Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 67 / 2 / a

Solution

∣ e_{t} ⟩ = ∣ ψ_{t} ⟩ \otimes ∣ c_{t} ⟩, ∣ c_{t} ⟩ = ∣1 ⟩^{\otimes t} ∣0 ⟩^{\otimes (T - t)}, t = 0, \dots, T,

where

ψ_{t} = U_{t} \dots U_{1} ψ_{0}

. Orthogonality follows from the distinct unary quantum clock strings, regardless of overlaps between the computational states.

The initialization term acts diagonally:

H_{init} ∣ e_{0} ⟩ = 0, H_{init} ∣ e_{t} ⟩ = ∣ e_{t} ⟩ (t \geq 1) .

Each propagation term only couples the two clock strings adjacent to its step. The gate and its adjoint give

H_{t} ∣ e_{t - 1} ⟩ = \frac{1}{2} (∣ e_{t - 1} ⟩ - ∣ e_{t} ⟩), H_{t} ∣ e_{t} ⟩ = \frac{1}{2} (∣ e_{t} ⟩ - ∣ e_{t - 1} ⟩),

and

H_{t} ∣ e_{j} ⟩ = 0

for

j \in / {t - 1, t}

. The same calculation covers the two endpoint terms.

Thus both Hamiltonians preserve $L$ , and in the history basis their restrictions are

D := H_{init} ∣_{L} = I - ∣ e_{0} ⟩ ⟨ e_{0} ∣, E := H_{final} ∣_{L} = \frac{1}{2} t = 1 \sum T (∣ e_{t} ⟩ - ∣ e_{t - 1} ⟩) (⟨ e_{t} ∣ - ⟨ e_{t - 1} ∣) .

1/2

. The specified two-clock endpoint formulas assume

T \geq 2

; a shorter circuit can first be padded with identity gates.

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