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Examination Questions and Answer Hints

This chapter collects representative examination questions drawn from all parts of the course, pitched at the M.Sc. level. Each question is followed by a structured Answer Hint — not a model answer, but a map of the concepts, logical steps, and equations an examiner would expect to see. Working through these before an examination is most effective when you attempt the question before reading the hint.

The questions are grouped by chapter and marked with an estimated difficulty: [S] short answer (5–10 min), [M] medium (15–20 min), [L] long / proof-based (30+ min).

Tip

Note on prerequisites. Q1.3 (variational He atom) and Q2.1 (HK proof) both rely on the Rayleigh–Ritz variational principle. These questions assume familiarity with the variational theorem: for any normalised trial state \(|\tilde\Psi\rangle\), the expectation value of the Hamiltonian satisfies \(\langle\tilde\Psi|\hat{H}|\tilde\Psi\rangle \geq E_0\), with equality iff \(|\tilde\Psi\rangle\) is the true ground state. This result underpins both the justification for DFT energy minimisation and the proof of the HK theorems, and is a thread that runs through the entire course.

Part I: Theoretical Foundations

Chapter 1 — The Many-Body Problem

Q1.1 [S] Write down the full many-body Hamiltonian for a system of \(N\) electrons and \(M\) nuclei. Identify each term and state which terms are neglected under the Born–Oppenheimer approximation. Why is this approximation justified?

**Answer Hint.** The Hamiltonian has five terms: nuclear kinetic energy \\(\hat{T}_n\\), electronic kinetic energy \\(\hat{T}_e\\), electron–electron repulsion \\(\hat{V}_{ee}\\), nuclear–nuclear repulsion \\(\hat{V}_{nn}\\), and electron–nuclear attraction \\(\hat{V}_{en}\\). Under Born–Oppenheimer, \\(\hat{T}_n\\) is neglected and \\(\hat{V}_{nn}\\) becomes a constant. Justification: nuclei are \\(\sim 10^3\\)–\\(10^5\\) times heavier than electrons, so they move on a far slower timescale. Electrons adiabatically follow the nuclear configuration, decoupling the two equations.

Q1.2 [S] Why does the number of variables in \(\Psi(r_1, r_2, \ldots, r_N)\) scale as \(3N\) rather than 3? What practical consequence does this have for \(N = 100\)?

**Answer Hint.** Each electron has three spatial coordinates; the wavefunction lives in \\(3N\\)-dimensional configuration space. For \\(N=100\\), storing \\(\Psi\\) on a grid of even 10 points per dimension requires \\(10^{300}\\) numbers — vastly more than the estimated \\(10^{80}\\) atoms in the observable universe. This exponential wall motivates the move to density-based (DFT) or other reduced descriptions.

Q1.3 [M] Use a hydrogenic trial wavefunction \(\phi(\mathbf{r}) = (\alpha^3/\pi)^{1/2} e^{-\alpha r}\) to estimate the ground-state energy of helium variationally. Find the optimal \(\alpha_{\rm opt}\), compute \(E(\alpha_{\rm opt})\), and compare to the Hartree–Fock result (\(-77.9\) eV) and experiment (\(-79.0\) eV). What physics is missing from this trial function?

**Answer Hint.**

Setup: Write the He Hamiltonian as \(\hat{H} = \hat{h}1 + \hat{h}2 + \hat{V}{ee}\), where \(\hat{h}i = -\tfrac{1}{2}\nabla_i^2 - Z/r_i\) (\(Z=2\)) and \(\hat{V}{ee} = 1/r{12}\). Use the product ansatz \(\Phi(\mathbf{r}_1, \mathbf{r}_2) = \phi(\mathbf{r}_1)\phi(\mathbf{r}_2)\) — two electrons in the same hydrogenic orbital with variational exponent \(\alpha\).

Three expectation values:

  1. Kinetic energy per electron: \(\langle\phi|-\tfrac{1}{2}\nabla^2|\phi\rangle = \alpha^2/2\).

  2. Electron–nuclear attraction: \(\langle\phi|-Z/r|\phi\rangle = -Z\alpha\).

  3. Electron–electron repulsion (the key integral): for two hydrogenic densities, the standard result is \(\langle \hat{V}_{ee}\rangle = 5\alpha/8\).

Total energy:

\[ E(\alpha) = \alpha^2 - 2Z\alpha + \frac{5\alpha}{8} = \alpha^2 - \left(2Z - \frac{5}{8}\right)\alpha. \]

Minimisation: \(\partial E/\partial\alpha = 0\) gives

\[ \alpha_{\rm opt} = Z - \frac{5}{16} = 2 - \frac{5}{16} = \frac{27}{16} \approx 1.6875. \]

The physical interpretation: \(\alpha_{\rm opt} < Z = 2\) because each electron partially screens the nucleus from the other, reducing the effective nuclear charge it feels.

Optimal energy:

\[ E(\alpha_{\rm opt}) = -\left(Z - \frac{5}{16}\right)^2 = -\left(\frac{27}{16}\right)^2 \approx -2.848 \text{ Ha} \approx -77.5 \text{ eV}. \]

Comparison: This is above the HF limit (\(-77.9\) eV) and experiment (\(-79.0\) eV), consistent with the variational principle — no approximation can go below the true ground state. The remaining \(\sim 1.1\) eV gap from experiment is electron correlation: the instantaneous \(r_{12}\)-dependence of the wavefunction, absent in any single-product ansatz. This is precisely the correlation energy \(E_c\) that DFT’s \(E_{\rm xc}\) must capture.


Chapter 2 — Hohenberg–Kohn Theorems

Q2.1 [L] State and prove the First Hohenberg–Kohn Theorem. Your proof must clearly identify where the Rayleigh–Ritz variational principle is used.

