Problem 1:

The potential energy of the nuclei of a diatomic molecule as a function of their separation r is given by
U(r) = -2D[a₀/r - a₀²/r²].
Here D is a constant with units of energy.
Approximate this potential energy function near its minimum by a harmonic oscillator potential energy function and determine the vibrational energy levels of the molecule with zero angular momentum.

Solution:

• Concepts:
  Motion of a fictitious particle of reduced mass μ in a spherically symmetric potential
• Reasoning:
  We have a spherically symmetric potential energy function U(r).
  The wave function ψ_klm(r,θ,φ) = R_kl(r)Y_lm(θ,φ) = [u_kl(r)/r]Y_lm(θ,φ) is a product of a radial function R_kl(r) and the spherical harmonic Y_lm(θ,φ).  The differential equation for u_kl(r) is
  [-(ħ²/(2μ))(∂²/∂r²) + ħ²l(l + 1)/(2μr²) + U(r)]u_kl(r) = E_klu_kl(r).
  When l = 0 we have [-(ħ²/(2μ))(∂²/∂r²) + U(r)]u_kl(r) = E_klu_kl(r).
• Details of the calculation:
  Find the location of the extrema:  dU/dr = 0,  -2D[-a₀/r² + 2a₀²/r³] = 0,  r = 2a₀ = r₀.
  There is only one extremum.  As r --> 0 U(r) --> ∞ and as r  --> ∞ U(r) --> 0.  There is only one extremum, it must be a minimum.
  Near r₀ we write  U(r) ≈ U(r₀) + ½ d²U(r)dr²|_r0(r - r₀)²
  d²U(r)dr²|_r0 =  - 2D[2a₀/(2a₀)³ - 6a₀²/(2a₀)⁴] = -¼D/a₀².
  U(r) = -U₀ + ½ μω²r^'2, where r' = r - 2a₀, U₀ = D/2, and ω² =  D/(4μa₀²).
  The vibrational energy levels of the molecule with l = 0 are E_ν = -U₀ + (ν + ½)ħω.
  E_ν = -D/2 + (ν + ½)ħ(D/(4μa₀²)^½,  ν = 0, 1, 2, ... .

Problem 2:

Find the rotational energy levels of a diatomic molecule with atoms of mass m₁ and m₂.  Take r₁ and r₂ to be the distances from atoms 1 and 2 to the center of mass.

Solution:

• Concepts:
  The rigid rotator
• Reasoning:
  We model the system as a rigid rotator.
• Details of the calculation:
  Considering only the rotational motion about the CM, the Hamiltonian of the molecule is L²/(2I) where I is its moment of inertia,
  I = m₁r₁² + m₂r₂² = μr²,  μ = m₁m₂/(m₁ + m₂). r = |r₁ - r₂|.
  The eigenfunctions of H are the eigenfunctions of L².
  These are the spherical harmonics, Y_lm(θ,φ).
  The eigenvalues are E_l = l(l + 1)ħ²/(2I), l = 1, 2, 3, ... .  Each energy eigenvalue is 2l + 1 fold degenerate
  The rotational energy levels of the molecule are E_l = l(l + 1)ħ²/(2I). 

Problem 3:

The operators a and a^† when acting on the energy eigenstates of the harmonic oscillator, denoted by |n>, have the property
a^†|n> = (n + 1)^½|n + 1>,  a|n> = n^½|n - 1>.
We have x = (a + a^†)/(2α),  p = -i(a - a^†)/(2β), where α =√(mω/(2ħ)),  β =1/√(2mωħ).
Find the mean value and root mean square deviation of p², when the oscillator is in the energy eigenstate |n>.

Solution:

• Concepts:
  Fundamental assumptions of QM, the eigenstates of the harmonic oscillator Hamiltonian, the raising and lowering operators
• Reasoning:
  For any operator A we have ∆A = (<(A - <A>)²>)^½ = (<A²> - <A>²)^½.
• Details of the calculation:
  p² = -(mħω/2)(a^†a^† - a^†a - aa^† + aa).
  <p²> = -(mħω/2)<n|(a^†a^† - a^†a - aa^†  + aa)|n> = (mħω/2)(2n + 1).
  The mean value is <p²> = (mħω/2)(2n + 1).
  
  p⁴ = (mħω/2)²(a^†a^† - a^†a - aa^† + aa) (a^†a^† - a^†a - aa^† + aa).
  <p⁴> = (mħω/2)²<n|(a^†a^† aa + a^†a a^†a + a^†a aa^† + aa^† a^†a + aa^† aa^†  + aaa^†a^†)|n>
  = (mħω/2)²(6n² + 6n + 3).
  <p⁴> - <p²>² = (mħω/2)²(2n² + 2n + 2).
  ∆p² = (mħω/2)( 2n² + 2n + 2)^½ is root mean square deviation of p².

