Suppose we have a quantum field theory (QFT) for a scalar field $\phi$ with vacuum state $|\Omega\rangle$. Then, in units where $\hbar = 1$, we postulate that the vacuum expectation value (VEV) of any operator $\hat O$ is given by $$ \langle\hat O\rangle \equiv \langle{\Omega}|\hat O|\Omega\rangle = \frac{\int \mathcal{D}\phi\, \hat O e^{i S[\phi]}}{\int \mathcal{D}\phi\,e^{i S[\phi]}}, $$ where $S$ is the action of the theory and $\int \mathcal{D} \phi \,e^{i S[\phi]}$ is a path integral over all field configurations with boundary conditions such that $\phi$ is equal to the vacuum in the infinite past and the infinite future. If we try to evaluate $\int \mathcal{D} \phi\, e^{i S[\phi]}$ directly by discretizing and taking the limit, then the answer is formally infinite because of an infinite prefactor that arises (the functional determinant). The fact that the VEV of $\hat O$ is defined as a quotient of two path integrals allows us to avoid this problem.
In the path integral formulation of quantum mechanics, we postulate that the transition amplitude $$K(x_f, t_f; x_i, t_i) \equiv \langle{x_f, t_f}|x_i, t_i\rangle$$ is given by $$ \langle{x_f, t_f}|x_i, t_i\rangle = \frac{1}{Z}\int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D}x\,e^{i S[x]}, $$ where $Z$ is a formally infinite constant, $S$ is the action of the theory, and $\int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D} x \,e^{i S[x]}$ is a path integral over all paths with boundary conditions such that $x(t_i) = x_i$ and $x(t_f) = x_f$. If we try to evaluate $\int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D} x \,e^{i S[x]}$ directly by discretizing and taking the limit, then the answer is formally infinite because of an infinite prefactor that arises (the functional determinant), so formally, we need to divide by $Z$ in order to prevent this.
I know that the normalization of the propagator in quantum mechanics can be fixed in an ad hoc manner (see e.g., this & this post) by requiring that $$ 1 = \int dx_0' \int dx_1 K^\ast(x_1,t;x_0',t_0) K(x_1,t;x_0,t_0). $$ My question is: is there a way to define $K(x_f, t_f; x_i, t_i)$ as a quotient of path integrals in order to avoid this ad hoc normalization requirement?
For example, in QFT, as shown above, the quotient of path integrals naturally arises in the expression for physical observables, allowing us to avoid the ad hoc normalization. I tried to achieve something similar for the quantum mechanical propagator by defining $K(x_f, t_f; x_i, t_i)$ as $$ K(x_f, t_f; x_i, t_i) = \frac{\int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D}x\,e^{i S[x]}}{\sqrt{\int dx_0' \int dx_1 \left(\int_{x(t_i)=x_0'}^{x(t_f)=x_1} \mathcal{D}x\,e^{i S[x]}\right)^*\int_{x(t_i)=x_0}^{x(t_f)=x_1} \mathcal{D}x\,e^{i S[x]}}}. $$ When I tried evaluating this expression for the free particle by discretizing the path integrals in the numerator and the denominator into $N$ different steps and taking the limit of the quotient as $N \to \infty$ (while using the $i\epsilon$ prescription to ensure convergence), this nearly worked, except that I had a left over factor in the numerator that went like $(i)^{N / 2}$.
