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=⟨xi|eiℏ∫Hdt|xf⟩
Splitting into infinitesimally many δt:
A=⟨xi|∏jeiℏHδtj|xf⟩
Inserting infinitely many (N−1↗∞) identity operators of the form ∫|xtk⟩⟨xtk|xtk
A=⟨xi|eiℏHδtj(N−1∏j=1∫|xtj⟩⟨xtj|dxtjeiℏHδtj)|xf⟩
Now, we stop at this step and pull out a random xtk:
A=∫∫⟨xi|eiℏHδtj(∏j≠k∫|xtj⟩⟨xtj|dxtjeiℏHδtj)|xf⟩∫⟨xtk−1|eiHδtk|xtk⟩⟨xtk|eiHδtk|xtk+1⟩dxtkdxtk−1dxtk+1
Adding over all possible tk's
NA=∑k∫∫⟨xi|eiℏHδtj(∏j≠k∫|xtj⟩⟨xtj|dxtjeiℏHδtj)|xf⟩∫⟨xtk−1|eiHδtk|xtk⟩⟨xtk|eiHδtk|xtk+1⟩dxtkdxtk−1dxtk+1
Since, k is just a dummy index we remove them (with δtl=δtm where l and m are arbitrary) and proceed to use Nδt=tf−ti=T (where tf is the final time and ti is the initial time):
TA=∑k∫∫(⟨xi|eiℏHδtj(∏j≠k∫|xtj⟩⟨xtj|dxtjeiℏHδtj)|xf⟩∫⟨xtk−1|eiHδtk|xtk⟩⟨xtk|eiHδtk|xtk+1⟩dxtk)δtkdxtk−1dxtk+1
We make the redefinition:
(∏j≠k∫|xtj⟩⟨xtj|dxtjeiℏHδtj)=ak
Now, swapping the summation and the 2 integrals
TA=∫∫N−1∑k=1ak∫⟨xtk−1|eiHδtk|xtk⟩⟨xtk|eiHδtk|xtk+1⟩dxtkδtkdxtk−1dxtk+1
Evaluating the δtk integral first using this:
TA=∫∫∞∑k=1lims→1akks×1ζ(s)∫∫tfti⟨xtk−1|eiHδtk|xtk⟩⟨xtk|eiHδtk|xtk+1⟩δtkdxtkdxtk−1dxtk+1