Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 3 / a / Solution

Using the printed measurement-vector convention, the unitary operator is

U (θ) = \frac{1}{2} (11 e^{- i θ} - e^{- i θ}) = H diag (1, e^{- i θ}) = J (- θ) .

Thus it is a phase gate followed by a Hadamard gate, with angle sign opposite to the parameter in the catalogued J gate. Its rows are the conjugate transposes of the orthonormal measurement vectors, so it is unitary. Right multiplication by the Pauli Z gate changes the sign of the second column, which is exactly the effect of left multiplication by the Pauli X gate, exchanging the two rows. Explicitly,

U (θ) Z = \frac{1}{2} (11 - e^{- i θ} e^{- i θ}) = X U (θ) .

Similarly, right multiplication by the Pauli X gate exchanges columns, giving

U (θ) X = \frac{1}{2} (e^{- i θ} - e^{- i θ} 11) = e^{- i θ} Z \frac{1}{2} (11 e^{i θ} - e^{i θ}) .

Therefore the exact identities, retaining the global phase in the second one, are

U (θ) Z = X U (θ), U (θ) X = e^{- i θ} Z U (- θ) .

These supply Pauli-frame propagation along a measurement wire.

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