level all intro standard hard
kind all mcq numeric predict short-answer
topic all basis bit-tricks bitmask ci competitive-programming complexity contest-craft correlation cpp debugging deliberate-practice density determinants digit-dp dynamic-programming fenwick full-ci hartree-fock integrals koopmans lambdas lazy move-semantics mp2 prefix-sums quantum-chemistry raii range-queries roothaan-hall rule-of-five scf second-quantization segment-tree smart-pointers sparse-table spin state-design strategy stress-testing templates tree-dp uhf value-categories variational
91 problems
hard Multiple choice
In a lazy-propagation segment tree, what does a node's lazy tag represent, and what invariant must hold?
○ An update already folded into the node's own aggregate, but not yet pushed down to its children. ○ An update that has not yet been applied to anything, including the node itself. ○ The node's current aggregate value, cached for speed. ○ The count of queries still waiting to touch this node.
hard Multiple choice
You need, for EVERY node of a tree, the sum of distances from that node to all other nodes. A separate DFS from each node is O(n²) and too slow for n = 10⁵. Which technique computes all n answers in O(n)?
○ Memoizing the per-node DFS. ○ Rerooting: solve for one root, then slide the root along each edge with an O(1) update. ○ Running Dijkstra from every node. ○ Floyd-Warshall all-pairs shortest paths.
hard Multiple choice
CISD keeps single and double excitations from the HF reference. Given Brillouin's theorem, what role do the singles actually play in the CISD ground-state energy?
○ Almost none directly — singles don't couple to HF (Brillouin), so they enter the ground state only indirectly, through their coupling to the doubles; the doubles carry the correlation. ○ Singles provide the bulk of the correlation energy. ○ Singles are excluded from CISD entirely. ○ Singles double-count the HF energy.
hard Multiple choice
For TWO non-interacting H₂ molecules, FCI gives exactly 2× one H₂ but CISD gives a higher energy. The missing piece is one specific determinant. Which?
○ The simultaneous double-on-both-monomers — a quadruple excitation from the supersystem reference, which singles-and-doubles cannot reach. ○ A single excitation on one monomer. ○ The HF determinant itself. ○ A charge-transfer determinant between the two monomers.
hard Multiple choice
Stretching H₂ toward dissociation makes the correlation problem qualitatively harder than it is at equilibrium. Which kind of correlation dominates there?
○ Static correlation: two determinants become nearly degenerate, and no single determinant is even qualitatively right. ○ Dynamic correlation: the electrons' short-range avoidance of each other intensifies. ○ Relativistic effects from the heavier reduced mass. ○ Basis-set superposition error.
hard Multiple choice
UHF rescues the H₂ dissociation curve by letting the α and β electrons take different spatial orbitals. What does it pay for this?
○ Spin contamination: the wavefunction stops being an eigenfunction of S², so a 'singlet' picks up triplet character and ⟨S²⟩ drifts above 0. ○ It violates the variational principle. ○ Its energy is always higher than RHF's. ○ Charge is no longer conserved.
hard Multiple choice
Mulliken charges are famously basis-set sensitive — add diffuse functions and the numbers can change wildly. What is the structural reason?
○ The scheme assigns density to whichever atom's basis function carries it; a diffuse function 'on' atom A can describe density physically sitting on atom B, and overlap population is split 50/50 by fiat. ○ Diffuse functions break the normalization of the density matrix. ○ Mulliken charges require orthogonal basis sets. ○ The Boys function loses precision for diffuse exponents.
hard Multiple choice
For a closed-shell system with two electrons in one spatial orbital φ (opposite spins), the exchange integral between them…
○ vanishes, because the two electrons have opposite spin and ⟨α|β⟩ = 0. ○ equals the Coulomb integral exactly. ○ is small but nonzero. ○ is negative and cancels the kinetic energy.
hard Multiple choice
Is the total Hartree-Fock energy equal to the sum of the occupied orbital energies Σεᵢ?
