Quantum Computing 3.6 -- Triple CNOT

Computer Science Level 4

A qubit pair has the initial state ∣ ψ ⟩ = α ∣ 00 ⟩ + β ∣ 01 ⟩ + γ ∣ 10 ⟩ + δ ∣ 11 ⟩ ≡ ( α β γ δ ) ∈ C 4 |\psi\rangle = \alpha |00\rangle + \beta |01\rangle + \gamma |10\rangle + \delta |11\rangle \equiv \left( \begin{array}{c} \alpha \\ \beta \\ \gamma\\ \delta \end{array} \right) \in \mathbb{C}^4 Three controlled NOT gates are applied on this initial state: ∣ ψ ′ ⟩ = CNOT 12 ⋅ CNOT 21 ⋅ CNOT 12 ∣ ψ ⟩ |\psi'\rangle = \text{CNOT}_{12} \cdot \text{CNOT}_{21} \cdot \text{CNOT}_{12} |\psi\rangle What is the result of the state vector of the final state ∣ ψ ′ ⟩ |\psi'\rangle ?

Details: The gates CNOT 12 _{12} and CNOT 21 _{21} differ in that control and target qubits are reversed. The application of these gates to the basis states results ∣ ψ ⟩ CNOT 12 ∣ ψ ⟩ CNOT 21 ∣ ψ ⟩ ∣ 00 ⟩ ∣ 00 ⟩ ∣ 00 ⟩ ∣ 01 ⟩ ∣ 01 ⟩ ∣ 11 ⟩ ∣ 10 ⟩ ∣ 11 ⟩ ∣ 10 ⟩ ∣ 11 ⟩ ∣ 10 ⟩ ∣ 01 ⟩ \begin{array}{c|cc} |\psi\rangle & \text{CNOT}_{12} |\psi\rangle & \text{CNOT}_{21} |\psi\rangle\\ \hline |00\rangle & |00\rangle & |00\rangle \\ |01\rangle & |01\rangle & |11\rangle \\ |10\rangle & |11\rangle & |10\rangle \\ |11\rangle & |10\rangle & |01\rangle \end{array}

( δ β γ α ) \left( \begin{array}{c} \delta \\ \beta \\ \gamma \\ \alpha \end{array}\right) ( α δ γ β ) \left( \begin{array}{c} \alpha \\ \delta \\ \gamma \\ \beta \end{array}\right) ( α γ β δ ) \left( \begin{array}{c} \alpha \\ \gamma \\ \beta \\ \delta \end{array}\right) ( δ γ β α ) \left( \begin{array}{c} \delta \\ \gamma \\ \beta \\ \alpha \end{array}\right) ( α β δ γ ) \left( \begin{array}{c} \alpha \\ \beta \\ \delta \\ \gamma \end{array}\right)

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jul 12, 2018

From the table we can read the matrix representation of the operators. The columns of the matrix correspond exactly to the vectors CNOT ∣ i j ⟩ \text{CNOT}|ij\rangle : CNOT 12 = ( CNOT 12 ∣ 00 ⟩ , CNOT 12 ∣ 01 ⟩ , CNOT 12 ∣ 10 ⟩ , CNOT 12 ∣ 11 ⟩ ) = ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) CNOT 21 = ( CNOT 21 ∣ 00 ⟩ , CNOT 21 ∣ 01 ⟩ , CNOT 21 ∣ 10 ⟩ , CNOT 21 ∣ 11 ⟩ ) = ( 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 ) \begin{aligned} \text{CNOT}_{12} &= \left( \text{CNOT}_{12}|00\rangle, \text{CNOT}_{12}|01\rangle, \text{CNOT}_{12}|10\rangle, \text{CNOT}_{12}|11\rangle \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right)\\ \text{CNOT}_{21} &= \left( \text{CNOT}_{21}|00\rangle, \text{CNOT}_{21}|01\rangle, \text{CNOT}_{21}|10\rangle, \text{CNOT}_{21}|11\rangle \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \end{aligned} The overall operator of the circuit is given by matrix multiplication U = CNOT 12 ⋅ CNOT 21 ⋅ CNOT 12 = ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) ⋅ ( 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 ) ⋅ ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) = ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) ⋅ ( 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 ) = ( 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ) = SWAP \begin{aligned} U &= \text{CNOT}_{12} \cdot \text{CNOT}_{21} \cdot \text{CNOT}_{12} \\ &= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) \cdot \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \cdot \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) \\ &= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) \cdot \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array} \right) \\ &= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)\\ &= \text{SWAP} \end{aligned} The result is the SWAP gate which, when applied to a separable state, swaps the individual states: SWAP ∣ ψ 1 ⟩ ⊗ ∣ ψ 2 ⟩ = ∣ ψ 2 ⟩ ⊗ ∣ ψ 1 ⟩ \text{SWAP}|\psi_1 \rangle \otimes |\psi_2\rangle = |\psi_2\rangle \otimes |\psi_1\rangle For an general state vector we get the result SWAP ∣ ψ ⟩ = ( 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ) ( α β γ δ ) = ( α γ β δ ) \text{SWAP} |\psi\rangle = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} \alpha \\ \beta \\ \gamma \\ \delta \end{array} \right) = \left( \begin{array}{c} \alpha \\ \gamma \\ \beta \\ \delta \end{array} \right)

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