Consider a quantum particle of mass $$m$$ moving in an $$n$$-dimensional space in the presence of an external (scalar) potential $$V(\boldsymbol{x})$$. The Hamiltonian governing the motion is $H = \frac{\boldsymbol{P} \cdot \boldsymbol{P}}{2 m} + V(\boldsymbol{X}) \,,$ where $$\boldsymbol{X}$$ and $$\boldsymbol{P}$$ are the position and momentum operators, respectively. The time-dependent state of the particle $$| \Psi_t \rangle$$ satisfies the Schrödinger equation $i \hbar \frac{\partial | \Psi_t \rangle}{\partial t} = H | \Psi_t \rangle \,.$ In position representation, we have $\boldsymbol{X} = \boldsymbol{x} \,, \qquad \boldsymbol{P} = -i \hbar \frac{\partial}{\partial \boldsymbol{x}} \,,$ and so the Schrödinger equation has the following familiar form: $i \hbar \frac{\partial \psi(\boldsymbol{x},t)}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial}{\partial \boldsymbol{x}} \cdot \frac{\partial \psi(\boldsymbol{x},t)}{\partial \boldsymbol{x}} + V(\boldsymbol{x}) \psi(\boldsymbol{x},t) \,,$ where $\psi(\boldsymbol{x},t) = \langle \boldsymbol{x} | \Psi_t \rangle \,.$
Let us now rewrite the Schrödinger equation in momentum representation, in which $\boldsymbol{X} = i \hbar \frac{\partial}{\partial \boldsymbol{p}} \,, \qquad \boldsymbol{P} = \boldsymbol{p} \,.$ We have \begin{align} i \hbar \frac{\partial \langle \boldsymbol{p} | \Psi_t \rangle}{\partial t} &= \langle \boldsymbol{p} | H | \Psi_t \rangle \\[0.1cm] &= \frac{|\boldsymbol{p}|^2}{2 m} \langle \boldsymbol{p} | \Psi_t \rangle + \langle \boldsymbol{p} | V(\boldsymbol{X}) | \Psi_t \rangle \\[0.2cm] &= \frac{|\boldsymbol{p}|^2}{2 m} \langle \boldsymbol{p} | \Psi_t \rangle + \int_{\mathbb{R}^n} d^n\boldsymbol{x} \int_{\mathbb{R}^n} d^n\boldsymbol{p}' \, \langle \boldsymbol{p} | V(\boldsymbol{X}) | \boldsymbol{x} \rangle \langle \boldsymbol{x} | \boldsymbol{p}' \rangle \langle \boldsymbol{p}' | \Psi_t \rangle \\[0.2cm] &= \frac{|\boldsymbol{p}|^2}{2 m} \langle \boldsymbol{p} | \Psi_t \rangle + \int_{\mathbb{R}^n} d^n\boldsymbol{x} \int_{\mathbb{R}^n} d^n\boldsymbol{p}' \, \langle \boldsymbol{p} | \boldsymbol{x} \rangle V(\boldsymbol{x}) \langle \boldsymbol{x} | \boldsymbol{p}' \rangle \langle \boldsymbol{p}' | \Psi_t \rangle \,. \end{align} Denoting the momentum wave function by $\phi(\boldsymbol{p},t) = \langle \boldsymbol{p} | \Psi_t \rangle \,,$ and making use of the fact that $\langle \boldsymbol{x} | \boldsymbol{p}' \rangle = \frac{e^{i \boldsymbol{p}' \cdot \boldsymbol{x}}}{(2 \pi \hbar)^{n/2}} \,, \qquad \langle \boldsymbol{p} | \boldsymbol{x} \rangle = \frac{e^{-i \boldsymbol{p} \cdot \boldsymbol{x}}}{(2 \pi \hbar)^{n/2}} \,,$ we get $i \hbar \frac{\partial \phi(\boldsymbol{p},t)}{\partial t} = \frac{|\boldsymbol{p}|^2}{2 m} \phi(\boldsymbol{p},t) + \frac{1}{(2 \pi \hbar)^n} \int_{\mathbb{R}^n} d^n\boldsymbol{x} \int_{\mathbb{R}^n} d^n\boldsymbol{p}' \, e^{-i (\boldsymbol{p} - \boldsymbol{p}') \cdot \boldsymbol{x} / \hbar} V(\boldsymbol{x}) \phi(\boldsymbol{p}', t) \,.$ Finally, denoting the Fourier transform of $$V$$ by $$W$$, i.e. $W(\boldsymbol{p}) = \frac{1}{(2 \pi \hbar)^n} \int_{\mathbb{R}^n} d^n\boldsymbol{x} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{x} / \hbar} V(\boldsymbol{x}) \,,$ we obtain the sought momentum representation of the Schrödinger equation: $\boxed{ i \hbar \frac{\partial \phi(\boldsymbol{p},t)}{\partial t} = \frac{|\boldsymbol{p}|^2}{2 m} \phi(\boldsymbol{p},t) + \int_{\mathbb{R}^n} d^n\boldsymbol{p}' \, W(\boldsymbol{p}-\boldsymbol{p}') \phi(\boldsymbol{p}', t) }$