In this note, we first obtain Schrödinger equation in momentum space and then solve a bound state problem with delta potential in position space. This is based on what I learned from my friend and a physicist Khin Maung. Let us begin with eigenstate Schrödinger equation
$$\hat H|\psi\rangle=E|\psi\rangle$$
with
$$\hat H=\frac{\hat p^2}{2m}+\hat V$$
in position space. As we will see Schrödinger equation in momentum space becomes an integral equation, while Schrödinger equation in position space is a differential equation.
Before we continue, here is the list of a few things we need for our discussion.

• The completeness: $1=\int |p\rangle\langle p|dp$, $1=\int |x\rangle\langle x|dx$.
• Eigenvalue equations: \begin{align*} \hat p|p’\rangle&=p’|p’\rangle\\ \hat x|x’\rangle&=x’|x’\rangle \end{align*}
• $\langle p|p’\rangle=\delta(p’-p)$, $\langle x|x’\rangle=\delta(x’-x)$.
• $\langle p|x\rangle=\frac{e^{-\frac{i}{\hbar}px}}{\sqrt{2\pi\hbar}}$, $\langle x|p’\rangle=\frac{e^{\frac{i}{\hbar}p’x}}{\sqrt{2\pi\hbar}}$.

Details about these can be found in any standard quantum mechanics textbook, particularly in the chapter about quantum systems with continuous spectra.

Using the completeness, we have
$$\int\left(\frac{\hat p^2}{2m}+\hat V\right)|p’\rangle\langle p’|\psi\rangle dp’=E|\psi\rangle$$
and
\begin{equation}
\label{eq:seq}
\int\langle p|\left(\frac{\hat p^2}{2m}+\hat V\right)|p’\rangle\tilde\psi(p’)dp’=E\tilde\psi(p)
\end{equation}
where $\tilde\psi(p)=\langle p|\psi\rangle$. Since $\hat p|p’\rangle=p’|p’\rangle$, $\hat p^2|p’\rangle=p’^2|p’\rangle$. So, \eqref{eq:seq} can be written as
\begin{equation}
\label{eq:seq2}
\int\langle p|p’\rangle\frac{p’^2}{2m}\tilde\psi(p’)dp’+\int\langle p|\hat V|p’\rangle\tilde\psi(p’)dp’=E\tilde\psi(p)
\end{equation}
The LHS of \eqref{eq:seq2} is then
$$\int\delta(p’-p)\frac{p’^2}{2m}\tilde\psi(p’)dp’+\int\langle p|\hat V|p’\rangle\tilde\psi(p’)dp’=\frac{p^2}{2m}\tilde\psi(p)+\int\langle p|\hat V|p’\rangle\tilde\psi(p’)dp’$$
Hence, we arrive at the one-dimensional Schrödinger equation in momentum space (or $p$-space)
\begin{equation}
\label{eq:seq3}
\frac{p^2}{2m}\tilde\psi(p)+\int\langle p|\hat V|p’\rangle\tilde\psi(p’)dp’=E\tilde\psi(p)
\end{equation}
\begin{align*} \langle p|\hat V|p’\rangle&=\langle p|I|\hat V|I|p’\rangle\\ &=\langle p|\int|x\rangle\langle x|dx|\hat V|\int|x’\rangle\langle x’|dx’|p’\rangle\\ &=\int\langle p|x\rangle\langle x|\hat V|x’\rangle\langle x’|p’\rangle dxdx’ \end{align*}
Since $\hat V$ is a scalar potential,
\begin{align*} \langle x|\hat V|x’\rangle&=\langle x|x’\rangle\hat V\\ &=\delta(x-x’)V(x) \end{align*}
So, we have
\begin{align*} \langle p|\hat V|p’\rangle&=\int\langle p|x\rangle\delta(x-x’)V(x)\langle x’|p’\rangle dx’dx\\ &=\int\langle p|x\rangle V(x)\langle x|p’\rangle dx\\ &=\int \frac{e^{-ipx}}{\sqrt{2\pi}}V(x)\frac{e^{ip’x}}{\sqrt{2\pi}}dx\ [\mbox{assuming}\ \hbar=1]\\ &=\frac{1}{2\pi}\int V(x)e^{i(p’-p)x}dx \end{align*}
If we want to keep $\hbar$, we have
$$\langle p|\hat V|p’\rangle=\frac{1}{2\pi\hbar}\int V(x) e^{\frac{i}{\hbar}(p’-p)x}dx$$

Now, we consider a special case of a delta potential
$$V(x)=-\delta(x)V_0$$
\begin{align*} \langle p|\hat V|p’\rangle&=\frac{1}{2\pi}\int_{-\infty}^\infty[-\delta(x)V_0e^{i(p’-p)x}]dx\\ &=-\frac{V_0}{2\pi} \end{align*}
The Schrödinger equation \eqref{eq:seq3} then becomes
\begin{equation}
\label{eq:seq4}
\left(\frac{p^2}{2m}-E\right)\tilde\psi(p)=\frac{V_0}{2\pi}\int_{-\infty}^\infty\tilde\psi(p’)dp’
\end{equation}
Note here that the bound state energy must be negative, so we assume that $E<0$. Let
$$A=\int_{-\infty}^\infty\tilde\psi(p’)dp’$$
Then \eqref{eq:seq4} can be written as
$$\tilde\psi(p)=\frac{V_0}{2\pi}\frac{A}{\frac{p^2}{2m}-E}$$
Integrating this with respect to $p$, we obtain
\begin{align*} 1&=\frac{V_0}{2\pi}\int_{-\infty}^\infty\frac{dp}{\frac{p^2}{2m}-E}\\ &=\frac{V_0}{2}\sqrt{\frac{2m}{-E}} \end{align*}
Hence, we find the bound state energy
\begin{equation}
\label{eq:bde}
E=-\frac{V_0^2m}{2}
\end{equation}
For $\hbar\ne 1$, we have
$$E=-\frac{V_0^2m}{2\hbar^2}$$
Recall that $\tilde\psi(p)$ has $A$ in it. $A$ can be found by normalizing $\tilde\psi(p)$.
$$1=\int_{-\infty}^\infty |\tilde\psi(p)|^2dp$$
Since
$$|\tilde\psi(p)|^2=\frac{V_0^2}{4\pi^2}\frac{|A|^2}{\left(\frac{p^2}{2m}-E\right)^2},$$
we have
\begin{equation}
\label{eq:sqe5}
1=\frac{V_0^2}{4\pi^2}|A|^2\int_{-\infty}^\infty\frac{dp}{\left(\frac{p^2}{2m}-E\right)^2}
\end{equation}
Let
$$I=\int_{-\infty}^\infty\frac{dp}{\left(\frac{p^2}{2m}-E\right)^2}=\sqrt{2m}\pi(-E)^{-\frac{1}{2}}$$
Then
$$\frac{\partial I}{\partial E}=\int_{-\infty}^\infty\frac{dp}{\left(\frac{p^2}{2m}-E\right)^2}=\sqrt{\frac{m}{2}}\pi(-E)^{-\frac{3}{2}}$$
Therefore, from \eqref{eq:sqe5} with \eqref{eq:bde}, we obtain
$$A=\sqrt{2\pi mV_0}e^{i\theta}$$
for some $\theta$.