So starting with an up particle:
$$
\lvert \uparrow \rangle = \begin{bmatrix}
1 \\
0 \\
\end{bmatrix}
$$
My understanding is that you can measure $\lvert \uparrow \rangle$ in $X$ and have a 50% chance of getting $\lvert \leftarrow \rangle$ and a 50% chance of getting $\lvert \rightarrow \rangle$ where:
$$
\lvert \rightarrow \rangle = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}}
\end{bmatrix}
\text{ and }
\lvert \leftarrow \rangle = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{-1}{\sqrt{2}}
\end{bmatrix}
$$
but you could also operate in $X$ which would be the equivalent of passing it though a NOT
gate:
$$
X \cdot \lvert \uparrow \rangle =
\begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix}
\cdot
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix} = \begin{bmatrix}
0 \\
1 \\
\end{bmatrix} = \lvert \downarrow \rangle
$$
I've read that measuring qubit spin is more or less equivalent to measuring the orientation of a photon which can be done by passing it through polarized filters. If the photon is measured to be in one orientation and then is measured in a different orientation it has a certain probability of snapping to be in that other direction.
So in this example, that would be equivalent to measuring an up qubit in $X$. But how does operating in $X$ come into the equation? I understand the mathematical effect it has, but what does this mean for the physical qubit? Is it also being passed through a filter and how is that different from measuring?
Thanks!