This Quantum World/Feynman route/Free propagator

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The propagator as a path integral
Suppose that we make m intermediate position measurements at fixed intervals of duration $$\Delta t.$$ Each of these measurements is made with the help of an array of detectors monitoring n mutually disjoint regions $$R_k,$$ $$k=1,\dots,n.$$ Under the conditions stipulated by Rule B, the propagator $$\langle B|A\rangle$$ now equals the sum of amplitudes

\sum_{k_1=1}^n\cdots\sum_{k_m=1}^n\langle B|R_{k_m}\rangle\cdots \langle R_{k_2}|R_{k_1}\rangle\,\langle R_{k_1}|A\rangle. $$ It is not hard to see what happens in the double limit $$\Delta t\rightarrow 0$$ (which implies that $$m\rightarrow\infty$$) and $$n\rightarrow\infty.$$ The multiple sum $$\sum_{k_1=1}^n\cdots\sum_{k_m=1}^n$$ becomes an integral $$\int\!\mathcal{DC}$$ over continuous spacetime paths from A to B, and the amplitude $$\langle B|R_{k_m}\rangle\cdots\langle R_{k_1} |A \rangle$$ becomes a complex-valued functional $$Z[\mathcal{C}:A\rightarrow B]$$ — a complex function of continuous functions representing continuous spacetime paths from A to B:
 * $$\langle B|A\rangle=\int\!\mathcal{DC}\,Z[\mathcal{C}:A\rightarrow B]$$

The integral $$\int\!\mathcal{DC}$$ is not your standard Riemann integral $$\int_a^b dx\,f(x),$$ to which each infinitesimal interval $$dx$$ makes a contribution proportional to the value that $$f(x)$$ takes inside the interval, but a functional or path integral, to which each "bundle" of paths of infinitesimal width $$\mathcal{DC}$$ makes a contribution proportional to the value that $$Z[\mathcal{C}]$$ takes inside the bundle.

As it stands, the path integral $$\int\!\mathcal{DC}$$ is just the idea of an idea. Appropriate evaluation methods have to be devised on a more or less case-by-case basis.

A free particle
Now pick any path $$\mathcal{C}$$ from A to B, and then pick any infinitesimal segment $$d\mathcal{C}$$ of $$\mathcal{C}$$. Label the start and end points of $$d\mathcal{C}$$ by inertial coordinates $$t,x,y,z$$ and $$t+dt,x+dx,y+dy,z+dz,$$ respectively. In the general case, the amplitude $$Z(d\mathcal{C})$$ will be a function of $$t,x,y,z$$ and $$dt,dx,dy,dz.$$ In the case of a free particle, $$Z(d\mathcal{C})$$ depends neither on the position of $$d\mathcal{C}$$ in spacetime (given by $$t,x,y,z$$) nor on the spacetime orientiaton of $$d\mathcal{C}$$ (given by the four-velocity $$(c\,dt/ds,dx/ds,dy/ds,dz/ds)$$ but only on the proper time interval $$ds=\sqrt{dt^2-(dx^2+dy^2+dz^2)/c^2}.$$

(Because its norm equals the speed of light, the four-velocity depends on three rather than four independent parameters. Together with $$ds,$$ they contain the same information as the four independent numbers $$dt,dx,dy,dz.$$)

Thus for a free particle $$Z(d\mathcal{C})=Z(ds).$$ With this, the multiplicativity of successive propagators tells us that
 * $$\prod_j Z(ds_j)=Z\Bigl(\sum_j ds_j\Bigr)\longrightarrow Z\Bigl(\int_\mathcal{C}ds\Bigr)$$

It follows that there is a complex number $$z$$ such that $$Z[\mathcal{C}]=e^{z\,s[\mathcal{C}:A\rightarrow B]},$$ where the line integral $$s[\mathcal{C}:A\rightarrow B]= \int_\mathcal{C}ds$$ gives the time that passes on a clock as it travels from A to B via $$\mathcal{C}.$$

A free and stable particle
By integrating $$\bigl|\langle B|A\rangle\bigr|^2$$ (as a function of $$\mathbf{r}_B$$) over the whole of space, we obtain the probability of finding that a particle launched at the spacetime point $$t_A,\mathbf{r}_A$$ still exists at the time $$t_B.$$ For a stable particle this probability equals 1:

\int\!d^3r_B\left|\langle t_B,\mathbf{r}_B|t_A,\mathbf{r}_A\rangle\right|^2= \int\!d^3r_B\left|\int\!\mathcal{DC}\,e^{z\,s[\mathcal{C}:A\rightarrow B]}\right|^2=1 $$ If you contemplate this equation with a calm heart and an open mind, you will notice that if the complex number $$z=a+ib$$ had a real part $$a\neq0,$$ then the integral between the two equal signs would either blow up $$(a>0)$$ or drop off $$(a<0)$$ exponentially as a function of $$t_B$$, due to the exponential factor $$e^{a\,s[\mathcal{C}]}$$.

Meaning of mass
The propagator for a free and stable particle thus has a single "degree of freedom": it depends solely on the value of $$b.$$ If proper time is measured in seconds, then $$b$$ is measured in radians per second. We may think of $$e^{ib\,s},$$ with $$s$$ a proper-time parametrization of $$\mathcal{C},$$ as a clock carried by a particle that travels from A to B via $$\mathcal{C},$$ provided we keep in mind that we are thinking of an aspect of the mathematical formalism of quantum mechanics rather than an aspect of the real world.

It is customary The purpose of using the same letter $$b$$ everywhere is to emphasize that it denotes the same physical quantity, merely measured in different units. If we use natural units in which $$\hbar=c=1,$$ rather than conventional ones, the identity of the various $$b$$'s is immediately obvious.
 * to insert a minus (so the clock actually turns clockwise!): $$Z=e^{-ib\,s[\mathcal{C}]},$$
 * to multiply by $$2\pi$$ (so that we may think of $$b$$ as the rate at which the clock "ticks" — the number of cycles it completes each second): $$Z=e^{-i\,2\pi\,b\,s[\mathcal{C}]},$$
 * to divide by Planck's constant $$h$$ (so that $$b$$ is measured in energy units and called the rest energy of the particle): $$Z=e^{-i(2\pi/h)\,b\,s[\mathcal{C}]}=e^{-(i/\hbar)\,b\,s[\mathcal{C}]},$$
 * and to multiply by $$c^2$$ (so that $$b$$ is measured in mass units and called the particle's rest mass): $$Z=e^{-(i/\hbar)\,b\,c^2\,s[\mathcal{C}]}.$$

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