10C (Quantum Information and Computation)

Immediately before the final Hadamard, the state is

\frac{1}{2} (∣0 ⟩ ∣ ψ ⟩ - i ∣1 ⟩ U ∣ ψ ⟩) .

The outcome-one component after that Hadamard is

(∣ ψ ⟩ + i U ∣ ψ ⟩) /2

. Thus, with

z = ⟨ ψ ∣ U ∣ ψ ⟩

p (1) = \frac{1}{4} ∥ ∣ ψ ⟩ + i U ∣ ψ ⟩ ∥^{2} = \frac{1}{2} (1 - Im z) .

b

For every product vector,

(A \otimes B) (∣ x ⟩ \otimes ∣ y ⟩) = A ∣ x ⟩ \otimes B ∣ y ⟩ .

Product vectors span the two-register state space, so the single

A \otimes B

gate equals parallel application of

A

to the first register and

B

to the second. Unitarity guarantees that both descriptions are valid quantum circuits.

Both

Z \otimes Z

and

X \otimes I

are unitary and Hermitian, so their expectations are real. Run the real-part Hadamard test twice, once with controlled

Z \otimes Z

and once with controlled

X \otimes I

. If the respective ancilla outcome-one probabilities are

p_{ZZ}

and

p_{X I}

, then

⟨ ψ ∣ Z \otimes Z ∣ ψ ⟩ = 1 - 2 p_{ZZ}, ⟨ ψ ∣ X \otimes I ∣ ψ ⟩ = 1 - 2 p_{X I} .

The identity in part (b) implements each controlled tensor product by gates on its two target wires with the common control. By linearity,

⟨ ψ ∣ W ∣ ψ ⟩ = 2 - 2 (p_{ZZ} + p_{X I}) .

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