Justify the value of the product of the uncertainty in position and momentum
The commutator $[\hat{A}, \hat{B}]$ of two operators $\hat{A}$ and $\hat{B}$ is also an operator, so when using the Robertson uncertainty relation we have to take the expectation value of the commutator first and then take the absolute value of it.

But for $\hat{x}$ and $\hat{p}$ the commutator $[\hat{x}, \hat{p}]$ is simply a constant $\mathrm{i} \hbar$, so that taking the expectation value is trivial: $\langle [\hat{x}, \hat{p}] \rangle = \mathrm{i} \hbar$. Then we take the absolute value of this result which is $\hbar$ and divide it by two to get $\frac{1}{2} \hbar$ as expected.