Problem 1:

Calculate the probability current density for of a freely moving particle whose quantum state is described by a de Broglie plane wave.

Solution:

• Concepts:
  The Schroedinger equation, ρ(x,t) = |ψ(x,t)|².
• Reasoning:
  Using the Schroedinger equation we can find an expression for ∂ρ(r,t)/∂t in terms of space derivatives.
• Details of the calculation:
  Let the plane wave propagate in the positive x-direction.  Then we can work in one dimension.
  The probability per unit length of finding a particle described by the normalized wave function ψ(x,t) is given by dP(x,t)/dx = |ψ(x,t)|².
  The total probability of finding the particle at time t anywhere in space is ∫_{all space} |ψ(x,t)|²dx = <ψ|ψ> = 1.
  Local conservation of a classical quantity is usually expressed through the equation ∇∙j = -(∂/∂t)ρ.  Here ρ(r,t) is the volume density and j(r,t) is the current density.
  In one dimension this becomes ∂j(x,t)/∂x = = -(∂/∂t)ρ(x,t).
  Assuming local conservation of probability we write
  ∂ρ/∂t = ∂ψ^*ψ/∂t = ψ^*∂ψ/∂t + ψ∂ψ^*/∂t
  = (1/(iħ))[ψ^*((-ħ²/(2m))∂²ψ/∂x² + Vψ) - ((-ħ²/(2m))∂²ψ/∂x² + Vψ)^*ψ]
  = (-ħ/(2im)[ψ^*∂²ψ/∂x² - (∂²ψ/∂x²)^*ψ], since V is real.
  [We use iħ∂ψ/∂t = Hψ,  ∂ψ/∂t = (1/(iħ))Hψ,  ∂ψ^*/∂t = (-1/(iħ))(Hψ)^*.]
  ∂ρ/∂t = (-ħ/(2im)(∂/∂x)[ψ^*∂ψ/∂x - ψ∂ψ^*/∂x] = -∂j(x,t)/∂x.
  j(x,t) = (-ħ/(2im)[ψ^*∂ψ/∂x - ψ∂ψ^*/∂x] = (-ħ/(m) Im(ψ^*∂ψ/∂x).
  We interpret ρ(x,t) as the probability density and j(x,t) as the probability current density.
  For the plane wave wave function of a free particle ψ(x,t) = Aexp(ikx - ωt) we have 
  (ħ/i)∂ψ(x,t)/∂x = ħkψ(x,t) = p_xψ(x,t).
  ψ*∂ψ/∂x =  ik|ψ(x,t)|²,  ψ∂ψ*/∂x =  ik|ψ(x,t)|²,
  j(x,t) = (ħk/m)|ψ(x,t)|² = (ħk/m)|A|².
  Since the eigenfunctions of the operator p_x are not square integrable, we cannot normalize the wave function in the traditional way.

Problem 2:

Find <x> and ∆x for the nth stationary state of a free particle in one dimension restricted to the interval 0 < x < a.  Show that as n --> ∞ these become the classical values.

Solution:

• Concepts:
  The eigenfunctions of the infinite square well, the postulates of quantum mechanics
• Reasoning:
  The expression for the mean value of an observable x in the normalized state |ψ> is
  <x> = <ψ|x|ψ>.  The root mean square deviation ∆x is given by 
  ∆x = (<x²> - <x>²)^½.  Here ψ_n(x) = (2/a)^½sin(nπx/a),  E_n = n²π²ħ²/(2ma²).
• Details of the calculation:
  <x>_n = (2/a)∫₀^a x sin²(nπx/a)dx = a/2.
  <x²>_n = (2/a)∫₀^a x² sin²(nπx/a)dx = (2/a)(a/(nπ))³∫₀^nπ x² sin²(x)dx
  = (2a²/(nπ)³)(nπ)³/6 - (2a²/(nπ)³)nπ/4 = a²/3 - ½a²/(nπ)².
  ∆x_n = (a²/12 - ½a²/(nπ)²)^½  --> a/12^½ as n --> ∞.
  
  Classically, the particle is moving back and forth in the well.  It is equally likely to be found at any position.
  The average position x_avg is a/2.
  For x²_avg we have x²_avg = a²/3.  (∫₀^a P(x) x²dx  with P(x) = 1/a.)
  If we define ∆x = (x²_avg - x_avg²)^½ we have ∆x = a/12^½.

Problem 3:

Consider a system of N particles with only 3 possible energy levels separated by ε.  Let the ground state energy be zero.  The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at temperature T.  Ignore interactions between particles and assume Boltzmann statistics apply.

(a)  What is the partition function for a single particle in the system?
(b)  What is the average energy per particle?
(c)  What is the probability that the 2ε level is occupied in the high-temperature limit k_BT >> ε?
Explain your answer on physical grounds.
(d)  What is the average energy per particle in the high-temperature limit k_BT >> ε?
(e)  At what energy is the ground state 1.1 times as likely to be occupied as the 2ε level?
(f)  Find the heat capacity C_V of the system, analyze the low-T (k_BT >> ε) and high-T (k_BT >> ε) limit, and sketch CV as a function of T.