**Answer Hint.**

Statement: The external potential \(V_{\rm ext}(\mathbf{r})\) is determined uniquely (up to a constant) by the ground-state electron density \(\rho_0(\mathbf{r})\).

Proof structure (by contradiction): Assume two potentials \(V\) and \(V’\) (differing by more than a constant) give the same \(\rho_0\). They define two Hamiltonians \(\hat{H}\), \(\hat{H}‘\) with ground states \(\Psi_0\), \(\Psi_0’\) and energies \(E_0\), \(E_0’\). Apply the variational principle using \(\Psi_0’\) as a trial state for \(\hat{H}\):

\[ E_0 < \langle \Psi_0’ | \hat{H} | \Psi_0’ \rangle = E_0’ + \int [V - V’]\rho_0, d\mathbf{r}. \]

By symmetry (swap primed and unprimed):

\[ E_0’ < E_0 + \int [V’ - V]\rho_0, d\mathbf{r}. \]

Adding gives \(E_0 + E_0’ < E_0 + E_0’\) — a contradiction. Note: the strict inequality requires the ground states to be non-degenerate; address this subtlety if asked.

Q2.2 [M] State the Second Hohenberg–Kohn Theorem and prove it using the result of the First. What role does \(v\)-representability play?

**Answer Hint.**

Statement: The true ground-state density \(\rho_0\) minimises the total energy functional \(E[\rho] = F[\rho] + \int V_{\rm ext}\rho, d\mathbf{r}\) among all \(v\)-representable densities.

Proof: By HK1, every trial density \(\tilde\rho \neq \rho_0\) maps to a unique \(\tilde\Psi \neq \Psi_0\). Since \(\Psi_0\) is the true ground state, the variational principle gives \(E_0 = \langle\Psi_0|\hat{H}|\Psi_0\rangle \leq \langle\tilde\Psi|\hat{H}|\tilde\Psi\rangle = E[\tilde\rho]\), with equality iff \(\tilde\rho = \rho_0\).

\(v\)-representability: The proof requires \(\tilde\rho\) to be the ground-state density of some external potential. Not all non-negative, \(N\)-normalised densities satisfy this condition — it is not trivially obvious which densities are physically realisable as ground states. Lieb’s constrained-search (Legendre-transform) formulation defines \(F[\rho]\) via

\[ F[\rho] = \inf_{\Psi \to \rho} \langle\Psi|\hat{T}+\hat{V}_{ee}|\Psi\rangle, \]

a minimisation over all antisymmetric \(\Psi\) yielding density \(\rho\). This is well-defined for any \(N\)-representable \(\rho\) (a much larger class), sidestepping the \(v\)-representability restriction entirely. The KS scheme then requires the weaker condition of non-interacting \(v\)-representability: the ground-state density of the interacting system must also be achievable as the ground-state density of some non-interacting system in a local potential. This is assumed throughout KS theory and is believed to hold for all physical ground-state densities, but remains unproven in general.

Q2.3 [S] What is the universal functional \(F[\rho]\)? Why is it called “universal,” and why is computing it accurately the central challenge of DFT?

**Answer Hint.** \\(F[\rho] = \langle\Psi[\rho]|\hat{T}+\hat{V}_{ee}|\Psi[\rho]\rangle\\). It is "universal" because it depends only on the density, not on \\(V_{\rm ext}\\) — it is the same functional for all systems. Computing it is hard because it encodes all many-body kinetic and correlation effects. In particular, the kinetic energy \\(T[\rho]\\) as a functional of density alone (as in Thomas–Fermi theory) is highly inaccurate; this motivates the Kohn–Sham trick of introducing orbitals.

Q2.4 [S] A student claims: “The HK theorem tells us DFT is exact, so any DFT calculation gives the exact answer.” Identify two reasons why this claim is wrong.

**Answer Hint.** (i) The HK theorems guarantee *existence* of a universal functional, not its explicit form. The exchange-correlation functional \\(E_{\rm xc}[\rho]\\) must be approximated in every practical calculation; LDA, GGA etc. introduce uncontrolled errors. (ii) The HK theorems apply only to the *ground state* at zero temperature. Excited states, finite temperature, and time-dependent processes require extensions (TDDFT, finite-temperature DFT).

Chapter 3 — Kohn–Sham Equations

Q3.1 [M] Derive the Kohn–Sham equations from the variational minimisation of the total energy functional. Clearly define the Kohn–Sham effective potential \(V_{\rm eff}\).

**Answer Hint.** Start from

\[ E[\rho] = T_s[\rho] + E_{\rm H}[\rho] + E_{\rm xc}[\rho] + \int V_{\rm ext}\rho, d\mathbf{r}. \]

Introduce \(T_s\) as an explicit orbital functional: \(T_s = \sum_i f_i \langle\phi_i|-\tfrac{1}{2}\nabla^2|\phi_i\rangle\), with \(\rho(\mathbf{r}) = \sum_i f_i |\phi_i(\mathbf{r})|^2\). Impose orthonormality \(\langle\phi_i|\phi_j\rangle = \delta_{ij}\) via Lagrange multipliers \(\epsilon_{ij}\). The stationarity condition \(\delta\mathcal{L}/\delta\phi_i^* = 0\) gives

\[ -\frac{1}{2}\nabla^2\phi_i + \frac{\delta(E_{\rm H}+E_{\rm xc})}{\delta\rho},\phi_i + V_{\rm ext}\phi_i = \sum_j \epsilon_{ij}\phi_j. \]

The functional derivatives are: \(\delta E_{\rm H}/\delta\rho = V_{\rm H}(\mathbf{r}) = \int\rho(\mathbf{r}‘)/|\mathbf{r}-\mathbf{r}’|, d\mathbf{r}’\) and \(\delta E_{\rm xc}/\delta\rho = V_{\rm xc}(\mathbf{r})\). Choosing the unitary transformation that diagonalises the multiplier matrix \(\epsilon_{ij} \to \epsilon_i \delta_{ij}\) (canonical KS orbitals) yields the KS equations in their standard form:

\[ \left[-\frac{1}{2}\nabla^2 + V_{\rm eff}(\mathbf{r})\right]\phi_i(\mathbf{r}) = \epsilon_i,\phi_i(\mathbf{r}), \qquad V_{\rm eff} = V_{\rm ext} + V_{\rm H} + V_{\rm xc}. \]

Note: \(V_{\rm eff}\) depends on \(\rho\) which depends on the \(\phi_i\) — the equations are nonlinear and must be solved self-consistently.

