The Free-Particle Propagator: Difference between revisions
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<math>S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.</math> | <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. The action may then be written as | 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> | <math>S=\tfrac{1}{2}m\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{i})^2}{\delta t}.</math> | ||
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The kernel now becomes | The kernel now becomes | ||
<math>K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty} | <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> | |||
<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> | |||
<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> | |||
<math>y_{ | |||
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 }exp\left | <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> | |||
<math> | or, rewriting in terms of <math>x_N=x_f\!</math> and <math>x_0=x_i,\!</math> | ||
<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{m}{2\hbar i(t_f-t_i)}(x_f-x_i)^2\right ].</math> |
Revision as of 02:38, 17 January 2014
We will now evaluate the kernel for a free particle. In this case, the action is just
Note that we renamed to and to the reason for this will become clear shortly. Let us now discretize the path that the particle takes, so that the intermediate positions are We discretize the time axis similarly, with a spacing between two subsequent times, so that and so on. The action may then be written as
The kernel now becomes
We will now evaluate this integral. Let us first switch to the variables,
We then get
Although the multiple integral looks formidable, it is not. Let us begin by doing the integral over Considering just the part of the integrand that involves we get
Now let us evaluate the integral over Again considering just the part of the integrand that involves we get
We now continue to do this until all of the have been integrated out. At the step (i.e., integrating out ), the integral that we evaluate and the solution are
Combining all of these results together, we find that the kernel is
or, rewriting in terms of and
Since we divided the time interval up into equal amounts, we note that We may now take the limit, finally obtaining the free-particle propagator,
Failed to parse (syntax error): {\displaystyle K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar(t_f-t_i)}\exp\left [-\frac{m}{2\hbar i(t_f-t_i)}(x_f-x_i)^2\right ].}