○ No — Σεᵢ double-counts the electron-electron repulsion, so a correction is needed. ○ Yes — the total energy is exactly Σεᵢ. ○ Yes, but only for closed-shell molecules. ○ No — the total energy is always exactly twice Σεᵢ.
hard Multiple choice
Restricted Hartree-Fock (RHF) famously fails as H₂ is pulled apart. The reason is that…
○ a single closed-shell determinant forces the bonding orbital to be doubly occupied, mixing in ionic H⁺H⁻ even at large separation. ○ the basis set runs out of functions at large distances. ○ the nuclear repulsion energy diverges. ○ the SCF always fails to converge for stretched bonds.
hard Multiple choice
FCI is often called 'exact'. Exact in what sense?
○ It is the exact solution within the chosen basis set — basis-incompleteness error still remains. ○ It is the exact solution of the Schrödinger equation, with no remaining error. ○ It is exact only for one-electron systems. ○ It is exact only at the Hartree-Fock geometry.
hard Multiple choice
FCI is variational and size-consistent . Which statement about the FCI energy is therefore true?
○ E_FCI ≤ E_exact(basis), and it lies at or below every truncated CI energy in the same basis. ○ E_FCI can drop below the true ground-state energy of the real system. ○ E_FCI is always above the Hartree-Fock energy. ○ E_FCI depends on the order in which determinants are listed.
hard Multiple choice
Koopmans' theorem makes two crude approximations, yet its ionization potentials are often surprisingly decent. Why?
○ Its two errors push in opposite directions: ignoring orbital relaxation overestimates the IP, while ignoring correlation underestimates it — and they partially cancel. ○ Both approximations are individually negligible for most molecules. ○ Hartree-Fock orbitals are exact for cations. ○ The basis set automatically compensates for the missing physics.
hard Multiple choice
Møller-Plesset perturbation theory splits the Hamiltonian as H = H₀ + V. What is H₀, and what is the perturbation?
○ H₀ is the sum of Fock operators; the perturbation is the difference between the true electron repulsion and the mean field — the fluctuation HF averaged away. ○ H₀ is the kinetic energy; the perturbation is all of the potential energy. ○ H₀ is the one-electron core Hamiltonian; the perturbation is the full electron-electron repulsion. ○ H₀ is the exact Hamiltonian; the perturbation is the basis-set error.
hard Multiple choice
The MP2 energy is a sum of terms |coupling|² / (εᵢ + εⱼ − εₐ − εᵦ). Which systems should make you distrust it?
○ Small-gap systems (stretched bonds, near-degeneracies): the denominator approaches zero, the 'small' correction blows up, and the perturbation series itself stops converging. ○ Large molecules, because the sum has too many terms to be accurate. ○ Closed-shell molecules near equilibrium. ○ Any charged species.
hard Multiple choice
Two Gaussian primitives have exponents α (large) and β (small) on centres A and B. Their product Gaussian sits at P. Where is P?
○ Shifted toward A, the centre with the larger exponent. ○ Exactly midway between A and B. ○ Shifted toward B, the centre with the smaller exponent. ○ Outside the segment AB entirely.
hard Multiple choice
The Roothaan-Hall equations take the form FC = SCε (a generalized eigenvalue problem) rather than FC = Cε. The overlap matrix S appears because…
○ the basis functions are non-orthogonal, so S = ⟨gⱼ|gᵢ⟩ ≠ I. ○ the Fock operator is non-Hermitian. ○ the electrons are indistinguishable. ○ the molecular orbitals must be normalized to two electrons each.
hard Multiple choice
The converged SCF electronic energy is computed as E = ½ tr[P(h + F)] rather than tr[P h]. Why the factor of ½ and the (h + F)?
○ F already contains the two-electron terms, so averaging h and F and halving avoids double-counting the electron-electron repulsion. ○ It converts Hartree to electron-volts. ○ Half the electrons are spin-up. ○ It accounts for the nuclear repulsion.
hard Multiple choice
Why are the Slater-Condon rules said to 'fall out for free' in second quantization?
