As I understand it, you can transform the different Bell states into one another by applying various gates. Wikipedia has the Bell states written out as follows: enter image description here

And says that you can generate bell states 2, 3, and 4 from bell state 1 by applying Pauli gates as follows:

enter image description here

I understand the Z gate simply flips the sign in the middle, so I can see how applying that to bell state 1 generates bell state 2. However, I don't get how applying the X/CNOT gate to bell state 1 generates bell state 3. Shouldn't the output of that be $|00 \rangle + |10 \rangle$ instead of $|01 \rangle + |10 \rangle$, since the control qubit is 0 in the first half?


1 Answer 1


$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.

  • $\begingroup$ Ah, I see! Thank you, that was a very dumb error on my part! $\endgroup$
    – Yuerno
    Oct 20, 2019 at 18:34

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