Question

In quantum theory a certain method (the Born approximation) gives the (socalled) amplitude \(f(\theta)\) for the scattering of a particle of mass \(m\) through an angle \(\theta\) by a uniform potential well of depth \(V_{0}\) and radius \(b\) (i.e. the potential energy of the particle is \(-V_{0}\) within a sphere of radius \(b\) and zero elsewhere) as $$ f(\theta)=\frac{2 m V_{0}}{h^{2} K^{3}}(\sin K b-K b \cos K b) $$ Here \(\hbar\) is the Planck constant divided by \(2 \pi\), the energy of the particle is \(h^{2} k^{2} / 2 m\) and \(K\) is \(2 k \sin (\theta / 2)\) Use l'Hôpital's rule to evaluate the amplitude at low energies, i.e. when \(k\) and hence \(K\) tend to zero, and so determine the low-energy total cross- section. (Note: the differential cross-section is given by \(|f(\theta)|^{2}\) and the total cross-section by the integral of this over all solid angles, i.e. \(2 \pi \int_{0}^{\lambda}|f(\theta)|^{2} \sin \theta d \theta .\) )

Short answer

At low energies, the amplitude is \( \frac{2mV_0 b}{3 h^2} \) and the total cross-section is \( 4\pi ( \frac{2mV_0 b}{3 h^2} )^2 \).

Step-by-step solution

- Understand the given function.

The amplitude function is given by: \[ f(\theta)=\frac{2 m V_{0}}{h^{2} K^{3}}(\sin K b-K b \cos K b) \] where \( K = 2 k \sin (\theta / 2) \). At low energies, \( k \) and \( K \) tend to zero.

- Apply L'Hôpital's Rule.

Due to the limit involving the form \( \frac{0}{0} \), apply L'Hôpital's Rule to evaluate the limit of \( f(\theta) \) as \( K \rightarrow 0 \). Differentiate the numerator and the denominator with respect to \( K \).

- Differentiate the numerator.

Differentiate the numerator \( \sin Kb - Kb \cos Kb \) with respect to \( K \): \[ \frac{d}{dK}(\sin Kb - Kb \cos Kb) = b \cos Kb - b \cos Kb + Kb^2 \sin Kb. \]

- Differentiate the denominator.

Differentiate the denominator \( K^3 \) with respect to \( K \): \[ \frac{d}{dK}(K^3) = 3 K^2. \]

- Apply the result of L'Hôpital's Rule.

After applying L'Hôpital's Rule, the expression becomes: \[ f(\theta) = \lim_{K \to 0} \frac{2mV_0 ( bK^2 \sin Kb )}{h^2 ( 3K^2 )}. \]

- Evaluate the limit.

As \( K \rightarrow 0 \), \( \sin(Kb) \approx Kb \). The final simplified form is: \[ f(\theta) = \frac{2mV_0 b}{3 h^2}. \]

- Find the differential cross-section.

The differential cross-section is given by \( |f(\theta)|^2 \). Hence, \[ \frac{d\sigma}{d\Omega} = \left( \frac{2mV_0 b}{3 h^2} \right)^2. \]

- Integrate to find the total cross-section.

Integrate the differential cross-section over all solid angles: \[ \sigma_{total} = 2 \pi \int_{0}^{\pi} \left( \frac{2 m V_0 b}{3 h^2} \right)^2 \sin \theta d\theta = 2 \pi \left( \frac{2 m V_0 b}{3 h^2} \right)^2 \int_{0}^{\pi} \sin \theta d\theta. \]

- Solve the integral.

Compute the integral: \[ \int_{0}^{\pi} \sin \theta d\theta = 2. \] Therefore, \[ \sigma_{total} = 2 \pi \left( \frac{2 m V_0 b}{3 h^2} \right)^2 \cdot 2 = 4 \pi \left( \frac{2 m V_0 b}{3 h^2} \right)^2. \]

Key concepts

L'Hôpital's rule

L'Hôpital's rule is a handy mathematical tool used for finding limits that involve indeterminate forms, often in the shape of \( \frac{0}{0} \) or \( \frac{\text{∞}}{\text{∞}} \). The rule suggests that if the limit \( \frac{f(x)}{g(x)} \) as \( x \to c \) results in an indeterminate form, we can find this limit by differentiating the numerator and the denominator separately. Formally, if \( \frac{f(x)}{g(x)} \) approaches an indeterminate form as \( x \to c \), then the limit can be computed as:
\[ \text{lim}_{x \to c} \frac{f(x)}{g(x)} = \text{lim}_{x \to c} \frac{f'(x)}{g'(x)} \] This rule is particularly useful in problems like the one we are discussing where the function tends to zero, resulting in indeterminate forms. In the given exercise, this rule helps us evaluate the amplitude function \( f(\theta) \) as \( K \to 0 \). Differentiating the numerator and the denominator in this zero-over-zero case helps us find a proper limit and solve the problem.
L'Hôpital's rule simplifies complex derivatives and evaluations, ensuring we don't get stuck in indeterminate situations.

differential cross-section

The differential cross-section in quantum scattering describes how the scattering amplitude varies as a function of angle. It tells us the likelihood that a particle deflects into a particular direction following a scattering event.
The differential cross-section, often denoted as \( \frac{dσ}{dΩ} \), is given by the squared magnitude of the scattering amplitude \( f(θ) \):
\[ \frac{dσ}{dΩ} = |f(θ)|^2 \] Understanding this concept helps in visualizing how particles scatter and distribute over various angles. In low-energy quantum scattering, the differential cross-section becomes vital to explaining observed patterns.
This is especially useful when interpreting experimental or theoretical data about how particles behave under specific potentials. For example, in our exercise, we derive \( f(θ) \) at low energies and use that to find the corresponding differential cross-section.

low-energy quantum scattering

Low-energy quantum scattering describes scenarios where the kinetic energy of the particles involved is quite small. In this regime, the wavelengths of particles are relatively long. This usually simplifies the mathematical treatment of the problem, as we often use approximations valid for small parameters.
For the given problem, at low energies, \( k \) and thus \( K \) tend towards zero. This makes the calculations more manageable because complex trigonometric functions simplify into linear or near-linear forms. For instance, \( \text{sin}(Kb) \) approximately becomes \( Kb \), which is used in the given solution.
Low-energy regimes help us explore fundamental behavior without the complexities introduced by high kinetic energies. This provides essential insights into the inherent nature of particles and their interactions under weak potentials.

total cross-section calculation

The total cross-section combines all possible angles into a single quantity, providing an overall measure of scattering probability. Mathematically, it's the integral of the differential cross-section over all solid angles, usually denoted as:
\[ σ_{total} = 2π \bigg( \frac{2mV_0 b}{3 h^2} \bigg)^2 \bigg( \bigg)\integral_{0}^{\pi} \text{sin} \theta d\theta \bigg) \]
This expression takes into account the symmetry of the problem. The integral simplifies using properties of sine functions, just like in our step-by-step solution:
\[ \bigg( \frac{2mV_0 b}{3 h^2} \bigg)^2 \cdot 2π \bigg( 2\bigg) \] resulting in:
\[ 4π \bigg( \frac{2mV_0 b}{3 h^2} \bigg)^2 \]
By integrating over all angles, we ensure a comprehensive understanding of the scattering process, giving a full-picture insight into the behavior of particles under the potential. This total cross-section can be compared with experimental values to validate theoretical models.