# Schrödinger Equation in Momentum Representation

$$\langle q | \hat{H} | \psi \rangle = \, – \frac{\hbar^2}{2m} \frac{d^2}{dq^2} \psi(q) + V(q) \psi(q) \overset{!}{=} E \psi(q) = E \langle q | \psi \rangle$$

Next, we switch to a basis of momentum eigenstates $$\hat{p} |p\rangle = p |p\rangle$$ and make use of

$$\langle p | q \rangle = \frac{1}{\sqrt{2 \pi \hbar}} e^{-\frac{i}{\hbar} p q}$$

So we would like to express Schrödinger equation in momentum representation $$\langle p | \hat{H} | \psi \rangle$$, but what we know this far is it’s position representation. Therefore we can insert a suitable “one” of position basis states

$$\int d{q} \langle p | q \rangle \langle q | \hat{H} | \psi \rangle = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \left[ – \frac{\hbar^2}{2m} \frac{d^2}{dq^2} \psi(q) + V(q) \psi(q) \right]$$

$$\overset{!}{=} E \int d{q} \langle p | q \rangle \langle q |\psi \rangle = E \bar{\psi}(p)$$

The first term represents a Fourier transform of a second derivative. After partial integration for two times

\begin{aligned} & \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \frac{d^2}{dq^2} \psi(q) \\ &= \underbrace{e^{-\frac{i}{\hbar} pq} \frac{d}{dq} \psi(q) \Big|_{-\infty}^{\infty}}_{= 0} – \left( -\frac{i}{\hbar} p \right) \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \frac{d}{dq} \psi(q) \\ &= \underbrace{\frac{i}{\hbar} p e^{-\frac{i}{\hbar} pq} \psi(q) \Big|_{-\infty}^{\infty}}_{= 0} -\left( -\frac{i}{\hbar} p \right)^2 \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \psi(q) \end{aligned}

we obtain

$$– \frac{\hbar^2}{2m} \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \frac{d^2}{dq^2} \psi(q) = \frac{p^2}{2m} \bar{\psi}(p)$$

The second term represents a Fourier transform of a product of $$V(q)$$ and $$\psi(q)$$, whose individual Fourier transforms read $$\psi(p)$$ and

$$V(q) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{p^\prime} e^{\frac{i}{\hbar} p^\prime q} \, \bar{V}(p^\prime)$$

\begin{aligned} &\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{q} \, e^{-\frac{i}{\hbar} p q} \, V(q) \psi(q) \\ &= \frac{1}{(2 \pi \hbar)^{3/2}} \int_{-\infty}^{\infty} d{q} \, e^{\frac{i}{\hbar} p q} \iint_{-\infty}^{\infty} d{p^\prime} d{p^{\prime\prime}} e^{\frac{i}{\hbar} (p^\prime + p^{\prime\prime}) q} \, \bar{V}(p^\prime) \bar{\psi}(p^{\prime\prime}) \\ &= \frac{1}{\sqrt{2 \pi \hbar}} \iint_{-\infty}^{\infty} d{p^\prime} d{p^{\prime\prime}} \, \delta(p^\prime + p^{\prime\prime} – p) \bar{V}(p^\prime) \bar{\psi}(p^{\prime\prime}) \\ &= \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{p^{\prime\prime}} \, \bar{V}(p – p^{\prime\prime}) \bar{\psi}(p^{\prime\prime}) \end{aligned}

So the Fourier transform of a product of functions is a convolution. The resulting Schrödinger equation reads in momentum representation

$$\frac{p^2}{2m} \bar{\psi} + \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d{p^{\prime\prime}} \, \bar{V}(p – p^{\prime\prime}) \bar{\psi}(p^{\prime\prime}) = E \bar{\psi}(p)$$