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| Now, let's evaluate the path integral:
| | {{Quantum Mechanics A}} |
| <math>A=A(t)=\int_{y(0)=0}^{y(t)=0}D[y(t')]e^{\frac{i}{\hbar}\int_{0}^{t}(\frac{1}{2}my'^2-\frac{1}{2}ky^2)dt'}</math> <br/>
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| Note that the integrand is taken over all possible trajectory starting at point <math>x_0</math> at time <math>t'=0</math>,
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| ending at point <math>x</math> at time <math>t'=t</math>.
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| Expanding this integral,
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| <math>A(t)=\left(\frac{m}{2\pi i \hbar}\right)^{\frac{N}{2}}\int_{-\infty}^{\infty} dy_1\ldots dy_{N-1}
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| e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}y^2_{N-1}-
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| \frac{\Delta t}{2}ky^2_{N-1}\right)\right]}e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}(y_{N-1}-y_{N-2})^2-
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| \frac{\Delta t}{2}ky^2_{N-2}\right)\right]}\ldots
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| e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}y^2_{1}-
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| \frac{\Delta t}{2}ky^2_{1}\right)\right]}
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| </math>
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|
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| where <math>N\Delta t=t\!</math>.
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| Expanding the path trajectory in Fourier series, we have
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| <math>
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| y(t')=\sum_n a_n \sin\left(\frac{n\pi t'}{t}\right)
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| </math>
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|
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| we may express <math>A(t)\!</math> in the form
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| <math>A(t)=C\int_{-\infty}^{\infty} da_1\ldots da_{N-1}
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| \exp{\left[\sum_{n=1}^{N-1}\frac{im}{2\hbar}\left(\left(\frac{n\pi}{t}\right)^2-
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| \omega^2\right)a^2_n\right]}
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| </math>
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| where C is a constant independent of the frequency which comes from the Jacobian of the transformation.
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| The important point is that it does not depend on the frequency <math>\omega\!</math>.
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| Thus, evaluating the integral of,
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| <math>A(t)=C'\prod_{n=1}^{N-1}\left[\left(\frac{n\pi}{t}\right)^2-\omega^2\right]^{-\frac{1}{2}}=
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| C'\prod_{n=1}^{N-1}\left[\left(\frac{n\pi}{t}\right)^2\right]^{-\frac{1}{2}}
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| \prod_{n=1}^{N-1}\left[1-\left(\frac{\omega t}{n\pi}\right)^2\right]^{-\frac{1}{2}}
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| </math>
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|
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| where C' is a constant directly related to C and still independent of the frequency of motion.
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| Since the first product series in this final expression is also independent of the frequency of motion,
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| we can absorb it into our constant C' to have a new constant, C''. Simplifying further,
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| <math>A(t)=C''\sqrt{\frac{\omega t}{\sin(\omega t)}}
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| </math>
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| In the limit <math>\omega\rightarrow 0</math>, we already know that
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| <math>C''=\sqrt{\frac{m}{2\pi i \hbar t}}
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| </math>
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|
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| Thus,
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| <math>A(t)=\sqrt{\frac{m}{2\pi i \hbar t}}\sqrt{\frac{\omega t}{\sin(\omega t)}}=
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| \sqrt{\frac{m}{2\pi i \hbar \sin(\omega t)}}
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| </math>
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|
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| and
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| <math><x|\hat{U}(t,0)|x_0>=\sqrt{\frac{m}{2\pi i \hbar \sin(\omega t)}}
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| e^{\frac{i}{\hbar}\left(\frac{m\omega}{2sin(\omega t)}((x^2+x_0^2)cos(\omega t)-2xx_0)\right)}
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| </math>
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| For a more detailed evaluation of this problem, please see Barone, F. A.; Boschi-Filho, H.; Farina, C. 2002. "Three methods for calculating the Feynman propagator". American Association of Physics Teachers, 2003. Am. J. Phys. 71 (5), May 2003. pp 483-491.
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