Momentum Space Representation of the 1\r Operator
Momentum Space Representation of the \( \frac{1}{r} \) Operator
The operator
\( \frac{1}{r}, \quad r = |\mathbf{x}| \)
plays a central role in quantum mechanics, especially in Coulomb potentials, hydrogen-like atoms, and scattering theory.
In momentum space, multiplication by \( \frac{1}{r} \) in position space becomes a convolution with a well-defined kernel.
1. Position-Space Operator
For a wavefunction \( \psi(\mathbf{x}) \), the action of the operator \( \frac{1}{r} \) is
\left( \frac{1}{r} \psi \right) (\mathbf{x}) = \frac{1}{|\mathbf{x}|} \psi(\mathbf{x}).
\]
2. Fourier Transform Conventions
We adopt the standard physics convention for the Fourier transform:
\psi(\mathbf{x}) = \frac{1}{(2\pi)^{3/2}} \int \tilde{\psi}(\mathbf{k})\, e^{i \mathbf{k}\cdot \mathbf{x}} \, d^3k,
\]
\[
\tilde{\psi}(\mathbf{k}) = \frac{1}{(2\pi)^{3/2}} \int \psi(\mathbf{x})\, e^{-i \mathbf{k}\cdot \mathbf{x}} \, d^3x.
\]
The momentum-space operator \( \widetilde{V}(\mathbf{k},\mathbf{k}’) \) associated with
\( V(\mathbf{x}) = \frac{1}{r} \) is defined through
\widetilde{(V \psi)}(\mathbf{k})
= \frac{1}{(2\pi)^{3/2}} \int \frac{1}{r}\, \psi(\mathbf{x})\, e^{-i \mathbf{k}\cdot \mathbf{x}}\, d^3x.
\]
Substituting the inverse Fourier transform of \(\psi(\mathbf{x})\):
\widetilde{(V \psi)}(\mathbf{k})
= \frac{1}{(2\pi)^3} \int d^3x \, \frac{e^{-i \mathbf{k}\cdot \mathbf{x}}}{r}
\int d^3k’ \, \tilde{\psi}(\mathbf{k}’)\, e^{i \mathbf{k}’\cdot \mathbf{x}}.
\]
Reordering integrals gives:
\widetilde{(V \psi)}(\mathbf{k})
= \frac{1}{(2\pi)^3} \int d^3k’ \, \tilde{\psi}(\mathbf{k}’)
\left[ \int d^3x \, \frac{e^{i(\mathbf{k}’ – \mathbf{k})\cdot \mathbf{x}}}{r} \right].
\]
3. Fourier Transform of \( 1/r \)
The inner integral is the Fourier transform of \( 1/r \):
\int_{\mathbb{R}^3} \frac{e^{i \mathbf{q}\cdot \mathbf{x}}}{|\mathbf{x}|} \, d^3x,
\quad \mathbf{q} = \mathbf{k}’ – \mathbf{k}.
\]
This integral evaluates to
\int \frac{e^{i \mathbf{q}\cdot \mathbf{x}}}{|\mathbf{x}|} \, d^3x
= \frac{4\pi}{|\mathbf{q}|^2}.
\]
This result is obtained by switching to spherical coordinates and integrating over the angular and radial parts.
4. Final Momentum-Space Representation
Substituting the Fourier transform result back gives:
\widetilde{(V \psi)}(\mathbf{k})
= \frac{1}{(2\pi)^3} \int d^3k’ \, \tilde{\psi}(\mathbf{k}’)
\left[ \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2} \right].
\]
Thus, the momentum-space kernel of the \( \frac{1}{r} \) operator is
\langle \mathbf{k} | \frac{1}{r} | \mathbf{k}’ \rangle
= \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2} \frac{1}{(2\pi)^3}.
\]
Equivalently,
\left[\frac{1}{r}\tilde{\psi}\right](\mathbf{k})
= \int \frac{d^3k’}{(2\pi)^3} \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2}\, \tilde{\psi}(\mathbf{k}’).
\]
5. Special Case: Coulomb Potential
For a Coulomb potential
\( V(\mathbf{x}) = -\frac{e^2}{4\pi\epsilon_0}\frac{1}{r} \),
the momentum-space kernel is
\widetilde{V}(\mathbf{k}, \mathbf{k}’)
= – \frac{e^2}{4\pi\epsilon_0} \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2} \frac{1}{(2\pi)^3}
= – \frac{e^2}{\epsilon_0} \frac{1}{|\mathbf{k} – \mathbf{k}’|^2} \frac{1}{(2\pi)^3}.
\]
This form is central in solving the Schrödinger equation in momentum space for hydrogen and in the Born approximation in scattering theory.
6. Summary Table
| Quantity | Position Space | Momentum Space |
|---|---|---|
| Operator | \( V(\mathbf{x}) = \frac{1}{r} \) | Integral operator with kernel \( \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2(2\pi)^3} \) |
| Action on \( \psi \) | Multiply by \( \frac{1}{r} \) | Convolution with \( \frac{4\pi}{|\mathbf{k} – \mathbf{k}’|^2} \) |
| Key Fourier Transform | — | \( \mathcal{F}\{ 1/r \} = \frac{4\pi}{q^2} \) |