Solution:

• Concepts:
  Boltzmann statistics
• Reasoning:
  In thermodynamic equilibrium P(E_i) ∝ g_iexp(-E_i/(kT).
  Here the degeneracy g_i is 1 for all i.
• Details of the calculation:
  (a)  P(E) = Cexp(-E/(k_BT)).
  Σ_jP(E_j) = CΣ_jexp(-E_j/(k_BT)) = 1, C = 1/Σ_jexp(-E_j/(k_BT)) = 1/Z.
  Let β = 1/(k_BT)).
  Z = 1 + exp(-βε) + exp(-2βε).
  (b)  <ε> = (ε exp(-βε) + 2ε exp(-2βε))/Z
  = ε(exp(-βε) + 2exp(-2βε))/(1 + exp(-βε) + exp(-2βε)).
  (c)  P =  exp(-2βε)/(1 + exp(-βε) + exp(-2βε)) ~ (1 - 2βε)/( 1 + 1 - βε + 1 - 2βε) = 1/3.
  In the high energy limit the probability is the same for all three levels.
  (d)  <ε> = 0/3 + ε/3 + 2ε/3 = ε.
  (e)  exp(-2βε) = 1/1.1,  2βε ln(1.1),  T = 2ε/(k_B*ln(1.1)).
  (f)  C_V = dU/dT = Nd<ε>/dT = Nd<ε>/dβ)(dβ/dT).
  C_V = (Nε²/(k_BT²))[ exp(-βε) + 4 exp(-2βε) + exp(-3βε)]/[1 + exp(-βε) + exp(-2βε)]².
  Low temperature limit:  C_V = (Nε²/(k_BT²)) exp(-ε/(k_BT)).  (C_V --> 0 as T --> 0)
  High temperature limit:  C_V = (2/3)(Nε²/(k_BT²)).  ( C_V --> 0 as T --> ∞)
  
  

Problem 4:

A river of width D flows on the northern hemisphere at a geographical latitude φ toward the north with a certain flow speed v₀.  By which amount is the right bank higher that the left one?
First apply the equation of motion in a non-inertial frame to the problem at hand, and then use D = 2 km, v₀ = 5 km/h, and φ = 45^o to find the super-elevation of the river for the given parameters.

Solution:

• Concepts:
  Motion in an accelerating frame
• Reasoning:
  
  Assume an observer in the rotating frame at mid latitudes φ = 90^o - θ.
  The equations of motion in a rotating coordinate system contain fictitious forces.
  mdv/dt = F_inertial - mΩ × (Ω × r) - 2mΩ × v.
• Details of the calculation:
  (a)  Consider a volume element dV of the water with mass m = ρdV.
  For this volume element we have
  -mΩ × (Ω × r) = centrifugal force.
  (Ω × r) = ΩRsinθ j.  Ω × (Ω × r) = -Ω²Rsin²θ  k - Ω²Rsinθcosθ  i.
  -2mΩ × v = Coriolis force.
  Ω × v = -Ωvcosθ j.
  For each volume element dV of the water we want dv_y/dt = 0.
  We therefore need 0 = F_{y inertial}  + 2ρdVΩvcosθ.
  F_{y inertial}  = -2ρdVΩvcosθ.
  F_{y inertial} is a force on ρdV due to a horizontal pressure gradient, if the banks have different heights.
  [P(y + dy) - P(y)] dA = -2ρdAdyΩvcosθ.
  dP = -2ρdyΩvcosθ,  ρgdh = -2ρdyΩvcosθ,  dh/dy = -2Ωvcosθ/g. 
  (b)  Ω = 2π/(24*3600 s), |dh/dy| = 1.46*10^-5.
  Δh = 2000*1.46*10^-5 m ~ 3 cm.
  The right bank is 3 cm higher that the left one.

Problem 5:

Two equal mass particles are connected by a spring and are executing simple harmonic motion when they are placed in a horizontal, frictionless open pipe rotating at a constant angular velocity Ω about a vertical axis through its midpoint. The midpoint of the spring is initially at the midpoint of the pipe, but neither the particles nor the spring are attached to the pipe.
(a)  Find the maximum angular velocity Ω for which the particles stay in the pipe.
(b)  Assume (a) is satisfied and the particles stay in the pipe.  Find their positions as a function of time.

Solution:

• Concepts:
  Motion in a non-inertial frame
• Reasoning:
  For a uniformly rotating frame the equations of motion for a particle with mass m is
  mdv/dt = F_inertial - 2mΩ × v - mΩ × (Ω × r).
• Details of the calculation:
  (a)  Chose coordinate in the rotating frame.
  
  Let x₀ be the equilibrium length of the spring.  Let x₁ and x₂ be the coordinate of the particles in the rotating frame.
  The equations of motions for the particles constrained to move in the pipe are
  md²x₁/dt² = -k(x₁ - x₂ - x₀) + mΩ²x₁,
  md²x₂/dt² = k(x₁ - x₂ - x₀) + mΩ²x₂.
  Introduce new coordinates.
  y₁ = (x₁ + x₂)/2,  y₂ = (x₁ - x₂)/2.  Then
  d²y₁/dt² = Ω²y₁,  d²y₂/dt² = Ω²y₂ - (2k/m)y₂ + (2k/m)x₀.
  For the initial conditions x₁ = -x₂, , dx₁/dt = -dx₂/dt, we have that y₁(t) = 0, for all t, the midpoint of the spring stays at the midpoint of the pipe. (unstable equilibrium)
  Since x₁ = -x₂, the equation describing the motion of x₁ = y₂ is
  d²y₂/dt² = -(2k/m - Ω²)y₂ + (2k/m)x₀.
  If 2k/m > Ω², then y₂ = C cos(ωt + φ) + y₂₀,
  with ω² = 2k/m - Ω² and y₂₀ = (2k/m)x₀/(2k/m - Ω²) = x₀/(1 - ½mΩ²/k).
  The particle can stay in the pipe if Ω < (2k/m)^½, Ω = ±Ω k.
  We also need y₂₀ + C = x₀/(1 - ½mΩ²/k) + C < L, where L is the length of the pipe.
  (b)  x₁ = - x₂ = ½C cos(ωt + φ) + ½y₂₀.  The constants C and φ are determined by the initial conditions.