= Solution
Let $N_c=c^\dagger c$, $N_d=d^\dagger d$ and $P=cd+d^\dagger c^\dagger$. Substitution of the <Bogoliubov transformation>, with <normal ordering> using $dd^\dagger=1-d^\dagger d$ and $cc^\dagger=1-c^\dagger c$, gives
$$
\begin{aligned}
a^\dagger a+b^\dagger b
&=(u^2-v^2)(N_c+N_d)+2v^2-2uvP,\\
ab+b^\dagger a^\dagger
&=2uv(N_c+N_d)-2uv+(u^2-v^2)P.
\end{aligned}
$$
For example, the $v^2dd^\dagger$ term in $a^\dagger a$ contributes $v^2-v^2N_d$; these constants must be retained to obtain the <ground-state energy>. The <two-orbital fermionic pairing Hamiltonian> is consequently
$$
\begin{aligned}
H={}&[\varepsilon(u^2-v^2)+2\Delta uv](N_c+N_d)\\
&+[\Delta(u^2-v^2)-2\varepsilon uv]P
+2\varepsilon v^2-2\Delta uv.
\end{aligned}
$$
To eliminate the pairing term, choose a <canonical rotation of two fermionic modes> satisfying
$$
2\varepsilon uv=\Delta(u^2-v^2).
$$
With $u=\cos\theta$, $v=\sin\theta$, take $0<\theta<\pi/4$, so $\tan2\theta=\Delta/\varepsilon$. Setting $E=\sqrt{\varepsilon^2+\Delta^2}>0$ gives
$$
u^2-v^2=\frac\varepsilon E,\qquad 2uv=\frac\Delta E,\qquad
\boxed{u=\sqrt{\frac{E+\varepsilon}{2E}},\qquad
v=\sqrt{\frac{E-\varepsilon}{2E}}.}
$$
This positive choice fixes a convenient convention for the <vacuum state> in part (d). It yields
$$
\boxed{H=E(N_c+N_d)+\varepsilon-E,\qquad
\lambda=\sqrt{\varepsilon^2+\Delta^2},\quad
\mu=\varepsilon-\sqrt{\varepsilon^2+\Delta^2}.}
$$
Both <number operators> independently have <eigenvalues> $0,1$. Hence the <energy eigenvalues> and <state degeneracies> are
$$
\boxed{\begin{array}{c|c|c}
N_c+N_d&\text{energy}&\text{degeneracy}\\\hline
0&\varepsilon-E&1\\
1&\varepsilon&2\\
2&\varepsilon+E&1
\end{array}}
$$
There is one <fermionic Bogoliubov quasiparticle> of <energy> $E$ per occupied transformed orbital. The two <Nambu spinor> <matrix> <eigenvalues> $\pm E$ should not be mistaken for two independently occupiable levels of opposite <energy>; the <normal ordering> constant and the two positive <quasiparticle> excitation costs give the four physical <energy eigenvalues> above.
An independent <diagonalization of a matrix> checks both the pairing sign and the additive constant. In the ordered <orthonormal basis> $|0\rangle,|11\rangle$, the even-<fermion parity> block is
$$
H_{\mathrm{even}}=\begin{pmatrix}0&-\Delta\\-\Delta&2\varepsilon\end{pmatrix},
$$
because $ab|11\rangle=-|0\rangle$ and $b^\dagger a^\dagger|0\rangle=-|11\rangle$. Its <characteristic polynomial> is $x^2-2\varepsilon x-\Delta^2$, giving $x=\varepsilon\pm E$. The odd-<fermion parity> <Fock states> $a^\dagger|0\rangle,b^\dagger|0\rangle$ are unaffected by pairing and both have <energy> $\varepsilon$.
Equivalently, with the <Nambu spinor> $\Psi=(a,b^\dagger)^{\mathsf T}$,
$$
H=\Psi^\dagger h\Psi+\varepsilon,\qquad
h=\begin{pmatrix}\varepsilon&-\Delta\\-\Delta&-\varepsilon\end{pmatrix},\qquad
R^{\mathsf T}hR=\operatorname{diag}(E,-E).
$$
Replacing $\Psi$ by $R(c,d^\dagger)^{\mathsf T}$ gives $E c^\dagger c-Edd^\dagger+\varepsilon=E(N_c+N_d)+\varepsilon-E$, reproducing the result by <normal ordering>.
\Image[/past-exam-of-the-advanced-quantum-theory-path-integrals-and-many-particle-physics-course-of-the-university-of-oxford/2025-pairing-spectrum.png]
{title=Pairing splits the even-parity levels and mixes the original vacuum with the occupied pair}
{description=Original exact <energy eigenvalues> in units of $\varepsilon$, and the <probability> $v^2$ of finding both original orbitals occupied in the <two-mode fermionic paired vacuum>. The middle <energy level> remains twofold degenerate; both outside <energy levels> are nondegenerate.}
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