DarMM said:
No, the Minkowski space path integral (Feynman path integral) does not have a well defined measure.
I'm wondering if an integral recursion relation whose measure is well defined might help define the Feynman path integral measure. In my limited reading on the subject I've not noticed the use of a recursion relation in the discussion of the Feynman path integral measure, has anyone here seen such use?
I'm thinking of using a Chapman-Kolmogorov equation as a recursion relation,
\[\int_{ - \infty }^{ + \infty } {\left( {\frac{\lambda }{{2\pi \left( {t - s} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - \upsilon )^2 } \over {2\left( {t - s} \right)}}}} \left( {\frac{\lambda }{{2\pi \left( {s - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\upsilon - u)^2 } \over {2\left( {s - r} \right)}}}} {\rm{d}}\upsilon } = \left( {\frac{\lambda }{{2\pi \left( {t - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - u)^2 } \over {2\left( {t - r} \right)}}}} \]
If necessary I can show proof of this equation. Now this equation is well defined, it's a normal type integral with a lebesgue measure, right? What if lambda is imaginary, is it still a well defined integral? I don't see why not.
If we put limits on (s-r) and (t-s) so they each approach zero independently, then we have,
\[<br />
\mathop {\lim }\limits_{\left( {t - s} \right) \to 0} \left( {\frac{\lambda }{{2\pi \left( {t - s} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - \upsilon )^2 } \over {2\left( {t - s} \right)}}}} = {\rm{\delta (}}\omega - \upsilon )<br />
\]<br />
\[<br />
\mathop {\lim }\limits_{\left( {s - r} \right) \to 0} \left( {\frac{\lambda }{{2\pi \left( {s - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\upsilon - u)^2 } \over {2\left( {s - r} \right)}}}} = {\rm{\delta (}}\upsilon - u)<br />
\]<br />
and,
\[<br />
\mathop {\lim }\limits_{\left( {t - r} \right) \to 0} \left( {\frac{\lambda }{{2\pi \left( {t - r} \right)}}} \right)^{\frac{1}{2}} e^{ - {\textstyle{{\lambda (\omega - u)^2 } \over {2\left( {t - r} \right)}}}} = {\rm{\delta (}}\omega - u)<br />
\]<br />
So that,
\[<br />
\int_{ - \infty }^{ + \infty } {{\rm{\delta (}}\omega - \upsilon {\rm{)\delta (}}\upsilon - u){\rm{d}}\upsilon } = {\rm{\delta (}}\omega - u)<br />
\]<br />
Does placing limits as shown above change the nature of the measure? Is it still a well defined integral? I suppose it would be necessary to do the integration first before taking the limits of (t-s), (s-r), and (t-r), in the same way that you have to do the integral first and then take the limit when integrating a single delta that equals one. So if you do the limits later, then the integral must have the same measure as before without the limits. Then perhaps taking the limits turns the resulting gaussing distribution into a Dirac measure, which is the measure used when integrating the next iteration, right?
If we replace \[<br />
\omega <br />
\]<br /> with \[<br />
{\rm{x}}<br />
\]<br />, replace \[<br />
\upsilon <br />
\]<br /> with \[<br />
{{\rm{x}}_1 }<br />
\]<br />, and replace \[<br />
u<br />
\]<br /> with \[<br />
{{\rm{x}}_0 }<br />
\]<br />, then we have,
\[<br />
\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )<br />
\]<br />
This is a recursion relation that can be iterated any number of times to get the path integral when the gaussian form of the dirac delta is used. See post 21 in this thread.
Since we will always get \[<br />
{\rm{\delta (x - x}}_0 )<br />
\]<br /> no matter how many times the recursion relation is iterated, doesn't this prove that the path integral must be well defined too, since the delta on the right can be considered a well defined measure, and since the delta here is independent of the number of iterations even if iterated an infinite number of times?