More Problems

Problem 1:

A particle of mass m is moving in a one dimensional potential,
U(x) = ∞  for x < 0 and x > a,  U(x) = 0  for 0 < x < a/2,  U(x) = U₀  = π²ħ²/(4ma²) for a/2 < x < a.
(a)  Find the most general solution) of the eigenvalue equation HΦ₀(x) = E₀Φ₀(x) for the ground state wave function in all regions and apply boundary conditions.
(b)  Explain possible approaches to solving the resulting equation for the ground state energy, without actually carrying out any approaches.

Solution:

• Concepts:
  Square potentials
• Reasoning:
  This is a "square potential" problem.  We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions
• Details of the calculation:
  Φ₀(x) = 0 for x< 0 and x > a.
  region 1,  0 < x < a/2:  Φ₀(x) = A sin(k₁a). 
  region 2,  a/2 < x < a:  Φ₀(x) = B sin(k₂(a - x)).
  Boundary conditions:  Φ and ∂Φ/∂x  are continuous at x = a/2  This implies
  A sin(k₁a/2) = B sin(k₂(a/2)),  k₁A cos(k₁a/2) = k₂B cos(k₂a/2).
  k₁² = (2m/ħ²)E , k₂² = (2m/ħ²)(E - U₀) = k₁² - k₀² with k₀² = (2m/ħ²)U₀.
  tan(k₁a/2) = (1 - k₀²/k₁²)^½tan((k₁² - k₀²)^½a/2).
  This is one transcendental equation with one unknown.  If we do not want to solve it numerically we can start with the result from stationary perturbation theory, E = π²ħ²/(2ma²) + ½U₀ = 5π²ħ²/(4ma²),
  k₁² = 5π²/(2a²),  k₁a/2 = √(5/2) π/2, k₀a/2 = π/(2√2) and use these values in both sides if the equation.
  We can then change k₁ by a small amount ε, until we are satisfied with the accuracy and precision of the result.
  Once we have found k₁, we can solve for the ratio A/B. 

Problem 2:

A system is in a state described by ψ(r) = Nf(r)(5 Y₄₃(θ,φ) + Y₆₃(θ,φ) - 2 Y₆₀(θ,φ)).
(a)  If the square of the orbital angular momentum,L², is measured, what values can be obtained with what probabilities?
(b)  If L_z is measured, what values can be obtained with what probabilities?
(c)  What are the expectation values for L² and L_z for this system?

Solution:

• Concepts:
  The eigenfunctions of the orbital angular momentum operator, the spherical harmonics
• Reasoning:
  The common eigenfunctions of L² and L_z are the spherical harmonics. 
• Details of the calculation:
  (a)  ψ(r) = f(r)χ(θ). Normalize χ(θ).
  N_ang²∫dΩ [5 Y₄₃(θ,φ) + Y₆₃(θ,φ) - 2 Y₆₀(θ,φ)] [5 Y₄₃(θ,φ) + Y₆₃(θ,φ) - 2 Y₆₀(θ,φ)]* = 1. 
  N_ang² (25 + 1 + 4) = 1  N_ang = 1/√(30).
  Normalized χ(θ) = (5 Y₄₃(θ,φ) + Y₆₃(θ,φ) - 2 Y₆₀(θ,φ))/√(30).
  Values of L² that can be measured are l(l +1 )ħ².
  The probability of obtaining l = 4, L² = 20ħ² is 5/6.
  The probability of obtaining l = 6, L² = 42ħ² is 1/6.
  
  (b)  Values of L_z that can be measured are mħ.
  The probability of obtaining m = 3, L_z = 3ħ is 13/15.
  The probability of obtaining m = 0, L_z = 0 is 2/15.
  
  (c)  The expectation or average value for L² is (20ħ²)(5/6) + (42ħ²)(1/6) = 42ħ².
  The expectation or average value for L_z is (3ħ)(13/15) + 0*(2/15) = (13/5)ħ.

Problem 3:

A beam of neutral hydrogen atoms in the ground state, with speed v₀, passes through a series of two Stern-Gerlach (SG) devices whose magnetic field is directed along the z-axis -- we call them SGz devices. 
The first SGz device "transmits" (i.e., it lets pass through) particles with S_z = ħ/2 and filters out particles with S_z = -ħ/2.  The second SGz device transmits particles with S_z = -ħ/2 and filters out particles with S_z = ħ/2.  Between the two devices is a region of length l₀ in which there is a uniform magnetic field of magnitude B₀ pointing in the x-direction.   Determine the smallest value of l₀ such that exactly 25% of the particles transmitted by the first SGz device are transmitted by the second device. 
Express your result in terms of v₀ and of the Larmor frequency ω₀ = q_eB₀/m_e, where q_e is the magnitude of the electron charge. 

Note: 
Recall that the spin gyromagnetic factor for the electron is given by γ = -2μ_B/ħ, where μ_B = q_eħ/(2m_e) is the Bohr magneton.  For the electron μ = γS.   In this problem, neglect the magnetic moment of the proton.

Solution:

• Concepts:
  The two dimensional state space of a spin ½ particle, the evolution operator, the postulates of Quantum Mechanics
• Reasoning:
  In the region between the devices H = -μ_xB_x = -γS_xB₀ = (2μ_BB₀/ħ)S_x = ω₀S_x.
  The evolution operator is exp(-iHt/ħ).
• Details of the calculation:
  The matrix of S_x in the eigenbasis of S_z is
  

  S_x = (ħ/2)  

  	0	1
  	1	0

  .  

  
  

   

  The eigenvectors of S_x are |+>_x and |->_x.
  |+> = 2^-½(|+>_x + |->_x),  |-> = 2^-½(|+>_x - |->_x). 
  |ψ(0)> = |+>.
  |ψ(t)> = U(t,0)|+> = 2^-½U(t,0)(|+>_x + |->_x)
  = 2^-½(exp(-iω₀t/2)|+>_x + exp(iω₀t/2)|->_x)
  = ½[exp(-iω₀t/2)(|+> + |->) + exp(iω₀t/2)(|+> - |->)]
  = cos(ω₀t/2)|+> - isin(ω₀t/2)|->.
  We want P(S_z=-ħ/2,t) = sin²(ω₀t/2) =  0.25 for the smallest value of t = (l₀/v₀).
  Then ω₀t/2 = ω₀l₀/(2v₀) = π/6,  l₀ = πv₀/(3ω₀).