What is the effect of the Hadamard gate H on the basis state |0>?
A|1>
B(|0> - |1>)/sqrt(2)
C(|0> + |1>)/sqrt(2)
Di|0>
The Hadamard gate maps |0> to (|0> + |1>)/sqrt(2) = |+>, creating an equal superposition. It also maps |1> to (|0> - |1>)/sqrt(2) = |->. The Hadamard is its own inverse (H^2 = I), so applying it twice returns to the original state.
Question 2 True / False
All quantum gates must be reversible because they are represented by unitary matrices.
TTrue
FFalse
Answer: True
A unitary matrix U satisfies U*U^dagger = I, which means U^dagger is the inverse operation. Every quantum gate therefore has a well-defined reverse. This contrasts with classical gates like AND and OR, which are irreversible — you cannot recover the inputs from the output. Reversibility is a fundamental requirement of quantum mechanics, not just a design choice.
Question 3 Short Answer
What does the CNOT gate do to the two-qubit state |10>? What about |11>?
Think about your answer, then reveal below.
Model answer: CNOT|10> = |11> and CNOT|11> = |10>. The CNOT flips the target qubit (second qubit) when the control qubit (first qubit) is |1>, and leaves the target unchanged when the control is |0>.
The CNOT acts as a conditional NOT: it applies the X (bit-flip) gate to the target qubit conditioned on the control qubit being 1. In the computational basis: |00> -> |00>, |01> -> |01>, |10> -> |11>, |11> -> |10>. When the control is in superposition, the CNOT creates entanglement — applying H then CNOT to |00> produces the Bell state (|00> + |11>)/sqrt(2).
Question 4 Multiple Choice
The Pauli Z gate leaves |0> unchanged and maps |1> to -|1>. Since measurement probabilities depend on |amplitude|^2, does the Z gate have any observable effect?
ANo — since |-1|^2 = 1, Z has no physically observable consequence
BYes — the relative phase between |0> and |1> affects subsequent interference and thus future measurement outcomes
CNo — global phases and relative phases are both unobservable
DYes — the minus sign directly changes the probability of measuring |1>
While a global phase (multiplying the entire state by e^(i*theta)) is unobservable, a relative phase between basis states is physically meaningful. Z changes the relative phase between |0> and |1> in a superposition. If the state is alpha|0> + beta|1>, Z produces alpha|0> - beta|1>. On the Bloch sphere, this is a 180-degree rotation about the z-axis, changing where the state sits on the equator. Subsequent Hadamard gates or other operations will translate this phase difference into a probability difference.