Free Schroedinger equation How can one find and prove the general solution to the equation $\dfrac{\partial f(x,t)}{\partial t} =c^2i\dfrac{\partial^2f(x,t)}{\partial x^2}$ ?
I can find the solutions $Ae^{ikx-E_kt}$, so I expect linear sombinations of this to solve the equation, but can, and if so why, every solution be written as $\int a(k)e^{ikx-iE_kt}~dk$?
I think its two questions,
a) (Why) can every 'basis' solution be found by separation of variables?
b) (Why) is the general solution an integral and not a sum of all LI. solutions?
 A: Similar to how to solve $ {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$:
Let $f(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=ic^2X''(x)T(t)$
$\dfrac{T'(t)}{ic^2T(t)}=\dfrac{X''(x)}{X(x)}=-(K(k))^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-ic^2(K(k))^2\\X''(x)+(K(k))^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(k)e^{-ic^2t(K(k))^2}\\X(x)=\begin{cases}c_1(k)\sin((x-m)K(k))+c_2(k)\cos((x-m)K(k))&\text{when}~K(k)\neq0\\c_1x+c_2&\text{when}~K(k)=0\end{cases}\end{cases}$
$\therefore f(x,t)=C_1x+C_2+\int_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))~dk+\int_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))~dk$
or $C_1x+C_2+\sum\limits_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))+\sum\limits_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))$
The choice of integral or summation depends on number of boundary conditions. If there is only one boundary condition or no boundary conditions, we should choose integral. If there are two boundary conditions, we should choose summation.
Another brilliant method is called the power series method.
Similar to PDE - solution with power series:
Let $f(x,t)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^n}{n!}\dfrac{\partial^nu(a,t)}{\partial x^n}$ ,
Then $f(x,t)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(a,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(a,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{i^nc^{2n}(2n)!}\dfrac{\partial^nu(a,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}\dfrac{\partial^{n+1}(a,t)}{\partial t^n\partial x}=\sum\limits_{n=0}^\infty\dfrac{F^{(n)}(t)(x-a)^{2n}}{i^nc^{2n}(2n)!}+\sum\limits_{n=0}^\infty\dfrac{G^{(n)}(t)(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}=\sum\limits_{n=0}^\infty\dfrac{(-1)^nF^{(2n)}(t)(x-a)^{4n}}{c^{4n}(4n)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nF^{(2n+1)}(t)(x-a)^{4n+2}}{c^{4n+2}(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nG^{(2n)}(t)(x-a)^{4n+1}}{c^{4n}(4n+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nG^{(2n+1)}(t)(x-a)^{4n+3}}{c^{4n+2}(4n+3)!}$
