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Schrödinger equation in momentum space

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: \[\boxed{ i \hbar \frac{\partial \phi(\boldsymbol{p},t)}{\partial t} = \frac{|\boldsymbol{p}|^2}{2 m} \phi(\boldsymbol{p},t) + \int_{\mathbb{R}^n} d^n\boldsymbol{p}' \, W(\boldsymbol{p}-\boldsymbol{p}') \phi(\boldsymbol{p}', t) }\]