# 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 $\Delta t.$  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 evalutation 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:

$\int \!d^{3}r_{B}\left|\langle t_{B},\mathbf {r} _{B}|t_{A},\mathbf {r} _{A}\rangle \right|^{2}=\int \!d^{3}r_{B}\left|\int \!{\mathcal {DC}}\,e^{z\,s[{\mathcal {C}}:A\rightarrow B]}\right|^{2}=1$

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 $a\neq 0,$  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 $e^{a\,s[{\mathcal {C}}]}$ .

### 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

• 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): $Z=e^{-(i/\hbar )\,b\,c^{2}\,s[{\mathcal {C}}]}.$

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.