Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 324 / 3 / ii

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- Solution ii

Solution

The computational-basis projectors on the first qubit are

Π_{0} = ∣0 ⟩ ⟨ 0∣ \otimes I^{\otimes (n - 1)}, Π_{1} = ∣1 ⟩ ⟨ 1∣ \otimes I^{\otimes (n - 1)} .

The Pauli Z gate acts as

+ 1

on the first subspace and

- 1

on the second, so

Z_{1} = Π_{0} - Π_{1}

. By the Born rule,

p_{b} = ⟨ ψ ∣ Π_{b} ∣ ψ ⟩

. Therefore the expectation determines the output probabilities:

⟨ ψ ∣ Z_{1} ∣ ψ ⟩ = p_{0} - p_{1}, p_{0} = \frac{1 + ⟨ Z _{1} ⟩}{2}, p_{1} = \frac{1 - ⟨ Z _{1} ⟩}{2} .

The normalization supplies

p_{0} + p_{1} = 1

; no product-state assumption is needed for this identity.

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