= Solution
For real angles write $u_p=\cos\theta_p$ and $v_p=\sin\theta_p$. For one independent pair $p>0$, the transformation and its conjugate read
$$
c_p=u_p\alpha_p+i v_p\alpha_{-p}^\dagger,\qquad
c_{-p}=u_p\alpha_{-p}-i v_p\alpha_p^\dagger.
$$
Using the <canonical anticommutation relations> for the $\alpha$ modes gives
$$
\begin{aligned}
\{c_p,c_p^\dagger\}&=u_p^2+v_p^2=1,\\
\{c_{-p},c_{-p}^\dagger\}&=u_p^2+v_p^2=1,\\
\{c_p,c_{-p}\}&=-i u_pv_p+i v_pu_p=0,\\
\{c_p,c_{-p}^\dagger\}&=0.
\end{aligned}
$$
The remaining <anticommutators> vanish for independent <momentum> pairs. The displayed two-by-two <matrix> is <unitary> and its inverse is obtained by changing $i$ to $-i$.
When the same formula is indexed over both signs of <momentum>, its first row at $-p$ must agree with the conjugate of its second row at $p$. Thus a consistent global convention requires
$$
\boxed{u_{-p}=u_p,\qquad v_{-p}=-v_p,\qquad \theta_{-p}=-\theta_p\pmod{2\pi}.}
$$
This is the <odd-momentum fermionic Bogoliubov angle> convention. If one checks only the first-row formulas, their cross <anticommutator> is $i\sin(\theta_p+\theta_{-p})$; its vanishing gives the condition modulo $\pi$, with possible signs absorbable in the $\alpha$ modes. Requiring both rows to describe the same operators fixes the convention above. Merely checking a separate <unitary matrix> at each <momentum> misses this redundancy: choosing $\theta_p=\theta_{-p}=\pi/4$ in both first rows would give $\{c_p,c_{-p}\}=i$.
At $p=0$, the <annihilation operator> and its conjugate belong to one mode. The full displayed formula is consistent only with $v_0=0$; choose $\theta_0=0$ and $\alpha_0=c_0$. This mode is not a two-mode pair.
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