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Why does applying the following circuit on a $00$ state produce $|0\rangle \otimes |+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)$. Shouldn't it produce $ |+\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)$?

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luciano
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Yash Wath
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  • See also here: https://quantumcomputing.stackexchange.com/questions/8244/big-endian-vs-little-endian-in-qiskit – Cryoris Nov 02 '20 at 15:07

1 Answers1

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Qubit ordering in Qiskit: qubit 0 is the rightmost one, i.e., the least significant bit.

Yael Ben-Haim
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