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| {{Quantum Mechanics A}} | | {{Quantum Mechanics A}} |
| Although our heuristic analysis yielded an exact free-particle propagator, we will now repeat the calculation without any approximation to illustrate the partial integration.
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| Consider <math>U(x_{N}, t_{N}; x_{0}, t_{0})</math>. The peculiar labeling of the end points will be justified later. Our problem is to perform the path integral
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| <math>\int_{_{X0}}^{_{XN}}e\tfrac{tS\left [ x(t) \right ]}{h}D\left [ x(t) \right ]</math>
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| <br/>Where<br/> <br/>
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| <math>\int_{_{X0}}^{_{XN}}D\left [ x(t) \right ]</math>
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| is a symbolic way of saying "integrate over all paths connecting <math>{x_{0}}</math> and <math>{x_{N}}</math> (in the interval <math>{t_{0}}</math> and <math>{t_{N}}</math>). | | 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 |
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| | <math>S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.</math> |
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| | 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 |
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| | <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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| ." Now, a path <math>{x_{t}}</math> is fully specified by an infinity of numbers <math>{x(t_{0})}</math>,..., <math>{x(t)}</math>, ...,<math>{x(t_{N})}</math>, namely, the values of the function <math>{x(t)}</math> at every point <math>{t}</math> is the interval <math>{t_{0}}</math> to <math>{t_{N}}</math>.To sum over all paths, we must integrate over all possible values of these infinite variables, except of course <math>{x(t_{0})}</math> and <math>{x(t_{N})}</math>, which will be kept fixed at <math>{x_{0}}</math> and <math>{x_{N}}</math>, respectively. To tackle this problem,we trade the function <math>{x_{t}}</math> for a discrete approximation which agrees with <math>{x_{t}}</math> at the <math>{N+1}</math> points.agrees with x{t) at the N + 1 points <math>t_{n}=t_{0}+n\varepsilon</math>, n = 0,.. . , N, where <math>\varepsilon =\frac{t_{n}-t_{0}}{N}</math>. In this approximation each path is specified by N+ 1 numbers <math>x(t_{0}),x(t_{1}),...,x(t_{N})</math>. The gaps in the discrete function are interpolated by straight lines. We hope that if we take the limit <math>N\to \infty</math> 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
| | The kernel now becomes |
| <math>
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| S=\int_{t_{0}}^{t_N}\mathcal{L}(t)dt\int_{t_{0}}^{t_N}\frac{1}{2}m\dot{x}^2dt</math>
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| <math>S\int_{t_{0}}^{N-1}\frac{m}{2}[\frac{x_{i+1}-x_{i}}{\varepsilon }^2]\varepsilon</math> | | <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> |
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| | We will now evaluate this integral. Let us first switch to the variables, |
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| where <math>x_{i}=x(t_{i})</math>. We wish to calculate
| | <math>y_{i}=\sqrt{\frac{m}{2\hbar\,\delta t}}x_{i}.</math> |
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| <math>U(x_{N},t_{N};x_{0},t_{0})=\int_{x_{0}}^{x_N} exp\frac{i S[x(t)]}{\hbar}D[x(t)]=lim A\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}</math>
| | We then get |
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| It is implicit in the above that <math>x(t_{0})</math> and <math>x(t_{N})</math> 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 <math>N\to \infty</math> is taken.
| | <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> |
| Let us first switch to the variables
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| <math>y_{i}=(\frac{m}{2\hbar\varepsilon })^{1/2}x_{i}</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 |
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| We then want
| | <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> |
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| <math>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}</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 |
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| where
| | <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> |
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| <math>{A}'=(\frac{2\hbar\varepsilon }{m })^{\frac{N-1}{2}}</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 |
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| Although the multiple integral looks formidable, it is not. Let us begin by doing the
| | <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> |
| <math>y_{1}</math> integration. Considering just the part of the integrand that involves <math>y_{1}</math>, we get | |
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| | Combining all of these results together, we find that the kernel is |
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| <math>\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}</math> | | <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> |
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| Consider next the integration over yr. Bringing in the part of the integrand involving
| | or, rewriting in terms of <math>x_N=x_f\!</math> and <math>x_0=x_i,\!</math> |
| <math>y_{2}</math> and combining it with the result above we compute next | |
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| | <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> |
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| <math>(\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}=</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, |
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| <math>(\frac{(i\pi )^{2}}{3})^{\frac{1}{2}}e^\frac{-(y_{3}-y_{0})^2}{3i}</math> | | <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> |
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
![{\displaystyle 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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46838cdfb0ce6842b7ed9bd5f6628df98744c127)
We will now evaluate this integral. Let us first switch to the variables,
We then get
![{\displaystyle 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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b710ef31b779148b22b729092cdc66a96aa2df0)
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
![{\displaystyle \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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7c5d3fb42e8828f4bf0b9069de08f0d876eb4ca)
Now let us evaluate the integral over 
Again considering just the part of the integrand that involves 
we get
![{\displaystyle \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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4084fdbdd37a63f53503d4240d39a7fecbd249b)
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
![{\displaystyle \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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ffd88c5e79906715ec7f7c876da11843eb90943)
Combining all of these results together, we find that the kernel is
![{\displaystyle 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],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91aeb91d087f3b62743e7447a43d322b2a0d0348)
or, rewriting in terms of 
and 
![{\displaystyle 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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc3a3d5b5b03e4c282feeecea473e136fc742f30)
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,
![{\displaystyle 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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfb8fea2a7b20ff48d1bba3e07daede700b29556)
