More Problems

Problem 1:

Is each of the following sets a valid combination of quantum numbers (n, ℓ, m_ℓ, m_s) for the energy eigenstates of hydrogen?  If not, explain why not.
(2, 2, -1, ½)
(3, 1, +2, -½)
(3, 1, 0, ½)
(4, 1, 1, -3/2)
(2, -1, 1, ½)

Solution:

• Concepts:
  The hydrogen atom
• Reasoning:
  We are supposed to add the orbital and spin angular momentum of the electron in the hydrogen atom and check if the given combinations are valid.
• Details of the calculation:
  The combinations below are not valid.
  (2, 2,-1, ½)  The quantum number ℓ must be non-negative and smaller than the quantum number n.
  (3, 1, +2, -½)  The quantum number m_ℓ can take on all integer values between -ℓ and ℓ.
  (4, 1, 1, -3/2)  The quantum number m_s for the spin ½ electron is restricted to the values ±½.
  (2, -1, 1, ½)  The quantum number ℓ must be non-negative.

Problem 2:

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 3:

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 move 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 is 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.