Math behind photon stopwatch path integral in Feynman's QED book

Question

I want to connect the math and the exact concepts hidden behind the simplified picture of photon propagation provided in Feynman's QED book.

Background

Feynman's photon stopwatches

In his book QED - the strange theory of light and matter, Richard Feynman introduces a procedure to compute the probability of photons going from a point A to B, bouncing off a mirror for example (and neglecting polarization).

While introducing his famous path integral formalism, he succeeds in avoiding mathematical expression by simply attaching an imaginary stopwatch to each photon exploring a different path. The final position of the stopwatch arrow at B changes with the distance traveled between A and B. Or, if you want, the time of travel.

Then, the final probability of detecting photons at B is computed by adding together all arrows head-to-tails, and squaring the length of the resulting total arrow (from the tail of the first arrow to the head of the last arrow).

The mathematical expression for an arrow is actually a phasor, $e^{i\varphi}$. And the arrow is drawn in the complex plane.

All of this is nicely illustrated in the QED wikipedia article

Space and time events

So far, while exploring the literature on the subject (for example the related wikipedia article or Feynman&Hibbs book), I have only been introduced to Feynman path integral for two events defined in space-time, i.e. $(t_A, x_A)$ and $(t_B, x_B)$.

The procedure is to sum over all the probability amplitudes $\phi_i$ associated to each space-time path connecting A and B. The probability amplitude $\phi_i$ for a given path $q(t)$ is proportional to an exponential complex of the action $S[q]=\int_{t_A}^{t_B} L(q,\dot q) \, dt$ associated to that path, with $L(q,\dot q)$ the Lagrangian function. i.e. $$ \phi_i = A e^{iS[q]/\hbar} $$ The amplitude $A$ is constant and equal for every path.

The final probability is given by $$ \begin{align} P(A,B) &= \sum_i \phi_i\\ &\propto \int \exp\left(\frac i \hbar \int_{t_A}^{t_B} L(q, \dot q) \, dt\right) \mathcal D q \end{align} $$ where $\int \mathcal D q$ is the path integral.

Question

But in Feynman's QED book, the two events are fixed in space only. The duration between photon starting at position $x_A$ and going to $x_B$ is left free. So, how do we jump from space-time events to space events only?. (Q1)

There is another closely related question, but the answer doesn't seem complete.

Answer 1

The transition from (space,time) events to (space,energy) events is a result of a Fourier transformation $$ \langle {\bf r}_f,E_f | {\bf r}_i,E_i \rangle ~=~\int_{\mathbb{R}}\!\mathrm{d}t_f \int_{\mathbb{R}}\!\mathrm{d}t_i~e^{\frac{i}{\hbar}(E_ft_f-E_it_i)} \langle {\bf r}_f,t_f | {\bf r}_i,t_i \rangle. $$ For details, see my Phys.SE answer here, which also derives the Feynman's stopwatch hand/arrow picture from first principles, that is, QED.