○ Evaluating ⟨Φ|H|Φ′⟩ means pushing annihilation operators to the right via the anticommutators; a term survives only when operators pair up, which is exactly when the determinants differ in ≤2 orbitals — the three cases, signs, and spectator sums all forced by the algebra. ○ Because the Hamiltonian is diagonal in the determinant basis. ○ Because second quantization ignores the two-electron terms. ○ They don't — Slater-Condon must still be applied by hand.
hard Multiple choice
A single Slater determinant with one α and one β electron in different spatial orbitals (|φ₁α φ₂β|) is…
○ Not a spin eigenfunction at all — it is an equal mixture of the singlet and the Ms = 0 triplet. ○ A pure singlet, because the spins are opposite. ○ A pure triplet, because the orbitals are different. ○ An eigenfunction of S² with eigenvalue ½.
hard Multiple choice
For H₂'s σ→σ excited states, the triplet lies below* the singlet by exactly 2K_ov = 0.363 hartree (in STO-3G). What pays for the triplet's discount?
○ Exchange: same-spin electrons get the −K term, and the antisymmetry that produces it also keeps them spatially apart, lowering their repulsion. ○ The triplet has lower kinetic energy. ○ The triplet feels weaker nuclear attraction. ○ Magnetic dipole interactions between the parallel spins.
hard Multiple choice
Schwarz screening lets a code skip computing an ERI (ab|cd) when…
○ the charge distributions ab and cd are far apart, so the Gaussian-product prefactor makes the integral negligibly small. ○ the four basis functions are on the same atom. ○ the molecule has no symmetry. ○ the basis set is large.
hard Multiple choice
The Coulson-Fischer point of H₂ (≈2.1 bohr in this basis) marks the geometry where…
○ a spin-symmetry-broken UHF solution first drops below the RHF solution — paying spin purity for energy starts to be a good trade. ○ the SCF stops converging. ○ the molecule dissociates completely. ○ the overlap S₁₂ reaches exactly ½.
hard Multiple choice
UHF dissociates H₂ to the right energy (two free atoms) with the wrong wavefunction (⟨S²⟩ ≈ 1). For which prediction would this contamination actually bite?
○ Anything that distinguishes singlet from triplet — singlet-triplet gaps, magnetic properties, spin densities — and post-HF methods built on the contaminated reference (spin-contaminated MP2 is notoriously poor). ○ Nothing — energy is the only observable in quantum chemistry. ○ Only the nuclear repulsion energy. ○ The Mulliken charges, which become complex numbers.
intro Multiple choice
A local object is created on the stack inside a function. When does its destructor run?
○ When execution leaves the enclosing scope — by any path, including an exception. ○ When the garbage collector next runs. ○ Only if you call delete on it. ○ At program exit.
intro Multiple choice
In int x = 5; int& r = x;, which expression below is an lvalue ?
○ x ○ 5 ○ x + 1 ○ std::move(x)
intro Predict + reveal
Using the prefix-sum identity sum(l, r) = P[r+1] − P[l], compute sum(2, 4). Work it out from the arrays below before revealing.
a = [3, 1, 4, 1, 5] # indices 0..4
P = [0, 3, 4, 8, 9, 14] # P[i] = sum of a[0..i-1]
sum(2, 4) = ?
Reveal answer
intro Multiple choice
The variational principle says that for any trial wavefunction ψ, the Rayleigh quotient ⟨ψ|H|ψ⟩ / ⟨ψ|ψ⟩ is…
○ an upper bound on the ground-state energy E₀. ○ a lower bound on the ground-state energy E₀. ○ exactly equal to E₀ for any normalized ψ. ○ unrelated to E₀ unless ψ is an eigenstate.
standard Predict + reveal
Does this compile? If not, why? Predict before revealing.
int a = 3, b = 4;
int* p = &(a + b);
std::cout << *p;
Reveal answer
standard Predict + reveal
Three guard objects are constructed in order. What is the destruction order at the closing brace? Predict the full output.
struct Guard {
const char* name;
Guard(const char* n) : name(n) { std::cout << "ctor " << name << "\n"; }
~Guard() { std::cout << "dtor " << name << "\n"; }
};
int main() {
Guard a("A");
Guard b("B");
Guard c("C");
}
Reveal answer
standard Predict + reveal
What does this print? Look closely at how n is captured.
int n = 10;
auto f = [n]() { return n + 1; };
n = 100;
std::cout << f();
Reveal answer
standard Multiple choice
Under the hood, what is a C++ lambda expression?
