Gaussian Potential in the Born Approximation

This worked example computes the first Born scattering amplitude and cross section for a three-dimensional Gaussian potential

$$V(r) = V_0e^{-r^2/a^2}.$$

The sign of $V_0$ determines whether the potential is repulsive or attractive, but the leading Born differential cross section depends on $V_0^2$.

Gaussian Potential and Weak Scattering

The elementary Fourier transform makes this a useful first Born example. Accuracy additionally requires weak enough scattering; smoothness alone does not justify the approximation. Here $m$ is the particle mass for a fixed target or the reduced mass for relative motion, $a>0$, and $V_0$ is real.

With the scattering-amplitude convention used in this volume,

$$f_{\mathrm B}(\mathbf q) = - \frac{m}{2\pi\hbar^2} \int d^3r\, e^{-i\mathbf q\cdot\mathbf r} V(r),$$

where

$$\mathbf q=\mathbf k'-\mathbf k, \qquad q=2k\sin\frac{\theta}{2}$$

for elastic scattering. The vector $\mathbf q$ has units of inverse length; the projectile momentum change is $\hbar\mathbf q$.

The basic weak-scattering parameter is of order

$$\epsilon \sim \frac{m|V_0|a^2}{\hbar^2}.$$

The Born result should be checked against unitarity and, when possible, numerical phase shifts as $\epsilon$ grows.

Born Approximation Numerical Test performs that check for the repulsive Gaussian, with independent coupling, momentum, grid, matching-radius, and partial-wave sweeps.

Fourier Transform

Because $r^2=x^2+y^2+z^2$, the transform separates into three one-dimensional Gaussian integrals. Each contributes

$$\int_{-\infty}^{\infty} dx\,e^{-x^2/a^2-iq_xx} =\sqrt\pi a\,e^{-a^2q_x^2/4}.$$

Multiplying the three factors gives

$$\int d^3r\, e^{-i\mathbf q\cdot\mathbf r} e^{-r^2/a^2} = \pi^{3/2}a^3 e^{-q^2a^2/4}.$$

Thus

$$\int d^3r\, e^{-i\mathbf q\cdot\mathbf r} V_0e^{-r^2/a^2} = V_0\pi^{3/2}a^3 e^{-q^2a^2/4}.$$

Born Amplitude

Substituting into the Born formula gives

$$f_{\mathrm B}(q) = - \frac{mV_0\sqrt\pi\,a^3}{2\hbar^2} e^{-q^2a^2/4}.$$

It is useful to define

$$A = \frac{mV_0\sqrt\pi\,a^3}{2\hbar^2}.$$

Then

$$f_{\mathrm B}(q) = - A e^{-q^2a^2/4}.$$

For this real, inversion-symmetric potential, the leading Born amplitude is real. Unitarity effects in the imaginary forward amplitude appear at higher Born orders.

Differential Cross Section

The differential cross section is

$$\frac{d\sigma}{d\Omega} = |f_{\mathrm B}(q)|^2 = A^2e^{-q^2a^2/2}.$$

Using

$$q=2k\sin\frac{\theta}{2},$$

this becomes

$$\frac{d\sigma}{d\Omega} = \frac{m^2V_0^2\pi a^6}{4\hbar^4} \exp \left[ - 2k^2a^2 \sin^2\frac{\theta}{2} \right].$$

Large-angle scattering is exponentially suppressed when $ka$ is large because large angles require large momentum transfer.

Total Cross Section

Let

$$\beta=k^2a^2.$$

Since

$$q^2a^2/2 = \beta(1-\cos\theta),$$

the total Born cross section is

$$\sigma_{\mathrm B} = 2\pi A^2 \int_{-1}^{1} d\mu\, e^{-\beta(1-\mu)}.$$

Evaluating the integral,

$$\sigma_{\mathrm B} = 2\pi A^2 \frac{1-e^{-2\beta}}{\beta}.$$

Restoring $A$,

$$\sigma_{\mathrm B} = \frac{\pi^2m^2V_0^2a^6}{2\hbar^4} \frac{1-e^{-2k^2a^2}}{k^2a^2}.$$

In the low-energy limit $ka\ll1$,

$$\sigma_{\mathrm B} \to 4\pi A^2 = \frac{\pi^2m^2V_0^2a^6}{\hbar^4}.$$

Validity Discussion

The Born approximation is most reliable when the potential is weak enough that the scattered wave remains small inside the interaction region. The dimensionless estimate

$$\epsilon \sim \frac{m|V_0|a^2}{\hbar^2}$$

should be small.

At low energy, an attractive Gaussian can support or nearly support a shallow bound state as $|V_0|$ grows. Near such a threshold, the scattering length can become large and the first Born approximation fails even if the potential looks smooth.

At high energy, the angular distribution narrows because the Fourier transform suppresses momentum transfers $q\gtrsim1/a$.

Exercises

1. Show that the low-energy total Born cross section is $4\pi A^2$.

Solution

For $ka\ll1$,

$$e^{-q^2a^2/2}\approx1.$$

The differential cross section is approximately isotropic:

$$\frac{d\sigma}{d\Omega}\approx A^2.$$

Integrating over solid angle gives

$$\sigma\approx4\pi A^2.$$

2. Why does the leading Born differential cross section not depend on the sign of $V_0$?

Solution

The Born amplitude is proportional to $V_0$:

$$f_{\mathrm B}\propto V_0.$$

The differential cross section is the magnitude squared:

$$\frac{d\sigma}{d\Omega}=|f_{\mathrm B}|^2.$$

Therefore the leading cross section is proportional to $V_0^2$ and is insensitive to the sign. The sign affects phases, higher-order terms, and possible bound-state physics.

References

• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Dover, 2002.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.