What BPS and protected mean

“BPS” can describe a particle, a local operator, or a line observable. In each case some supersymmetry acts trivially, but the resulting constraint is different. A mass can obey an exact formula while changing with the vacuum. An operator can have a fixed scaling dimension while its correlator normalization changes. A supersymmetric loop can have a nontrivial expectation value.

Start with local operators and radial quantization. The common argument below needs no dimension-specific spinor formalism. The three-dimensional calculation additionally uses the three-dimensional algebra and R-symmetry lessons; the independent four-dimensional particle comparison uses the extended-algebra lesson. Read the specialization appropriate to your route. Neither specialization is a prerequisite for the other.

Start with a positive norm, not a label

Let $q$ be an odd conserved charge on a positive Hilbert space and $q^\dagger$ its adjoint in that Hilbert space. For any state $|v\rangle$ in their domains,

$$\langle v|\{q,q^\dagger\}|v\rangle =\|q|v\rangle\|^2+\|q^\dagger|v\rangle\|^2\geq0.$$

Suppose the algebra identifies this anticommutator with a Hermitian generator $\mathcal H_q$, and $\mathcal H_q|v\rangle=h|v\rangle$. For a unit-normalized state,

$$h=\|q|v\rangle\|^2+\|q^\dagger|v\rangle\|^2.$$

This gives both the bound $h\geq0$ and its saturation condition:

$$h=0\quad\Longleftrightarrow\quad q|v\rangle=q^\dagger|v\rangle=0.$$

Annihilation by $q$ alone is insufficient unless there is a separate reason that $q^\dagger$ also annihilates the state. The equation becomes a mass bound or a dimension bound only after we specify the Hilbert space, its adjoint, and the generators in $\mathcal H_q$.

A zero-norm descendant is called null. Positivity implies that it has zero inner product with every physical state, by the Cauchy–Schwarz inequality. Removing null states and their null descendants produces a short multiplet, with fewer independent states than a generic representation. A local-operator multiplet still contains infinitely many ordinary derivative descendants; “short” does not mean a finite set of cylinder states.

What superconformal symmetry adds

In a conformal field theory, the generators are translations $P_\mu$, rotations $M_{\mu\nu}$, dilations $D$, and special conformal transformations $K_\mu$. A superconformal field theory, or SCFT, also has Poincaré supercharges $Q$, conformal supercharges $S$, and the R-symmetry generators required by its superconformal algebra.

The structure, suppressing spinor and family labels, is

$$\{Q,Q\}\sim P,\qquad \{S,S\}\sim K,\qquad \{S,Q\}\sim D+M+R.$$

Here $Q$ denotes the collection of Poincaré families: this schematic notation does not say that every same-family anticommutator is nonzero. The coefficients and R-generators depend on the dimension and amount of supersymmetry. A flavor generator commutes with the supercharges and is not an interchangeable substitute for the R-generator in these relations.

In radial quantization, $D$ measures dimension, $P$ raises it by one, and a Poincaré supercharge raises it by one-half:

$$[D,Q]=\frac12Q,\qquad [D,S]=-\frac12S,\qquad P^\dagger=K,\qquad Q^\dagger=S,$$

with matching component labels understood. The adjoint is the radial one. It differs from the ordinary Lorentzian Poincaré adjoint, whose mixed anticommutator produces momentum. The conformal part of this distinction is established in Simmons-Duffin, §7.2, (108)–(116).

A superconformal primary $|\mathcal O\rangle$ is a lowest-dimension state of a superconformal multiplet: all conformal supercharges annihilate it, and $K_\mu|\mathcal O\rangle=0$. Its superdescendants are generated by $Q$ and ordinary derivatives. A conformal primary need only be annihilated by $K_\mu$. A nonzero $Q|\mathcal O\rangle$ can itself be a conformal primary, since $[K,Q]$ is proportional to $S$, even though it is a superdescendant.