○ A compiler-generated unnamed class (the closure type) with an overloaded operator() and the captures stored as data members. ○ A function pointer. ○ A preprocessor macro expanded inline. ○ Always a std::function.
standard Multiple choice
Why prefer std::make_unique<T>(args...) over std::unique_ptr<T>(new T(args...))?
○ It does the allocation and the wrapping in one step, so a raw owning pointer never exists to leak — and it keeps new out of your code. ○ It makes the resulting pointer copyable. ○ It allocates on the stack instead of the heap. ○ It disables the destructor for speed.
standard Predict + reveal
What does this print? Pay attention to the state of a after the move.
std::vector<int> a = {1, 2, 3};
std::vector<int> b = std::move(a);
std::cout << b.size() << " " << a.size();
Reveal answer
standard Short answer
Name the special member function that constructs a new object by stealing the resources of an expiring one. Two words.
standard Multiple choice
Why does wrapping a mutex lock in a stack guard (std::lock_guard) make it exception-safe, where a manual lock() / unlock() pair is not?
○ If an exception is thrown between lock and unlock, stack unwinding runs the guard's destructor, which unlocks. A manual unlock() is skipped by the throw. ○ lock_guard catches the exception for you. ○ lock_guard makes the critical section atomic. ○ Manual unlock() is slower, so it times out.
standard Multiple choice
The Rule of Five says that if you write one of a certain set of special member functions, you should think about all five. Which set?
○ Destructor, copy constructor, copy assignment, move constructor, move assignment. ○ Constructor, destructor, copy, move, and swap. ○ The five SOLID design principles. ○ operator new, delete, new[], delete[], and placement new.
standard Multiple choice
What does the Rule of Zero recommend, and when does it apply?
○ Design classes so you write none of the five — let resource-managing members (smart pointers, containers) supply correct copy/move/destroy for free. ○ Never write a destructor under any circumstances. ○ Always = default all five special members explicitly. ○ Avoid classes and use free functions only.
standard Predict + reveal
What does this print? The template parameter is taken by value .
template<class T> void bump(T x) { x += 1; }
int a = 5;
bump(a);
std::cout << a;
Reveal answer
standard Multiple choice
Given template<class T> T add(T a, T b) { return a + b; }, you call add(1, 2) and add(1.5, 2.5). How many concrete versions of add does the compiler generate?
○ Two — add<int> and add<double>, instantiated at compile time. ○ One generic version, reused for both at runtime. ○ Zero — templates are resolved by the linker. ○ One per call site, so two here but four if you call each twice.
standard Multiple choice
What happens here?
auto a = std::make_unique<int>(5);
std::unique_ptr<int> b = a; // <-- this line
○ Compile error — unique_ptr's copy constructor is deleted; it is move-only. ○ a and b end up sharing ownership of the same int. ○ b gets a deep copy — a separate int with value 5. ○ a silently becomes null and b takes ownership.
standard Multiple choice
What does std::move(x) actually do at runtime?
○ Nothing — it's a compile-time cast to an rvalue reference (static_cast<T&&>). ○ It moves the contents of x into a hidden temporary. ○ It frees the memory owned by x. ○ It zeroes out x.
standard Multiple choice
Bitmask DP for the traveling salesman runs in O(2ⁿ · n²). Under a typical contest budget of about 10⁸ basic operations, what is the largest n that fits comfortably?
○ about 10 ○ about 20 ○ about 40 ○ about 60
standard Multiple choice
In digit DP (building a number left-to-right under an upper bound), the 'tight' flag is true at a position exactly when...
○ the digits placed so far exactly match the prefix of the upper bound, so the next digit is capped at the bound's digit. ○ the number built so far is even. ○ the running accumulator has already reached its target value. ○ every remaining digit is forced to be 9.
standard Multiple choice
For "maximum sum of a subsequence with no two chosen elements adjacent" (house robber) over an array, what is a sufficient and minimal DP state?
○ (current index) alone. ○ (current index, whether the previous element was taken). ○ (current index, running sum so far, set of taken indices). ○ (the full subset of indices chosen so far).
standard Multiple choice
A contestant codes 0/1 knapsack with the state (item index, capacity used, current total value) and gets Memory Limit Exceeded. Which dimension is redundant and causing the blowup?
