Is the quantum NOT operation similar to the classical NOT operation? $\renewcommand{ket}[1]{\left| #1 \right\rangle}$
Classical NOT operation
Suppose I had an interval $S = [a,b]\in\Bbb{R}$, then $$\mathrm{NOT}(S) = (-\infty,a) \cup (b,\infty)$$
Quantum NOT operation
$$\ket{\downarrow} =
\begin{bmatrix}
0 \\ 1 \\
\end{bmatrix}$$
$$\ket{\uparrow} =
\begin{bmatrix}
1 \\ 0 \\
\end{bmatrix}$$
$$\begin{bmatrix}
0 & 1 \\ 1 & 0\\
\end{bmatrix}
\ket{\downarrow} = \ket{\uparrow}$$
See here for a source on how I defined the quantum NOT operation. 
My Confusion
If I consider
$$\ket{\downarrow} =
\begin{bmatrix}
0 \\ 1 \\
\end{bmatrix}$$
this is just one point of a set of all possible linear combinations of $a\ket{\uparrow} +b\ket{\downarrow}$. So then
$$\mathrm{NOT}(\ket{\downarrow})\neq \ket{\uparrow}$$ but instead would be a set of points.
My Question
Is the quantum NOT operation supposed to be similar to the classical NOT operation? Am I just misunderstanding the definition of the quantum NOT operation or is quantum logic somehow different from classical logic? 
 A: $\renewcommand{ket}[1]{\left| #1 \right\rangle}$
In my question I was trying to define the $\mathrm{NOT}$ operation using set theory. I got confused because I regarded $ \mathrm{NOT}$ as mapping from points in $[a,b]$ onto points in $(-\infty,a,)\cup (b,\infty)$. This is wrong. 
Let $X = [a,b]$ and let $Y = (-\infty,a,)\cup (b,\infty)$. Consider the set $S = \lbrace X,Y\rbrace$. $\mathrm{NOT}(X)$ maps from the element $X$ to the element $Y$. In other words, it is not a map from real numbers to real numbers but a map from elements of a set $S$ to elements of a set $S$. That is, $$\mathrm{NOT}: S \rightarrow S$$ So when $\mathrm{NOT}(X)$ acts on elements of $S$, it doesn't treat them like points in $\Bbb{R}$ but just as sets. So there is one and only one value for $\mathrm{NOT}(X)$ or $\mathrm{NOT}(Y)$ That is, $$\mathrm{NOT}(X) = Y$$
$$\mathrm{NOT}(Y)=X$$
So the key point here is, it isn't a map between points in $\Bbb{R}$ but instead a map between elements of $S$. 

Let me now comment on $\mathrm{QNOT}$. 
First, let us note $\mathrm{NOT}$ has two key properties 


*

*It is bijective.

*It is an involution.


I will demonstrate $\mathrm{QNOT}$ has these properties. 
Let $V$ be a complex vector space. Let $\alpha,\beta\in\Bbb{C}$. Let 
$$\ket{\downarrow} =
\begin{bmatrix}
0 \\ 1 \\
\end{bmatrix}$$
$$\ket{\uparrow} =
\begin{bmatrix}
1 \\ 0 \\
\end{bmatrix}$$
$$ \mathrm{QNOT}(\vec{v}):= \begin{bmatrix}
0 &1 \\ 1&0 \\
\end{bmatrix}\vec{v}$$ 
Claim 1: $\mathrm{QNOT}$ is an involution
Reasoning:
\begin{align*} \mathrm{QNOT}\left( \mathrm{QNOT}(\vec{v})\right)&= \begin{bmatrix}
0 &1 \\ 1&0 \\
\end{bmatrix} \begin{bmatrix}
0 &1 \\ 1&0 \\
\end{bmatrix}\vec{v}\\
&= \begin{bmatrix}
1 &0 \\ 0&1 \\
\end{bmatrix}\vec{v}\\
&=I_2 \vec{v}
\end{align*}
Claim 2: 
For any given $\ket{\psi} = \alpha \ket{\uparrow} + \beta\ket{\downarrow} \in V$,  $\mathrm{QNOT}\left(\ket{\psi}\right)$ is bijective.  
Reasoning:
As defined, $\mathrm{QNOT}$ is a linear operator acting on a vector $\ket{\psi} \in V$. In this case, from Claim 1, we can see $$\mathrm{QNOT} = \mathrm{QNOT}^{-1} = \begin{bmatrix}
0 &1 \\ 1&0 \\
\end{bmatrix} $$
Thus, $\mathrm{QNOT}$ is has both a right and left inverse. Any matrix with both a right and left inverse is called an isomorphism and is also a bijection. 
So the key point here is that, like $\mathrm{NOT}$, $\mathrm{QNOT}$ 


*

*maps to one and only one element 

*can be applied again and get the original vector $\ket{\psi}$


I am not a quantum information theory expert, but to me that seems like a sensible enough argument to show why $\mathrm{NOT}$ and $\mathrm{QNOT}$ have similar properties. 
