Codex @codex  0

Taylor expansion about any supersymmetric vacuum $v$ givesTherefore the fluctuation scalar potential isFor the nonzero vacuum $w=-2M/g$ in the shifted coordinate, $M_v=M+gw=-M$. Thus it is also $|-M\tilde{\phi }+g{\tilde{\phi }}^2/2{|}^2$. The cubic interaction changes sign compared with expansion at the other root, whereas the quadratic mass term is the same.

Use the usual normalization of the Super-Poincaré algebra, with ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ under the corresponding index convention:The supercharges are odd operators: they turn bosonic states into fermionic states and vice versa. If $\Pi =(-1{)}^F$ is fermion parity, this statement is $\Pi Q_{\alpha }{\Pi }^{-1}=-Q_{\alpha }$, soThe same relation holds for the conjugate supercharges.

For a finite-dimensional physical supermultiplet at fixed four-momentum with energy $E>0$, let $n_B,n_F$ count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation implySumming over the two spinor indices givesThis supertrace pairing at positive energy proves boson-fermion degeneracy in a supermultiplet for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement $E>0$ matters: a zero-energy supersymmetric vacuum can be a bosonic singlet without a paired fermionic vacuum.

If supersymmetry-breaking operators are explicitly added to the Lagrangian, the original supercharges generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical supermultiplets; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by $4E\Pi$ on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.

This is explicit versus spontaneous supersymmetry breaking. In spontaneous breaking the action still has conserved supercharges, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.

In the Polonyi model, the Kähler metric is $K_{z\bar{z}}=1$. The relevant Kähler covariant derivative of a superpotential isThe supergravity auxiliary field is $F^z=-e^{|z{|}^2/2}\overline{D_{z}W}$, up to an irrelevant common phase convention. Thus a constant vacuum preserves supersymmetry exactly when this auxiliary field vanishes. For $m=0$, the superpotential and scalar potential vanish identically, and every constant scalar value is a supersymmetric vacuum.

For $m\ne 0$, put $⟨z⟩=x+iy$. Its supersymmetry condition becomesSince $\beta >0$, it requires $y=0$ and $x^2+\beta x+1=0$. Hence the Polonyi supersymmetry branches areAt these points $W\ne 0$, so the supergravity F-term potential is negative, $V=-3e^K|W{|}^2$: these are supersymmetric Anti-de Sitter spacetime vacua, not zero-energy ones. The condition $D_{z}W=0$ also makes them stationary, as follows by differentiating the supergravity F-term potential.

For $0<\beta <2$ and $m\ne 0$, the auxiliary field cannot vanish anywhere, so any vacuum has supersymmetry breaking. For $\beta \ge 2$, a stationary vacuum at any other scalar value still breaks supersymmetry; the parameter condition alone does not determine which vacuum is selected. In particular, a nontrivial zero-energy vacuum cannot preserve supersymmetry: $D_{z}W=0$ and $V=0$ would also imply $W=0$, whereas $W=0$ gives $z=-\beta$ and $D_{z}W=m^2\ne 0$.

Use the minimal four-dimensional Super-Poincaré algebra, with $\{Q_{\alpha },Q_{\beta }\}=0$, $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\nu }{P}_{\nu }$ and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for $P^{\mu },Q_{\alpha },Q_{\beta }$ therefore givesFor example, with ${\sigma }^0=I$, $\mu =0$ and $\alpha =\beta =1$, the coefficient is a nonzero multiple of $A(P_1-i{P}_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently $[Q_{\alpha },P^2]=0$. If $|u⟩$ has a particular mass-shell condition, the nonzero state $Q_{\alpha }|u⟩$ has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.

For the O'Raifeartaigh model, write $X={\varphi }_1$, $Y={\varphi }_2$, $Z={\varphi }_3$. Field phases allow $g,M,m$ to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-g{m}^2$ could not vanish. Thus F-flatness is impossible. More precisely, with $Z=u+iv$,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere $F_X=-\overline{W_X}=g{m}^2$, while $F_Y=F_Z=0$. The Weyl spinor ${\psi }_X$ is the massless goldstino, and the complex scalar field $X$ is a pseudomodulus with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.

At the representative vacuum $x=0$, the chiral-superfield fermion mass matrix isIts physical masses are its singular values: $0,M,M$. The two massive Weyl spinors can be combined into one massive Dirac spinor. For $Y=(y_R+i{y}_I)/\sqrt{2}$, $Z=(z_R+i{z}_I)/\sqrt{2}$, the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the $Z$ masses displays supersymmetry breaking directly.

For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are $M^2(M^2+2ϵg^2{m}^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.

Define the physical mass supertrace byEach real scalar contributes once and each Weyl spinor twice with a minus sign. At $x=0$ this gives $4M^2-2(2M^2)=0$. At general $x$, ${\sum }_{ϵ,\pm }{b}_{ϵ,\pm }=4M^2+8g^2|x{|}^2=2(f_{+}+f_{-})$. Hence the tree-level mass supertrace vanishes despite broken supersymmetry:This is the tree-level supertrace mass sum rule, not a claim that the individual masses are equal.

