# Solution of the simple form of the Schrodinger equation

The following equation is given: $$\frac{i}{a}\psi'_t(t,x) = -\frac{1}{2}\psi''_{xx}(t,x)$$ with initial condition $\psi(0,x) = \psi_0(x)$. How to explicitely solve this equation for $\psi: [0, +\infty)\times\mathbb{R}\to\mathbb{C}$?

## 1 Answer

Here's a formal solution: $$\psi(t,x)=\sum_{k=0}^{+\infty}\left(i\frac{\alpha t}2\right)^k\frac1{k!}\psi_0^{(2k)}(x).$$

It is obtained directly from the formal solution: $$\psi(t,x)=\mathrm{e}^{\alpha i t/2\Delta}\psi_0(x),$$ where $\Delta$ is the (spatial) Laplacian, i.e., the “second derivative with respect to $x$”.

• Thank you for your answer! Sadly, this is not quite what I was looking for... You solved it using formal series representation for operator semigroup. Can we get an explicit solution for $\psi$? By explicit I mean without limits and approximations. – Glinka Feb 17 '16 at 17:14
• @Glinka: I doubt it. Maybe a good next move would be to express that in terms of Fourier transforms (which makes sense since, in the context of Schrödinger's equation, assuming $\psi_0\in L^2$ is natural). But that would add more integrals (and hence be less explicit?). Now, if you have $\psi_0$ explicitly, depending on its form, it might be possible to have $\psi$ explicitly. – gniourf_gniourf Feb 17 '16 at 17:48
• Okey, too bad... But the formula doesn't seem to work when $\psi \in L^2$, since you take derivatives of any high order, $\psi$ bounded to be at least in $C^\infty$ – Glinka Feb 17 '16 at 19:26
• @Glinka: with Fourier transforms you'll be able to (formally) use $\mathcal{F}(\psi_0^{(2k)})(\omega)=(-\omega^2)^k\mathcal{F}(\psi_0)(\omega)$ and the series will be happily convergent; so that's not a big deal, really. – gniourf_gniourf Feb 17 '16 at 19:29