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Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

A Majorana spinor equals its transform under charge conjugation, so its spinor components are not independent of their conjugates. In a real two-dimensional gamma matrix representation it can be taken real, with anticommuting Grassmann variables as components. Each target index labels a separate worldsheet Majorana fermion; it is not itself a target-space spinor.

Use worldsheet signature $(-,+)$ and the real gamma matricesTake $\bar{\psi }={\psi }^{T}C$ with $C={\gamma }^0$. Define the chirality matrix with the chosen worldsheet orientation byIt squares to one and anticommutes with each gamma matrix. Chirality is its eigenvalue, selected by the chiral projectors. Its overall sign is conventional.

Writing $\psi =({\psi }_{+},{\psi }_{-}{)}^T$, the fermion equation of motion ${\gamma }^{\mu }{\partial }_{\mu }\psi =0$ becomesHenceThe first profile travels toward increasing $\sigma$, and the second toward decreasing $\sigma$. Thus the choice of chirality orientation realizes the requested worldsheet chirality and propagation direction correspondence; reversing the orientation interchanges the labels.

For the rigid worldsheet supersymmetry variation, let $ϵ$ be a constant Grassmann-odd Majorana spinor. With the displayed conventions,The minus sign follows from $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. A second identity from the Majorana Grassmann bilinear interchange is $\bar{\chi }{\gamma }^{\nu }{\gamma }^{\mu }ϵ=\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }\chi$. Keeping these Grassmann signs is essential.

The bosonic kinetic variation is $2i{\partial }_{\mu }{X}_a\overline{ϵ}{\partial }^{\mu }{\psi }^a$. The two fermionic kinetic variations, using the two identities above, give the full resultFor the second equality use ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and symmetry of the second embedding derivative. No field equation has been used. Thus the variation is an off-shell total derivative:This is the rigid worldsheet supersymmetry boundary term. It vanishes when the boundary flux vanishes, for example on a closed worldsheet with compatible fields and variations. On a spatial circle, a constant supersymmetry parameter must also respect the chosen spin structure. Periodic fermions allow this rigid transformation; antiperiodic fermions and a constant parameter would give an antiperiodic $\delta X$ and fail to preserve a periodic bosonic embedding. The local action identity remains valid, but global symmetry requires compatible boundary data. This is the spin-structure obstruction to constant worldsheet supersymmetry.

For the spectrum, assume the usual critical RNS string, physical-state constraints, a nonzero null target momentum, and the supersymmetric GSO projection. Analyze just one chiral sector. Its matter central charge is $d+d/2$; the bc system and superconformal ghosts contribute $-26+11$. Therefore $3d/2-15=0$ gives the critical dimension of the RNS superstring $d=10$. Light-cone gauge in string theory leaves eight transverse bosonic and eight transverse fermionic string oscillator directions.

In the Neveu–Schwarz sector, the normal-ordering intercept is $a_{\mathrm{NS}}=1/2$. The massless level is $N=1/2$ and has statesThe GSO projection retains these eight transverse vector polarizations while removing the tachyonic ground state. They are spacetime bosons, although the creating string oscillator is a worldsheet fermion. In covariant language transversality and the longitudinal null-state quotient leave $10-2=8$ polarizations.

In the Ramond sector, bosonic and fermionic zero-point energies cancel, so $a_{\mathrm{R}}=0$ and the ground states are massless. The eight transverse fermion zero modes obeyThis Ramond zero-mode Clifford algebra acts on a sixteen-dimensional ground-state space. Its two chirality subspaces each have dimension eight. The GSO projection keeps one, an $8_s$ or $8_c$ spinor of $\mathrm{SO}(8)$, the rotation subgroup of the massless little group. These are spacetime fermions. Equivalently a ten-dimensional Majorana-Weyl spinor has sixteen real components before the massless Dirac equation reduces the physical polarization count to eight.

Consequently the massless chiral RNS spectrum after GSO projection satisfiesThe GSO projection is essential: without it, the transverse Ramond sector would retain both eight-dimensional spinor chiralities. These are counts for one chiral sector, not the tensor-product counts of the full closed-string spectrum.

