# Transforming the first Bell state into the other Bell states

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:

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

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?

$$\mathrm{X}$$ 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.