Question Bank

C1000-179 style. 20 questions, single best answer (A–D). No answers in this file — see solution-bank.md.

Q1. In Qiskit, Pauli('IZ') applies the Z operator to which qubit?

• A. Qubit 1, because 'Z' is the second character
• B. Qubit 0, because Pauli strings are little-endian (rightmost character = qubit 0)
• C. Both qubits
• D. Neither; 'IZ' is an invalid label

Q2. What is the diagonal of Pauli('IZ').to_matrix()?

• A. [1, 1, -1, -1]
• B. [1, -1, -1, 1]
• C. [1, -1, 1, -1]
• D. [-1, 1, -1, 1]

Q3. What does Pauli('-iXY').phase return?

• A. 0
• B. 1
• C. 2
• D. 3

Q4. What is the output of the following?

from qiskit.quantum_info import Pauli
print(Pauli('X').compose(Pauli('Y')))

• A. iZ
• B. Z
• C. -iZ
• D. XY

Q5. What does Pauli('X') @ Pauli('Z') evaluate to (@ is the dot product, X·Z)?

• A. iY
• B. -iY
• C. Y
• D. -Y

Q6. What does this print?

from qiskit.quantum_info import SparsePauliOp
op = SparsePauliOp.from_list([("XX", 1.0), ("XX", 1.0), ("ZZ", 0.0)])
print(op.simplify())

• A. SparsePauliOp(['XX', 'ZZ'], coeffs=[2.+0.j, 0.+0.j])
• B. SparsePauliOp(['XX'], coeffs=[2.+0.j])
• C. SparsePauliOp(['XX', 'XX', 'ZZ'], coeffs=[1.+0.j, 1.+0.j, 0.+0.j])
• D. SparsePauliOp(['XX'], coeffs=[1.+0.j])

Q7. Which Pauli string does this produce?

SparsePauliOp.from_sparse_list([("ZX", [1, 4], 2.0)], num_qubits=5)

• A. 'IZIIX'
• B. 'ZXIII'
• C. 'XIIZI'
• D. 'IZIXI'

Q8. After op = SparsePauliOp.from_list([("Z", 1.0)]), what is the dtype of op.coeffs?

• A. float64, since 1.0 is a float
• B. complex128, coefficients are always stored as complex
• C. int64
• D. object

Q9. Statevector.from_label('10') has its single nonzero amplitude at which index of the statevector array?

• A. 0
• B. 1
• C. 2
• D. 3

Q10. What does this print?

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector
qc = QuantumCircuit(2)
qc.x(0)
print(Statevector.from_instruction(qc).probabilities_dict())

• A. {'10': 1.0}
• B. {'01': 1.0}
• C. {'00': 1.0}
• D. {'11': 1.0}

Q11. For the Bell state (|00⟩ + |11⟩)/√2, what is expectation_value(SparsePauliOp(['ZI']))?

• A. 1
• B. -1
• C. 0.5
• D. 0

Q12. Which circuit prepares the Bell state (|00⟩ + |11⟩)/√2 from |00⟩?

• A. qc.cx(0, 1); qc.h(0)
• B. qc.h(0); qc.cx(0, 1)
• C. qc.h(0); qc.h(1)
• D. qc.x(0); qc.cx(0, 1)

Q13. Which matrix is the S gate?

• A. [[1, 0], [0, -1]]
• B. [[1, 0], [0, e^{iπ/4}]]
• C. [[1, 0], [0, i]]
• D. (1/√2)[[1, 1], [1, -1]]

Q14. Applying the T gate twice in a row is exactly equal (not just up to phase) to which single gate?

• A. S
• B. Z
• C. X
• D. H

Q15. The circuit h(0); z(0); h(0) implements which single-qubit gate?

• A. Z
• B. Y
• C. H
• D. X

Q16. Given:

import numpy as np
from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator, Pauli
qc = QuantumCircuit(1)
qc.rz(np.pi, 0)
a = Operator(qc) == Operator(Pauli('Z'))
b = Operator(qc).equiv(Operator(Pauli('Z')))

What are a and b?

• A. a = True, b = True
• B. a = False, b = True
• C. a = True, b = False
• D. a = False, b = False

Q17. Which pair of operators commutes?

• A. Pauli('X') and Pauli('Z')
• B. Pauli('XI') and Pauli('ZI')
• C. Pauli('XX') and Pauli('ZZ')
• D. None of the above

Q18. Applying Z to the state |+⟩ produces |−⟩. What is the effect on the outcome probabilities of an immediate computational-basis (Z-basis) measurement?

• A. '0' becomes certain
• B. '1' becomes certain
• C. Probabilities are unchanged (still 50/50); only the relative phase changed
• D. The state acquires only a global phase, so nothing about the state changed at all

Q19. For bell = Statevector of (|00⟩ + |11⟩)/√2, what is partial_trace(bell, [1])?

• A. The pure state |0⟩⟨0|
• B. The maximally mixed state I/2, with purity 0.5
• C. The pure state |+⟩⟨+|
• D. An error — you cannot trace out part of an entangled state

Q20. What does this print (up to numerical precision)?

from qiskit.quantum_info import Statevector, Pauli
sv = Statevector.from_label('+0')
print(sv.expectation_value(Pauli('XI')), sv.expectation_value(Pauli('IX')))

• A. 1.0 0.0
• B. 0.0 1.0
• C. 1.0 1.0
• D. 0.0 0.0