○ item index ○ capacity used ○ current total value ○ none — all three are genuinely needed
standard Predict + reveal
A Fenwick query walks i down to 0 by repeatedly subtracting the lowest set bit: i -= i & (-i), summing bit[i] each step. Which indices does query(11) visit? (11 = 1011 in binary.) List them before revealing.
long long query(int i) {
long long s = 0;
for (; i > 0; i -= i & -i)
s += bit[i];
return s;
}
// query(11): which i values?
Reveal answer
standard Multiple choice
For a positive integer i in two's-complement, what does the expression i & (-i) compute?
○ i with its lowest set bit cleared. ○ The lowest set bit of i, as a power of two (this is "lowbit"). ○ The highest set bit of i. ○ The number of set bits (popcount) of i.
standard Multiple choice
Which practice habit most efficiently raises your competitive-programming rating over a season?
○ Solving a large volume of easy problems you already know how to do. ○ Upsolving the problems you couldn't finish in contests, then drilling those themes until they're reflex. ○ Reading editorials before attempting, to learn the solutions faster. ○ Competing constantly and never practicing between contests.
standard Multiple choice
You must apply many range updates of the form "add v to every a[i] for i in [l, r]" and read the array only once, at the end. Using a difference array d (1-indexed, sized n+1, all zero), what is the correct O(1) action per update?
○ d[l] += v; d[r] -= v ○ d[l] += v; d[r+1] -= v ○ d[l-1] += v; d[r] -= v ○ d[l] += v; d[r] += v
standard Multiple choice
A problem has n ≤ 2·10⁵ and a 1-second time limit. Which intended time complexity should you assume when designing the solution?
○ O(n²) — about 4·10¹⁰ operations. ○ O(n log n) — sort, segment tree, or binary search. ○ O(2ⁿ) — a bitmask over all subsets. ○ O(n³) — three nested loops.
standard Multiple choice
Both a Fenwick tree and a segment tree do point update + range query in O(log n). Which task specifically needs a segment tree, because a plain Fenwick tree cannot do it directly?
○ Range sum with point updates. ○ Prefix sum queries on a static array. ○ Range minimum with point updates. ○ Adding a constant to a single element.
standard Multiple choice
A sparse table answers range minimum in O(1) by combining two power-of-two blocks that overlap in the middle. The very same structure does NOT work for range sum. Why?
○ Sum is not associative, so the blocks cannot be combined. ○ The two blocks overlap, and sum would count the overlapping elements twice; min is idempotent, so overlap is harmless. ○ Sparse tables require the array to be sorted, and sums need it unsorted. ○ Sum changes under updates, but a sparse table only stores minima.
standard Multiple choice
Your clever solution gets Wrong Answer on a hidden test, somewhere among 10⁵ cases you cannot see. What is the most reliable way to find the failing case?
○ Re-read your code line by line until you spot the bug. ○ Generate random SMALL inputs, run both your solution and a trusted brute force, and diff their outputs until they disagree. ○ Resubmit a few more times in case the judge was wrong. ○ Add print statements and inspect the judge's hidden tests.
standard Multiple choice
In a stress-testing harness, why should the random generator be bounded to produce SMALL inputs (e.g. n ≤ 8) rather than full-size ones?
○ Small inputs run faster, so you can try more of them. ○ The first mismatch is then a minimal counterexample you can trace by hand; the same bug on a huge input is undebuggable. ○ The brute force only works on small inputs anyway. ○ Large inputs cannot trigger logic bugs, only performance bugs.
standard Multiple choice
Forty minutes into a contest you have spent all of it on problem B with no working idea, and problems C and D are still unread. What is the best move?
○ Keep going on B — you've already invested forty minutes, so quitting wastes them. ○ Switch now: read C and D, bank whichever is easiest, and come back to B with a fresh head. ○ Resubmit your best guess on B to see if it passes. ○ Give up on the contest.
standard Numeric
Rerooting sum-of-distances uses ans[v] = ans[u] + (n − 2·sz[v]) when moving the root from u to an adjacent child v. A tree has n = 6 nodes; the subtree rooted at v has sz[v] = 2; and ans[u] = 9. Compute ans[v].
standard Multiple choice
Why does quantum chemistry build orbitals out of Gaussians rather than the more physically-accurate Slater functions?
