Problem 1:

The Schroedinger equation for a particle in an isotropic harmonic oscillator potential in coordinate-space representation is

-[ħ²/(2m)]∇²ψ(r,t) + ½mω²r²ψ(r,t) = iħ(∂/∂t)ψ(r,t).

(a)  What is the corresponding Schroedinger equation in momentum-space representation?
(b)  Write down the ground state wave function ψ(r,t) for the particle.
(c)  Find the ground state wave function of the particle Φ(r,t) in momentum space.

Solution:

• Concepts:
  The r and p representation, the 3D isotropic harmonic oscillator
• Reasoning:
  The Hamiltonian operator is H = P²/2m + ½mω²R², where R and P are the position and momentum operator, respectively.
• Details of the calculation:
  (a)  <r|R|ψ> = rψ(r), <r|P|ψ> = (ħ/i)∇ψ(r).
  <p|P|ψ> = pΦ(p), <p|R|ψ> = (iħ)∇_pΦ(p).
  ∇_p = i ∂/∂p_x + j ∂/∂p_y + k ∂/∂p_z.
  The given Schroedinger equation in momentum-space representation therefore is
  [P²/(2m)]Φ(p,t) - ½ħ²mω²∇_p²Φ(p,t) = iħ(∂/∂t)Φ(p,t).
  (b)  ψ(r,t) = N exp(-mωr²/(2ħ)) exp(iE₀t/ħ).  N = (mω/(πħ))^3/4.  E₀ = (3/2)ħω.
   ψ(r,t) = (mω/(πħ))^3/2 exp(-mωr²/(2ħ)) exp(i3ωt/2).
  (c)  Fourier transform:  ψ(r) =  ψ(r) = N exp(-mωr²/(2ħ)).
  Φ(p) = Φ(p) = (2πħ)^-3/2∫d³r ψ(r)exp(-ip∙r/ħ).
  Φ(p) = N(2πħ)^-3/2∫dx exp(-ax²)exp(-ip_xx/ħ)∫dy exp(-ay²)exp(-ip_yy/ħ)∫dz exp(-az²)exp(-ip_zz/ħ)
  ∫dx exp(-ax²)exp(-ik_xx)
  =∫dx exp(-(ax² + ik_xx - k_x²/(4a) + k_x²/(4a)))      (completing the square)
  = exp(-k_x²/(4a))∫_-∞^+∞ exp(-(ax² + ik_xx  - k_x²/(4a))dx
  = a^-1/2exp(-k_x²/(4a))∫_-∞^+∞ exp(-x'²) dx' = (π/a)^1/2exp(-k_x²/(4a)).
  Therefore Φ(p) = (mω/(πħ))^3/4(2πħ)^-3/2(2ħπ/mω)^3/2exp(-ħk²/(2mω)).
  Φ(p,t) = (πħmω)^-3/4exp(-p²/(2ħmω))exp(i3ωt/2) is the ground space wave function in momentum space representation.

Problem 2:

(a)  Find the energy eigenfunctions and energy levels for a spinless particle confined to a two-dimensional rectangular box, with |x| < a and |y| < b.
Do not just write down your answer, but derive it.  Justify each step.
(b)  Make a diagram or table showing the 5 lowest energy levels and the degeneracies for b = a and for b = 2a.

Solution:

• Concepts:
  The particle in a 2-dimensional box, separation of variables
• Reasoning:
  We are asked to find the eigenfunctions and eigenvalues of the Hamiltonian of a particle in a 2D box.
• Details of the calculation:
  Let x' = x + a and y ' = y + b.
  U(x',y') = 0 if 0 < x' < 2a AND 0 < y' < 2b,  U(x',y') = infinite otherwise.
  In Cartesian coordinate the 2D time-independent Schroedinger equation is 
  (-ħ²/(2m))[∂²Φ(x',y')/∂x'² + ∂²Φ(x',y')/∂y'²] + U(x',y')Φ(x',y')  = EΦ(x',y').

  We usually try to solve such equations by a technique called separation of variables.
  For a particle trapped in a rectangular "infinite well" the potential is 0 inside the well and infinite outside the well.  In the region where the potential is zero we solve the Schroedinger equation by trying a solution of the form Φ(x',y') = X(x')Y(y').
  Then
  (-ħ²/(2m))[Y(y')∂²X(x')/∂x'² + X(x') ∂²Y(y')/∂y'²]  = E X(x')Y(y'). 
  X(x')^-1∂² X(x')/∂x'² + Y(y')^-1∂²Y(y')/∂y'² + (2m/ħ²)E = 0.
  The first term is a function of x' only and the second term is a function of y' only.  Both terms must be equal to constants and the sum of these constants must be equal to -(2m/ħ²)E.
  Write E = E_x' + E_y'.
  ∂²X(x')/∂x'² + (2m/ħ²)E_x'X(x') = 0.
  ∂²Y(y')/∂y'² + (2m/ħ²)E_y'Y(y') = 0.
  The boundary conditions are
  X(x') = 0 at x' = 0 and x' = 2a.
  Y(y') = 0 at y' = 0 and y' = 2b.
  The normalized solutions to the differential equations are
  X(x') = (1/a)^½sin(n_xπx'/(2a)),  Y(y') = (1/b)^½sin(n_yπy'/(2b)),
  n_x = 1, 2, 3, ...  ,  n_y = 1, 2, 3, ...  . 
  E_nx = n_x²π²ħ²/(8ma²),  E_ny = n_y²π²ħ²/(8mb²).