Q3.2 [S] The KS scheme introduces a fictitious non-interacting system. In what precise sense is this system “fictitious,” and what physical quantity does it share with the true interacting system?

**Answer Hint.** The non-interacting system has no electron–electron interaction; it moves in an effective one-body potential \\(V_{\rm eff}\\) that is designed so that its ground-state density equals that of the true interacting system. The *density* \\(\rho(\mathbf{r})\\) is exact (by construction); the individual KS eigenvalues \\(\epsilon_i\\) and orbitals \\(\phi_i\\) are *not* observables of the real system (with the notable exception of the highest occupied eigenvalue, which equals the negative of the ionisation energy under the exact XC functional).

Q3.3 [S] Write out the self-consistent field (SCF) cycle for solving the KS equations. What quantity is converged, and why is iterative solution necessary?

**Answer Hint.** The SCF loop: (1) Start with an initial guess \\(\rho^{(0)}\\). (2) Construct \\(V_{\rm eff}[\rho]\\). (3) Solve the KS eigenvalue problem to get \\(\{\phi_i, \epsilon_i\}\\). (4) Compute the output density \\(\rho^{\rm out} = \sum_i f_i |\phi_i|^2\\). (5) Mix \\(\rho^{\rm in}\\) and \\(\rho^{\rm out}\\) to form the new input. Repeat until \\(|\rho^{\rm out} - \rho^{\rm in}| \lt \epsilon_{\rm tol}\\). Iteration is necessary because \\(V_{\rm eff}\\) depends on \\(\rho\\), which in turn depends on the eigenstates of \\(V_{\rm eff}\\) — a nonlinear self-consistency condition.

Chapter 4 — Exchange-Correlation Functionals

Q4.1 [S] Write down the LDA expression for \(E_{\rm xc}\) and state the physical system from which \(\varepsilon_{\rm xc}^{\rm UEG}(\rho)\) is derived.

**Answer Hint.**

\[ E_{\rm xc}^{\rm LDA}[\rho] = \int \varepsilon_{\rm xc}^{\rm UEG}(\rho(\mathbf{r})),\rho(\mathbf{r}), d\mathbf{r}. \]

\(\varepsilon_{\rm xc}^{\rm UEG}\) is the XC energy per particle of the uniform electron gas (UEG), a model of interacting electrons in a uniform positive background. The exchange part is analytic (Dirac): \(\varepsilon_{\rm x}^{\rm UEG} = -(3/4)(3/\pi)^{1/3}\rho^{1/3}\). The correlation part is obtained from quantum Monte Carlo simulations (Ceperley–Alder data, parameterised by Perdew–Wang or Vosko–Wilk–Nusair).

Q4.2 [M] Explain why LDA often gives surprisingly good results despite being based on the uniform electron gas. What is the role of the XC hole sum rule?

**Answer Hint.** The XC energy depends on the *spherically averaged* XC hole \\(\bar{n}_{\rm xc}(r)\\), not its angular details. The LDA hole (borrowed from the UEG) satisfies the exact sum rule \\(\int n_{\rm xc}(\mathbf{r},\mathbf{r}')\, d\mathbf{r}' = -1\\) by construction, so its spherical average integrates correctly. Even if the shape of the hole is wrong in detail, this constraint produces systematic error cancellation. The overbinding tendency of LDA (too negative \\(E_{\rm xc}\\)) is a known and quantifiable error rather than random noise.

Q4.3 [M] Describe the GGA improvement over LDA. Why does the raw gradient expansion approximation (GEA) fail, and how does PBE correct for this?

**Answer Hint.** GEA adds a correction proportional to \\(|\nabla\rho|^2/\rho^{4/3}\\) (the dimensionless reduced gradient \\(s^2\\)), but it violates the XC hole sum rule in the density tails (where \\(s \to \infty\\)) — the hole becomes positive and the \\(-1\\) sum rule is broken.

PBE uses an enhancement factor form \(E_{\rm xc}^{\rm GGA} = \int \varepsilon_{\rm xc}^{\rm UEG}(\rho) F_{\rm xc}(\rho, s)\rho, d\mathbf{r}\) and determines \(F_{\rm xc}\) by imposing exact constraints: recovery of LDA at \(s=0\), correct GEA gradient coefficient near \(s=0\), saturation at large \(s\) to restore the sum rule, and the Lieb–Oxford bound. PBE has zero empirical parameters.

Q4.4 [S] What is Jacob’s Ladder in the context of XC functionals? Name one functional at each of the first four rungs and the key new ingredient added at each rung.

**Answer Hint.**
RungIngredient addedExample
LDAUEG energy density \(\rho\)PW92
GGAReduced gradient \(s =\nabla\rho
meta-GGAKinetic energy density \(\tau\) or \(\nabla^2\rho\)SCAN
HybridFraction of exact (HF) exchangePBE0, HSE06

Double hybrids add perturbative correlation (GL2). Each rung adds non-locality of a different kind.

