Past exam of the field theories and collective phenomena in condensed matter course of the University of Oxford / 2025 / 3

The canonical anticommutation relations imply $a^2=(a^{†}{)}^2=0$. Thus the number operator satisfiesIt is a orthogonal projection, so its eigenvalues obey $n_a^2=n_a$ and are $\boxed{n_a\in \{0,1\}}$. The same argument gives $\boxed{n_b\in \{0,1\}}$. This is the Pauli exclusion principle for each orbital. The two number operators commute, so they have simultaneous eigenstates.

Take the vacuum state to have unit norm and fix the double-occupation convention $|11⟩=a^{†}{b}^{†}|0⟩$. An orthonormal basis of the four-dimensional fermionic Fock space isEach individual number operator has two-dimensional eigenspaces: for $N_a$ these are spanned by $\{|00⟩,|01⟩\}$ and $\{|10⟩,|11⟩\}$; for $N_b$ they are spanned by $\{|00⟩,|10⟩\}$ and $\{|01⟩,|11⟩\}$. Higher occupation numbers vanish because the square of a fermionic creation operator is zero. Reversing the order of the two fermionic creation operators changes only the sign convention for $|11⟩$.

The Bogoliubov transformation and its Hermitian conjugate readUsing the canonical anticommutation relations of $c,d$, the nonzero same-mode anticommutators areThe equal-type anticommutators satisfy $\{a,a\}=\{b,b\}=0$, since every contributing equal-type or mixed-mode anticommutator of $c,d$ vanishes. The potentially nontrivial cross term cancels:Also $\{a,b^{†}\}=0$, and the other cross relations follow by Hermitian conjugation. Therefore the necessary and sufficient condition isThe real transformation matrix is then an orthogonal matrix, with inverseThis preservation of the canonical anticommutation relations makes it a canonical Bogoliubov transformation.

Write $Q=c^{†}c+d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Expanding the number operators and restoring the displayed order with the canonical anticommutation relations givesThe minus sign in the first line follows from $c^{†}{d}^{†}+dc=-(cd+d^{†}{c}^{†})$. Substitution into the Hamiltonian operator yieldsLet $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$. A convenient real Bogoliubov transformation hasThese obey $u^2+v^2=1$, $u^2-v^2=\epsilon /E$ and $2uv=\Delta /E$. Equivalently $u=\cos \theta$, $v=\sin \theta$ with $0<\theta <\pi /4$ and $\tan (2\theta )=\Delta /\epsilon$. The coefficient of $P$ vanishes, while the coefficient of $Q$ is $E$. The constant isThus the diagonalized two-mode fermion pairing Hamiltonian isThe two transformed fermionic occupation numbers independently take values zero or one. The energy eigenvalues and their eigenspace dimensions are thereforeThe positive quasiparticle energy $E$ makes the transformed vacuum state the unique ground state. The constant $\mu$ is essential for the correct ground-state energy.

As a separate sign and spectrum check, use the original fermionic Fock space basis. Both singly occupied states are energy eigenstates with energy $\epsilon$, since the pairing terms annihilate them. In the ordered even-fermion parity basis $(|0⟩,a^{†}{b}^{†}|0⟩)$,The off-diagonal signs follow from $b^{†}{a}^{†}=-a^{†}{b}^{†}$ and $ab{a}^{†}{b}^{†}|0⟩=-|0⟩$. Its characteristic polynomial is $z^2-2\epsilon z-{\Delta }^2$, giving the remaining two eigenvalues $\epsilon \pm E$, in agreement with the Bogoliubov transformation.

The inverse Bogoliubov transformation gives $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$. Because $(a^{†}{b}^{†}{)}^2=0$ by the canonical anticommutation relations, the exponential truncates:The fermionic signs are fixed byIt follows thatBoth annihilation conditions hold precisely when $B=v/u$. Normalizing the paired fermionic vacuum, using the orthogonality of its two Fock states, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result isEquivalently,Its positive pair coefficient follows from the specified order $ab+b^{†}{a}^{†}$ in the Hamiltonian operator. The state has even fermion parity and is a quantum superposition of zero and two original particles. In the limit $\Delta \to 0^{+}$, $A\to 1$ and $B\to 0$, so it reduces to the original vacuum state, as expected for $\epsilon >0$.