Harmonic Oscillator: Integration Over Fluctuations

Now, let's evaluate the path integral: $A=A(t)=\int _{y(0)=0}^{y(t)=0}D[y(t')]e^{{\frac {i}{\hbar }}\int _{0}^{t}({\frac {1}{2}}my'^{2}-{\frac {1}{2}}ky^{2})dt'}$

Note that the integrand is taken over all possible trajectory starting at point $x_{0}$ at time $t'=0$, ending at point $x$ at time $t'=t$.

Expanding this integral,

$A(t)=\left({\frac {m}{2\pi i\hbar }}\right)^{\frac {N}{2}}\int _{-\infty }^{\infty }dy_{1}\ldots dy_{N-1}e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{N-1}^{2}-{\frac {\Delta t}{2}}ky_{N-1}^{2}\right)\right]}e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}(y_{N-1}-y_{N-2})^{2}-{\frac {\Delta t}{2}}ky_{N-2}^{2}\right)\right]}\ldots e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{1}^{2}-{\frac {\Delta t}{2}}ky_{1}^{2}\right)\right]}$

where $N\Delta t=t\!$.

Expanding the path trajectory in Fourier series, we have $y(t')=\sum _{n}a_{n}\sin \left({\frac {n\pi t'}{t}}\right)$

we may express $A(t)\!$ in the form

$A(t)=C\int _{-\infty }^{\infty }da_{1}\ldots da_{N-1}\exp {\left[\sum _{n=1}^{N-1}{\frac {im}{2\hbar }}\left(\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right)a_{n}^{2}\right]}$

where C is a constant independent of the frequency which comes from the Jacobian of the transformation. The important point is that it does not depend on the frequency $\omega \!$. Thus, evaluating the integral of,

$A(t)=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right]^{-{\frac {1}{2}}}=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}\right]^{-{\frac {1}{2}}}\prod _{n=1}^{N-1}\left[1-\left({\frac {\omega t}{n\pi }}\right)^{2}\right]^{-{\frac {1}{2}}}$

where C' is a constant directly related to C and still independent of the frequency of motion. Since the first product series in this final expression is also independent of the frequency of motion, we can absorb it into our constant C' to have a new constant, C. Simplifying further,

$A(t)=C''{\sqrt {\frac {\omega t}{\sin(\omega t)}}}$

In the limit $\omega \rightarrow 0$, we already know that

$C''={\sqrt {\frac {m}{2\pi i\hbar t}}}$

Thus,

$A(t)={\sqrt {\frac {m}{2\pi i\hbar t}}}{\sqrt {\frac {\omega t}{\sin(\omega t)}}}={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}$

and

$<x|{\hat {U}}(t,0)|x_{0}>={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}e^{{\frac {i}{\hbar }}\left({\frac {m\omega }{2sin(\omega t)}}((x^{2}+x_{0}^{2})cos(\omega t)-2xx_{0})\right)}$

Reference

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.