Q4.5 [S] What is the self-interaction error (SIE) in DFT, and why does it not appear in Hartree–Fock theory?

**Answer Hint.** In exact theory, the Hartree self-repulsion of each electron \\(E_{\rm H}[\rho_i]\\) is exactly cancelled by a corresponding exchange term. In DFT with approximate \\(E_{\rm xc}\\), this cancellation is incomplete, leaving a spurious self-repulsion. Consequence: delocalisation error, incorrect dissociation of \\(H_2^+\\), underestimated barriers. In Hartree–Fock, the exact exchange integral \\(\langle ij|ji\rangle\\) cancels the self-interaction in \\(E_{\rm H}\\) exactly for each orbital, by construction.

Q4.6 [S] A DFT+PBE calculation of a van der Waals bonded molecular crystal (e.g. graphite or a layered organic solid) predicts a near-zero interlayer binding energy. Explain precisely why PBE fails here, and describe two physically motivated remedies.

**Answer Hint.** London dispersion forces arise from *non-local, correlated density fluctuations* at large inter-fragment separations \\(R\\), producing a \\(-C_6/R^6\\) attraction. PBE is semi-local: \\(E_{\rm xc}^{\rm PBE}\\) at point \\(\mathbf{r}\\) depends only on \\(\rho\\) and \\(\nabla\rho\\) at that same point. It cannot describe the instantaneous dipole–dipole correlations between distant fragments; the interaction energy decays exponentially rather than as \\(R^{-6}\\). The GEA and its constrained resummations cannot cure this — it is a fundamental non-locality, not a gradient correction problem.

Remedies:

  1. DFT-D3(BJ) (Grimme et al., 2010): Add a pairwise atom–atom correction \(-C_6^{AB}/r_{AB}^6 - C_8^{AB}/r_{AB}^8\) with Becke–Johnson damping. Cheap, widely implemented, recommended default for molecular crystals. Invoke in VASP via IVDW = 12.

  2. Non-local correlation functional (vdW-DF / VV10): Replace \(E_c^{\rm GGA}\) with a functional that includes a double spatial integral coupling densities at different points \(\mathbf{r}\) and \(\mathbf{r}’\). Physically includes the \(C_6/R^6\) tail without empirical atom parameters. Computationally more expensive but more transferable, especially for heterogeneous systems.

The key point for the examiner: the failure is not due to missing gradient information (adding \(\nabla\rho\) does not help) but due to missing non-local correlation.

Q4.7 [S] For each of the following materials, state the most appropriate class of XC functional and justify your choice: (a) bcc iron; (b) a hybrid organic–inorganic perovskite with a band gap of \(\sim 1.6\) eV; (c) a weakly correlated simple metal (Al); (d) a Mott insulator (VO\(_2\)).

**Answer Hint.**

(a) bcc Fe — spin-polarised GGA (PBE or PBEsol): Fe is a transition metal with itinerant magnetism; GGA captures the exchange splitting adequately, and the \(3d\) states are not as strongly correlated as in oxides. DFT+U is sometimes added for oxides but is not standard for pure Fe.

(b) Hybrid perovskite — range-separated hybrid (HSE06): Band gap is in the semiconductor regime where GGA underestimates by \(\sim 40%\). PBE0 is too expensive for large unit cells; HSE06 provides similar accuracy with tractable cost by limiting exact exchange to short range. DFT-D3 should also be added for vdW stacking in layered perovskites.

(c) Al — LDA or GGA: Al is a nearly-free-electron metal; its electron density is close to uniform, so LDA’s errors cancel well. PBE slightly overbinds but both work. Hybrids add cost with no accuracy benefit for nearly free-electron metals.

(d) VO\(_2\) — DFT+U or hybrid (HSE06): V \(3d\) states are localised and GGA fails to reproduce the Mott gap. DFT+U with \(U_{\rm eff} \sim 3\)–\(4\) eV on V \(d\) is the standard workhorse; HSE06 is an alternative if a parameter-free approach is needed but at higher cost.


Chapter 5 — Basis Sets and Pseudopotentials

Q5.1 [S] What is the kinetic energy cutoff \(E_{\rm cut}\) in a plane-wave calculation? Write the condition that a reciprocal lattice vector \(\mathbf{G}\) must satisfy to be included in the basis.

**Answer Hint.** The plane-wave basis includes all \\(\mathbf{G}\\) satisfying

\[ \frac{1}{2}|\mathbf{k}+\mathbf{G}|^2 \leq E_{\rm cut}. \]

(atomic units; in SI replace \(1/2\) with \(\hbar^2/2m_e\)). Increasing \(E_{\rm cut}\) adds more plane waves and systematically improves the basis — a key advantage over atom-centred bases. VASP quotes \(E_{\rm cut}\) in eV; Quantum ESPRESSO in Ry (1 Ry = 13.6 eV).

Q5.2 [M] Compare norm-conserving pseudopotentials (NCPP), ultrasoft pseudopotentials (USPP), and the PAW method. What physical quantity does norm conservation enforce, and what does relaxing it require?

**Answer Hint.**
  • NCPP: The pseudo-wavefunction is normalised identically to the all-electron wavefunction inside \(r_c\): \(\int_0^{r_c}|\tilde\phi|^2, dr = \int_0^{r_c}|\phi|^2, dr\). This ensures correct scattering properties (logarithmic derivatives). Requires \(E_{\rm cut} \sim 60\)–\(100\) Ry.

  • USPP: Relaxes norm conservation, allowing smoother pseudo-wavefunctions. Missing charge is compensated by augmentation charges added to the density. Requires only \(\sim 25\)–\(40\) Ry.

