Chapter 19 Quiz: The Variational Principle

Instructions: This quiz covers the core concepts from Chapter 19. 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 variational theorem states that for any normalized trial state $|\psi\rangle$ and Hamiltonian $\hat{H}$ with ground state energy $E_0$:

(a) $\langle \psi | \hat{H} | \psi \rangle = E_0$ (b) $\langle \psi | \hat{H} | \psi \rangle \leq E_0$ (c) $\langle \psi | \hat{H} | \psi \rangle \geq E_0$ (d) $\langle \psi | \hat{H} | \psi \rangle \geq E_1$ (the first excited state energy)

Q2. In the proof of the variational theorem, the key mathematical step is:

(a) Using the completeness of energy eigenstates and $E_n \geq E_0$ (b) Applying the uncertainty principle (c) Using time-dependent perturbation theory (d) Employing the WKB approximation

Q3. A Gaussian trial function $\psi \propto e^{-\alpha r^2}$ applied to the hydrogen atom gives a variational energy approximately:

(a) Exactly $-13.6$ eV (the exact ground state energy) (b) About 15% above the exact ground state energy (c) About 15% below the exact ground state energy (d) About 50% above the exact ground state energy

Q4. In the variational treatment of helium with trial function $\psi \propto e^{-Z'(r_1 + r_2)}$, the optimal effective nuclear charge is $Z'_{\text{opt}} = 27/16 \approx 1.69$. The physical interpretation is:

(a) The nucleus has been partially neutralized by virtual photon exchange (b) Each electron partially screens the nuclear charge from the other electron (c) Relativistic effects reduce the effective nuclear charge (d) The electron-electron repulsion pushes both electrons away equally

Q5. In the LCAO treatment of H₂⁺, the exchange integral $K$ is responsible for:

(a) The electrostatic attraction between the electron and the nearest proton (b) The quantum mechanical splitting between bonding and antibonding states (c) The nuclear-nuclear repulsion energy (d) The kinetic energy of the electron

Q6. The Ritz (linear variational) method reduces the Schrödinger equation to:

(a) A set of coupled differential equations (b) A generalized matrix eigenvalue problem $\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$ (c) A single nonlinear algebraic equation (d) A path integral

Q7. According to the Hylleraas-Undheim-MacDonald theorem, the $k$-th eigenvalue of the Ritz method provides:

(a) An exact value of the $k$-th true eigenvalue (b) A lower bound on the $k$-th true eigenvalue (c) An upper bound on the $k$-th true eigenvalue (d) An estimate with no guaranteed relationship to the true eigenvalue

Q8. In variational Monte Carlo (VMC), the local energy $E_L(\mathbf{r})$ is defined as:

(a) $\hat{H}\psi(\mathbf{r})$ (b) $\psi(\mathbf{r})/\hat{H}\psi(\mathbf{r})$ (c) $\hat{H}\psi(\mathbf{r})/\psi(\mathbf{r})$ (d) $|\psi(\mathbf{r})|^2 \hat{H}$

Q9. The "zero-variance principle" in VMC states that:

(a) The energy estimate always has zero statistical error (b) The variance of the local energy vanishes when the trial function is an exact eigenstate (c) Adding more Metropolis samples always reduces the variance to zero (d) The Metropolis acceptance rate should be tuned to give zero rejected moves

Q10. Adding more variational parameters to a trial function will:

(a) Always decrease the optimized energy (b) Never decrease the optimized energy (c) Decrease or leave unchanged the optimized energy (never increase it) (d) Sometimes increase the optimized energy if the parameters are poorly chosen

True/False (4 questions)

Q11. True or False: If a variational calculation gives an energy that is very close to the exact ground state energy, then the trial wavefunction must be a very accurate approximation to the true ground state wavefunction.

Q12. True or False: The variational method can only provide upper bounds on the ground state energy; it cannot be directly applied to estimate excited state energies.

Q13. True or False: In the Metropolis algorithm used for VMC, the acceptance probability depends on the normalization of the trial wavefunction.

