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

Question

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?

Answer 1

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 $\alpha = -i$.

Global phase is considered irrelevant in quantum theory as it is undetectable. Any measurement protocol you apply to one state will give the exact same probabilities for the other state.