Self-Assessment Quiz: Quantum Gates

Twenty questions on the Pauli group, Hadamard, phase gates, CNOT, Toffoli, universality, and gate identities. Aim for 16+.

Question 1

$X|0\rangle$ equals:

A) $|0\rangle$ B) $|1\rangle$ C) $|+\rangle$ D) $-|0\rangle$

Question 2

$Z|+\rangle$ equals:

A) $|+\rangle$ B) $|-\rangle$ C) $|0\rangle$ D) $|1\rangle$

Question 3

$HZH$ equals:

A) $X$ B) $Y$ C) $Z$ D) $I$

Question 4

$S^2$ equals:

A) $I$ B) $Z$ C) $X$ D) $T$

Question 5

$T^2$ equals:

A) $I$ B) $S$ C) $Z$ D) $H$

Question 6

Which gates are Hermitian (equal to their own inverse)?

A) $X, Y, Z, H$ B) $S, T$ C) All gates D) Only $I$

Question 7

CNOT with control $|+\rangle$ and target $|0\rangle$ produces:

A) $|+0\rangle$ B) A Bell state C) $|10\rangle$ D) $|++\rangle$

Question 8

The Toffoli (CCX) gate flips its target when:

A) Either control is 1 B) Both controls are 1 C) Both controls are 0 D) Always

Question 9

The Toffoli gate is significant because:

A) It is the fastest two-qubit gate B) It is universal for classical reversible computation C) It cannot be decomposed D) It is not unitary

Question 10

A universal gate set for quantum computation is:

A) $\{X, Z\}$ B) $\{H, T, \text{CNOT}\}$ C) $\{H, S, \text{CNOT}\}$ D) $\{\text{CNOT}\}$

Question 11

$\{H, S, \text{CNOT}\}$ generates the Clifford group, which is:

A) Universal B) Efficiently simulable classically (Gottesman–Knill) C) Not physically realizable D) Sufficient for Shor's algorithm

Question 12

The gate that promotes Clifford to universal is typically:

A) $X$ B) $T$ C) $Z$ D) SWAP

Question 13

SWAP can be decomposed into:

A) One CNOT B) Two CNOTs C) Three CNOTs D) It cannot be decomposed

Question 14

Any single-qubit unitary can be written (up to global phase) as:

A) $R_z(\gamma)R_y(\beta)R_z(\alpha)$ B) $R_x(\alpha)$ alone C) $H^n$ for some $n$ D) A product of Paulis only

Question 15

The Solovay–Kitaev theorem guarantees:

A) Exact synthesis of any unitary from a finite gate set B) Approximation to accuracy $\varepsilon$ using $O(\log^c(1/\varepsilon))$ gates C) That $T$ gates are free D) That all circuits can be made depth-1

Question 16

True or false: A global phase applied by a gate has observable consequences.

Question 17

True or false: CNOT can create entanglement from a product state.

Question 18

True or false: Every two-qubit unitary can be built from at most three CNOTs plus single-qubit gates.

Question 19

Short answer. Why is $T$ expensive in fault-tolerant architectures while $H$ and CNOT are cheap?

Question 20

Short answer. Explain why quantum gates must be reversible, and what that implies for implementing a classical AND.

Answer Key

Q Ans Note
1 B $X$ is the quantum NOT in the computational basis.
2 B $Z$ flips the sign of the $|1\rangle$ component, turning $|+\rangle$ into $|-\rangle$. $Z$ is a "bit flip" in the $X$ basis — the duality is the point.
3 A $H$ conjugation swaps $X$ and $Z$.
4 B $S = \sqrt Z$, so $S^2 = Z$.
5 B $T = \sqrt S$, so $T^2 = S$ and $T^4 = Z$, $T^8 = I$.
6 A Each squares to $I$. $S$ and $T$ do not — their inverses are $S^\dagger$ and $T^\dagger$.
7 B Control in superposition entangles: $(|00\rangle + |11\rangle)/\sqrt2$. This is the standard Bell recipe.
8 B AND of the controls.
9 B With ancillas, Toffoli implements any classical reversible circuit — the bridge between classical and quantum computation.
10 B Clifford + $T$. Option C is Clifford only, and hence not universal.
11 B Gottesman–Knill: Clifford circuits on stabilizer states are polynomially simulable classically. Entanglement alone therefore does not imply quantum advantage.
12 B Any non-Clifford gate suffices; $T$ is the conventional choice.
13 C Three alternating CNOTs. Because most hardware lacks a native SWAP, routing costs are counted in CNOTs — a major source of overhead on limited-connectivity devices.
14 A The Euler ZYZ decomposition.
15 B Efficient approximation, not exact synthesis — with polylogarithmic gate count in $1/\varepsilon$.
16 False Global phase is unobservable. It matters only when the gate is controlled, because then the global phase becomes a relative phase between control branches — the mechanism behind phase kickback (Ch. 11).
17 True That is exactly what Q7 demonstrates; entangling capability is what distinguishes two-qubit gates from products of single-qubit ones.
18 True The KAK/Cartan decomposition gives an exact three-CNOT upper bound for arbitrary two-qubit unitaries.
19 — Clifford gates (including $H$ and CNOT) can be implemented transversally on many codes — fault-tolerantly, with no extra machinery. $T$ cannot be (Eastin–Knill theorem), so it requires magic-state distillation, which consumes many physical qubits and many rounds. Consequently the $T$-count, not the total gate count, is the dominant cost metric in fault-tolerant resource estimates (Ch. 25).
20 — Gates are unitary, and unitaries are invertible, so no information may be destroyed. Classical AND is irreversible: output 0 does not determine the inputs. It is therefore implemented as a Toffoli, $(a, b, 0) \mapsto (a, b, a\wedge b)$, which keeps the inputs so the map stays invertible — at the cost of an ancilla, and of "garbage" that must be uncomputed to avoid destroying interference.