Chapter 38 Quiz: Capstone — Hydrogen Atom from First Principles
Instructions: This quiz covers the core concepts from Chapter 38. For multiple choice, select the single best answer. For true/false, provide a brief justification (1-2 sentences). For short answer, aim for 3-5 sentences. For applied scenarios, show your work.
Multiple Choice (10 questions)
Q1. The "accidental" degeneracy of the hydrogen atom — the fact that states with the same $n$ but different $l$ have the same energy — arises from:
(a) The spherical symmetry of the Coulomb potential (b) A hidden $SO(4)$ symmetry related to the conserved Runge-Lenz vector (c) The spin-orbit coupling canceling the relativistic correction (d) The fact that $l$ can only take integer values
Q2. The fine-structure constant $\alpha \approx 1/137$ appears in the fine-structure correction as $E_{\text{fs}} \sim \alpha^2 E_n$. If $\alpha$ were twice as large ($\alpha \approx 1/68.5$), the fine-structure splitting would be:
(a) Twice as large (b) Four times as large (c) Half as large (d) Unchanged — $\alpha$ does not affect the splitting
Q3. The Darwin term in the fine-structure Hamiltonian is nonzero only for $l = 0$ states because:
(a) Only $s$-states have nodes (b) Only $s$-states have nonzero probability density at the nucleus (c) The centrifugal barrier prevents $l > 0$ states from existing (d) The Darwin term depends on $\hat{L} \cdot \hat{S}$, which vanishes for $l = 0$
Q4. A Gaussian trial wavefunction $\psi = Ae^{-\alpha r^2}$ gives a variational ground-state energy of $-11.5$ eV for hydrogen, compared to the exact $-13.6$ eV. This means:
(a) The Gaussian wavefunction is 85% correct (b) The true ground-state energy lies between $-13.6$ eV and $-11.5$ eV (c) The true ground-state energy is at most $-11.5$ eV (more negative) (d) The true ground-state energy is at most $-11.5$ eV (less negative)
Q5. The hydrogen $2S_{1/2}$ state is metastable because:
(a) It is the lowest energy state, so it cannot decay further (b) The $2S \to 1S$ electric dipole transition violates $\Delta l = \pm 1$ (c) Spin-orbit coupling prevents the transition (d) The energy difference is too small to produce a real photon
Q6. The Lamb shift ($2S_{1/2}$ vs. $2P_{1/2}$ splitting of about 1058 MHz) is caused by:
(a) The spin-orbit interaction between the electron and proton (b) The finite size of the proton (c) Quantum electrodynamic effects — the electron's interaction with virtual photons (d) The magnetic interaction between electron spin and proton spin
Q7. When solving the hydrogen radial equation numerically using finite differences, a logarithmic grid ($r_i = r_{\min}e^{i\Delta s}$) is preferred over a uniform grid ($r_i = i\Delta r$) because:
(a) The logarithmic grid avoids the singularity at $r = 0$ (b) The logarithmic grid places more points near the nucleus where the wavefunction varies rapidly (c) The uniform grid cannot represent exponential functions (d) The logarithmic grid is computationally faster
Q8. The hydrogen 21-cm line corresponds to:
(a) The $2P \to 1S$ Lyman-$\alpha$ transition (b) The fine-structure splitting of the $n = 2$ level (c) The hyperfine splitting of the $1S$ ground state ($F = 1 \to F = 0$) (d) The $3D \to 2P$ Balmer-$\alpha$ transition
Q9. In the hierarchy of hydrogen energy corrections, the ordering from largest to smallest is:
(a) Gross structure > Lamb shift > fine structure > hyperfine structure (b) Gross structure > fine structure > hyperfine structure > Lamb shift (c) Gross structure > fine structure > Lamb shift $\approx$ hyperfine structure (d) Fine structure > gross structure > Lamb shift > hyperfine structure
Q10. The complete fine-structure correction for hydrogen depends on the quantum numbers:
(a) $n$ and $l$ only (b) $n$ and $j$ only (c) $n$, $l$, and $j$ (d) $n$, $l$, $j$, and $m_j$
True/False (4 questions)
Q11. True or false: The variational method always gives an energy that is greater than or equal to the true ground-state energy.
Q12. True or false: For the hydrogen atom, the shooting method can find the exact ground-state energy (to machine precision), while the finite-difference matrix method always has a finite discretization error.
Q13. True or false: The spin-orbit coupling term $\hat{H}_{\text{SO}} \propto \hat{\mathbf{L}} \cdot \hat{\mathbf{S}}/r^3$ contributes to the fine structure of $s$-states ($l = 0$).
Q14. True or false: The expectation value $\langle 1/r \rangle$ for a hydrogen eigenstate depends on both $n$ and $l$.
Short Answer (4 questions)
Q15. Explain why Slater-type orbitals (STOs) converge faster than Gaussian-type orbitals (GTOs) for the hydrogen atom, yet GTOs dominate computational chemistry.
Q16. The experimental ionization energy of hydrogen ($13.5984$ eV) differs from the Coulomb model prediction ($13.6057$ eV) by about $0.05\%$. List all the effects that contribute to this discrepancy, in order of importance.
Q17. Describe the physical meaning of the Kramers relation and explain how it is useful in computing perturbation corrections for hydrogen.
Q18. The selection rule $\Delta l = \pm 1$ for electric dipole transitions has a deep physical origin in the properties of the photon. Explain this connection.
Applied Scenarios (2 questions)
Q19. You are given numerical wavefunctions for the $2S$ and $2P$ states of hydrogen from a finite-difference calculation on a 200-point grid. You want to compute the fine-structure splitting. Describe the procedure step by step: - What expectation values do you need to compute? - How do you handle the $l = 0$ (Darwin) and $l = 1$ (spin-orbit) terms differently? - How do you estimate the error in your numerical fine-structure splitting? - What grid refinement study would you perform to verify convergence?
Q20. A colleague has built a variational solver for hydrogen that consistently returns a ground-state energy of $-14.2$ eV — below the exact value of $-13.6$ eV. This appears to violate the variational principle. Diagnose three possible bugs in their code and explain how you would test for each one.
Answer Key Notes
(For instructor use)
- Q1: (b) — The $SO(4)$ symmetry is the key; spherical symmetry alone gives only $m$-degeneracy.
- Q4: (c) — The variational principle gives an upper bound: $E_0 \leq E_{\text{trial}} = -11.5$ eV, meaning the true energy is more negative (lower).
- Q10: (b) — The remarkable cancellation makes the fine-structure formula depend only on $n$ and $j$.
- Q11: True — This is the variational theorem, valid for the ground state of any Hamiltonian bounded below.
- Q12: True — The shooting method adjusts $E$ continuously (limited only by floating-point precision), while finite differences have inherent grid error.
- Q13: False — For $l = 0$, $\hat{\mathbf{L}} \cdot \hat{\mathbf{S}} = 0$, and the $1/r^3$ expectation value diverges. The Darwin term handles $l = 0$ instead.
- Q14: False — $\langle 1/r \rangle_{nl} = 1/(n^2 a_0)$ depends only on $n$.
- Q20: Likely bugs: (1) Normalization error — the trial function is not properly normalized, making $\langle H \rangle$ meaningless. (2) Sign error in the potential — the Coulomb potential must be negative ($-e^2/r$). (3) Numerical integration error — insufficient quadrature points causing inaccurate matrix elements.