Question

For a non-relativistic free particle, the ratio of phase velocity to group velocity is

• A. 2
• B. \(\frac{1}{2}\)
• C. 1
• D. \(\frac{1}{4}\)

Hint:

Remember the fundamental definitions: \(v_p = \omega/k\) and \(v_g = d\omega/dk\). For any power-law dispersion relation of the form \(\omega \propto k^n\), the ratio is \(\frac{v_p}{v_g} = \frac{\omega/k}{d\omega/dk} = \frac{Ak^n/k}{nAk^{n-1}} = \frac{Ak^{n-1}}{nAk^{n-1}} = \frac{1}{n}\). For a non-relativistic particle, \(n=2\), so the ratio is \(1/2\). For light in a vacuum, \(n=1\), and the ratio is 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a quantum mechanical particle, we associate a wave packet. The phase velocity (\(v_p\)) is the speed of an individual wave crest within the packet, while the group velocity (\(v_g\)) is the speed of the overall envelope of the wave packet. The group velocity corresponds to the classical particle velocity. We need to find the relationship between these two velocities for a non-relativistic free particle.
Step 2: Key Formula or Approach:
The key relationships are:

Energy of a non-relativistic free particle: \(E = \frac{p^2}{2m}\), where \(p\) is the momentum and \(m\) is the mass.
De Broglie relations: \(E = \hbar \omega\) and \(p = \hbar k\), where \(\omega\) is the angular frequency and \(k\) is the wave number.
Phase velocity: \(v_p = \frac{\omega}{k}\)
Group velocity: \(v_g = \frac{d\omega}{dk}\)
Step 3: Detailed Explanation:
1. Find the dispersion relation (\(\omega\) as a function of \(k\)): Start with the energy-momentum relation for a non-relativistic free particle: \[ E = \frac{p^2}{2m} \] Substitute the De Broglie relations \(E = \hbar \omega\) and \(p = \hbar k\): \[ \hbar \omega = \frac{(\hbar k)^2}{2m} = \frac{\hbar^2 k^2}{2m} \] Solving for \(\omega\), we get the dispersion relation: \[ \omega(k) = \frac{\hbar k^2}{2m} \] 2. Calculate the phase velocity (\(v_p\)): \[ v_p = \frac{\omega}{k} = \frac{1}{k} \left( \frac{\hbar k^2}{2m} \right) = \frac{\hbar k}{2m} \] 3. Calculate the group velocity (\(v_g\)): \[ v_g = \frac{d\omega}{dk} = \frac{d}{dk} \left( \frac{\hbar k^2}{2m} \right) = \frac{\hbar}{2m} (2k) = \frac{\hbar k}{m} \] 4. Find the ratio of phase velocity to group velocity: \[ \frac{v_p}{v_g} = \frac{\frac{\hbar k}{2m}}{\frac{\hbar k}{m}} = \frac{\hbar k}{2m} \cdot \frac{m}{\hbar k} = \frac{1}{2} \] Alternative approach using classical velocity (v): The classical velocity of the particle is \(v = \frac{p}{m}\). The group velocity is equal to the classical particle velocity: \(v_g = \frac{dE}{dp} = \frac{d}{dp}(\frac{p^2}{2m}) = \frac{2p}{2m} = \frac{p}{m} = v\). The phase velocity is \(v_p = \frac{E}{p} = \frac{p^2/2m}{p} = \frac{p}{2m} = \frac{v}{2}\). The ratio is \(\frac{v_p}{v_g} = \frac{v/2}{v} = \frac{1}{2}\). Step 4: Final Answer:
The ratio of phase velocity to group velocity for a non-relativistic free particle is \(\frac{1}{2}\). Therefore, option (B) is correct.