Q14. True or False: The variational method with a trial function $\psi(r; \alpha) = N e^{-\alpha r}$ applied to the hydrogen atom recovers the exact ground state energy because this functional form includes the true ground state.

Short Answer (4 questions)

Q15. Explain why the variational method is said to give results that are "second-order accurate." Specifically, if the trial function deviates from the true ground state by a small amount $\epsilon$, how does the energy error scale with $\epsilon$? Derive this scaling.

Q16. Describe the key differences between the nonlinear variational method (e.g., optimizing $Z'$ in the helium calculation) and the linear variational (Ritz) method. When would you prefer one over the other?

Q17. A student uses the variational method on the hydrogen atom with the trial function $\psi(r) = N e^{-\alpha r^3}$. Without doing any calculation, explain why this is a poor choice of trial function. What physical features does it get wrong?

Q18. Why is Monte Carlo integration essential for applying the variational method to many-electron systems? What is the "curse of dimensionality" and how does Monte Carlo overcome it?

Applied Scenarios (2 questions)

Q19. Consider the one-dimensional harmonic oscillator $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2 x^2$.

You use the trial function $\psi(x; a) = N/(x^2 + a^2)$ (a Lorentzian).

(a) This trial function has the wrong asymptotic behavior for the harmonic oscillator. What behavior does it have for large $|x|$, and what behavior should the correct ground state have?

(b) Despite this, the variational theorem guarantees that $E(a) \geq E_0 = \frac{1}{2}\hbar\omega$. Using the fact that $\int_{-\infty}^{\infty} \frac{dx}{(x^2 + a^2)^2} = \frac{\pi}{2a^3}$ and $\int_{-\infty}^{\infty} \frac{x^2 \, dx}{(x^2 + a^2)^2} = \frac{\pi}{4a}$, compute the normalization constant $N$ and then $\langle T \rangle$ and $\langle V \rangle$ as functions of $a$.

(c) Find $a_{\text{opt}}$ and $E_{\min}$. Express the result as $E_{\min} = \kappa \cdot \frac{1}{2}\hbar\omega$ and find $\kappa$. How far above the exact ground state energy is the Lorentzian bound?

Q20. You want to estimate the ground state energy of the lithium atom ($Z = 3$, three electrons) using a variational approach.

(a) The simplest trial function would be a product of hydrogen-like 1s orbitals for all three electrons: $\psi = \phi_{1s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)\phi_{1s}(\mathbf{r}_3)$. Why is this trial function forbidden by the Pauli exclusion principle? (Hint: think about the spin state.)

(b) A valid simple trial function places two electrons in a 1s orbital (with opposite spins) and one in a 2s orbital: $\psi = \mathcal{A}[\phi_{1s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)\phi_{2s}(\mathbf{r}_3)]$ where $\mathcal{A}$ is the antisymmetrizer (Slater determinant). If you use hydrogen-like orbitals with effective charges $Z_1$ (for the 1s electrons) and $Z_2$ (for the 2s electron) as variational parameters, would you expect $Z_1 > Z_2$ or $Z_1 < Z_2$? Explain physically.

(c) The experimental ground state energy of lithium is $E_0 = -7.478$ hartree. A simple two-parameter variational calculation gives $E \approx -7.34$ hartree. What is the percentage error? Would you expect the error to be larger or smaller than the helium variational error (1.9%), and why?

Answer Key

Q1: (c) — The variational theorem provides an upper bound on $E_0$.

Q2: (a) — Expanding in energy eigenstates and using $E_n \geq E_0$ is the essential step.

Q3: (b) — The Gaussian gives $E \approx -0.424$ hartree vs. exact $-0.500$ hartree, about 15% too high.

Q4: (b) — Each electron screens part of the nuclear charge from the other.

Q5: (b) — The exchange integral $K$ creates the energy splitting between bonding and antibonding orbitals.

