1. A single qubit is prepared in the state
\[
|\psi\rangle=\frac{3}{5}|+\rangle...
Sep 28, 2024
Solution by Steps
step 1
For question 1(a), the possible outcomes in the computational basis are ∣0⟩ and ∣1⟩. The probabilities are calculated as follows: P(0)=∣⟨0∣ψ⟩∣2 and P(1)=∣⟨1∣ψ⟩∣2
step 2
The state ∣ψ⟩=53∣+⟩−54i∣−⟩ can be expressed in the computational basis as ∣ψ⟩=5321(∣0⟩+∣1⟩)−54i21(∣0⟩−∣1⟩)
step 3
Simplifying this gives ∣ψ⟩=(523−524i)∣0⟩+(523+524i)∣1⟩
step 4
The probabilities are then calculated as P(0)=523−4i2 and P(1)=523+4i2
step 5
For question 2, the probability of measuring the qubit in state ∣1⟩ is given by P(1)=232=43
step 6
For question 3, we verify that P0+P1=E by showing that the projection operators P0=∣0⟩⟨0∣ and P1=∣1⟩⟨1∣ sum to the identity matrix E=∣0⟩⟨0∣+∣1⟩⟨1∣
step 7
For question 4(a), applying the Hadamard gate H to the first qubit of ∣01⟩ results in H∣0⟩=21(∣0⟩+∣1⟩), giving the state 21(∣0⟩+∣1⟩)∣1⟩
step 8
For question 4(b), applying the Hadamard gate H to the second qubit results in ∣0⟩H∣1⟩=∣0⟩21(∣0⟩−∣1⟩), giving the state ∣0⟩21(∣0⟩−∣1⟩)
[1] Answer
C
[2] Answer
C
[3] Answer
A
[4] Answer
A
Key Concept
Quantum Measurement and Probability
Explanation
The probabilities of measuring a qubit in a specific state are derived from the squared magnitudes of the coefficients in its state representation. The Hadamard gate transforms qubit states, affecting measurement outcomes.