Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2018 / 1

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

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 readUsing the canonical anticommutation relations for the $\alpha$ modes givesThe 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 requiresThis 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.

Group the nonzero modes into independent pairs $p>0$. For one pair set $\epsilon =ϵ(p)=p^2/(2m)+\mu$, and introduce the Nambu spinor ${\Psi }_p=(c_p,c_{-p}^{†}{)}^{\mathsf{T}}$. Normal ordering using $c_{-p}{c}_{-p}^{†}=1-c_{-p}^{†}{c}_{-p}$ givesThis 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 ${\mathcal{A}}_p=({\alpha }_p,{\alpha }_{-p}^{†}{)}^{\mathsf{T}}$ and $U_p$ is the Bogoliubov transformation matrix. Direct matrix multiplication yieldsChoose the diagonalizing branchThen 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 asHere $u_p$ 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 givesThe zero-momentum term was never paired and is simply $\mu {\alpha }_0^{†}{\alpha }_0$. HencewithThe constant is the ground-state energy. At fixed $L$, its ultraviolet tail converges: $ϵ(p)-E(p)=-{\Delta }^2/[ϵ(p)+E(p)]\sim -m{\Delta }^2/p^2$ in one dimension. Replacing $E(0)$ by $\sqrt{{\mu }^2+{\Delta }^2}$ would incorrectly pair a spinless mode with itself.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.