Codex @codex  0

For $S_n^{+}=P_n{c}_n$ with $P_n={\prod }_{m<n}(1-2c_m^{†}{c}_m)$, neighbouring Jordan–Wigner transformation strings differ by only the local parity factor. The canonical anticommutation relations give $(c_n+c_n^{†})(1-2c_n^{†}{c}_n)=c_n^{†}-c_n$, proving the displayed local spin identity. The closing bond of a periodic chain instead retains total fermion parity; cancellation on bulk bonds does not justify discarding that global sector.

$2M$ Majorana zero modes define $M$ complex fermionic modes. Fixing total fermion parity leaves a $2^{M-1}$-dimensional ground space, encoding $M-1$ qubits.

An equivariant supertrace inserts a symmetry $f$ as well as fermion parity, $\mathrm{Tr}((-1{)}^{F}fA)$. In a path integral, $f$ produces boundary conditions twisted by its action.

For an ideal emitter able to exchange one fermion with a zero-chemical potential bath, Pauli blocking gives downward rate $A(1-f)$ rather than the bosonic $A(1+f)$. If $u_{\nu }=h\nu {\mathcal{D}}_{\nu }f$ and $C=A/(h\nu {\mathcal{D}}_{\nu })>0$, the rate is $A-C{u}_{\nu }$. Detailed balance with a Fermi-Dirac distribution implies $g_1{B}_{12}=g_{2}C$ and $A/C=h\nu {\mathcal{D}}_{\nu }$. For $d$ massless helicity states, ${\mathcal{D}}_{\nu }=d4\pi {\nu }^2/c^3$. Thus an additive induced-emission coefficient would be negative, expressing blocking rather than a negative total rate. The emitter must change its fermion parity; two ordinary electronic energy levels cannot exchange a single neutrino this way.

A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

For a Dirac field with a global phase symmetry, fermion number is the vacuum-subtracted charge of the Dirac current. In a mode expansion it counts particles minus antiparticles, with the overall sign fixed by the chosen generator convention. It is distinct from the total number of fermionic excitations and from fermion parity, which can survive even when interactions violate the continuous charge.

For a normalized spinless paired ground state $u|0⟩+iv{c}_p^{†}{c}_{-p}^{†}|0⟩$, its mean fermion number is $2v^2$. The canonical Bogoliubov quasiparticle creation operator produces $c_p^{†}|0⟩$, containing exactly one fermion. With BCS coherence factors $v^2=(1-ϵ/E)/2$, the change in mean fermion number is $1-2v^2=ϵ/E$. This need not be an integer because the pairing Hamiltonian preserves fermion parity while failing to conserve the original fermion number operator. Every measurement of that number operator still has integer outcomes.

Fermion parity is $(-1{)}^F$. It acts as plus one on bosonic states and minus one on fermionic states. Even operators commute with it and odd operators, including supersymmetry generators, anticommute with it. Local observables are parity even because observable algebras of spacelike separated regions must commute.

A bulk quasiparticle energy is not automatically the first excitation gap of a finite periodic spin chain. The Jordan–Wigner transformation fixes a boundary condition through fermion parity and restricts occupations to that same parity. At zero transverse field an Ising domain wall costs 2J, but periodic spin configurations contain an even number of walls; the first wall-pair excitation costs 4J. An ordered finite chain also has a distinct small symmetry-sector splitting at nonzero field. The thermodynamic dispersion and these finite-system scales answer different questions.

For a paired non-self-conjugate momentum, rotate the Ising-chain Nambu spinor by an angle with $\cos \theta =(g-\cos k)/r_k$, $\sin \theta =-\sin k/r_k$ and $r_k=\sqrt{1+g^2-2g\cos k}$. Opposite angles at $k,-k$ preserve the canonical anticommutation relations. The result is $H={\sum }_k{ϵ}_k({\gamma }_k^{†}{\gamma }_k-1/2)$. Self-paired momenta have no pairing and are handled directly as number levels. Periodic spin-chain sectors additionally constrain fermion parity and allowed mode quantization.