Problem 4:

An exotic atom consists of a Helium nucleus (Z = 2) and an electron and an antiproton p(bar) both in n = 2 states.  Take the mass of the p(bar) to be 2000 electron masses and that of the helium nucleus to be 8000 m_e.  For an electron in the n = 1 state of hydrogen E = -13.6 eV.
(a)  How much energy is required to remove the electron from this atom?
(b)  How much energy is required to remove the p(bar) from this atom?
(c)  Assume both the p(bar) and the electron are in 2p states.  Then each can de-excite to their ground state.  It is observed that radiation always accompanies those transitions when the electron jumps first, but when the p(bar) jumps first there is often no photon emitted.  Explain!

Solution:

• Concepts:
  Hydrogenic atoms
• Reasoning:
  To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom a₀ by
  a₀' = ħ²/(μ'Ze²) = a₀(μ/μ')(1/Z),
  and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_I by
  E_I' = μ'Z²e⁴/(2ħ²) = E_I(μ'/μ)Z².
  Here μ is the reduced mass for the hydrogen atom.
• Details of the calculation:
  (a)  For the p(bar) we have μ' = 2000*8000/10000 m_e = 1600 m_e.
  For the p(bar) we have a₀' = a₀/3200.
  The p(bar) is most likely found very close to the nucleus.
  The electron therefore moves in a field that is approximately that of a nucleus with Z = 1.
  E₂ = -13.6 eV*/4 = -3.4 eV.  3.4 eV is required to remove the electron.
  (b)  The electron is most likely found very far away from the nucleus compared to the p(bar).  In the 2p state the probability of finding the electron mear the nucleus id very low.  The p(bar) therefore move in a field that is approximately that of a nucleus with Z = 2.
  For the p(bar) we have μ' = 1600, Z = 2.
  E₂ = -13.6 eV*4*1600/4 = -21760 eV.  21760 eV is required to remove the p(bar).
  (c)  When the electron jumps first, (13.6 - 3.4) = 10.2 eV of energy is released.  This energy can only be given to a photon, since at least 13.6*4*1600(1/4 - 1/9) eV = 12088 eV is needed to excite the p(bar) to the n = 3 level.
  When the p(bar) jumps first, enough energy is released to remove the electron from the atom.  Radiation-less transitions, in which the energy is given to the electron, are now possible.

Problem 5:

An electron in the hydrogen atom occupies the combined position and spin state

R₂₁(r)[√⅓ Y₁₀(θ,φ)χ₊ + √⅔ Y₁₁(θ,φ)χ_-].

(a)  If you measure L², what value(s) might you get, and with what probability(ies)?
(b)  If you measure L_z, what value(s) might you get, and with what probability(ies)?
(c)  If you measure S², what value(s) might you get, and with what probability(ies)?
(d)  If you measure S_z, what value(s) might you get, and with what probability(ies)?
(e)  If you measured the position of the electron, what is the probability density for finding the electron at r, θ, φ in terms of the variables given above.
(f)  If you measured both S_z and the distance of the electron from the proton, what is the probability per unit length for finding the particle with spin up a distance r from the proton in terms of the variables given above?

Useful integral:  ∫₀^πsinθ dθ∫₀^2πdφ |Y_lm(θ,φ))|² = 1.

Solution:

• Concepts:
  The eigenfunctions of the orbital and spin angular momentum operators, the spherical harmonics, tensor product states
• Reasoning:
  The common eigenfunctions of L² and L_z are the spherical harmonics.
• Details of the calculation:
  (a)  L² = 2ħ²,  probability 1.
  (b)  L_z = 0,  probability ⅓,  L_z = ħ,  probability ⅔.
  (c)  S² = (3/4)ħ²,  probability 1.
  (d)  S_z = ħ/2,  probability ⅓,  S_z = -ħ/2,  probability ⅔.
  (e)  Let P(r,θ,φ) the probability per unit volume of for finding the electron at  r, θ, φ.  
  P(r,θ,φ) = <ψ|P_r,θ,φ|ψ>, where the projector
  P_r,θ,φ = ∑_i(|r,θ,φ |χ_i>)(<r,θ,φ <χ_i|), and i = +, -.
  P(r,θ,φ)
  = (√⅓<210|<χ₊| + √⅔<211|<χ_-|) (|r,θ,φ |χ₊>)(<r,θ,φ|<χ₊|)(√⅓|210>|χ₊> + √⅔|211>|χ_->)
  + (√⅓<210|<χ₊| + √⅔<211|<χ_-|) (|r,θ,> |χ_->)(<r,θ,φ|<χ_-|)(√⅓|210>|χ₊> + √⅔|211>|χ_->)
  = ⅓|<210|r,θ,φ>|² + ⅔|<211|r,θ,φ>|², since <χ_i|χ_j|> = δ_ij.
  P(r,θ,φ) = |R₂₁r)|²[⅓|Y₁⁰(θ,φ |² + ⅔|Y₁¹(θ,φ |²].
  (f)  P₊(r,θ,φ) = ⅓|R₂₁(r)|²|Y₁⁰(θ,φ)|².
  Probability per unit length P₊(r) = ⅓r²|R₂₁(r)|².