Quantization means to replace $p$ (the momentum) in the expressions of classical physical quantities with $-i\hbar\nabla$, so we get an operator belonging to each physical quantity. However, an expression (which defines a physical quantity) can be written in different forms, due to some identities, e.g. $$(x+y)\cdot p=x\cdot p+y\cdot p\tag 1$$ (where $x$ is a vector). In this example, there isn't any problem, because $(x+y)\cdot\nabla=x\cdot\nabla + y\cdot\nabla$, so it doesn't matter if we quantize the right side or the left side of (1). However, there are vector identities which cease to be identities when we quantize them, e.g. $$a\cdot (p\times a)=p(a\cdot a)-a(a\cdot p).\tag 2$$ This is the so-called "bac-cab identity" in the case of $c=a$, $b=p$, but the quantized version of it is $$a\cdot (\nabla\times a)=\frac{1}{2}\nabla(a\cdot a)-(a\cdot \nabla)a\tag 3$$ (see here) so we will get different equations when we quantize the LHS or the RHS of (2).
This issue is unlikely to arise in the practice of conventional quantum mechanics, but I wonder how we could theoretically rule out such contradictions. Especially, since in a tricky way, using such kind of ambiguity of quantization, W. F. Eberlein showed that transforming a bit the expression of the Hamiltonian of the electron in magnetic field, then performing the quantization starting from an alternative form of the classical expression of the energy, the resulting Schrödinger equation excludes the presence of the magnetic field, and it can be corrected so that we replace the complex wave function to a pair of such functions, and if we do that, then instead of the Schrödinger equation, we will get the Pauli-equation.
