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:
$$T A = \int \int \sum_{k=1}^{\infty} \lim_{s \to 1} \frac{a_k}{k^s} \times \frac{1}{\zeta(s)} \int \int_{t_i}^{t_f} \langle x_{t_{k-1}}| e^{i H \delta t_k} | x_{t_{k}} \rangle \langle x_{t_{k}}| e^{i H \delta t_k} | x_{t_{k+1}} \rangle \delta t_k dx_{t_{k}} d x_{t_{k-1}} d x_{t_{k+1}} $$
Q&A (4901)
