= Solution
Since every <quasiparticle> <energy> is positive, the <ground state> is the <vacuum> of all the $\alpha_p$ <annihilation operators>, including $\alpha_0=c_0$. Its zero-momentum mode is therefore empty. For each positive <momentum>, the inverse <Bogoliubov transformation> gives
$$
\alpha_p=u_pc_p-i v_pc_{-p}^\dagger,\qquad
\alpha_{-p}=u_pc_{-p}+i v_pc_p^\dagger.
$$
Both annihilate the normalized pair state $(u_p+i v_pc_p^\dagger c_{-p}^\dagger)|0\rangle$. For example, applying $\alpha_p$ produces the two cancelling terms $i u_pv_pc_{-p}^\dagger|0\rangle$. Thus
$$
\boxed{|\Omega\rangle=\prod_{p>0}\left(u_p+i v_pc_p^\dagger c_{-p}^\dagger\right)|0\rangle,\qquad c_0|\Omega\rangle=0.}
$$
The factors have even <fermion parity> and commute for disjoint <momentum> pairs. Each normalized factor is a <superposition> of zero and two <fermions>, whose pair-occupation <probability> is $v_p^2$. Therefore
$$
\boxed{\langle\Omega|\widehat N|\Omega\rangle=2\sum_{p>0}v_p^2=\sum_{p>0}\left(1-\frac{\epsilon(p)}{E_p}\right).}
$$
There is no zero-mode contribution. For fixed $L$, the <sum> converges since $v_p^2=O(p^{-4})$.
A normalized simple excited state is $|k\rangle=\alpha_k^\dagger|\Omega\rangle$ for $k>0$. Its <excitation energy> is $E_k$, and its total <momentum> is $k$: both terms in $\alpha_k^\dagger=u_kc_k^\dagger+i v_kc_{-k}$ add <momentum> $k$. Acting on the $k,-k$ pair gives
$$
\alpha_k^\dagger\left(u_k+i v_kc_k^\dagger c_{-k}^\dagger\right)|0\rangle
=(u_k^2+v_k^2)c_k^\dagger|0\rangle=c_k^\dagger|0\rangle.
$$
The affected pair now contains exactly one <fermion> instead of mean number $2v_k^2$; all other pairs are unchanged. Consequently
$$
\boxed{\langle k|\widehat N|k\rangle=\langle\widehat N\rangle_\Omega+1-2v_k^2
=\langle\widehat N\rangle_\Omega+\frac{\epsilon(k)}{E_k}.}
$$
This is the <fermion number of a Bogoliubov excitation>. Its change in mean <fermion number> lies between zero and one for $\mu>0$, with the value one at $\Delta=0$. The <Bogoliubov quasiparticle> coherently mixes adding a <fermion> and removing the opposite-momentum <fermion>; it is not an eigen-excitation of the original <fermion number operator>. Indeed,
$$
[\widehat N,H]=2i\Delta\sum_{p>0}\left(c_p^\dagger c_{-p}^\dagger+c_{-p}c_p\right),
$$
so <fermion number> is not conserved when $\Delta\ne0$, while <fermion parity> is conserved. The generally noninteger change is an <expectation value> in a coherent paired background, rather than a fractional <eigenvalue> of the <fermion number operator>. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added <fermion>.
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