This Quantum World/Feynman route/Free propagator

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[edit] Propagator for a free and stable particle

[edit] The propagator as a path integral

Suppose that we make m intermediate position measurements at fixed intervals of duration Δt. Each of these measurements is made with the help of an array of detectors monitoring n mutually disjoint regions Rk, 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.

[edit] 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}.

[edit] 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 tB. 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 tB, due to the exponential factor e^{a\,s[\mathcal{C}]}.

[edit] 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 2π (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 c2 (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.

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