Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2025 / 2

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular $\{a,b^{†}\}=\{a^{†},b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by $⟨0|0⟩=1$ and fix the orderingThe canonical anticommutation relations give $a^2=(a^{†}{)}^2=0$ and $a{a}^{†}=1-a^{†}a$. Consequently the number operator is an orthogonal projector:The same holds for $N_b$. Hence each eigenvalue obeys $n^2=n$ and is either $0$ or $1$, expressing the Pauli exclusion principle. Also $[N_a,N_b]=0$, so the two number operators have a simultaneous orthonormal basis of Fock states:Each number operator has eigenvalues $0,1$, each with state degeneracy two. For $N_a$, the zero eigenspace is spanned by $|00⟩,|01⟩$ and the unit eigenspace by $|10⟩,|11⟩$; for $N_b$, these are respectively $|00⟩,|10⟩$ and $|01⟩,|11⟩$. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below: $b^{†}{a}^{†}|0⟩=-|11⟩$ and

Writing the Bogoliubov transformation componentwise, together with its Hermitian adjoint, givesThe full canonical anticommutation relations for $c,d$ implyEqual-type anticommutators vanish because $c^2=d^2=0$ and $\{c,d^{†}\}=\{c^{†},d\}=0$:For the mixed anticommutators, the only potentially nonzero terms cancel:while $\{a,b^{†}\}=0$ follows from $\{c,c\}=\{d^{†},d^{†}\}=\{c,d^{†}\}=0$. Taking the Hermitian adjoint supplies the remaining relations. Thus the necessary and sufficient condition isEquivalently, the matrix $R=\left(\begin{array}{cc}u& v\\ -v& u\end{array}\right)$ is an orthogonal matrix, $R^{\mathsf{T}}R=I$, giving a canonical rotation of two fermionic modes. Its inverse is therefore

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere 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 $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

The canonical anticommutation relations make $(a^{†}{b}^{†}{)}^2=0$, so the proposed exponential function terminates exactly:Use the inverse Bogoliubov transformation $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$, and the actions $a|11⟩=b^{†}|0⟩$ and $b|11⟩=-a^{†}|0⟩$. The two vacuum state conditions becomeBoth therefore require $B=v/u$. Normalizing this quantum state gives $1=A^2(1+B^2)$. Choosing positive $A$, which fixes an irrelevant overall global phase, yields $A=u$ because $u^2+v^2=1$. Thus the two-mode fermionic paired vacuum isThe other sign for $A$ describes the same physical quantum state. The positive relative coefficient is fixed by the paper's pairing order and the convention $|11⟩=a^{†}{b}^{†}|0⟩$.

As a direct eigenvector check, $\Delta v=(E-\epsilon )u$ and $\Delta u=(E+\epsilon )v$ implyIt is therefore the unique ground state, as also follows from positive quasiparticle excitation energy $E$. It has definite even fermion parity but is not an eigenstate of the original total number operator: the empty-pair and occupied-pair probabilities are $u^2$ and $v^2$. In particular $⟨N_a⟩=⟨N_b⟩=v^2$. The pair amplitude is $⟨ab⟩=-uv$, consistently with $ab|11⟩=-|0⟩$.

For the limiting check $\Delta \to 0^{+}$ at fixed positive $\epsilon$, $A\to 1$, $B\to 0$, so $|\tilde{0}⟩\to |0⟩$. For $\epsilon \to 0^{+}$ at fixed positive $\Delta$, $A\to 1/\sqrt{2}$, $B\to 1$, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the Bogoliubov transformation produces the ground state.