2
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

$\mathrm{X}$ 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.