○ Gaussians reproduce the nuclear cusp and long-range tail more accurately. ○ Products of Gaussians collapse to a single Gaussian, making multi-centre integrals analytic. ○ Gaussians require fewer functions per atom than Slater orbitals. ○ Slater orbitals cannot represent s-type orbitals.
standard Multiple choice
The Gaussian product theorem states that the product of two s-type Gaussians centred at A and B is…
○ a single Gaussian centred at the weighted point P = (αA + βB)/(α+β), times a constant. ○ a sum of two Gaussians, one at A and one at B. ○ a Gaussian centred exactly midway between A and B regardless of exponents. ○ zero unless A = B.
standard Multiple choice
In an STO-3G calculation, the three contraction coefficients that combine the primitive Gaussians into one basis function are…
○ fixed, tabulated values (fitted once to a Slater orbital) that don't change during the calculation. ○ optimized during each SCF iteration. ○ the unknowns the Roothaan-Hall equations solve for. ○ recomputed whenever the molecular geometry changes.
standard Multiple choice
What physical feature of a true atomic orbital is lost when you approximate it with a sum of Gaussians?
○ The sharp cusp at the nucleus and the correct exponential long-range tail. ○ The number of radial nodes. ○ The total electron count. ○ The spin of the electron.
standard Multiple choice
The Davidson correction ΔE_Q ≈ (1 − c₀²)(E_CISD − E_HF) estimates the missing quadruples. What does a SMALL c₀ (HF-coefficient) signal?
○ Strong double-excitation character in the wavefunction — exactly the regime where disconnected quadruples (a double atop a double) matter most, so the correction grows. ○ That the SCF failed to converge. ○ That the basis set is too small. ○ That the molecule is at its equilibrium geometry with little correlation.
standard Multiple choice
The density matrix P = 2C_occ C_occᵀ compresses the wavefunction. What does its trace with the overlap matrix, tr(PS), always equal?
○ The number of electrons. ○ The total energy. ○ The number of basis functions. ○ Zero, by orthonormality.
standard Multiple choice
HeH⁺'s dipole moment is 0.889 a.u. with the origin at He . For H₂O the origin never needs stating. What's the difference?
○ An origin shift changes the dipole by (total charge) × (shift) — zero for neutral water, but ±1·d for the cation, so charged-species dipoles are origin-dependent bookkeeping. ○ Water's symmetry fixes the origin automatically. ○ Cations have no electronic contribution to the dipole. ○ It is a basis-set artifact of STO-3G.
standard Multiple choice
Why is the Hartree product ψ(r₁,r₂) = φ(r₁)φ(r₂) an unacceptable wavefunction for two electrons?
○ It is symmetric under exchange of the two electrons, but fermions require an antisymmetric wavefunction. ○ It is not normalizable. ○ It puts the two electrons in different orbitals. ○ It violates conservation of energy.
standard Multiple choice
A Slater determinant automatically encodes the Pauli exclusion principle because…
○ swapping two rows (electrons) flips its sign, and two identical columns (spin-orbitals) make it vanish. ○ it normalizes to one. ○ it is built only from s-type orbitals. ○ it assigns every electron the same spin.
standard Multiple choice
The Slater–Condon rules are useful because they…
○ reduce the energy of a Slater determinant to a short sum of one- and two-electron integrals over the occupied orbitals. ○ prove that the Hartree product and Slater determinant give the same energy. ○ are only valid for a single electron. ○ eliminate the need for any two-electron integrals.
standard Multiple choice
What does it mean that Hartree-Fock is a mean-field theory?
○ Each electron moves in the averaged field of all the others, not in their instantaneous positions. ○ Each electron is treated as completely independent, with no electron-electron interaction at all. ○ The electrons are smeared into a single classical charge cloud with no orbitals. ○ All electrons share one wavefunction with no antisymmetry.
standard Multiple choice
The correlation energy is defined as…
○ E_exact − E_HF: the energy Hartree-Fock misses by using a mean field. ○ E_HF − E_exact, a positive quantity HF overshoots by. ○ the sum of the occupied orbital energies. ○ the nuclear repulsion energy.
standard Multiple choice
What is the Full Configuration Interaction (FCI) wavefunction?