  • PAW: Formally equivalent to an all-electron calculation via a linear transformation \(\hat{\mathcal{T}}\) that maps smooth pseudo-wavefunctions onto full oscillatory all-electron wavefunctions inside augmentation spheres. Provides access to the full all-electron density. \(E_{\rm cut} \sim 400\)–\(600\) eV for GGA+PAW. Default in VASP.

Q5.3 [S] A student runs a calculation doubling the plane-wave cutoff but finds no change in the total energy. What does this confirm, and what should be checked next?

**Answer Hint.** It confirms that the total energy is converged with respect to \\(E_{\rm cut}\\) — the basis is complete enough for this property. Next convergence checks: (1) \\(k\\)-point mesh density (especially for metals); (2) supercell size if periodic images of a defect or molecule are a concern; (3) SCF convergence threshold to ensure the self-consistency loop has actually converged; (4) force convergence for structural relaxation.

Q5.4 [S] What is basis set superposition error (BSSE), and in which type of basis does it arise? How can it be corrected?

**Answer Hint.** BSSE arises in **atom-centred (localised) basis sets**: when two atoms approach, each atom borrows basis functions from the other's basis, artificially lowering the total energy and overestimating the binding. It does *not* occur in plane-wave bases because the basis is not atom-centred. Correction: the **Boys–Bernardi counterpoise correction** — compute the energy of each fragment using the full dimer basis and subtract the fragment-only energies.

Chapter 6 — Spin-Polarised DFT

Q6.1 [S] Why must the electron density be spin-resolved \(\rho_\uparrow(\mathbf{r}), \rho_\downarrow(\mathbf{r})\) for magnetic systems? What goes wrong if spin polarisation is neglected for, say, iron?

**Answer Hint.** In a magnetic system, spin-up and spin-down electrons experience different effective potentials due to exchange splitting. A spin-unpolarised calculation forces \\(\rho_\uparrow = \rho_\downarrow\\) at every point, preventing the system from lowering its energy by spin polarisation — it constraints the variational search to a subspace that excludes the magnetic ground state. For bcc Fe, the spin-unpolarised calculation gives the wrong crystal structure (predicts fcc, because GGA without magnetism favours the more closely-packed fcc minimum) and zero magnetic moment; spin-polarised LDA/GGA correctly gives bcc with \\(\sim 2.2\,\mu_B\\) per atom and the correct cohesive energy.

Q6.2 [S] Define the spin magnetisation density \(m(\mathbf{r})\) and the local spin density approximation (LSDA). How does LSDA reduce to LDA in the non-magnetic limit?

**Answer Hint.** \\(m(\mathbf{r}) = \rho_\uparrow(\mathbf{r}) - \rho_\downarrow(\mathbf{r})\\). LSDA:

\[ E_{\rm xc}^{\rm LSDA}[\rho_\uparrow,\rho_\downarrow] = \int \varepsilon_{\rm xc}^{\rm UEG}(\rho_\uparrow(\mathbf{r}),\rho_\downarrow(\mathbf{r})),\rho(\mathbf{r}), d\mathbf{r}, \]

where \(\varepsilon_{\rm xc}^{\rm UEG}(\rho_\uparrow, \rho_\downarrow)\) is the XC energy per particle of the spin-polarised UEG, parameterised from quantum Monte Carlo (Ceperley–Alder). The spin-polarisation enters through the relative spin polarisation \(\zeta = (\rho_\uparrow - \rho_\downarrow)/\rho\), which interpolates between the unpolarised (\(\zeta=0\)) and fully polarised (\(\zeta=1\)) limits via a spin-scaling relation due to von Barth and Hedin. When \(\rho_\uparrow = \rho_\downarrow = \rho/2\) (non-magnetic limit), \(\zeta = 0\) and \(\varepsilon_{\rm xc}^{\rm UEG}\) reduces to the unpolarised UEG value — recovering LDA exactly.

Q6.3 [M] Derive the spin-dependent Kohn–Sham equations for collinear spin-polarised DFT. Show explicitly that the exchange splitting of eigenvalues arises from \(V_{\rm xc}^\uparrow \neq V_{\rm xc}^\downarrow\), and write down what \(V_{\rm xc}^\sigma\) is in LSDA.

**Answer Hint.**

Starting point: In collinear SDFT, the total energy is a functional of both spin densities:

\[ E[\rho_\uparrow,\rho_\downarrow] = T_s[\rho_\uparrow,\rho_\downarrow] + E_{\rm H}[\rho] + E_{\rm xc}[\rho_\uparrow,\rho_\downarrow] + \int V_{\rm ext}(\mathbf{r}),\rho(\mathbf{r}),d\mathbf{r}. \]

Variational minimisation with respect to \(\phi_{i\sigma}^\)* (subject to orthonormality in each spin channel) gives two independent sets of KS equations — one per spin channel \(\sigma \in {\uparrow,\downarrow}\):

\[ \left[-\frac{1}{2}\nabla^2 + V_{\rm eff}^\sigma(\mathbf{r})\right]\phi_{i\sigma}(\mathbf{r}) = \epsilon_{i\sigma},\phi_{i\sigma}(\mathbf{r}), \]

with the spin-dependent effective potential

\[ V_{\rm eff}^\sigma = V_{\rm ext} + V_{\rm H}[\rho] + V_{\rm xc}^\sigma[\rho_\uparrow, \rho_\downarrow], \]

where \(V_{\rm H}\) depends only on the total density \(\rho = \rho_\uparrow + \rho_\downarrow\) (same for both spins), and

\[ V_{\rm xc}^\sigma(\mathbf{r}) = \frac{\delta E_{\rm xc}[\rho_\uparrow, \rho_\downarrow]}{\delta \rho_\sigma(\mathbf{r})}. \]

