Schrödinger Equation in Momentum Representation

We start with the 1D Schrödinger equation in position 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

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

leading to

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

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

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

This leads to

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

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

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

Further Reading

Grundkurs Theoretische Physik 5/1. Quantenmechanik – Grundlagen. Wolfgang Nolting. 2009 Springer Verlag, Exercise 2.3.9