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Dec 21, 2019 at 15:38 comment added met927 It works in the same way as the one qubit case, only now your basis states are in terms of 2 qubits. This means the coefficients correspond to $\vert 00 \rangle,\vert 01 \rangle, \vert 10\rangle, \vert 11 \rangle$ respectively. The order of this is simply counting up in binary,
Dec 21, 2019 at 15:01 comment added marissalianam Thank you so much! Now everything makes sense. I am trying to follow the same procedure with 2qubits gate (CNOT) like this: qc.cx(q[0], q[1]) but I always receive this result : {'counts': {'0x0': 1}, 'statevector': [[1.0, 0.0], [0.0, 0.0], [0.0, 0.0], [0.0, 0.0]]} How can I read this?
Dec 21, 2019 at 11:42 vote accept marissalianam
Dec 21, 2019 at 9:38 history edited met927 CC BY-SA 4.0
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Dec 20, 2019 at 23:43 history answered met927 CC BY-SA 4.0