Exchange splitting: In LSDA, \(\varepsilon_{\rm xc}^{\rm UEG}\) is an asymmetric function of \((\rho_\uparrow, \rho_\downarrow)\) whenever \(\rho_\uparrow \neq \rho_\downarrow\). Therefore \(\delta E_{\rm xc}/\delta\rho_\uparrow \neq \delta E_{\rm xc}/\delta\rho_\downarrow\), i.e., \(V_{\rm xc}^\uparrow \neq V_{\rm xc}^\downarrow\). This spin-asymmetric potential shifts the spin-up eigenvalues \(\epsilon_{i\uparrow}\) relative to spin-down \(\epsilon_{i\downarrow}\) — the exchange splitting \(\Delta_{\rm xc} = V_{\rm xc}^\uparrow - V_{\rm xc}^\downarrow\). In LSDA explicitly:

\[ V_{\rm xc}^{\sigma,\rm LSDA}(\mathbf{r}) = \varepsilon_{\rm xc}^{\rm UEG}(\rho_\uparrow,\rho_\downarrow) + \rho,\frac{\partial \varepsilon_{\rm xc}^{\rm UEG}}{\partial \rho_\sigma}\bigg|_{\mathbf{r}}. \]

The two equations are coupled only through the shared \(V_{\rm H}[\rho]\) and the LSDA XC potential — they are solved self-consistently together within the same SCF loop.


Chapter 7 — DFT+U

Q7.1 [M] Explain the self-interaction error in the context of strongly correlated systems. Why does standard GGA predict NiO to be a metal rather than a Mott insulator?

**Answer Hint.** In NiO, the Ni \\(3d\\) states are narrow and strongly localised. Standard GGA treats the on-site Coulomb repulsion \\(U\\) among \\(d\\) electrons at the mean-field level, which underestimates the Mott gap. Because the self-interaction error in GGA artificially delocalises the \\(d\\) electrons, the predicted density of states shows \\(d\\) bands straddling the Fermi level — i.e., metallic behaviour — even though experiment shows a gap of \\(\sim 4\\) eV. The DFT+U correction adds an explicit penalty for fractional occupation of the \\(d\\) subspace, driving the occupations toward integer values and opening the gap.

Q7.2 [S] Write the Dudarev DFT+U energy correction and explain the role of the occupation matrix \(\mathbf{n}^{I\sigma}\). What does the term \(\mathrm{Tr}[\mathbf{n}(1-\mathbf{n})]\) penalise?

**Answer Hint.**

\[ E_U^{\rm Dur} = \frac{U_{\rm eff}}{2}\sum_{I,\sigma}\mathrm{Tr}!\left[\mathbf{n}^{I\sigma}(\mathbf{1}-\mathbf{n}^{I\sigma})\right], \]

where \((n^{I\sigma}){mm’} = \sum_i f_i \langle\phi_i|\tilde{p}m^I\rangle\langle\tilde{p}{m’}^I|\phi_i\rangle\) is the occupation of the correlated \(d\) (or \(f\)) subspace on atom \(I\). The trace \(\mathrm{Tr}[\mathbf{n}(1-\mathbf{n})]\) is zero when all eigenvalues of \(\mathbf{n}\) are 0 or 1 (integer occupations) and positive otherwise — so the correction penalises fractional occupations, driving localisation. Only one parameter \(U{\rm eff} = U - J\) is needed, making this formulation simpler than the Liechtenstein approach.

Q7.3 [S] What is the double-counting correction in DFT+U, and why is it necessary? Name the two most common schemes.

**Answer Hint.** The standard GGA already partially includes Coulomb interactions among the correlated electrons through \\(E_{\rm H}\\) and \\(E_{\rm xc}\\). Adding \\(E_U\\) directly would count these interactions twice. The double-counting correction \\(E_{\rm dc}\\) is subtracted to avoid this. Two common schemes: (1) **Fully localised limit (FLL)**, appropriate for strongly correlated insulators; (2) **Around mean-field (AMF)**, appropriate for weakly correlated metals. Unfortunately, the exact form of \\(E_{\rm dc}\\) is not known; this ambiguity is a fundamental limitation of DFT+U.

Part II: Numerical Implementation


Chapter 8 — SCF Convergence and Density Mixing

Q8.1 [S] What is charge sloshing in a metallic DFT calculation? Identify the physical origin of the instability and state which mixing scheme was designed to suppress it.

Answer Hint. In metals, a small perturbation of the density at long wavelength (small \\(\mathbf{G}\\)) produces a large response (diverging dielectric screening as \\(G \to 0\\)). Simple linear mixing \\(\rho^{\rm in}\_{n+1} = \rho^{\rm in}\_n + \alpha(\rho^{\rm out}\_n - \rho^{\rm in}\_n)\\) amplifies these long-wavelength modes: the correction at small \\(G\\) is too large, causing the density to oscillate between iterations. **Kerker preconditioning** suppresses this by multiplying the residual by \\(G^2/(G^2 + k\_0^2)\\), damping corrections at \\(G \lt k\_0\\), where \\(k_0\\) is the Thomas–Fermi screening wavevector.

Q8.2 [M] Describe the Pulay (DIIS) mixing scheme. What quantity is minimised, and how does it use information from previous iterations?

**Answer Hint.** DIIS (Direct Inversion in the Iterative Subspace) stores the last \\(m\\) input densities \\(\{\rho_n^{\rm in}\}\\) and residuals \\(\{R_n = \rho_n^{\rm out} - \rho_n^{\rm in}\}\\). The next input is a linear combination

\[ \rho^{\rm in}{n+1} = \sum{i=1}^m \alpha_i \rho_i^{\rm in} \]

where the coefficients \({\alpha_i}\) minimise \(|\sum_i \alpha_i R_i|^2\) subject to \(\sum_i \alpha_i = 1\). This is solved by a small linear system. The method is quasi-Newton: it builds a model of the Jacobian of the residual from the stored history, achieving superlinear convergence near the fixed point, compared to linear convergence for simple mixing.