○ A linear combination of all Slater determinants you can form from the orbitals: |Ψ⟩ = Σ_I c_I |Φ_I⟩. ○ The single best Slater determinant, optimized variationally. ○ A product of one-electron orbitals with no antisymmetry. ○ The Hartree-Fock determinant with a correlation correction added perturbatively.
standard Multiple choice
Why is FCI limited to small systems (roughly 10–20 electrons in modest bases)?
○ The number of determinants grows combinatorially (≈ exponentially) with electrons and basis size. ○ The two-electron integrals stop being computable. ○ The variational principle no longer holds. ○ Floating-point precision runs out.
standard Numeric
H₂ in the 6-31G basis has M = 4 spatial orbitals, one α electron, and one β electron. How many determinants does the FCI expansion contain? (Use N = C(M, Nα) · C(M, Nβ).)
standard Multiple choice
According to Koopmans' theorem, the ionization potential of a molecule is approximately…
○ −ε_HOMO: the negative of the highest occupied orbital energy, assuming the remaining orbitals don't relax. ○ +ε_HOMO, the highest occupied orbital energy itself. ○ ε_LUMO − ε_HOMO, the orbital energy gap. ○ The total Hartree-Fock energy divided by the number of electrons.
standard Multiple choice
For H₂/STO-3G the MP2 correction is −13.2 mHa, computed from a single excitation. What made the minimal basis collapse the whole MP2 sum to one term?
○ There is exactly one occupied and one virtual orbital, so (ia|jb) has only one index choice: the σ→σ* double excitation. ○ All other terms cancel by spin symmetry. ○ The Boys function vanishes for the other terms. ○ MP2 always has one term for closed-shell molecules.
standard Multiple choice
The Boys function F₀(x) shows up in which integrals when using a Gaussian basis?
○ Any integral involving the Coulomb operator — including one-electron nuclear-attraction integrals. ○ Only the two-electron repulsion integrals. ○ Only the overlap integrals. ○ Only the kinetic-energy integrals.
standard Multiple choice
Why do the overlap and kinetic-energy integrals between two Gaussians on different centres have simple closed forms?
○ The Gaussian product theorem collapses the two-centre product to one Gaussian, which integrates analytically. ○ They are computed by numerical quadrature on a grid. ○ They are always zero by symmetry. ○ They require the Boys function to evaluate.
standard Multiple choice
Diagonalizing the Roothaan-Hall equations in an N-function basis returns N orbitals. The N/2 with the lowest energies are occupied. What are the rest?
○ Virtual (unoccupied) orbitals — a discrete image of the excited spectrum, essential for post-HF correlation methods. ○ Numerical artifacts that should be discarded. ○ Duplicate copies of the occupied orbitals. ○ Core orbitals frozen out of the calculation.
standard Multiple choice
In practice, how do you solve the generalized eigenvalue problem FC = SCε for a non-orthogonal Gaussian basis?
○ Hand it directly to a generalized eigensolver (e.g. scipy.linalg.eigh(F, S)). ○ It cannot be solved; you must use a different basis. ○ Invert the Fock matrix and read off the orbitals. ○ Only by iterating the SCF — there is no direct solve.
standard Multiple choice
Why must Hartree-Fock be solved by iteration (the SCF cycle) rather than in one shot?
○ The Fock matrix depends on the orbitals (through the density), but the orbitals come from diagonalizing the Fock matrix — a circular dependence. ○ The basis set changes every step. ○ Floating-point error forces repeated refinement. ○ The nuclei move during the calculation.
standard Multiple choice
The SCF procedure is best described as…
○ a fixed-point iteration: keep applying the map P → F[P] → C → P′ until the density stops changing. ○ a single matrix inversion. ○ a Monte Carlo sampling of electron positions. ○ a minimization that needs no starting guess.
standard Multiple choice
A common SCF starting guess is the core Hamiltonian guess. What does it ignore in the first step?
