Recall the derivation of the Schr\"odinger equation (8.5.8) starting from Eq.\ (8.5.4). Note that although we chose the argument of V to be the midpoint x+x′/2, it did not matter very much: any choice x+aη, (where η=x′−x) for 0≤a≤1 would have given the same result since the difference between the choices is of order ηε≃ε3/2. All this was thanks to the factor ε multiplying V in Eq.\ (8.5.4) and the fact that ∣η∣≃ε1/2, as per Eq.\ (8.5.6).
Consider now the case of a vector potential which will bring in a factor
exp[ℏciqεεx−x′A(x+aη)]≡exp[−ℏciqεεηA(x+aη)]
to the propagator for one time slice. (We should really be using vectors for position and the vector potential, but the one-dimensional version will suffice for making the point here.) Note that ε now gets canceled, in contrast to the scalar potential case. Thus, going to order ε to derive the Schr\"odinger equation means going to order η2 in expanding the exponential. This will not only bring in an A2 term, but will also make the answer sensitive to the argument of A in the linear term. Choose a=1/2 and verify that you get the one-dimensional version of Eq.\ (4.3.7). Along the way you will see that changing a makes an order ε difference to ψ(x,ε) so that we have no choice but to use a=1/2, i.e., use the midpoint prescription.