Q8.3 [S] A VASP calculation for a metallic slab fails to converge after 200 iterations with AMIX = 0.4. Suggest three modifications to the INCAR and explain the physical reasoning for each.

**Answer Hint.** (1) **Reduce `AMIX`** (e.g. to 0.02–0.1): smaller mixing step damps the sloshing, at the cost of slower convergence per iteration. (2) **Set `BMIX = 1.0`–`3.0`**: enables Kerker preconditioning, directly targeting the long-wavelength instability in the metal. (3) **Increase `MAXMIX`** (e.g. to 40–60): gives Broyden/DIIS a larger history to build a better Jacobian model, which is important for slabs that have many degrees of freedom in the charge distribution. Optionally: switch to `ALGO = All` to use subspace rotation.

Chapter 9 — Brillouin Zone Integration and Smearing

Q9.1 [S] Why do metals require special treatment in Brillouin zone integration? Write the BZ integral for the electron density and explain the origin of the numerical difficulty.

**Answer Hint.** Physical observables involve integrals of the form

\[ \langle A \rangle = \frac{\Omega}{(2\pi)^3}\sum_n \int_{\rm BZ} A_{n\mathbf{k}}, f_{n\mathbf{k}}, d\mathbf{k}, \]

where \(f_{n\mathbf{k}} = \theta(\mu - \epsilon_{n\mathbf{k}})\) is the Fermi–Dirac occupation at \(T=0\). The Heaviside function is discontinuous at the Fermi surface. On a discrete \(k\)-mesh, the Fermi surface may pass between grid points, causing the integral to oscillate wildly with mesh density. This is why insulators converge far faster than metals: there is no Fermi surface discontinuity.

Q9.2 [M] Compare Gaussian smearing, Methfessel–Paxton smearing, and the linear tetrahedron method. For which systems and properties is each recommended?

**Answer Hint.**
  • Gaussian: Replaces \(\theta(\mu-\epsilon)\) by a Gaussian broadened step. Simple but introduces a \(\sigma\)-dependent error in the free energy; the \(\sigma \to 0\) extrapolation can be unreliable. Use: only for quick tests.

  • Methfessel–Paxton (MP): Expands the step function in a Hermite polynomial series; the \(N=1\) variant gives a broadened occupation with negative oscillations that cancel the smearing error to higher order. The \(\sigma\to 0\) energy extrapolation is \(E(\sigma) = E_0 + \mathcal{O}(\sigma^{2N+2})\). Use: structural relaxations and total energies of metals with moderate \(\sigma \sim 0.1\)–\(0.2\) eV.

  • Tetrahedron method (ISMEAR = -5 in VASP): Linear interpolation of band energies between \(k\)-points within each tetrahedron. Exact integration in the limit of a fine mesh; no \(\sigma\) parameter. Use: DOS, optical spectra, and properties requiring high accuracy in the Fermi surface topology. Do not use for structural relaxations (forces are not analytic).


Chapter 10 — Eigenvalue Solvers

Q10.1 [S] Why is the full diagonalisation of the KS Hamiltonian impractical for large systems? What is the computational scaling of full diagonalisation, and how do iterative methods improve on this?

**Answer Hint.** The KS Hamiltonian has dimension \\(N_{\rm PW} \times N_{\rm PW}\\), where \\(N_{\rm PW}\\) grows with the system size and \\(E_{\rm cut}\\). Full diagonalisation (e.g., LAPACK) scales as \\(\mathcal{O}(N_{\rm PW}^3)\\), which becomes prohibitive for \\(N_{\rm PW} \sim 10^5\\)–\\(10^6\\). Iterative methods (Davidson, RMM-DIIS) exploit the fact that only the \\(N_{\rm occ} \ll N_{\rm PW}\\) lowest eigenpairs are needed. They build a small Krylov/Rayleigh–Ritz subspace, diagonalise within it, and refine — achieving \\(\mathcal{O}(N_{\rm PW} \times N_{\rm occ})\\) per iteration.

Q10.2 [M] Describe the Davidson diagonalisation algorithm. What is the subspace expansion step, and how does it avoid recomputing the full Hamiltonian?

**Answer Hint.** Starting from a trial subspace \\(\{|\psi_i\rangle\}\\): (1) Compute the Rayleigh quotient \\(\epsilon_i = \langle\psi_i|\hat{H}|\psi_i\rangle/\langle\psi_i|\psi_i\rangle\\) — only matrix-vector products \\(\hat{H}|\psi_i\rangle\\) are needed, not matrix elements. In a plane-wave code this is done via FFT in \\(\mathcal{O}(N_{\rm PW}\log N_{\rm PW})\\). (2) Compute residuals \\(|r_i\rangle = (\hat{H} - \epsilon_i)|\psi_i\rangle\\). (3) Apply a **preconditioner** \\(\hat{D}^{-1}\\) (diagonal in \\(\mathbf{G}\\)-space: \\(D_{\mathbf{G}} = |\mathbf{k}+\mathbf{G}|^2/2 + \epsilon_0\\)) to obtain search directions that are steepest-descent steps in kinetic-energy-weighted distance. (4) Orthogonalise and add to the subspace; diagonalise the small projected Hamiltonian; iterate. Convergence is quadratic near the solution.

Cross-Cutting Conceptual Questions

Q11.1 [M] A colleague applies DFT+GGA to a cobalt oxide (CoO) surface and reports a magnetic moment of 0 \(\mu_B\) per Co, which contradicts experiment. Walk through the chain of possible failures, from theory choice to numerical implementation.