○ Electron-electron repulsion — it keeps only kinetic energy and nuclear attraction, then the iteration adds repulsion back. ○ The nuclear attraction term. ○ The kinetic energy term. ○ The basis set.
standard Multiple choice
In second quantization, what does the number operator a_p† a_p return when applied to a determinant?
○ 1 if spin-orbital p is occupied in that determinant, 0 if it is empty — it counts occupation. ○ The energy of orbital p. ○ Always 1, since it creates and then annihilates. ○ The total number of electrons.
standard Multiple choice
The anticommutator {a_p†, a_q†} = 0. Setting p = q gives (a_p†)² = 0. Which physical law is this?
○ The Pauli exclusion principle — you cannot create two electrons in the same spin-orbital. ○ Conservation of charge. ○ The variational principle. ○ Koopmans' theorem.
standard Multiple choice
⟨S²⟩ for a UHF wavefunction is computed as Ms(Ms+1) + Nβ − Σ|⟨φᵢα|φⱼβ⟩|². When does it reduce to the exact singlet value of 0?
○ When every β orbital exactly coincides with an α orbital — i.e. when UHF has collapsed back to RHF. ○ Whenever Nα = Nβ. ○ Only in a complete basis set. ○ Never — UHF always has ⟨S²⟩ > 0.
standard Multiple choice
How does the number of two-electron repulsion integrals (ERIs) scale with the number of basis functions N?
○ O(N⁴) — they carry four basis-function indices (ab|cd). ○ O(N²). ○ O(N³). ○ O(N!).
standard Multiple choice
Real-orbital ERIs obey an 8-fold permutation symmetry, e.g. (ab|cd) = (ba|cd) = (cd|ab) = …. Practically, this means a code can…
○ compute only about 1/8 of the integrals and recover the rest by symmetry. ○ skip the two-electron integrals entirely. ○ reduce the formal scaling from O(N⁴) to O(N²). ○ avoid ever building the Fock matrix.
standard Multiple choice
In chemist notation, what does the two-electron integral (ab|cd) represent?
○ The Coulomb repulsion between charge distribution a(r₁)b(r₁) for electron 1 and c(r₂)d(r₂) for electron 2. ○ The overlap of four basis functions at one point. ○ A one-electron kinetic-energy integral over four centres. ○ The product of two orbital energies.
standard Multiple choice
Pople-Nesbet UHF builds two Fock matrices. In Fα, the exchange term subtracts only Pα — never Pβ. Why?
○ Exchange exists only between same-spin electrons, so an α electron exchanges with the α density alone — while the Coulomb term uses the total density. ○ Pβ is already counted in the nuclear attraction. ○ It is an approximation to halve the cost. ○ Pα and Pβ are identical, so the choice is arbitrary.
standard Multiple choice
For hydrogen, the trial orbital ψ(r) = e^{−ζ|r|} with ζ as a parameter makes the variational calculation a…
○ linear variational problem (a matrix eigenvalue problem). ○ nonlinear variational problem, because the energy depends nonlinearly on ζ. ○ problem with no minimum, since any ζ works equally well. ○ problem that requires the exact analytic wavefunction to start.
standard Multiple choice
You compute the energy of a deliberately crude trial wavefunction and get −0.40 Hartree for hydrogen (exact: −0.50). What can you conclude?
○ Nothing — a crude guess gives no usable information. ○ The true ground state lies at or below −0.40 Hartree, and a better guess can only lower the estimate. ○ The calculation must be wrong, since −0.40 is above −0.50. ○ The true ground state must be above −0.40 Hartree.
standard Multiple choice
Optimizing ζ in the trial ψ = e^{−ζ|r|} for hydrogen gives E = −0.5 hartree — the exact ground-state energy, exactly. Why does the variational method land on the truth here?
○ The trial family happens to contain the exact wavefunction (ζ = 1 is the hydrogen 1s orbital), so the minimum finds it. ○ The variational method is always exact for one-parameter families. ○ Hydrogen has no electron-electron repulsion, so any trial function gives the exact energy. ○ It's a numerical coincidence specific to atomic units.