Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 66 / 2 / b

Solution

Write the arbitrary joint pure state of the input

A

and an external reference

R

∣Ω ⟩_{A R} = ∣0 ⟩_{A} ∣ r_{0} ⟩_{R} + ∣1 ⟩_{A} ∣ r_{1} ⟩_{R}, ⟨ r_{0} ∣ r_{0} ⟩ + ⟨ r_{1} ∣ r_{1} ⟩ = 1.

The reference vectors need not be normalized or orthogonal. Share

∣ Φ^{+} ⟩_{a B}

. Label Alice's Bell states by

∣ β_{rs} ⟩_{A a} = \frac{1}{2} \sum_{j = 0}^{1} (- 1)^{r j} ∣ j ⟩_{A} ∣ j \oplus s ⟩_{a}, r, s \in {0, 1} .

Projecting onto this Bell basis gives the unnormalized state

_{A a} ⟨ β_{rs} ∣ (∣Ω ⟩_{A R} ∣ Φ^{+} ⟩_{a B}) = \frac{1}{2} (X_{B}^{s} Z_{B}^{r} \otimes I_{R}) ∣Ω ⟩_{BR} .

Each outcome has probability

1/4

. Alice sends

(r, s)

and Bob applies $Z_{B}^{r} X_{B}^{s}$ , reversing the two Pauli gates. The final state is exactly

∣Ω ⟩_{BR}

for every outcome. In particular all input entanglement and other correlations with

R

now belong to

B

Conditioned on the record, Alice's two measured carriers are in

∣ β_{rs} ⟩

and factor from

BR

. Thus

A

no longer retains its original correlations with

R

. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

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