Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 66 / 4 / a

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- Solution a

Solution

All time-dependent Hamiltonian operators are multiples of the same fixed operator, so they commute and their time-ordered exponential reduces to an ordinary exponential. Let

P_{1} = (I - Z_{S}) /2

. With

ℏ = 1

and the given unit coupling integral,

U = exp [- \frac{iπ}{2} P_{1} \otimes (I - X_{D})] = P_{0} \otimes I + P_{1} \otimes e^{- iπ (I - X_{D}) /2} .

Since

X_{D}^{2} = I

,

e^{- iπ (I - X_{D}) /2} = e^{- iπ /2} [cos (π /2) I + i sin (π /2) X_{D}] = X_{D} .

Thus

U = P_{0} \otimes I + P_{1} \otimes X_{D}, U ∣ s, d ⟩ = ∣ s, d \oplus s ⟩ .

It is exactly the CNOT gate with

S

controlling

D

, with no residual relative phase. Acting on

(c_{0} ∣0 ⟩ + c_{1} ∣1 ⟩)_{S} ∣0 ⟩_{D}

correlates the system basis label with the apparatus label, producing

c_{0} ∣00 ⟩ + c_{1} ∣11 ⟩

as required.

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