Mute said:
As for your question as to whether or not there are any functions that are their own Fourier transform, the answer is yes. If we take our Fourier transform convention to be
\hat{y}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t),
(which is the unitary version of the Fourier transform), where \hat{y}(\omega) is the Fourier transform of y(t), then y(t) = \exp(-t^2/2) and y(t) = \mbox{sech}(\sqrt{\pi/2}t) are two examples of functions which solve the integral equation
y(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t).
(Note the absence of the hat on the left hand side to indicate that it is the same function y(t) that occurs in the integrand).
Since you seem to have a genuine interest in these matters, let me tell you what I'm up to.
I've stumbled across a novel approach to derive Feynman's path integral from the Dirac delta function. The Dirac delta function has the following properties:
1)
\int_{ - \infty }^{ + \infty } {{\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}){\rm{d}}{{\rm{x}}_1}} = 1
and also,
\int_{ - \infty }^{ + \infty } {{\rm{f(}}{{\rm{x}}_1}{\rm{)\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}){\rm{d}}{{\rm{x}}_1}} = {\rm{f(}}{{\rm{x}}_0})
where if we let {\rm{f(}}{{\rm{x}}_1}) = {\rm{\delta (x - }}{{\rm{x}}_1}), we get,
2)
\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})
This is on the wikepedia.org site for the Dirac delta function about half way down under section 4.3 for Translation. I've also seen this in Lowell S. Brown's book, Quantum Field Theory, page 30.Yet the above can be iterated to get,
\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_2}){\rm{\delta (}}{{\rm{x}}_2}{\rm{ - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} d{x_2} = } \int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_2}){\rm{\delta (}}{{\rm{x}}_2}{\rm{ - }}{{\rm{x}}_0})d{x_2}} = {\rm{\delta (x - }}{{\rm{x}}_0})
and if iterated an infinite number of times one gets,
3)
\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - }}{{\rm{x}}_n}){\rm{\delta (}}{{\rm{x}}_n}{\rm{ - }}{{\rm{x}}_{n - 1}}) \cdot \cdot \cdot {\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1} = {\rm{\delta (x - }}{{\rm{x}}_0})
This is explicitly written out in Prof. Hagen Kleinert's book, Path Integrals i Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, page 91.
Then if we use the gaussian form of the Dirac delta function,
4)
{\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{1/2}}}}{e^{ - {{({x_1} - {x_0})}^2}/{\Delta ^2}}}
with
5)
{\Delta ^2} = \frac{{2i\hbar }}{m}({t_1} - {t_0})
the dirac deltas become
\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im{{({x_1} - {x_0})}^2}}}{{2\hbar ({t_1} - {t_0})}}} \right]
which can be manipulated to
\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})}^2}({t_1} - {t_0})} \right] = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {\left[ {\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{\Delta {x_{1,0}}}}{{\Delta {t_{1,0}}}})}^2}\Delta {t_{1,0}}} \right]
or,
6)
\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}
When this is substituted for each of the dirac deltas in 3) above we get
7)
\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta {t_{,n}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{,n}})}^2}\Delta {t_{,n}}}}{{(\frac{m}{{2\pi i\hbar \Delta {t_{n,n - 1}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{n,n - 1}})}^2}\Delta {t_{n,n - 1}}}} \cdot \cdot \cdot {{(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}
with the appropriate limits implied.
Then since the exponents add up, 7) above becomes
8)
\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta t}})}^{n/2}}{e^{\,\,{\textstyle{-i \over \hbar }}\int_0^t {\frac{m}{2}{{(\dot x)}^2}dt} }}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}
Which is Feynman's path integral for a free particle.
This is very curious. I have speculative theories as to why the dirac delta would be fundamental to deriving physics. But the math I've presented here seems pretty straight forward.
However, the use of the gaussian form of the dirac delta seems arbitrary in 4), along with the use of the complex number for the standard deviation in 5). So I am looking for some more fundamental reasons why the complex gaussian would be preferred in this situation. And I wonder if there is not something in the algebra of 2) that dictates the use of the complex gaussian for each of the deltas.
So when I look at 2)
2)
\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})
and consider that
9)
{e^{\frac{{-im}}{{2\hbar }}{{(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})}^2}({t_1} - {t_0})}} = {e^{\frac{{-im}}{{2\hbar }}(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})({x_1} - {x_0})}} = {e^{-ip({x_1} - {x_0})/2\hbar }}
It seems that 2) can be written as
10)
\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = \int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){{(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})}^{1/2}}{e^{ - ip({x_1} - {x_0})/2\hbar }}d{x_1}}
which seems suspiciously close to the Fourier transform of the dirac delta. Although I,m not sure whether the factor in front of the exponential or the 2\hbar in the exponent can be made to fits in the format of the Fourier transform. This brings up questions as to whether 2) is even a Linear, Translationally Invariant (LTI) system, and if so, does that mean the Fourier transform is automatically implied. And since 3) is equivalent to 2), it would seem that I'm looking for an equation for the deltas such that any number of transforms are equal to one transform. I'm not sure what that means. Any insight anyone might have into these questions would help greatly. Thank you.
