View source for The Free-Particle Propagator

{{Quantum Mechanics A}}

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

<math>S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.</math>

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

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

The kernel now becomes

<math>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 ].</math>
 
We will now evaluate this integral.  Let us first switch to the variables, 

<math>y_{i}=\sqrt{\frac{m}{2\hbar\,\delta t}}x_{i}.</math> 

We then get

<math>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 ].</math> 

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

<math>\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 ].</math>

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

<math>\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 ].</math>

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

<math>\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 ].</math>

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

<math>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 ],</math>

or, rewriting in terms of <math>x_N=x_f\!</math> and <math>x_0=x_i,\!</math>

<math>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 ].</math>

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

<math>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 ].</math>