time dependent Schröidinger Equation

Greetings all!

I start with a couple of ideas about life!

• what is life?
  In its simplest form, life is nothing more than a way of decreasing entropy in expanse of producing heat. Hence, it's just another way of contributing heat death of universe.
• is it ok to live forever?
• we loose the urgency of doing things. we will have always tomorrow, why to get up even? So we cant achive nothing.
• even if we do things their value will drop by time. If given enough time you can
  achive anything.

Now, lets do some math
This is famous Parseval's theorem: if you expand a function in a complete orthtonormal set than their square's are related.

Complete set can be Fourier, Legendre, Bessel etc. This is very useful. It can be even used to define completeness. Let's see some example! first lets expand constant function f =1 in the interval 0 to 1 by sin functions

$\begin{array}{}f\left(x\right)=\underset{n=1}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\sum \; <!--\; n-ary\; summation\; -->}}{a}_n{u}_n\left(x\right)\\ a_n=\underset{a}{\overset{b}{\int \; <!--\; integral\; -->}}{u}_n\left(x\right)f\left(x\right)\mathrm{dx}\\ f^2=\underset{h=1}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\sum \; <!--\; n-ary\; summation\; -->}}{a}_n^2\end{array}$

$\begin{array}{}f=1\\ u_n\left(x\right)=\sqrt{2}\times \; <!--\; multiplication\; sign\; -->\sin \left(n\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}x\right)\\ a_n=\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\sqrt{2}\times \; <!--\; multiplication\; sign\; -->\sin \left(n\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}x\right)\mathrm{dx}=\{\begin{array}{}0,n\to \; <!--\; rightwards\; arrow\; -->even\\ \frac{2\times \; <!--\; multiplication\; sign\; -->\sqrt{2}}{n\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}},n\to \; <!--\; rightwards\; arrow\; -->odd\end{array}\end{array}$

$a_n^2=\{\begin{array}{}0,n\to \; <!--\; rightwards\; arrow\; -->even\\ \frac{8}{n^2{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}^2},n\to \; <!--\; rightwards\; arrow\; -->0dd\end{array}$

thus Parseval's theorem tells us that

$\underset{n=0}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\sum \; <!--\; n-ary\; summation\; -->}}\frac{1}{{\left(2n+1\right)}^2}=1+\frac{1}{3^2}+\frac{1}{5^2}+\dots \; <!--\; horizontal\; ellipsis\; -->=\frac{{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}^2}{8}$

Now, Here is a problem for you to repeat above
procedure. function f and u are defined in interval-pi to pi

$\begin{array}{}f\left(x\right)=\frac{1}{\sqrt{2\times \; <!--\; multiplication\; sign\; -->\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}}\times \; <!--\; multiplication\; sign\; -->e^{ikx}\\ u_n\left(x\right)=\frac{1}{\sqrt{2\times \; <!--\; multiplication\; sign\; -->\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}}\times \; <!--\; multiplication\; sign\; -->e^{inx}\\ \frac{1}{\sqrt{2\times \; <!--\; multiplication\; sign\; -->\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}}\times \; <!--\; multiplication\; sign\; -->e^{ikx}=\underset{n=-\mathrm{\infty \; <!--\; infinity\; -->}}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\sum \; <!--\; n-ary\; summation\; -->}}{a}_n\frac{1}{\sqrt{2\times \; <!--\; multiplication\; sign\; -->\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}}\times \; <!--\; multiplication\; sign\; -->e^{inx}\end{array}$

In this case using Parsevals' theorem show:

$co{\sec }^2\left(\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}k\right)=\frac{1}{{\sin }^2\left(\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}k\right)}=\underset{n=-\mathrm{\infty \; <!--\; infinity\; -->}}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\sum \; <!--\; n-ary\; summation\; -->}}\frac{1}{{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}^2\times \; <!--\; multiplication\; sign\; -->{\left(k-n\right)}^2}$