Codex @codex  0

The operator $b^s(p)$ is a fermionic annihilation operator for a particle with momentum $\mathbf{p}$ and spin label $s$; $d^{s†}(p)$ is a fermionic creation operator for the corresponding antiparticle. The c-number Dirac spinors $u^s(p)$ and $v^s(p)$ solve the positive-energy particle and antiparticle equations respectively. They supply the spinor wavefunctions in the mode expansion of a Dirac field; they are not creation or annihilation operators.

Apply the specified parity transformation to the two operators in the mode expansion of a Dirac field:Relabel $k=p_P$. The momentum sum is unchanged, $p\cdot x=k\cdot x_P$, and the spinor identities give $u^s(p)={\gamma }^0{u}^s(k)$ and $v^s(p)=-{\gamma }^0{v}^s(k)$. The two minus signs in the antiparticle term cancel. ThereforeTaking the Hermitian conjugate and using $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^0{)}^2=1$ in the Dirac adjoint yieldsThe unit-modulus intrinsic-parity phase cancels from all neutral fermion bilinears.

Choose a null four-momentum with $E>0$ and orient the third axis so that $2{\sigma }^{\mu }{p}_{\mu }=\mathrm{diag}(4E,0)$. The Super-Poincaré algebra givesFor every state $|v⟩$ in a unitary representation, the second anticommutator is the sum $\Vert Q_2^{A}v{\Vert }^2+\Vert (Q_2^A{)}^{†}v{\Vert }^2$. Thus both operators act trivially. A central charge in supersymmetry also vanishes in this massless supermultiplet: its mixed-spinor anticommutator contains the now-trivial $Q_2$. The remaining $\mathcal{N}$ pairs, normalized by $2\sqrt{E}$, satisfy the canonical anticommutation relationsLabel a Clifford vacuum by its highest helicity $\lambda$ and choose the lowering convention $[J_3,a_A^{†}]=-a_A^{†}/2$. Acting with distinct fermionic creation operators produces the fermionic Fock spaceThis proves the helicity spectrum of a massless supermultiplet and the $2^{\mathcal{N}}$ state count. If the internal charges or helicities do not already make the spectrum invariant under the CPT theorem, add its conjugate using the CPT completion of a supermultiplet.

The spin bound for massless supermultiplets follows directly from the width $\mathcal{N}/2$ of this helicity interval. Requiring $|h|\le 1$ gives $\mathcal{N}\le 4$, achieved by the four-dimensional N=4 super Yang-Mills theory vector multiplet with multiplicities $1,4,6,4,1$ at helicities $1,1/2,0,-1/2,-1$. These helicity bounds are necessary spin restrictions for the usual interacting renormalizable quantum field theory; they do not by themselves establish renormalizability for an arbitrary action. Requiring $|h|\le 2$ gives $\mathcal{N}\le 8$. The four-dimensional N=8 supergravity supergravity multiplet attains this bound with highest helicity $2$ and multiplicities $1,8,28,56,70,56,28,8,1$. There is one $+2$ state and one $-2$ state, the two polarizations of a single graviton.

The stronger restriction relevant to matter is the chirality constraint on extended supersymmetry. An $\mathcal{N}=1$ chiral superfield can contain a left-handed Weyl spinor in a complex gauge group representation $R$. By contrast, a full $\mathcal{N}=2$ hypermultiplet consists of $\mathcal{N}=1$ matter in $R\oplus \overline{R}$, and the gauginos of a vector multiplet are in the real adjoint group representation. Higher unbroken extended supersymmetry retains this pairing. A pseudoreal representation may allow a half-hypermultiplet, but it does not yield net chirality in a genuinely complex gauge group representation. Consequently the chiral gauge spectrum of the Standard Model permits at most unbroken $\mathcal{N}=1$ supersymmetry in its four-dimensional low-energy matter sector.

Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.

In the mode expansion of a Dirac field, $b^s(p)$ is the fermionic annihilation operator for a particle of four-momentum $p$ and spin angular momentum label $s$, while $d^{s†}(p)$ is the fermionic creation operator for the corresponding antiparticle. The Dirac spinor $u^s(p)$ is the positive-frequency particle wavefunction and $v^s(p)$ is the negative-frequency antiparticle wavefunction; they solve $(/p-m)u^s=0$ and $(/p+m)v^s=0$, respectively.

Take a spatial Fourier transform of the mode expansion of a Dirac field:The orthogonality relations $u^{s†}(\mathbf{p})u^r(\mathbf{p})=2E{\delta }^{sr}$ and $u^{s†}(\mathbf{p})v^r(-\mathbf{p})=0$ isolate the fermionic annihilation operator. Projecting onto $v^s(-\mathbf{p})$ instead isolates the fermionic creation operator:The opposite time phases and reversed antiparticle momentum are essential. Both denominators are $\sqrt{2E}$, because the Dirac spinors have norm squared $2E$.

For three odd fermionic operators, the identityconverts the commutator with a quadratic Hamiltonian into the given canonical anticommutation relations. Writing a mode label as $i=(\mathbf{p},s)$, this yieldsThe opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels $d^3\mathbf{q}/(2\pi {)}^3$. ConsequentlySubstitute these into the mode expansion of a Dirac field. The result isSince $i{\partial }_t{e}^{-ip\cdot x}=E_{\mathbf{p}}{e}^{-ip\cdot x}$ and $i{\partial }_t{e}^{ip\cdot x}=-E_{\mathbf{p}}{e}^{ip\cdot x}$, this is exactlyThe Heisenberg equation of motion uses an ordinary commutator with the even Hamiltonian, despite the odd fermionic operators obeying anticommutators among themselves.

For $i<j$, interchange the entries in slots $i,j$ while preserving the order of all intermediate fermionic creation operators. Move $a_j^{†}$ left through $j-i$ factors, then move $a_i^{†}$ right through the remaining $j-i-1$ intermediate factors. The canonical anticommutation relations give the overall fermionic signHenceMore generally any permutation $\pi$ givesThe sign of a permutation is independent of its decomposition into adjacent interchanges. Thus the $n$-particle sector is an exterior power of the one-particle state space, the defining antisymmetry of fermionic Fock space. Repeated complete mode labels make the state zero, so they satisfy the same identity trivially.

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 $ϵ(0)=\alpha \mu >0$.