  The eigenfunctions for the problem therefore are
  Φ_nx,ny(x',y') = (1/(ab))^½sin(n_xπx'/(2a)) sin(n_yπy'/(2b)),
  with the corresponding eigenvalues E_nx,ny = [π²ħ²/(8m)][(n_x/a)² + (n_y/b)²].
  In terms of x and y the eigenfunctions are
  Φ_nx,ny(x,y) = (1/(ab))^½sin(n_xπx/(2a) + n_xπ/2) sin(n_yπy/(2b) + n_yπ/2).

  (b)  If b = a then E_nx,ny = [π²ħ²/(8ma²)][n_x² +  n_y²].

  level	nx	ny	nx2 + ny2	degeneracy
  E1	1	1	2	1
  E2	1 2	2 1	5	2
  E3	2	2	8	1
  E4	1 3	3 1	10	2
  E5	2 3	3 2	13	2

  
  If b = 2a then E_nx,ny = [π²ħ²/(32ma²)][4n_x² + n_y²].
   

  level	nx	ny	4nx2 + ny2	degeneracy
  E1	1	1	5	1
  E2	2	2	8	1
  E3	1	3	13	1
  E4	2	1	17	1
  E5	2 1	2 4	20	2

Problem 3:

The Rydberg constant, R_H = 109737.568525/cm is one of the most accurately known fundamental constants.
(a)  Find the wave number of the Balmer alpha line (n = 3 to n' = 2) in atomic hydrogen.  Neglect fine structure.
(b)  Is the Balmer alpha line in atomic deuterium shifted towards the blue or towards the red compared to normal hydrogen?
(c)  Calculate the shift in wave number between deuterium and hydrogen.

Solution:

• Concepts:
  The hydrogen atoms, reduced mass
• Reasoning:
  The deuterium is a hydrogen atom with a slightly different reduced mass than ordinary hydrogen.
• Details of the calculation:
  (a)  For ordinary hydrogen:  E_n = -μe⁴/(2ħ²n²) with μ = m_pm_e/(m_p + m_e).
  E_n - E_n' = [μe⁴/(2ħ²)](1/n'² - 1/n²) = hν = hc/λ = hcR_H(1/n'² - 1/n²).
  So R_H = [2πμe⁴/(2ħ³c)].
  For the wave number we have 1/λ = R_H(1/3² - ½²) = 15241.328952 cm^-1.
  [Note: There are unfortunately two different definitions of the wave number, 1/λ or 2π/λ.  Clearly state which definition you are using.]
  (d)  For deuterium:  E_n = -μ'e⁴/(2ħ²n²) with μ' = (m_p + m_n)m_e/(m_p + m_n + m_e).
  R_D = R_H μ'/μ > R_H.
  So 1/λ increases, λ decreases, the line is blue-shifted.
  (c)  1/λ' - 1/λ  = (R_D - R_H)(1/3² - ½²) = (μ'/μ - 1)R_H(1/3² - ½²) = 4.15 cm^-1.

Problem 4:

A negative K meson with mass m = 1000 electron masses is captured into a circular Bohr orbit around a lead nucleus (Z = 82).  Assume it starts with principal quantum number n = 10 and then cascades down through n = 9, 8, 7, ... etc.
(a)  What is the energy of the photon emitted in the n = 10 to n = 9 transition?
(b)  What is the approximate radius of the lead nucleus if no further quanta are observed after the n = 4 to n = 3 transition (because of nuclear absorption of the K meson)?

Solution:

• Concepts:
  The hydrogenic atom, the ground-state energy of the hydrogen atom, the Bohr radius
• Reasoning:
  The lead nucleus and the meson form a hydrogenic atom.  The energy levels and the average distance of the meson from the nucleus can be determined from the ground-state binding energy of the hydrogen atom and the Bohr radius.  The photon emitted in a transition has an energy equal to the energy difference between the initial and the final state.  When the average distance of the meson from the nucleus in the final state becomes of the order of the radius of the nucleus, then a nuclear reaction becomes likely, and the hydrogenic atom will be destroyed.
• Details of the calculation:
  (a)  We have a hydrogen like system with Z = 82.
  E_n = E_I'/n²,  E_I' = E_I(μ'/μ)Z².
  μ = 1000 m_em_nucleus/(1000 m_e + m_nucleus) ≈ 1000 m_e.  μ'/μ = 1000.  E_I = 13.6 eV.
  E₁₀ - E₉ = 82²*13.6*1000* (1/9² - 1/10²) eV = 2.2*10⁵ eV.
  
  (b)  The radius of the Bohr orbit is r_n = n²a₀' = n²(a₀/Z)(μ/μ') = 9a₀/(82*1000) = 5*10^-15 m.
  This is the approximate radius of the lead nucleus.