= Solution
Group the nonzero modes into independent pairs $p>0$. For one pair set $\varepsilon=\epsilon(p)=p^2/(2m)+\mu$, and introduce the <Nambu spinor> $\Psi_p=(c_p,c_{-p}^\dagger)^{\mathsf T}$. <Normal ordering> using $c_{-p}c_{-p}^\dagger=1-c_{-p}^\dagger c_{-p}$ gives
$$
H_p=\Psi_p^\dagger\begin{pmatrix}\varepsilon&i\Delta\\-i\Delta&-\varepsilon\end{pmatrix}\Psi_p+\varepsilon.
$$
This exactly contains both <kinetic terms> and the pairing terms once; there is no additional factor of one half because only $p>0$ is summed.
Insert $\Psi_p=U_p\mathcal A_p$, where $\mathcal A_p=(\alpha_p,\alpha_{-p}^\dagger)^{\mathsf T}$ and $U_p$ is the <Bogoliubov transformation> <matrix>. Direct <matrix multiplication> yields
$$
U_p^\dagger h_pU_p=\begin{pmatrix}
\varepsilon\cos2\theta_p-\Delta\sin2\theta_p&i(\varepsilon\sin2\theta_p+\Delta\cos2\theta_p)\\
-i(\varepsilon\sin2\theta_p+\Delta\cos2\theta_p)&-\varepsilon\cos2\theta_p+\Delta\sin2\theta_p
\end{pmatrix}.
$$
Choose the diagonalizing branch
$$
\boxed{\cos2\theta_p=\frac{\varepsilon}{E_p},\qquad \sin2\theta_p=-\frac\Delta{E_p},\qquad E_p=\sqrt{\varepsilon^2+\Delta^2},\quad p>0.}
$$
Then the off-diagonal terms vanish and the upper diagonal entry is $E_p>0$. Extend the angle to negative <momenta> as in part b; the sine relation just written applies to the positive representative of each pair. The <BCS coherence factors> can be chosen as
$$
u_p=\sqrt{\frac{E_p+\varepsilon}{2E_p}},\qquad
v_p=-\frac{\Delta}{\sqrt{2E_p(E_p+\varepsilon)}}.
$$
Here $u_p$ denotes $\cos\theta_p$, and $v_p$ denotes $\sin\theta_p$; these expressions are nonsingular because $\mu>0$ and $m>0$. The minus sign of $v_p$ for positive $\Delta$ is fixed by the pairing <phase> in this <Hamiltonian>.
A second use of the <canonical anticommutation relations> gives
$$
H_p=E_p\left(\alpha_p^\dagger\alpha_p-\alpha_{-p}\alpha_{-p}^\dagger\right)+\varepsilon
=E_p\left(\alpha_p^\dagger\alpha_p+\alpha_{-p}^\dagger\alpha_{-p}\right)+\varepsilon-E_p.
$$
The zero-momentum term was never paired and is simply $\mu\alpha_0^\dagger\alpha_0$. Hence
$$
\boxed{H=\sum_pE(p)\alpha_p^\dagger\alpha_p+C,\qquad
C=\sum_{p>0}\bigl[\epsilon(p)-\sqrt{\epsilon(p)^2+\Delta^2}\bigr],}
$$
with
$$
\boxed{E(0)=\mu,\qquad E(p)=\sqrt{\epsilon(p)^2+\Delta^2}\quad(p\ne0).}
$$
The constant is the <ground-state energy>. At fixed $L$, its ultraviolet tail converges: $\epsilon(p)-E(p)=-\Delta^2/[\epsilon(p)+E(p)]\sim-m\Delta^2/p^2$ in one dimension. Replacing $E(0)$ by $\sqrt{\mu^2+\Delta^2}$ would incorrectly pair a spinless mode with itself.
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