Q6: (b) — The Ritz method yields the generalized eigenvalue problem $\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$.

Q7: (c) — Each Ritz eigenvalue is an upper bound on the corresponding true eigenvalue.

Q8: (c) — $E_L = \hat{H}\psi/\psi$ is the local energy.

Q9: (b) — When $\psi$ is exact, $\hat{H}\psi = E_0\psi$, so $E_L = E_0$ everywhere (constant, zero variance).

Q10: (c) — More parameters expand the variational manifold, so the optimized energy can only decrease or stay the same.

Q11: False — The energy is a second-order functional of the wavefunction. A 0.1% energy error can correspond to a ~3% wavefunction error.

Q12: False — The variational method can bound excited state energies if the trial function is constrained to be orthogonal to lower states (or uses symmetry).

Q13: False — The Metropolis acceptance ratio $|\psi(\mathbf{r}')|^2/|\psi(\mathbf{r})|^2$ involves a ratio, so the normalization cancels.

Q14: True — The family $\{N e^{-\alpha r}\}$ contains the exact ground state wavefunction (at $\alpha = 1/a_0$), so the variational method finds it exactly.

Q15: Write $|\psi\rangle = |0\rangle + \epsilon|\delta\rangle$. Then $\langle\psi|\hat{H}|\psi\rangle = E_0 + \epsilon^2 \langle\delta|(\hat{H} - E_0)|\delta\rangle + O(\epsilon^3)$. The first-order term vanishes because $\langle 0|\hat{H}|\delta\rangle = E_0\langle 0|\delta\rangle$ cancels. The energy error is $O(\epsilon^2)$ — second order in the wavefunction error.

Q16: Nonlinear: parameters enter the exponents or functional form; requires numerical optimization (scipy.optimize); good for capturing specific physics (screening, correlation). Linear (Ritz): coefficients multiply fixed basis functions; reduces to matrix diagonalization; systematically improvable by enlarging basis; gives all eigenvalues simultaneously. Prefer nonlinear when you have strong physical intuition; prefer Ritz when you have a good basis set and want systematic convergence.

Q17: The trial function $e^{-\alpha r^3}$ decays too rapidly at large $r$ (faster than any exponential) and has the wrong cusp behavior at $r = 0$ (derivative $\propto r^2$ instead of constant). It does not satisfy the Kato cusp condition. While the variational theorem still guarantees a valid upper bound, the bound will be very loose.

Q18: For $N$ electrons, the energy integral is $3N$-dimensional. Grid-based quadrature requires $M^{3N}$ evaluations (exponential in $N$) — the curse of dimensionality. Monte Carlo integration converges as $1/\sqrt{M}$ regardless of dimension, making it feasible for $N \gg 3$.

Q19: (a) Lorentzian decays as $1/x^2$ (power law); true ground state decays as $e^{-m\omega x^2/(2\hbar)}$ (Gaussian). (b) $N^2 = 2a^3/\pi$; $\langle T \rangle = \hbar^2/(4ma^2)$; $\langle V \rangle = m\omega^2 a^2/4$. (c) $a_{\text{opt}}^2 = \hbar/(m\omega)$, $E_{\min} = \hbar\omega/\sqrt{2} \approx 0.707\hbar\omega$, so $\kappa = \sqrt{2} \approx 1.414$. The Lorentzian bound is 41% above the exact ground state energy.

Q20: (a) Three electrons in 1s with only two possible spin states ($m_s = \pm 1/2$) violates Pauli — at least two electrons must have the same spin, but then they cannot both be in 1s. The antisymmetrized product of three identical spatial functions vanishes. (b) $Z_1 > Z_2$: the 1s electrons are closer to the nucleus and experience less screening, while the 2s electron is farther out and screened by both 1s electrons. (c) Error: $(7.478 - 7.34)/7.478 = 1.8\%$. This is comparable to helium's 1.9%. With three electrons, screening is more complex but the same physical idea applies. More electrons generally require more parameters for comparable accuracy.