Let's go step-by-step for each sub-question: --- ### **a) Velocity Before and After Collision in Wall's Frame** Let: - Particle velocity before collision in lab frame: \( v_1 \) - Wall moves with velocity \( u \) (assume \( u > 0 \), to right, and \( u \ll v_0 \)) - Particle velocity before collision in wall's frame: \( v_1' = v_1 - u \) - After **elastic** collision, in wall's frame: \( v_2' = -v_1' \) - Back to lab frame: \( v_2 = v_2' + u \) **Summary:** - Before: \( v_1' = v_1 - u \) - After: \( v_2' = -v_1' = -(v_1-u) = -v_1 + u \) --- ### **b) Change in Velocity Over the Collision** In the **wall's frame**: \[ \Delta v' = v_2' - v_1' = [-(v_1-u)] - (v_1-u) = -2(v_1-u) \] In the **lab frame**: \[ \Delta v = v_2 - v_1 = [(-v_1 + u) + u] - v_1 = -v_1 + 2u - v_1 = -2v_1 + 2u \] --- ### **c) What is \( v_2 \) in the Lab Frame?** From above: \[ v_2 = v_2' + u = (-v_1 + u) + u = -v_1 + 2u \] If initially \( v_1 = v_0 \): \[ v_2 = -v_0 + 2u \] --- ### **d) Next Collision with Stationary Wall** After bouncing off the moving wall, the particle has \( v_2 \) in the lab frame. It travels to the stationary wall and elastically bounces back: - Before collision: \( v_2 \) - After collision: \( v_3 = -v_2 \) (since the wall is at rest and collision is elastic) So, \[ v_3 = -v_2 = -(-v_0 + 2u) = v_0 - 2u \] --- ### **e) Average Acceleration Over the Cycle** The change in velocity over the cycle is: - Starts with \( v_0 \) - After one round trip (collides with moving wall, then stationary wall): ends with \( v_0 - 2u \) - Change in velocity: \( \Delta v = (v_0 - 2u) - v_0 = -2u \) Time for the cycle: \( \Delta t \approx \frac{2L}{v_0} \) (since \( u \ll v_0 \), length doesn't change much) So average acceleration: \[ a = \frac{\Delta v}{\Delta t} = \frac{-2u}{2L/v_0} = -\frac{u v_0}{L} \] --- ### **f) Adiabatic Invariant \( L^2 E \) in Classical and Quantum Cases** **Classical:** \( E = \frac{1}{2} m v^2 \) We want to show \( \frac{d}{dt}(L^2 E) \approx 0 \) if \( \dot{L} \ll v_0 \). Let’s compute: \[ \frac{d}{dt}(L^2 E) = 2L \dot{L} E + L^2 \frac{dE}{dt} \] During slow wall motion, the changes in \( E \) and \( L \) compensate so that \( L^2 E \) is (approximately) conserved—this is the classical adiabatic invariant for a slowly-varying 1D box. **Quantum:** Energy levels: \[ E_n = \frac{n^2 \pi^2 \hbar^2}{2m L^2} \] So, \[ L^2 E_n = L^2 \cdot \frac{n^2 \pi^2 \hbar^2}{2m L^2} = \frac{n^2 \pi^2 \hbar^2}{2m} \] This is **constant** for a given quantum number \( n \). --- ## **Summary of Answers** **a)** - Before: \( v_1' = v_1 - u \) - After: \( v_2' = -v_1' = -v_1 + u \) **b)** - \( \Delta v = v_2 - v_1 = -2v_1 + 2u \) **c)** - \( v_2 = -v_0 + 2u \) **d)** - \( v_3 = v_0 - 2u \) **e)** - Average acceleration: \( a = -\frac{u v_0}{L} \) **f)** - \( L^2 E \) is approximately conserved (adiabatic invariant) for slow wall motion. - In quantum case, \( L^2 E_n = \frac{n^2 \pi^2 \hbar^2}{2m} \) is constant for each quantum state \( n \). --- Let me know if you'd like clarification or further explanation on any step!