Apply the Grassmann Gaussian integral to a regulated finite collection of field components. Up to a field-independent measure normalization and phase,In the continuum this is a formal functional determinant, including spinor and spacetime indices. Its meaningful definition requires a regulator and boundary conditions. A complex Dirac field supplies a determinant, not the inverse square root obtained for a real commuting field.

The comparison is clean after Wick rotation. A free real scalar field with positive Euclidean operator $P_E=-{\partial }_E^2+m^2$ hasTaking a logarithm gives $-\frac{1}{2}\mathrm{Tr}\log P_E$ for the scalar and $+\mathrm{Tr}\log K_E$ for the Dirac field. The opposite statistics sign is the determinant counterpart of the minus sign for a closed fermion loop. The magnitude also differs because a Dirac field has several independent spin and antiparticle degrees of freedom.

The vacuum energy sign of a fermionic oscillator makes the comparison explicit. A real scalar mode contributes $+E/2$; each independent fermionic oscillator contributes $-E/2$. There is one oscillator per scalar momentum, but a massive Dirac field has two particle and two antiparticle oscillators. Thus the vacuum energy densities are formallyEquivalently, the large Euclidean-time vacuum functional behaves as $\log Z_E\sim -\beta V{\rho }_{\mathrm{vac}}$. Both displayed vacuum energies are ultraviolet divergent; a common regularization in quantum field theory and the appropriate renormalization are needed before comparing them. The zero-point vacuum energies of bosons and fermions have opposite signs, but do not cancel without matching masses and degrees of freedom. Normal ordering removes an additive vacuum constant in nongravitating flat-space theory. Coupling to the metric in general relativity makes that constant contribute to the cosmological constant, so it cannot simply be discarded without a renormalization condition.

For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is $(\square -m^2)\varphi =0$, or $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.

Resolve the free field into positive- and negative-frequency plane waves. With $E_{\mathbf{k}}=\sqrt{|\mathbf{k}{|}^2+m^2}$ and $k\cdot x=-E_{\mathbf{k}}t+\mathbf{k}\cdot \mathbf{x}$,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in $[\varphi ,\pi ]$ each supply half the Dirac delta function. Thus the normalization $1/\sqrt{2E_{\mathbf{k}}}$ is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with $a$ and $a^{†}$ its annihilation operator and creation operator.

Choose the Fock vacuum by $a(\mathbf{k})|0⟩=0$. Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume $V$ can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, $[H,a^{†}(\mathbf{k})]=E_{\mathbf{k}}{a}^{†}(\mathbf{k})$ and $[\mathbf{P},a^{†}(\mathbf{k})]=\mathbf{k}{a}^{†}(\mathbf{k})$. A one-particle state therefore has positive energy, momentum $\mathbf{k}$, and invariant mass $m$. A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.

A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase $e^{-i{E}_{\mathbf{k}}t}$; a narrow packet has group velocity $\mathbf{v}=\mathbf{k}/E_{\mathbf{k}}$, whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.

Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so $[\varphi (x),\varphi (y)]=0$ for spacelike $x-y$. Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.

Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are $(a^{†}{)}^n|0⟩/\sqrt{n!}$ with $n=0,1,2,\dots$: any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.

For occupation much smaller than $2S$, expand the square root in the Holstein–Primakoff transformation. To leading order $S_n^{+}=\sqrt{2S}{a}_n+O(S^{-1/2})$, $S_n^{-}=\sqrt{2S}{a}_n^{†}+O(S^{-1/2})$ and $S_n^z=S-a_n^{†}{a}_n$. ConsequentlyPeriodic boundaries count each occupation in two bonds. The linear spin-wave approximation is thereforeThe displayed fluctuation Hamiltonian is bilinear and conserves boson number. The omitted terms are magnon interactions, quartic at the first omitted order. With occupation held of order one, they are smaller by $1/S$ than the order-$S$ quadratic terms. The polarized vacuum has no zero-point depletion in this ferromagnetic expansion.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.

