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Please answer all parts ASAP In each of the problems 1-3, I have listed a quantum circuit and input qubit(s) (on the left of the

Please answer all parts ASAP

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In each of the problems 1-3, I have listed a quantum circuit and input qubit(s) (on the left of the circuit). In each case, calculate the output (i.e., calculate what comes out on the right). X, Y, and Z are, respectively, the Pauli ox, Oy, and oz matrices: As you work through these examples, keep in mind that quantum circuits differ from classical logic circuits in more than just the definition of the gates. In particular, when qubits become entangled, you can't compute the state change for each qubit separately from the others, because in this case an individual qubit doesn't have a well-defined state. So in this case you have to work with tensor-product states. 1. (9 points) 10) -H Z 10) [Hint: We know what the 2-qubit state vector looks like after the two Hadamards have been applied to the input qubits: this is an instance of a basic QC construction, discussed in the lectures and the notes. See, for example, section 2.5 of the "Quantum Computing Model" notes in module 5.] 2. (9 points) a |0) + b|1) 10) X 10) 3. (10 points) 10) -H Y (1) -Z 10) X7. It is possible to define different versions of the CNOT gate, depending on whether the control qubit is the first or the second qubit, and whether the gate acts trivially when the control qubit is set to |0) or |1). (A gate acts 'trivially" if it has the effect of the identity operator.) Consider the following version, which we'll call Co: Co flips the second qubit if the first is set to |0). Thus Co has the following effect on the 2-qubit basis vectors: ID) ID) > ID) I1) IO) I1) > ID) IO) |1>|0> > |1>|0) |1>|1)>|1>|1) We will represent C9 like this in our quantum circuits: 1 i. (4 points) Write out the 4 x 4 matrix for C0 ii. (6 points) Construct a circuit, using only CNOT and X gates, that implements C0. The Hadamard gate (H for short) plays a fundamental role in quantum computing Definition of H: HO) = * ( 10 ) + 1 1 ) ) H1) = 2 (10) - 1))

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