I understand from several heuristic arguments that in one dimension, the quantum-mechanical operator $\hat{p} = -i\hbar\,\partial_x$ corresponds to the classical momentum $p$, in the sense that a particle described by the wavefunction $\Psi(x,t)$ has expected momentum $$ \langle p \rangle = \int_{-\infty}^\infty \Psi^* \hat{p} \Psi \,\mathrm{d}x. $$ Why does it then follow that the quantum-mechanical operator $\hat{p}^2 = \hat{p} \circ \hat{p}$ corresponds to the square of the classical momentum, $p^2$? In general,does the square of a classical quantity $q$ always correspond to the composition of its quantum-mechanical operator, $\hat{q}$, with itself? Does the same apply to higher powers of $q$, i.e. does the $n$-fold composition $\hat{q} \circ \hat{q} \circ \cdots \circ \hat{q}$ form the operator corresponding to $q^n$?