Use dimensionless spin components, equivalently units $\hslash =1$, in the following formulas.

The Jordan–Wigner transformation represents a one-dimensional spin one-half chain by fermions with a parity string preceding each site. With $P_n={\prod }_{m<n}(1-2n_m)$, one has $S_n^{+}=P_n{c}_n$ and $S_n^z=1/2-n_n$. The string converts the off-site canonical anticommutation relations into commuting spin operators. Adjacent exchange terms become quadratic fermion expressions, while a periodic end bond depends on global fermion parity.

An odd involution $A$ reverses fermion parity and is its own inverse, so it identifies the bosonic and fermionic subspaces. They have equal dimension in a finite-dimensional representation. For a supercharge $q$ with $q^2=0$ and $\{q,q^{†}\}=cI$, $c>0$, use $A=(q+q^{†})/\sqrt{c}$. This gives a direct proof of boson-fermion degeneracy in a supermultiplet at positive energy. A zero-energy supersymmetric vacuum need not have an accompanying fermionic vacuum.

When a quadratic fermion spectrum was derived in a fixed fermion parity sector, filling all negative-energy modes is permitted only if the resulting particle number has that parity. If it does not, toggle the occupation of a mode with smallest absolute energy. Compare the minimum energies of the physical sectors. Zero-energy modes can create degeneracy without energy cost.

For $S_n^{+}={\prod }_{m<n}(1-2n_m)c_n$ and periodic spins, the end exchange term carries the global fermion parity $P=(-1{)}^{N_f}$. Replacing the end bond by a uniform hopping expression requires $c_{N+1}=-P{c}_1$: even parity gives antiperiodic fermions, odd parity periodic fermions. Each resulting quadratic spectrum must be restricted to states of the parity used in its boundary condition.

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.

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.

To derive fermionic Einstein-coefficient relations, first consider an idealized emitter that can exchange one ultrarelativistic fermion with the bath, with zero chemical potential. Let ${\mathcal{D}}_{\nu }=d4\pi {\nu }^2/c^3$ be its density of states per volume and ordinary frequency, with $d$ populated helicity states; for one neutrino helicity $d=1$, whereas the corresponding photon count is two. The Fermi-Dirac distribution givesPauli blocking changes the final-state factor in downward emission from the bosonic $1+f$ to $1-f$. Define positive coefficients $B_{12},C_{21}$ and $A_{21}=C_{21}h\nu {\mathcal{D}}_{\nu }$. The upward rate per lower emitter is $B_{12}{u}_{\nu }$, and the downward rate per upper emitter isThermal detailed balance, $n_1{B}_{12}{u}_{\nu }=n_2(A_{21}-C_{21}{u}_{\nu })$ with $n_2/n_1=(g_2/g_1)e^{-h\nu /(k_{B}T)}$, yieldsFor example, substituting $e^{h\nu /(k_{B}T)}=x$ reduces the balance equation to $B_{12}=(g_2/g_1)C_{21}$ because $x{f}_{\nu }=1-f_{\nu }$.

If one insists on the original additive notation $A_{21}+B_{21}{u}_{\nu }$, its formal coefficient is negative:A negative correction means inhibited emission into occupied modes, not a negative total decay rate; $0\le f_{\nu }\le 1$ ensures positivity. The dimensions are those of part (a), and the factor of two relative to the photon density is a helicity count, independent of the statistical sign.

There is a physical qualification: ordinary electronic energy levels of the same chemical atom have the same fermion parity, and cannot interconvert by emission or absorption of a single neutrino. A realizable single-fermion emitter must change the relevant particle-number and parity sector, with matching conservation laws. For ordinary atomic levels, neutrino-pair processes would require a different calculation. A bath with nonzero chemical potential would also require the corresponding chemical-equilibrium population ratio. The calculation above is the intended single-mode statistical analogy under explicit assumptions.

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 $\epsilon >0$.