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How can I construct the below 2-qubit state using suitable single qubit gates (maximum 3) and one CNOT gate starting with state $|00\rangle$?

$$ |\omega\rangle=\frac{1}{3}(2|00\rangle+|01\rangle+2|11\rangle) $$

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  • $\begingroup$ Would 2 CNOTs do? $\endgroup$
    – KAJ226
    Nov 7 '20 at 17:03
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Using only one CNOT gate and 3 single qubit gates:

enter image description here

Where: $$ \mathrm{U3}(\theta,\phi,\lambda)= \begin{pmatrix} \cos(\theta/2) & -\mathrm{e}^{i\lambda}\sin(\theta/2) \\ \mathrm{e}^{i\phi}\sin(\theta/2) & \mathrm{e}^{i(\phi+\lambda)}\cos(\theta/2) \end{pmatrix} $$


To prepare a state, you can use Qiskit's QuantumCircuit.initialize. But when I used this method, the constructed circuit contains 2 CNOTs even with maximum optimization_level. Qiskit provides another method, QuantumCircuit.iso, which yields the circuit shown above.

Code:

# Using QuantumCircuit.initialize
qc = QuantumCircuit(2)
qc.initialize([2/3, 1/3, 0, 2/3], [0, 1])

qc_by_initialize = transpile(qc, basis_gates=['u3','cx'], optimization_level = 3)
qc_by_initialize.draw('mpl')


# Using QuantumCircuit.iso
qc = QuantumCircuit(2)
qc.iso([2/3, 1/3, 0, 2/3], [0, 1], [])

qc_by_isometry = transpile(qc, basis_gates=['u3','cx'], optimization_level = 3)
qc_by_isometry.draw('mpl')
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Below is a circuit that can generate the state $|w\rangle$ that you wanted. However, It is NOT what you are looking for as it is using 2 CNOTs gate.

enter image description here

and if you look at the state vector output, you can see that it is indeed the correct state:

enter image description here


Hopefully someone will be able give you an answer that you are looking for :)

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  • $\begingroup$ would you also please explain how did you come up with this circuit/ gate combination? $\endgroup$
    – 22_J
    Nov 7 '20 at 18:24
  • $\begingroup$ I used Qiskit state preparation called Custom... qiskit.org/documentation/stubs/… ... But it seems like another person from the other answer found a better method. $\endgroup$
    – KAJ226
    Nov 7 '20 at 23:53
  • $\begingroup$ I see! Thanks for the link. $\endgroup$
    – 22_J
    Nov 8 '20 at 12:21
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If you want to understand how to do this, think of the Schmidt decomposition. This shows you that, up to unitaries on the two qubits, the state can be written in the form $\alpha|00\rangle+\beta|11\rangle$. Now, can you answer your own question for this state?

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  • $\begingroup$ Nice hint! This method will give this circuit $\endgroup$
    – tsgeorgios
    Nov 8 '20 at 19:24

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