# Vector math of applying an X-gate on an $|i\rangle$ basis state

It is well known that the X-gate will apply a rotation about the x-axis on the bloch sphere. Knowing this, the $$|i\rangle$$ state should be converted to the $$|-i\rangle$$ state on the application of this gate.

To be clear I define these states as: $$|i\rangle$$ = $${1 \over \sqrt{2}}(|0\rangle + i|1\rangle)$$ and $$|-i\rangle$$ = $${1 \over \sqrt{2}}(|0\rangle - i|1\rangle)$$

When trying to do the vector math with $$X|i\rangle$$ I get: $$\begin{bmatrix}0&1\\1&0\end{bmatrix}1 \over \sqrt{2}\begin{bmatrix}1\\i\end{bmatrix}$$ = $$1 \over \sqrt{2}\begin{bmatrix}i\\1\end{bmatrix}$$

But I expect to be getting the $$|-i\rangle$$ state: $$1 \over \sqrt{2}\begin{bmatrix}1\\-i\end{bmatrix}$$

What am I doing wrong, am I missing some intrinsic property of quantum theory?

In your calculations you are getting the state $$|\psi \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} i \\ 1 \end{pmatrix}$$ instead of what you are expecting $$|\phi \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}.$$
Well it turns out that in quantum theory these two states are considered the same! This is because they only differ up to a global phase. That is there is an $$\alpha \in \mathbb{C}$$ with $$|\alpha|=1$$ such that $$|\phi\rangle = \alpha |\psi\rangle$$, In this case $$\alpha = -i$$.