Ex5 9

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 5.9}$

How can the circuit

be used to measure the qubits $b_0$ and $b_1$ for equality without learning anything else about the state of $b_0$ and $b_1$? (Hint: you are free to chose any initial state on the register consisting of qubits $a_0$ and $a_1$.)

Transfer correlation to a0, a1

Anders B Madsen 07 Dec 2021 20:29

Chose $a_0, a_1$ to have the same initial state.

If $a_0, a_1$ come out in the same state, then either they both stayed the same or both flipped, meaning that $b_0, b_1$ are the same state.

If $a_0, a_1$ come out as different states, then $b_0, b_1$ are also different, without knowing the exact states of $b_0, b_1$.