Feynman path integral in an EM field

Question

I'm studying Feynman and Hibbs, Quantum Mechanics and Path Integrals

In problem 4-2, the book says for a particle of charge $e$ in an EM field the Lagrangian is

$$L=\frac{m}{2}\dot{\boldsymbol{x}}^2+\frac{e}{c}\dot{\boldsymbol{x}}\cdot \boldsymbol{A}(x,t)-e\phi(\boldsymbol{x},t),\tag{4-17}$$

and asks the reader to show that the corresponding Schrodinger's equation is

$$i\hbar\frac{\partial\psi}{\partial t}=\frac{1}{2m}\biggl(\frac{\hbar}{i}\nabla-\frac{e}{c}\boldsymbol{A}\biggr)^2\psi+e\phi\psi.\tag{4-18}$$

My question is, in the definition of path integrals, there is a normalization factor $$\biggl(\frac{m}{2\pi i\hbar\delta t}\biggr)^{\frac{1}{2}},\tag{4-8}$$ is the normalization factor still the same in this case? Or should it be a normalization factor involving $\boldsymbol{A}$ as well? Because I'm having difficulty proving this using the standard normalization.

Alternatively, is there a place where this derivation is done in more detail?

Answer 1

In principle the Feynman fudge factor (4-8) (which comes from Gaussian momentum integrations of the Hamiltonian phase space path integral, cf. e.g. this Phys.SE post) should not be affected by the E&M field (because the E&M field does not change the Hessian in the momentum sector.)