Certainly! Let's go through each sub-question carefully with detailed explanations, step-by-step calculations, and final answers. --- ### **Problem Context Recap:** - A particle in a 1D box of length \( L \). - The box walls are perfectly elastic; when the particle hits a wall, its velocity reverses direction, conserving magnitude. - At \( t=0 \), the particle has velocity \( v_0 \). - The right wall starts moving very slowly to the right with velocity \( u \) (where \( u \ll v_0 \)), so the box length becomes time-dependent: \( L(t) \). --- ### **a) Particle’s velocity in the moving wall's frame before and after collision** **Goal:** Find the particle's velocity relative to the moving wall **before** and **after** collision. --- **Step 1:** Define velocities: - Particle's velocity in the **lab frame**: \( v_1 \) (initially \( v_0 \)) - Wall's velocity in the lab frame: \( u \) (to the right) --- **Step 2:** Convert to the wall's frame: \[ v'_1 = v_1 - u \] - The particle's velocity relative to the moving wall **before** collision is \( v'_1 \). --- **Step 3:** Elastic collision in the wall's frame: - Since the collision is elastic and the wall is massive (or infinite mass), the particle's velocity reverses in the wall's frame: \[ v'_2 = -v'_1 \] --- **Step 4:** Convert back to lab frame: \[ v_2 = v'_2 + u = -v'_1 + u = - (v_1 - u) + u = -v_1 + u + u = -v_1 + 2u \] **Answer:** - **Before collision:** \( v'_1 = v_1 - u \) - **After collision:** \( v_2 = -v_1 + 2u \) --- ### **b) Change in the particle's velocity over the collision** \[ \Delta v = v_2 - v_1 \] Using the above: \[ \Delta v = (-v_1 + 2u) - v_1 = -2v_1 + 2u \] If the initial velocity is \( v_1 = v_0 \): \[ \boxed{ \Delta v = -2v_0 + 2u } \] **This represents the change in velocity of the particle after the collision.** --- ### **c) Velocity \( v_y \) of the particle in the lab frame** If initially the particle has velocity \( v_0 \) in the lab frame, then **after** the collision: \[ v_y = v_2 = -v_0 + 2u \] --- ### **d) Velocity after hitting the stationary wall** **Sequence:** - The particle hits the moving wall, velocity becomes: \[ v_{\text{after moving wall}} = v_2 = -v_0 + 2u \] - It then travels to the **stationary wall**, located at the other end; during this travel, velocity remains \( v_2 \). - Upon collision with the stationary wall, the particle reverses velocity elastically: \[ v_{\text{after stationary wall}} = -v_2 \] Substituting \( v_2 \): \[ v_{\text{after stationary wall}} = -(-v_0 + 2u) = v_0 - 2u \] **Final velocity after the full cycle:** \[ \boxed{ v_{f} = v_0 - 2u } \] --- ### **e) Average acceleration \( a = \frac{\Delta v}{\Delta t} \) over the cycle** **Step 1:** Initial velocity: \( v_0 \) **Step 2:** Final velocity after a full cycle: \[ v_f = v_0 - 2u \] **Step 3:** Change in velocity: \[ \Delta v = v_f - v_0 = -2u \] **Step 4:** Time taken for one cycle: - The particle travels a distance \( 2L \) (from one wall to the other and back). - Speed during travel is approximately \( v_0 \). \[ \Delta t \approx \frac{2L}{v_0} \] **Step 5:** Average acceleration: \[ a = \frac{\Delta v}{\Delta t} = \frac{-2u}{2L/v_0} = -\frac{2u v_0}{2L} = -\frac{u v_0}{L} \] **Answer:** \[ \boxed{ a = - \frac{u v_0}{L} } \] This indicates the particle's velocity decreases on average when the wall moves slowly to the right. --- ### **f) Adiabatic invariant \( L^2 E \) in classical and quantum cases** --- **Classical case:** - Particle's kinetic energy: \[ E = \frac{1}{2} m v^2 \] - The quantity: \[ L^2 E \] - **Claim:** This quantity remains approximately constant during slow (adiabatic) changes of \( L \). **Justification:** - When \( L \) changes slowly (\( \dot{L} \ll v \)), the action variable associated with the system is conserved. - For the particle-in-box, the classical adiabatic invariant is proportional to \( L \times p \), or more precisely, \( L^2 E \) remains constant because the energy scales as \( 1/L^2 \). --- **Quantum case:** - Quantized energy levels: \[ E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2} \] - Multiply by \( L^2 \): \[ L^2 E_n = \frac{n^2 \pi^2 \hbar^2}{2 m} \] which **does not depend on \( L \)**; it depends only on quantum number \( n \). **Conclusion:** \[ \boxed{ L^2 E_n = \text{constant for fixed } n } \] This demonstrates that in quantum mechanics, \( L^2 E \) is an **exact invariant** for the adiabatic change of the box size. --- ## **Summary of Final Answers** | Sub-question | Final Answer | |----------------|--------------| | **a)** | Before: \( v'_1 = v_1 - u \); After: \( v_2 = -v_1 + 2u \) | | **b)** | \( \Delta v = -2v_0 + 2u \) | | **c)** | \( v_y = -v_0 + 2u \) | | **d)** | Velocity after hitting stationary wall: \( v_f = v_0 - 2u \) | | **e)** | Average acceleration: \(\boxed{a = - \frac{u v_0}{L}}\) | | **f)** | \( L^2 E \) is approximately conserved; in quantum case, \( L^2 E_n = \frac{n^2 \pi^2 \hbar^2}{2 m} \) (constant for fixed \( n \)). | --- If you'd like further clarification on any step, feel free to ask!