Quantum computing circuit for qubits equality

  • #1
Hill
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Homework Statement
How can the circuit shown below be used to measure the qubits b0 and b1 for equality without learning anything else about the state of b0 and b1? (Hint: you are free to choose any initial state on the register consisting of qubits a0 and a1.)
Relevant Equations
Cnot = |0⟩⟨0|⊗I + |1⟩⟨1| ⊗X
This is the given circuit:

1697639388267.png


I think to add another Cnot on the right with a1 as control and a0 as target, to set initial states of a0 an a1 both |0⟩, and to measure the a0. If a0=|0⟩ then b0=b1, and vice versa.
Is it correct?
 
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  • #2
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