# Divergent solution in time-dependent Schrödinger equation

if I transform the time-dependent Schrödinger equation without a potential I get:

$$- \hbar \omega \psi(\omega,x) = \frac{- \hbar^2}{2m} \frac{\partial^2 \psi(\omega,x)}{\partial x^2}$$

The solutions is clealy: $$\psi(\omega,x)={\it C1}\,{{\rm e}^{{\frac {\sqrt {2\omega m}x}{\sqrt {\hbar}}}} }+{\it C2}\,{{\rm e}^{{-\frac {\sqrt {2\omega m}x}{\sqrt {\hbar}}}} }$$

I don't really understand this result. The problem is, that if I want to transform back, the Fourier-transform will be divergent, so what does this mean regarding my solution? Is there a work-around to get rid of this divergence? Why did this Fourier-transform fail?

(Should I have used the Laplace-transform?)

• How did you get that first equation? – Kyle Kanos Jun 22 '14 at 2:36
• Just like @KyleKanos, how come you still have derivative w.r.t. $x$ in the transformed equation? – 299792458 Jun 22 '14 at 5:41
• Your solutions do not satisfy your equation! Pay attention to signs... – Valter Moretti Jun 22 '14 at 7:19
• @KyleKanos I did the Fourier transform with respect to $t$ to the equation $i \hbar \partial_t \psi(x,t) = -\frac{\hbar^2}{2m} \partial_x^2 \psi(x,t)$ – Xin Wang Jun 22 '14 at 9:17
• @V.Moretti sorry, had a minus sign too much in the original equation. – Xin Wang Jun 22 '14 at 9:18

To find the solution you proceed in a very standard way as follows. Consider the operator $-\Delta$: it is self-adjoint on a dense domain of $L^2(\mathbb{R}^d)$ (assuming you are in $d$ dimensions), and thus to it can be associated an unitary one-parameter group $\exp(it\Delta)$ (I am assuming here $m$ and $\hbar$ to be one-half and one respectively, for simplicity). This unitary one parameter group is defined for all functions of $L^2$, so given $\psi(t_0,x)=\psi_0(x)\in L^2(\mathbb{R}^d)$, the solution of your Cauchy problem is $$\psi(t,x)=e^{i(t-t_0)\Delta}\psi_0(x)\; .$$ If you take the spatial fourier transform (in $x$), that is another unitary transformation on $L^2$, you obtain the maybe more explicit formula $$\hat{\psi}(t,k)=e^{-i(t-t_0)k^2}\hat{\psi}_0(k)$$ where $\hat{\psi}$ is the Fourier transform of $\psi$.