Getting 0 solving Schrodinger equation with Dirac delta by Fourier transform I am attempting to solve the Schrödinger equation with the potential $V = - \delta (x)$. This leads to a differential equation
$$
\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0
$$
where
$$
\alpha \equiv \frac{\hbar^2}{2m}
$$
My instinct to solve the equation was to use an integral transform as I didn't know what else to do with the delta function. Since I am concerned with all space, I chose the Fourier transform. To get the Fourier transform, I transform each term, $\alpha \psi''(x)$, $E \psi(x)$, $\delta(x) \psi(x)$, and $0$. I get these transforms:
\begin{align}
&\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\
&\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\
&\mathcal{F}\{ \delta(x)\psi(x) \} = \hat{\psi}(p) \\
&\mathcal{F} \{0 \} = 0
\end{align}
Where $\hat{\psi}(p)$ is the Fourier transform of $\psi(x)$. Unfortunately, when I combine this with the differential equation above, I get
$$
(1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = 0
$$
which results in the $\hat{\psi}(p)$ term going away and leaving me with a function only of $p$, so I am not able to solve for $\psi(x)$ via an inverse Fourier transform like I expected. I did replace the right hand side with an arbitrary function and work through it, but that results in $\psi(x) = 0$ is that function is $0$.
Is there something that I have done incorrectly in my approach? Is there a better approach to solving this equation? My professor and book both solved this question by solving on $(-\infty, 0)$ and $(0, \infty)$ and determining the condition at the origin by integrating the Schrödinger equation over $(-\epsilon, \epsilon)$, but I was hoping to be able to solve it in a more general way and plan to move to a Dirac comb next, which should be very similar using Fourier transformations.

From the Schrodinger equation
$$\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0$$
and
\begin{align}
&\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\
&\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\
&\mathcal{F}\{ \delta(x)\psi(x) \} = \psi(0) \\
&\mathcal{F} \{0 \} = 0
\end{align}
then it is seen that:
$$(1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = \psi(0).$$
Now, since $\hat{\psi}$ is on $(-\infty, \infty)$ then the only way to determine $\psi(0)$ is to consider the two regions, $p < 0$ and $p > 0$, and use $\psi(0)$ as the matching condition. In general the delta function acts like a discontinuity and needs to be treated as a discontinuity. 
