This Quantum World/Feynman route/Free propagator

Propagator for a free and stable particle

The propagator as a path integral

Suppose that we make m intermediate position measurements at fixed intervals of duration $\mathrm{\Delta }t.$ 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 [[../two slits|propagator]] $⟨B|A⟩$ now equals the sum of amplitudes

$\sum _{k_1=1}^n\cdots \sum _{k_m=1}^n⟨B|R_{k_m}⟩\cdots ⟨R_{k_2}|R_{k_1}⟩⟨R_{k_1}|A⟩.$

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

$⟨B|A⟩=\int 𝒟𝒞Z[𝒞:A\to B]$

The integral $\int 𝒟𝒞$ is not your standard Riemann integral $\underset{a}{\overset{b}{\int }}dxf(x),$ 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 $𝒟𝒞$ makes a contribution proportional to the value that $Z[𝒞]$ takes inside the bundle.

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

(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 $Z(d𝒞)=Z(ds).$ With this, the [[../two_slits#Why_product.3F|multiplicativity of successive propagators]] tells us that

$\prod _{j}Z(d{s}_j)=Z\left(\sum _{j}d{s}_j\right)⟶Z\left(\underset{𝒞}{\int }ds\right)$

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

A free and stable particle

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

$\int d^3{r}_B{|⟨t_B,{𝐫}_B|t_A,{𝐫}_A⟩|}^2=\int d^3{r}_B{|\int 𝒟𝒞e^{zs[𝒞:A\to B]}|}^2=1$

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 $a\ne 0,$ 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 $e^{as[𝒞]}$.

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 $e^{ibs},$ with $s$ a proper-time parametrization of $𝒞,$ as a clock carried by a particle that travels from A to B via $𝒞,$ 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

• to insert a minus (so the clock actually turns clockwise!): $Z=e^{-ibs[𝒞]},$
• 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): $Z=e^{-i2\pi bs[𝒞]},$
• to divide by Planck's constant $h$ (so that $b$ is measured in energy units and called the rest energy of the particle): $Z=e^{-i(2\pi /h)bs[𝒞]}=e^{-(i/\hslash )bs[𝒞]},$
• and to multiply by $c^2$ (so that $b$ is measured in mass units and called the particle's rest mass): $Z=e^{-(i/\hslash )b{c}^{2}s[𝒞]}.$

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 $\hslash =c=1,$ rather than conventional ones, the identity of the various $b$'s is immediately obvious.