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Put $d_k=\sqrt{(g-\cos k{)}^2+{\sin }^{2}k}$. The matrix in the Ising-chain Nambu spinor has eigenvalues $\pm d_k$. A canonical Bogoliubov transformation, with paired angles satisfying $\cos {\vartheta }_k=(g-\cos k)/d_k$ and $\sin {\vartheta }_k=-\sin k/d_k$, diagonalizes it. Accounting for full-zone fermionic Nambu energy counting givesThus the bulk Ising-chain ground-state energy isExact finite-chain states additionally obey their fermion parity sector.

The Ising-chain quasiparticle dispersion has bulk gap $2J|1-g|$, closing at the Ising-chain quantum critical point. Finite-system parity splittings and the bulk gap are distinct scales. The three requested limits are illustrated below.

For a bulk bond, $K_{n+1}=K_n(1-2n_n)$ and $c_n(1-2n_n)=-c_n$. Reordering the remaining different-site operators then givesThe periodic spin boundary condition requires separate attention. Set $P={\prod }_{n=1}^N(1-2n_n)=(-1{)}^{N_f}$. The end bond becomes $S_N^{+}{S}_1^{-}=-P{c}_1^{†}{c}_N$, with the conjugate relation for its adjoint. Consequently the exact finite-ring Hamiltonian isfermion parity is conserved, so diagonalize in a sector $P=\pm 1$. There the end bond can be written like every other bond provided $c_{N+1}=-P{c}_1$. This parity-dependent boundary condition of the Jordan-Wigner chain is antiperiodic for even fermion number and periodic for odd fermion number. Simply assuming periodic fermions in both sectors changes the finite-size spectrum.

The occupation formula from part (b) givesConsequentlyThe independent occupation numbers $n_1,n_2$ are each zero or one. The Hamiltonian is independent of $n_1$, so its four states haveEach energy admits $n_1=0$ or $n_1=1$; these states differ by occupancy of the zero-energy mode. The ground energy is $-|t|$ for either sign of $t$. These degeneracies refer to the full two-mode fermionic Fock space. If one separately restricts to a fixed global fermion parity, only one state at each energy remains in that parity sector.

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.

A supersymmetry generator is an odd Weyl spinor generator of the Super-Poincaré algebra. Acting on a physical state reverses fermion parity while preserving its four-momentum.

A supertrace is a trace weighted by the grading of a Hilbert space. For fermion parity $(-1{)}^F$, it is $\mathrm{Str}A=\mathrm{Tr}[(-1{)}^{F}A]$: even states contribute with a plus sign and odd states with a minus sign.

In supersymmetric quantum mechanics, paired positive-energy states have opposite fermion parity and cancel in the supertrace of $e^{-\beta H}$. Unpaired zero modes then contribute to the Witten index. This cancellation requires a well-defined trace or a consistent regulator; a continuum reaching zero energy can invalidate a naive term-by-term argument. The supertrace is therefore useful for counting protected differences between sectors, while an ordinary thermal trace adds both sectors and counts a different quantity.

For a finite physical supermultiplet at fixed positive energy, fermion parity anticommutes with each odd supercharge. Trace cyclicity makes $\mathrm{Tr}[(-1{)}^F\{Q,Q^{†}\}]=0$. Summing the Super-Poincaré algebra diagonal entries replaces the anticommutator by $4E$, giving equal bosonic and fermionic state counts. A zero-energy vacuum is an exception to the division by $E$, so it need not have a partner. The argument requires an invariant fixed-momentum representation, which is the step lost under explicit supersymmetry breaking.

The normalized vacuum state annihilated by $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$, for positive $u,v$ with $u^2+v^2=1$. Since $(a^{†}{b}^{†}{)}^2=0$, it can also be written $u\exp [(v/u)a^{†}{b}^{†}]|0⟩$. Acting with either annihilation operator makes the two terms cancel. Its original occupation numbers each have expected value $v^2$ and its pair amplitude is $⟨ab⟩=-uv$. It has even fermion parity although the original total number operator is not sharp. For the two-orbital fermionic pairing Hamiltonian, it is the unique ground state of energy $\epsilon -\sqrt{{\epsilon }^2+{\Delta }^2}$.

For distinct fermionic orbitals obeying the full canonical anticommutation relations, fix the Fock state order $|11⟩=a^{†}{b}^{†}|0⟩$. With real $\epsilon ,\Delta >0$, the even-fermion parity matrix is $\left(\begin{array}{cc}0& -\Delta \\ -\Delta & 2\epsilon \end{array}\right)$, since $ab|11⟩=-|0⟩$. The two odd-fermion parity energy eigenvalues are both $\epsilon$. The even-fermion parity energy eigenvalues are $\epsilon \pm \sqrt{{\epsilon }^2+{\Delta }^2}$. This finite model isolates the relative sign and constant shift required in a fermionic Bogoliubov transformation without any momentum sum or particle-hole doubling factor.

With a well-defined regulated supertrace, positive-energy states of supersymmetric quantum mechanics are paired by an odd invertible charge. Their opposite fermion parities cancel, leaving $I=n_B^0-n_F^0$, independent of $\beta$. A nonzero index proves the existence of a supersymmetric vacuum; zero does not prove its absence. Discrete-spectrum or suitable infrared conditions matter, because continuum thresholds and states escaping to infinity can invalidate naive trace manipulations or deformation invariance.