Why does charge-conjugation operator anti-commute with fermion number operator? On pp.6 right-hand side of this paper beneath eqn (27) it is stated that the charge-conjugation operator $\hat{C}$ anti-commutes with the fermion number operator $\hat{N}$.  The algebra is fine but I wish to have a physically intuitive explanation of this commutation relation.  Is it because the total number = the total charge?  Then can we use "number operator" and "charge operator" interchangeably?
 A: Yes: the number operator $N$ is essentially the same operator as the charge operator $Q$:
$$
Q=eN\tag{1}
$$
where $e$ is some electric charge (say, the electron charge). By definition,
$$
\begin{align}
N|\text{particle}\rangle&=+|\text{particle}\rangle\\
N|\text{anti-particle}\rangle&=-|\text{anti-particle}\rangle
\end{align}\tag{2}
$$
and a similar relation for $Q$.
Let $\mathscr C$ be the charge-conjugation operator:
$$
\mathscr C|\text{particle}\rangle\overset{\text{def}}=|\text{anti-particle}\rangle\tag{3}
$$
which satisfies
$$
\mathscr C^2=1\tag{4}
$$
Here we will try to explain the intuitive reason for the fact that $N\mathscr C+\mathscr CN\equiv0$.
If we act on both sides of $(3)$ with $N$, we get
$$
\begin{aligned}
N\mathscr C|\text{particle}\rangle&=N|\text{anti-particle}\rangle\\
&\overset{(2)}=-|\text{anti-particle}\rangle\\
&\overset{(3,4)}=-\mathscr C|\text{particle}\rangle\\
&\overset{(2)}=-\mathscr CN|\text{particle}\rangle
\end{aligned}\tag{5}
$$
that is,
$$
(N\mathscr C+\mathscr CN)|\text{particle}\rangle=0\tag{6}
$$
Finally, assuming that the vacuum is neutral ($N|0\rangle=0$) and charge-invariant ($\mathscr C|0\rangle=|0\rangle$), we also get
$$
(N\mathscr C+\mathscr CN)|0\rangle=0\tag{7}
$$
As the particle states -- together with the vacuum -- are a basis, we obtain the operator equation
$$
N\mathscr C+\mathscr CN=0\tag{8}
$$
as required.
Moreover, as $N\propto Q$, we also obtain
$$
Q\mathscr C+\mathscr CQ\overset{(1,8)}=0\tag{9}
$$
as a trivial corollary.
