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 $\Delta t.$$\Delta t.$ Each of these measurements is made with the help of an array of detectors monitoring n mutually disjoint regions $R_{k},$$R_k,$ $k=1,\dots ,n.$$k=1,\dots,n.$ Under the conditions stipulated by Rule B, the propagator $\langle B|A\rangle$$\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 .$$\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$$\Delta t\rightarrow 0$ (which implies that $m\rightarrow \infty$$m\rightarrow\infty$) and $n\rightarrow \infty .$$n\rightarrow\infty.$ The multiple sum $\sum _{k_{1}=1}^{n}\cdots \sum _{k_{m}=1}^{n}$$\sum_{k_1=1}^n\cdots\sum_{k_m=1}^n$ becomes an integral $\int \!{\mathcal {DC}}$$\int\!\mathcal{DC}$ over continuous spacetime paths from A to B, and the amplitude $\langle B|R_{k_{m}}\rangle \cdots \langle R_{k_{1}}|A\rangle$$\langle B|R_{k_m}\rangle\cdots\langle R_{k_1} |A \rangle$ becomes a complex-valued functional $Z[{\mathcal {C}}:A\rightarrow B]$$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]$$\langle B|A\rangle=\int\!\mathcal{DC}\,Z[\mathcal{C}:A\rightarrow B]$

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

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

A free particle

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

Thus for a free particle $Z(d{\mathcal {C}})=Z(ds).$$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 )}$$\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$$z$ such that $Z[{\mathcal {C}}]=e^{z\,s[{\mathcal {C}}:A\rightarrow B]},$$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$$s[\mathcal{C}:A\rightarrow B]= \int_\mathcal{C}ds$ gives the time that passes on a clock as it travels from A to B via ${\mathcal {C}}.$$\mathcal{C}.$

A free and stable particle

By integrating ${\bigl |}\langle B|A\rangle {\bigr |}^{2}$$\bigl|\langle B|A\rangle\bigr|^2$ (as a function of $\mathbf {r} _{B}$$\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}$$t_A,\mathbf{r}_A$ still exists at the time $t_{B}.$$t_B.$ For a stable particle this probability equals 1:

$\int \!d^{3}r_{B}\left|\langle t_{B},\mathbf {r} _{B}|t_{A},\mathbf {r} _{A}\rangle \right|^{2}=\int \!d^{3}r_{B}\left|\int \!{\mathcal {DC}}\,e^{z\,s[{\mathcal {C}}:A\rightarrow B]}\right|^{2}=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$$z=a+ib$ had a real part $a\neq 0,$$a\neq0,$ then the integral between the two equal signs would either blow up $(a>0)$$(a>0)$ or drop off $(a<0)$$(a<0)$ exponentially as a function of $t_{B}$$t_B$, due to the exponential factor $e^{a\,s[{\mathcal {C}}]}$$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.$$b.$ If proper time is measured in seconds, then $b$$b$ is measured in radians per second. We may think of $e^{ib\,s},$$e^{ib\,s},$ with $s$$s$ a proper-time parametrization of ${\mathcal {C}},$$\mathcal{C},$ as a clock carried by a particle that travels from A to B via ${\mathcal {C}},$$\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

• to insert a minus (so the clock actually turns clockwise!): $Z=e^{-ib\,s[{\mathcal {C}}]},$$Z=e^{-ib\,s[\mathcal{C}]},$
• to multiply by $2\pi$$2\pi$ (so that we may think of $b$$b$ as the rate at which the clock "ticks" — the number of cycles it completes each second): $Z=e^{-i\,2\pi \,b\,s[{\mathcal {C}}]},$$Z=e^{-i\,2\pi\,b\,s[\mathcal{C}]},$
• to divide by Planck's constant $h$$h$ (so that $b$$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}}]},$$Z=e^{-i(2\pi/h)\,b\,s[\mathcal{C}]}=e^{-(i/\hbar)\,b\,s[\mathcal{C}]},$
• and to multiply by $c^{2}$$c^2$ (so that $b$$b$ is measured in mass units and called the particle's rest mass): $Z=e^{-(i/\hbar )\,b\,c^{2}\,s[{\mathcal {C}}]}.$$Z=e^{-(i/\hbar)\,b\,c^2\,s[\mathcal{C}]}.$

The purpose of using the same letter $b$$b$ everywhere is to emphasize that it denotes the same physical quantity, merely measured in different units. If we use natural units in which $\hbar =c=1,$$\hbar=c=1,$ rather than conventional ones, the identity of the various $b$$b$'s is immediately obvious.