**Answer Hint.** Work from first principles outward: (1) *Spin polarisation:* Is `ISPIN = 2` set? A spin-unpolarised calculation cannot produce a magnetic moment by construction. (2) *Initial magnetic moments:* Were reasonable initial moments provided (e.g., `MAGMOM`)? DFT can converge to a local minimum (non-magnetic) if started too close to zero. (3) *GGA self-interaction error:* For Co \\(3d\\) states, GGA may underestimate localisation. Try DFT+U with reasonable \\(U_{\rm eff} \sim 3\\)–\\(5\\) eV for Co. (4) *k\\)-mesh:* A coarse mesh for a metallic oxide surface may smear out the magnetic splitting. (5) *Surface vs. bulk:* Surface CoO may have quenched moments; compare to bulk CoO first.

Q11.2 [S] Rank the following functionals in increasing computational cost for a periodic solid with 100 atoms: LDA, PBE, SCAN, PBE0, HSE06. Briefly justify each step.

**Answer Hint.**

LDA \(<\) PBE \(\approx\) SCAN \(<\) HSE06 \(<\) PBE0.

  • LDA and PBE are semi-local: same cost, \(\mathcal{O}(N^3)\) dominated by diagonalisation.
  • SCAN is meta-GGA; needs \(\tau(\mathbf{r})\), slightly more expensive but same scaling.
  • HSE06 includes short-range exact exchange only: the Fock operator is localised in real space (exponential decay), so its evaluation scales \(\sim\mathcal{O}(N)\) for large systems and is tractable for 100-atom cells.
  • PBE0 includes long-range exact exchange: the Fock operator is non-local, requiring 4-index integrals over all reciprocal space. Scales \(\mathcal{O}(N^3)\) with a large prefactor, impractical for 100-atom periodic cells.

Q11.3 [L] A student is computing the vacancy formation energy in rutile TiO\(_2\) using VASP. List all the numerical convergence parameters that must be checked, the order in which to check them, and a physically motivated reason for each.

**Answer Hint.** In order:
  1. \(E_{\rm cut}\): Plane-wave basis completeness. For TiO\(_2\) with PAW, converge to \(\sim 1\) meV/atom; typical value \(\sim 500\)–\(600\) eV.
  2. \(k\)-mesh (bulk and supercell): BZ sampling. Use a \(\Gamma\)-centred mesh; the supercell mesh is coarser than bulk but must be tested independently.
  3. Supercell size: Vacancy–vacancy interaction through periodic images. Converge the formation energy with cell size (\(2\times2\times2\), \(3\times3\times2\), etc.); electrostatic correction schemes (Freysoldt–Neugebauer–Van de Walle) may be needed for charged vacancies.
  4. SCF convergence (EDIFF): Set to \(10^{-6}\)–\(10^{-7}\) eV for energy differences; forces need \(10^{-7}\) eV or tighter.
  5. Ionic relaxation (EDIFFG): Forces below \(0.01\)–\(0.02\) eV/Å; the vacancy geometry relaxes significantly.
  6. Smearing and \(\sigma\): TiO\(_2\) is a semiconductor; use ISMEAR = 0 (Gaussian) or ISMEAR = -5 (tetrahedron). Check that EENTRO \(< 1\) meV/atom.
  7. Spin polarisation: Ti \(3d^0\) nominally; but oxygen vacancies can create \(d^1\) Ti, which is magnetic. Run spin-polarised (ISPIN = 2) to be safe.
  8. (Optional) DFT+U: GGA underestimates the TiO\(2\) band gap (\(\sim 1.8\) vs \(3.0\) eV experimental); a \(U{\rm eff}\) on Ti \(d\) states may be needed for reliable defect levels.

Quick-Reference Formula Sheet

The following identities are cited frequently in examination answers. You should be able to derive each from first principles.

QuantityExpression
Electron density from KS orbitals\(\rho(\mathbf{r}) = \sum_i f_i |\phi_i(\mathbf{r})|^2\)
KS effective potential\(V_{\rm eff} = V_{\rm ext} + V_{\rm H} + V_{\rm xc}\)
Spin-dependent \(V_{\rm eff}\)\(V_{\rm eff}^\sigma = V_{\rm ext} + V_{\rm H}[\rho] + V_{\rm xc}^\sigma[\rho_\uparrow,\rho_\downarrow]\)
Hartree potential\(V_{\rm H}(\mathbf{r}) = \int \rho(\mathbf{r}’)/
XC potential\(V_{\rm xc}(\mathbf{r}) = \delta E_{\rm xc}[\rho]/\delta\rho(\mathbf{r})\)
Variational He trial energy\(E(\alpha) = \alpha^2 - (2Z - 5/8)\alpha\), minimum at \(\alpha_{\rm opt} = Z - 5/16\)
LDA exchange (Dirac)\(\varepsilon_{\rm x}^{\rm UEG} = -(3/4)(3/\pi)^{1/3}\rho^{1/3}\)
XC hole sum rule\(\int n_{\rm xc}(\mathbf{r},\mathbf{r}‘), d\mathbf{r}’ = -1\)
Dudarev DFT+U correction\(E_U = \frac{U_{\rm eff}}{2}\sum_{I,\sigma}\mathrm{Tr}[\mathbf{n}^{I\sigma}(\mathbf{1}-\mathbf{n}^{I\sigma})]\)
Plane-wave cutoff condition\(\frac{1}{2}
Kerker preconditioner\(G^2/(G^2 + k_0^2)\)
PAW density\(\rho = \tilde{\rho} + \rho^1 - \tilde{\rho}^1\)