Problem 25 - Physics for Scientists and Engineers with Modern Physics

Question

(II) For a particle in a box with rigid walls, determine whether our results for the ground state are consistent with the uncertainty principle by calculating the product \(\Delta p \Delta x\). Take \(\Delta x \approx \ell\), since the particle is somewhere within the box. For \(\Delta p,\) note that although \(p\) is known \((=\hbar k),\) the direction of \(\overrightarrow{\mathbf{p}}\) is not known, so the \(x\) component could vary from \(-p\) to \(+p ;\) hence take \(\Delta p \approx 2 p\)

Step-by-Step Solution

Answer

The product \(\Delta p \Delta x = 2\hbar \pi\) is consistent with the uncertainty principle.

Step 1: Define the Solution Parameters

Given a particle in a box, the ground state energy is characterized by the wavelength equal to twice the box length, \(\lambda = 2\ell\). Moreover, the momentum \(p\) is defined as \(p = \hbar k\), where \(k = \frac{2\pi}{\lambda}\). Here, \(\Delta x \approx \ell\) and \(\Delta p \approx 2p\).

Step 2: Calculate Momentum \\(p\\)

For a particle in a one-dimensional box, the wavenumber \(k\) in the ground state corresponds to \(k = \frac{\pi}{\ell}\), because the wavefunction forms half a sine wave within the box. Therefore, \(p = \hbar \frac{\pi}{\ell}\).

Step 3: Determine \\(\Delta p\\)

Given that \(\Delta p \approx 2p\), substitute the expression of \(p\) from the previous step to obtain \(\Delta p = 2 \left( \frac{\hbar \pi}{\ell} \right) = \frac{2\hbar \pi}{\ell}\).

Step 4: Calculate the Product \\(\Delta p \Delta x\\)

Substitute \(\Delta x \approx \ell\) and \(\Delta p = \frac{2\hbar \pi}{\ell}\) into the expression for the product: \(\Delta p \Delta x = \left(\frac{2\hbar \pi}{\ell}\right)(\ell) = 2\hbar \pi\).

Step 5: Verify Against the Uncertainty Principle

The Heisenberg uncertainty principle states \(\Delta p \Delta x \geq \frac{\hbar}{2}\). Since \(2\hbar \pi> \frac{\hbar}{2}\), our result is consistent with the uncertainty principle.

Key Concepts

Uncertainty Principle

The uncertainty principle is a fundamental concept in quantum mechanics, introduced by physicist Werner Heisenberg. It states that certain pairs of physical properties, like position and momentum, cannot be known simultaneously to arbitrary precision. This means if we have a very precise knowledge of a particle's position, its momentum becomes very uncertain, and vice versa.

For a particle in a box, this principle suggests that even though we can estimate where the particle might be (within the box), the exact momentum of the particle remains uncertain. In the exercise involving a particle in a box with rigid walls, we used the concepts of uncertainty by calculating \(\Delta p \Delta x\) to ensure it aligns with the uncertainty principle.

• \(\Delta x\) represents the uncertainty in the particle's position, which in this scenario is approximated as the length of the box, \(\ell\).
• \(\Delta p\), the uncertainty in momentum, was derived considering that the momentum could vary from \(-p\) to \(+p\), leading to \(\Delta p \approx 2p\).

By calculating \(\Delta p \Delta x = 2\hbar \pi\), we verify that our understanding adheres to \(\Delta p \Delta x \geq \frac{\hbar}{2}\), as posited by the uncertainty principle.

Ground State Energy

In quantum mechanics, the concept of ground state energy refers to the lowest energy state of a quantum mechanical system. For a particle in a box, this ground state energy is linked to the wave characteristics of the particle.

The wavelength \(\lambda\) of the particle in its ground state is twice the length of the box. This relationship arises because the particle's wave function must constructively interfere within the box, forming a half sine wave that meets the boundary conditions of the box's walls.

• Because \(\lambda = 2\ell\), it follows that the wavenumber \(k = \frac{\pi}{\ell}\).
• The momentum of the particle in its ground state is then calculated as \(p = \hbar k = \frac{\hbar \pi}{\ell}\).

This description helps us understand the quantum nature of particles trapped in potential wells, like particles in a box, and explains why they cannot have zero energy (as they do classically).

Momentum Calculation

Computing the momentum of a particle in a quantum box involves understanding its wave-like properties. For particles, momentum depends on the wavelength of their associated wave function.

In the scenario of a particle in a one-dimensional box, the wave number \(k\) describes how many waves fit into the spatial confines of the box. For the ground state, the wavenumber is \(k = \frac{\pi}{\ell}\), matching the condition of having a half-wavelength filling the box length.

• This leads to a momentum defined as \(p = \hbar \frac{\pi}{\ell}\), using the relationship between wavenumber and wavelength.
• The wavelike nature requires that as the uncertainty of momentum \(\Delta p\) includes the full range of possible momentum directions, we write \(\Delta p \approx 2p\).

Understanding these calculations underscores the link between quantum mechanical principles and the macroscopic phenomena we observe, bridging our intuitive sense of particle behavior with the less intuitive nature of quantum objects.