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As I understood, the X gate flips the state around : $X(|0\rangle) = |1\rangle$. It can also be visualized with a $\pi$ rotation around the $x$ axis in the Bloch sphere. I have no problem with that.

The problem is with the minus state. The Bloch sphere visualization of the Hadamard gate is a $\pi$ rotation around the $z$ axis and a $\pi/2$ rotation around the $y$ axis. This makes it that when we use the H gate on the basis state we obtain $H(|0\rangle) = \frac {|0\rangle+|1\rangle} {\sqrt 2} = |+\rangle$ and $H(|1\rangle) = \frac {|0\rangle-|1\rangle} {\sqrt 2} = |-\rangle$. I understand that without any problem.

These two states are on the $x$ axis, this means that applying an X gate and therefor create a rotation shouldn't change anything. This is the case with the $|+\rangle$ state since $X(|+\rangle) = |+\rangle$. However. I simply do not understand how the case of the $|-\rangle$ state can be explained $X(|-\rangle) = -|-\rangle$ using the Bloch sphere since a rotation of $\pi$ around the $x$ axis shouldn't change anything.

Thanks for reading, and I hope you can explain

PS : I do understand the other demonstration $X(|-\rangle) = X(\frac {1}{\sqrt 2}|0\rangle-\frac {1}{\sqrt 2}|1\rangle)=\frac {1}{\sqrt 2}|\rangle-\frac {1}{\sqrt 2}|0\rangle=-|-\rangle$

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You are correct on the part that rotation about the X axis should not cause any change in the Bloch Sphere representation. In fact, $\left|-\right>$ and $-\left|-\right>$ do have the same Bloch sphere representation. Since $-\left|-\right>=e^{i\pi}\left|-\right>$, the two states differ only by a global phase factor. The the Bloch sphere representation is independent of the global phase as it has 'no observable effect' on the state.
Section 1.2 of Nielson-Chuang (mainly page 15 in the 10th anniversary edition) can be a good reference!

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  • $\begingroup$ Thanks a lot, I kind of doubted it was the same thing, but I sure didn't consider there could be a global phase. Thanks ! $\endgroup$ – Jonathcraft May 9 at 15:05

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