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| <math>S=S(t,x,x_0)=\frac{m\omega}{2\sin(\omega t)}((x^2+x_0^2)\cos(\omega t)-2xx_0)</math><br/> | | <math>S=S(t,x,x_0)=\frac{m\omega}{2\sin(\omega t)}((x^2+x_0^2)\cos(\omega t)-2xx_0)</math><br/> |
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| | Explicit evaluation of the path integral for the harmonic oscillator can be found here [[Image:FeynmanHibbs_H_O_Amplitude.pdf]] |
Revision as of 01:53, 17 January 2014
The classical action
can be evaluated as follows:
Where
is the kinetic engergy and
is the potential energy.
Equation of motion for harmonic oscillator:

and
are constants.
At
(starting point),
.
At
(final point),
. </math>
is a symbolic way of saying "integrate over all paths connecting
and
(in the interval
and
)." Now, a path
is fully specified by an infinity of numbers
,...,
, ...,
, namely, the values of the function
at every point
is the interval
to
. To sum over all paths, we must integrate over all possible values of these infinite variables, except of course
and
, which will be kept fixed at
and
, respectively. To tackle this problem, we follow this idea that was used in section 1.10: we trade the function
for a discrete approximation which agrees with
at the
points.
Substitute:
![{\displaystyle KE={\frac {1}{2}}m\left({\frac {dx_{cl}}{dt}}\right)^{2}={\frac {1}{2}}m\left[-\omega x_{0}\sin(\omega t')+\omega {\frac {x-x_{0}\cos(\omega t)}{\sin(\omega t)}}\cos(\omega t')\right]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6a285c0b9345614b34749f61c0b488e501361cb)
Substituting, integrating from time 0 to time
and simplifying, we get:

Explicit evaluation of the path integral for the harmonic oscillator can be found here File:FeynmanHibbs H O Amplitude.pdf