# Effect of Pauli X gate on minus state using bloch sphere

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 $$X(|-\rangle) = X(\frac {1}{\sqrt 2}|0\rangle-\frac {1}{\sqrt 2}|1\rangle)=\frac {1}{\sqrt 2}|\rangle-\frac {1}{\sqrt 2}|0\rangle=-|-\rangle$$

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.