For direct breaking in the Minimal supersymmetric Standard Model, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant superpotential term needs a gauge-singlet chiral superfield, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral F-term breaking, without D-term mass shifts, the MSSM tree-level sfermion mass constraint applies separately to conserved charge sectors. For one electron pair it gives $m_{{\tilde{e}}_1}^2+m_{{\tilde{e}}_2}^2=2m_e^2$: both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total supertrace alone. A separate hidden supersymmetry-breaking sector communicating effective soft supersymmetry breaking through loops or suppressed operators avoids the direct canonical tree-level obstruction.

Vary $F_i$ and $F_i^{\ast }$ independently. The auxiliary fields have algebraic, rather than differential, Euler-Lagrange equations:Alternatively complete the square:Substitution therefore yields the on-shell auxiliary fields and nonnegative scalar potentialThe sign in the Lagrangian is minus the F-term scalar potential; changing the metric convention does not change this algebraic result. A supersymmetric vacuum requires all $W_i=0$, namely F-flatness. For a noncanonical positive Kähler metric $K_{i\bar{j}}$, the purely bosonic elimination instead gives $V_F=K^{i\bar{j}}{W}_i\overline{W_j}$; fermionic connection terms must also be included in a complete nonlinear component action.

On shell, a massless $\mathcal{N}=1$ vector multiplet contains a gauge boson with two transverse polarizations and a gaugino with two fermionic states. A chiral superfield contains a complex scalar field, hence two real bosonic states, and a Weyl spinor with two fermionic states. Their total is $4$ bosonic and $4$ fermionic states.

Suppose charged scalar fields acquire vacuum expectation values at a supersymmetric vacuum, where F-flatness and D-flatness hold. For each broken generator of the gauge symmetry, the supersymmetric Higgs mechanism combines one massless vector multiplet and one chiral superfield into a massive N=1 vector multiplet. The gauge boson absorbs one real Goldstone boson, acquiring its third polarization. The other real scalar field remains physical. The gaugino and the relevant higgsino mix into the four fermionic states. Thus the spin content isUnbroken supersymmetry gives all these states the same mass through supersymmetric mass degeneracy. Remaining massive chiral superfields instead have spin content $(1/2,0,0)$, with two fermionic and two bosonic states of a common mass. The auxiliary fields ensure matching off shell but are not additional propagating particles in these counts.

The super-Higgs mechanism occurs when local supersymmetry is spontaneously broken in supergravity. Its spin-$3/2$ gauge field, the gravitino, absorbs the spin-$1/2$ goldstino:The extra states have helicities $\pm 1/2$, supplementing $\pm 3/2$. The ordinary supersymmetric Higgs mechanism can preserve supersymmetry while breaking the internal gauge symmetry; the super-Higgs mechanism breaks the local supersymmetry itself. Spontaneously broken global supersymmetry leaves a physical massless goldstino, because there is no dynamical gravitino to absorb it.

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.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

Forthe three F-terms areThe equation $W_Z=0$ forces $X=0$, after which $W_Y=\alpha \ne 0$; hence no supersymmetric vacuum exists. Minimizing over $Z$ sets $Z=-2\beta XY/\gamma$, leavingThe two real quadratic eigenvalues about $X=0$ are $|\gamma {|}^2\pm 2|\alpha \beta |$, so the stated hierarchy makes $X=0$ stable. The vacua areThis is an O'Raifeartaigh model: supersymmetry is spontaneously broken and $Y$ is a classical pseudomodulus.

For $W=X^{2}Z$, the F-term scalar potential isBoth terms vanish exactly when $X=0$, while $Z$ is arbitrary. Thus the supersymmetric vacua form one complex line,

Suppose $\zeta >0$ and all $m_i$ are distinct. A nonzero ${\varphi }_j$ forces $x=m_j$. Distinctness then makes every other charged field vanish, and ${\sum }_i{\tilde{\varphi }}_i{\varphi }_i=0$ forces ${\tilde{\varphi }}_j=0$. The D-flatness equation fixes $|{\varphi }_j{|}^2=\zeta$, while its phase is removed by the gauge group. There is therefore one isolated supersymmetric vacuum for each flavor,for a total of $N$ vacua. Negative $\zeta$ gives the analogous vacua with ${\tilde{\varphi }}_j$ nonzero.

For the Polonyi model with canonical Kähler potential and $W=m^2(z+\beta )$, $\beta >0$, a constant supersymmetric vacuum with $m\ne 0$ requires real $z=(-\beta \pm \sqrt{{\beta }^2-4})/2$ and exists exactly for $\beta \ge 2$. Its supergravity F-term potential is negative because $W\ne 0$. For $0<\beta <2$, the supergravity auxiliary field cannot vanish. If $m=0$, the model is flat with unbroken supersymmetry. This parameter-dependent statement is sharper than asserting that every linear superpotential breaks supersymmetry in supergravity.

In an unbroken supersymmetric vacuum, physical boson and fermion partners have equal masses because supercharges commute with four-momentum. For one canonical chiral superfield, fluctuations about $W^{\prime }(v)=0$ have mass $|W^{\prime \prime }(v)|$ for both the complex scalar field and Weyl spinor.

In the small-field global approximation, the waterfall fluctuations on $u=v=0$ have the displayed masses. One becomes negative when $|S{|}^2<|\zeta |$, driving the fields towards the supersymmetric vacuum branch. In the canonical supergravity model, the exact masses follow from expanding the exponential and Kähler-covariant derivatives; their Planck-suppressed corrections should not be discarded unless the corresponding field and symmetry-breaking scales are small.

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.