The Free-Particle Propagator

Although our heuristic analysis yielded an exact free-particle propagator, we will now repeat the calculation without any approximation to illustrate the partial integration. Consider $U(x_{N},t_{N};x_{0},t_{0})$. The peculiar labeling of the end points will be justified later. Our problem is to perform the path integral

$\int _{_{X0}}^{_{XN}}e{\tfrac {tS\left[x(t)\right]}{h}}D\left[x(t)\right]$
Where

$\int _{_{X0}}^{_{XN}}D\left[x(t)\right]$

is a symbolic way of saying "integrate over all paths connecting ${x_{0}}$ and ${x_{N}}$ (in the interval ${t_{0}}$ and ${t_{N}}$).

." Now, a path ${x_{t}}$ is fully specified by an infinity of numbers ${x(t_{0})}$,..., ${x(t)}$, ...,${x(t_{N})}$, namely, the values of the function ${x(t)}$ at every point ${t}$ is the interval ${t_{0}}$ to ${t_{N}}$.To sum over all paths, we must integrate over all possible values of these infinite variables, except of course ${x(t_{0})}$ and ${x(t_{N})}$, which will be kept fixed at ${x_{0}}$ and ${x_{N}}$, respectively. To tackle this problem,we trade the function ${x_{t}}$ for a discrete approximation which agrees with ${x_{t}}$ at the ${N+1}$ points.agrees with x{t) at the N + 1 points $t_{n}=t_{0}+n\varepsilon$, n = 0,.. . , N, where $\varepsilon ={\frac {t_{n}-t_{0}}{N}}$. In this approximation each path is specified by N+ 1 numbers $x(t_{0}),x(t_{1}),...,x(t_{N})$. The gaps in the discrete function are interpolated by straight lines. We hope that if we take the limit $N\to \infty$ at the end we will get a result that is insensitive to these approximations.t Now that the paths have been discretized, we must also do the same approximations.paths discretized, we must also do the same to the action integral. We replace the continuous path definition $S=\int _{t_{0}}^{t_{N}}{\mathcal {L}}(t)dt\int _{t_{0}}^{t_{N}}{\frac {1}{2}}m{\dot {x}}^{2}dt$

$S\int _{t_{0}}^{N-1}{\frac {m}{2}}[{\frac {x_{i+1}-x_{i}}{\varepsilon }}^{2}]\varepsilon$

where $x_{i}=x(t_{i})$. We wish to calculate

$U(x_{N},t_{N};x_{0},t_{0})=\int _{x_{0}}^{x_{N}}exp{\frac {iS[x(t)]}{\hbar }}D[x(t)]=limA\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }...\int _{-\infty }^{\infty }exp({\frac {i}{\hbar }}{\frac {m}{2}}\sum _{i=0}^{N-1}{\frac {(x_{i+1}-x_{i})^{2}}{\varepsilon }})dx_{1}...dx_{N-1}$

It is implicit in the above that $x(t_{0})$ and $x(t_{N})$ have the values we have chosen at the outset. The factor A in the front is to be chosen at the end such that we get the correct scale for U when the limit $N\to \infty$ is taken. Let us first switch to the variables

$y_{i}=({\frac {m}{2\hbar \varepsilon }})^{1/2}x_{i}$

We then want

$lim{A}'\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }...\int _{-\infty }^{\infty }exp(-\sum _{i=0}^{N-1}{\frac {(y_{i+1}-y_{i})^{2}}{i}}dy_{1}...dy_{N-1}$

where

${A}'=({\frac {2\hbar \varepsilon }{m}})^{\frac {N-1}{2}}$

Although the multiple integral looks formidable, it is not. Let us begin by doing the $y_{1}$ integration. Considering just the part of the integrand that involves $y_{1}$, we get

$\int _{-\infty }^{\infty }exp\left({\frac {-1}{i}}(y_{2}-y_{1})^{2}+(y_{1}-y_{0})^{2}\right)dy_{1}=({\frac {i\pi }{2}})^{1/2}e^{\frac {-(y_{2}-y_{0})^{2}}{2i}}$

Consider next the integration over yr. Bringing in the part of the integrand involving $y_{2}$ and combining it with the result above we compute next

$({\frac {i\pi }{2}})^{1/2}\int _{-\infty }^{\infty }e^{\frac {-(y_{3}-y_{2})^{2}}{<}}math>{i}e^{\frac {-(y_{2}-y_{0})^{2}}{2i}}dy_{2}=$

$({\frac {(i\pi )^{2}}{3}})^{\frac {1}{2}}e^{\frac {-(y_{3}-y_{0})^{2}}{3i}}$