**Question**

The following is an attempt to reformulate Feynman's path integral formulation. Is this correct? Can this be made rigorous? What is the radius of convergence?

**Proof**

Let's think about Feynman's Path Integral Formulation of Quantum Mechanics.

$$ A = \langle x_i|e^{\frac{i}{\hbar} \int H dt}| x_f\rangle $$

Splitting into infinitesimally many $\delta t$:

$$ A = \langle x_i|\prod_j e^{\frac{i}{\hbar} H \delta t_j}| x_f\rangle $$

Inserting infinitely many ($N -1 \nearrow \infty$) identity operators of the form $\int | x_{t_k}\rangle\langle x_{t_k}| x_{t_k}$

$$ A = \langle x_i|e^{\frac{i}{\hbar} H \delta t_j} \Big(\prod_{j=1}^{N-1} \int |x_{t_j}\rangle \langle x_{t_j}| d x_{t_j}e^{\frac{i}{\hbar} H \delta t_j}\Big)| x_f\rangle $$

Now, we stop at this step and pull out a random $x_{t_k}$:

$$ A = \int \int \langle x_i|e^{\frac{i}{\hbar} H \delta t_j} \Big(\prod_{j\neq k} \int|x_{t_j}\rangle \langle x_{t_j}| d x_{t_j}e^{\frac{i}{\hbar} H \delta t_j}\Big)| x_f\rangle \int \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle dx_{t_{k}} d x_{t_{k-1}} d x_{t_{k+1}} $$

Adding over all possible $t_k$'s

$$N A = \sum_{k} \int \int \langle x_i|e^{\frac{i}{\hbar} H \delta t_j} \Big(\prod_{j\neq k} \int|x_{t_j}\rangle \langle x_{t_j}| d x_{t_j}e^{\frac{i}{\hbar} H \delta t_j}\Big)| x_f\rangle \int \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle dx_{t_{k}} d x_{t_{k-1}} d x_{t_{k+1}} $$

Since, $k$ is just a dummy index we remove them (with $\delta t_l = \delta t_m$ where $l$ and $m$ are arbitrary) and proceed to use $N \delta t= t_f -t_i =T$ (where $t_f$ is the final time and $t_i$ is the initial time):

$$T A = \sum_{k} \int \int \Big( \langle x_i|e^{\frac{i}{\hbar} H \delta t_j} \Big(\prod_{j\neq k} \int|x_{t_j}\rangle \langle x_{t_j}| d x_{t_j}e^{\frac{i}{\hbar} H \delta t_j}\Big)| x_f\rangle \int \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle dx_{t_{k}} \Big ) \delta t_k d x_{t_{k-1}} d x_{t_{k+1}} $$

We make the redefinition:

$$ \Big(\prod_{j\neq k} \int|x_{t_j}\rangle \langle x_{t_j}| d x_{t_j}e^{\frac{i}{\hbar} H \delta t_j}\Big) = a_k$$

Now, swapping the summation and the $2$ integrals

$$T A = \int \int \sum_{k=1}^{N-1} a_k \int \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle dx_{t_{k}} \delta t_k d x_{t_{k-1}} d x_{t_{k+1}} $$

Evaluating the $\delta t_k$ integral first using this:

$$T A = \int \int \sum_{k=1}^{\infty} \lim_{s \to 1} \frac{a_k}{k^s} \times \frac{1}{\zeta(s)} \int \int_{t_i}^{t_f} \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle \delta t_k dx_{t_{k}} d x_{t_{k-1}} d x_{t_{k+1}} $$