 Solution

In the position representation, $\hat{x}$ multiplies by $x$ and ${\hat{p}}_x=-i\hslash \mathrm{d}/\mathrm{d}x$. Acting on an arbitrary differentiable test function $f$ shows the commutator explicitly:Therefore $[\hat{x},{\hat{p}}_x]=i\hslash \hat{1}$, whose expectation value is $i\hslash$ in every normalized state in the relevant domains. The Robertson uncertainty principle givesThis is a lower bound, not a claim that every state saturates it.