 Solution

ID: past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2017/5/b/solution

Before the gates, write the two Bell states as

∣ Φ^{+} ⟩_{12} ∣ Φ^{+} ⟩_{34} = \frac{1}{2} \sum_{a, b = 0}^{1} ∣ a ⟩_{1} ∣ a ⟩_{2} ∣ b ⟩_{3} ∣ b ⟩_{4} .

The CNOT gate sends the last bit to

b \oplus a

. Projection onto outcome 0 on qubit 4 therefore requires

b = a

. After the Hadamard gate on qubit 2, projection onto its outcome 0 contributes

1/ 2

for either remaining value of

a

. The unnormalized surviving state is

\frac{∣00 ⟩ _{13} + ∣11 ⟩ _{13}}{2 2} = \frac{1}{2} ∣ Φ^{+} ⟩_{13} .

Thus

∣ ψ ⟩_{13} = ∣ Φ^{+} ⟩_{13}, P (0 0_{24}) = \frac{1}{4} .

This is entanglement swapping: photons 1 and 3 become entangled even though the joint measurement acts only on their original partners.

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