A Majorana spinor equals its charge conjugation, ${\psi }^c=C{\bar{\psi }}^T=\psi$, with conventional phase choices absorbed into $C$. The two-component worldsheet Majorana fermion therefore has no independent complex conjugate components; in a Majorana representation its components can be real Grassmann variables. The same condition applies to the constant supersymmetry parameter. Grassmann statistics matter in the variation: replacing the spinors by commuting numerical vectors would give the wrong bilinear interchange signs.

The displayed rigid transformation is understood in flat conformal gauge. Take $h_{\mu \nu }={\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$, $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, and chooseHere $C={\gamma }^0$ is a valid invariant charge-conjugation form: $C^T=-C$ and $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. These concrete matrices make all signs checkable; equivalent Majorana conventions give the same result with their consistently transformed bars.

Let $A=\sqrt{{\alpha }^{\prime }/2}$ and $B=1/\sqrt{2{\alpha }^{\prime }}$, so ${\alpha }^{\prime }B=A$. Suppressing the common factor $-1/(4\pi {\alpha }^{\prime })$ in the action, the flat kinetic density isThe rigid variations have $\delta X^a=iA\overline{ϵ}{\psi }^a$ and $\delta {\psi }^a=B{\gamma }^{\nu }{\partial }_{\nu }{X}^aϵ$. The supersymmetry variation is even, so its product rule has no extra graded sign. From the Majorana Grassmann bilinear interchange identities,For example the last identity follows by moving the Grassmann-odd parameter through $\psi$, then using $C^T=-C$ and the transpose relations twice. No equation of motion has been used.

The two kinetic variations areThe Clifford algebra turns the first fermionic term plus the bosonic variation into $iA{\partial }_{\nu }{X}_a\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }{\partial }_{\mu }{\psi }^a$. In the second term, commute the partial derivatives and use the same algebra: the antisymmetric gamma product drops out, leaving $iA\square X_a\overline{ϵ}{\psi }^a$. Together they are precisely the rigid worldsheet supersymmetry boundary term:Therefore $\delta S=-(4\pi {\alpha }^{\prime }{)}^{-1}{\int }_{\partial \Sigma }d{\Sigma }_{\mu }{J}^{\mu }$. It vanishes on a closed or periodic worldsheet, for compactly supported changes, or under compatible supersymmetric endpoint conditions. The local total-derivative identity is off shell; the boundary assumptions are needed to call the integrated action invariant.

The flat-gauge qualification is substantive. On an arbitrary curved worldsheet, spinors need a zweibein and spin connection and a constant spinor parameter need not exist. A fully covariant locally supersymmetric action also involves the worldsheet gravitino; the isolated ordinary-derivative expression does not prove rigid invariance for arbitrary $h_{\mu \nu }$. We have proved the intended rigid symmetry of its flat gauge-fixed action, with the fermionic term inside the same integral and overall normalization. If the displayed last term were read as outside the integral, it would not even define an action.

For phenomenology, bosonic string theory has no spacetime fermions and its usual vacuum contains a tachyon. The spinning string has a Ramond sector, whose fermionic zero modes form a spacetime Clifford algebra and give spacetime spinor states, as well as a Neveu–Schwarz sector. In suitable consistent theories the GSO projection removes the tachyonic NS ground state and keeps the appropriate Ramond chirality. The critical dimension of the RNS superstring is 10 rather than 26: $D$ bosons and $D$ real fermions have matter central charge $3D/2$ per chiral sector, while the reparameterization and superconformal ghosts contribute $-26+11=-15$, so the total anomaly cancels at $D=10$.

Suitable GSO-projected superstrings admit spacetime fermions and a tachyon-free spectrum, making them more promising for particle physics than the bosonic string. Rigid worldsheet supersymmetry alone does not establish a tachyon-free spacetime theory or realistic phenomenology. The projection, consistent sectors and compactification in string theory are additional input; spacetime supersymmetry and a realistic four-dimensional spectrum are not automatic consequences of this classical action.

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.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.