The Free-Particle Propagator

We will now evaluate the kernel $K(x_f,t_f;x_i,t_i)\!$ for a free particle. In this case, the action is just

$S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.$

Note that we renamed $t_i\!$ to $t_0\!$ and $t_f\!$ to $t_N;\!$ the reason for this will become clear shortly. Let us now discretize the path that the particle takes, so that the intermediate positions are $x_1,\,x_2,\,\ldots,\,x_{N-1}.$ We discretize the time axis similarly, with a spacing $\delta t\!$ between two subsequent times, so that $x(t_1)=x_1,\,x(t_2)=x_2,\!$ and so on. The action may then be written as

$S=\tfrac{1}{2}m\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{i})^2}{\delta t}.$

The kernel now becomes

$K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty}\left (\frac{m}{2\pi i\hbar\,\delta t}\right )^{N/2}\int_{-\infty }^{\infty }\cdots\int_{-\infty }^{\infty }dx_{1}\ldots dx_{N-1}\,\exp\left [\frac{im}{2\hbar}\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{{i}})^2}{\delta t}\right ].$

We will now evaluate this integral. Let us first switch to the variables,

$y_{i}=\sqrt{\frac{m}{2\hbar\,\delta t}}x_{i}.$

We then get

$K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\lim_{N\to\infty}\left (-\frac{i}{\pi}\right )^{(N-1)/2}\int_{-\infty }^{\infty }\cdots\int_{-\infty }^{\infty }dy_{1}\ldots dy_{N-1}\,\exp\left [-\sum_{i=0}^{N-1}\frac{(y_{i+1}-y_{{i}})^2}{i}\right ].$

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

$\int_{-\infty }^{\infty}dy_1\,\exp\left [\frac{-(y_{2}-y_{1})^2-(y_{1}-y_{0})^2}{i}\right ]=\sqrt{\frac{i\pi }{2}}\exp\left [-\frac{(y_{2}-y_{0})^2}{2i}\right ].$

Now let us evaluate the integral over $y_{2}.\!$ Again considering just the part of the integrand that involves $y_{2},\!$ we get

$\int_{-\infty }^{\infty}dy_2\,\exp\left [-\frac{(y_{3}-y_{2})^2}{i}-\frac{(y_{2}-y_{0})^2}{2i}\right ]=\sqrt{\frac{2i\pi }{3}}\exp\left [-\frac{(y_{3}-y_{0})^2}{3i}\right ].$

We now continue to do this until all of the $y_i\!$ have been integrated out. At the $k^{\text{th}}\!$ step (i.e., integrating out $y_k\!$), the integral that we evaluate and the solution are

$\int_{-\infty }^{\infty}dy_k\,\exp\left [-\frac{(y_{k+1}-y_{k})^2}{i}-\frac{(y_{k}-y_{0})^2}{ki}\right ]=\sqrt{\frac{ki\pi }{k+1}}\exp\left [-\frac{(y_{k+1}-y_{0})^2}{(k+1)i}\right ].$

Combining all of these results together, we find that the kernel is

$K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\lim_{N\to\infty}\frac{1}{\sqrt{N}}\exp\left [-\frac{(y_{N}-y_{0})^2}{Ni}\right ],$

or, rewriting in terms of $x_N=x_f\!$ and $x_0=x_i,\!$

$K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty}\sqrt{\frac{m}{2\pi i\hbar N\,\delta t}}\exp\left [-\frac{m}{2\hbar iN\,\delta t}(x_f-x_i)^2\right ].$

Since we divided the time interval up into equal amounts, we note that $N\,\delta t=T=t_f-t_i.\!$ We may now take the limit, finally obtaining the free-particle propagator,

$K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar(t_f-t_i)}}\exp\left [\frac{im}{2\hbar}\frac{(x_f-x_i)^2}{t_f-t_i}\right ].$