (A-\langle A\rangle)^2\rangle = \langle A^2\rangle - 2 \langle A\langle A\rangle \rangle + \langle A\rangle^2 = \langle A^2\rangle - \langle A\rangle^2 $$
This manipulation justifies this thing.
$\langle\hat A\rangle$ is the expectation value of $\hat A$.
$\langle\hat A\rangle^2$ is the square of item 1.
$\langle\hat A^2\rangle$ is the expectation value of $\hat A^2=\hat A \hat A$.
Item 2 and 3 do not have to be equal.
If $\hat A$ is selfadjoint, then it is possible to show
that the expectation value $\langle\hat A\rangle~\in~\mathbb{R}$ is real, and
that $\langle\hat A^2\rangle ~\geq~ \langle\hat A\rangle^2$ (because of CS ineq.).