Thus one superconformal multiplet generally contains several conformal families. The generator structure and this organization are described in Córdova, Dumitrescu, and Intriligator v1, §1.2. Counting the full algebra also matters: three-dimensional $\mathcal N=2$ has four real Poincaré and four real conformal supercharges; four-dimensional $\mathcal N=4$ has sixteen of each. Choosing one $q$ for a norm calculation does not change either theory’s full supersymmetry.

Three-dimensional shortening

This section uses the three-dimensional algebra and R-symmetry versus flavor symmetry. Work in a unitary three-dimensional $\mathcal N=2$ SCFT, in its conformally invariant vacuum. Local states are those on the unit $S^2$ of radial quantization. In a gauge theory, the local operators considered here must be gauge invariant. No massive-particle central charge or dimensionful deformation is present.

Write the eigenvalues of $D$, the superconformal R-generator, and one Hermitian rotation generator $J_3$ as $\Delta,r,m$. For a sphere of radius $L$, the cylinder energy gap is $\Delta/L$; $\Delta$ itself is dimensionless.

Choose a component $q$ of the R-charge-$+1$ Poincaré family, with charges

$$[D,q]=\frac12q,\qquad [R,q]=q,\qquad [J_3,q]=-\frac12q.$$

It belongs to the family called $\widetilde Q$ in the earlier component lessons, not the R-charge-$-1$ family $Q$. The family signs agree with the text after P v3, (1.2). We choose the normalization of this radial component so that

$$\boxed{\{q,q^\dagger\}=D-R-J_3.}$$

This is Kapustin and Willett v1, (2.2), with their energy generator written as $D$. Its relative coefficients pass a quick check: $D-R-J_3$ commutes with $q$, since $1/2-1+1/2=0$.

Now take a unit-normalized scalar superconformal primary. It has $m=0$ and $q^\dagger|\mathcal O\rangle=0$, because the radial adjoint is a conformal supercharge. Therefore

$$\begin{aligned} \|q|\mathcal O\rangle\|^2 &=\langle\mathcal O|q^\dagger q|\mathcal O\rangle\\ &=\langle\mathcal O|\{q^\dagger,q\}|\mathcal O\rangle\\ &=\Delta-r. \end{aligned}$$

Rotating the spinor component gives the same norm for its other component on a scalar state. The opposite Poincaré family gives $\Delta+r$. In a basis with unit component normalization, the two first-level Gram matrices are consequently

$$G_{\widetilde Q}=(\Delta-r)I_2,\qquad G_Q=(\Delta+r)I_2.$$

These matrices are necessary positivity tests. They are not the complete classification of superconformal representations: higher descendants impose further conditions.

If the scalar primary is chiral, by definition both components of $\widetilde Q$ annihilate it. The norm equation gives

$$\boxed{\Delta=r\quad\text{for a scalar chiral superconformal primary}.}$$

An antichiral primary is annihilated by the opposite family and has $\Delta=-r$. The ordinary conformal scalar bound adds $\Delta\geq1/2$ for a nonidentity scalar in three dimensions. Thus a nonzero, nonidentity chiral primary must have $r\geq1/2$. The identity has $\Delta=r=0$ and is a separate case.

The full representation constraints agree with these scalar chiral relations; see CDI v1, §2.1.2, (2.5) and tables 2–4. We are deriving the chiral relation, not treating the weaker first-level inequality $\Delta\geq|r|$ as sufficient to construct an arbitrary multiplet.

Check the free chiral and a trial R-charge

For the gauge-neutral free chiral scalar $\phi$, the two-point function is

$$\langle\phi(x)\phi^\dagger(0)\rangle =\frac1{4\pi|x|}.$$

It has $\Delta=1/2$ and superconformal charge $r=1/2$. Hence $G_{\widetilde Q}=0$ and $G_Q=I_2$: the chiral family annihilates the primary, while the opposite family produces its fermionic descendants. The additional ordinary conformal null relation is $P^2|\phi\rangle=0$, the separated-point free equation.

This theory also has a flavor generator $F$ with $F\phi=\phi$ and $[F,q]=0$. The infinitesimal trial generator

$$R_t=R+tF$$

still assigns charge $+1$ to $q$. But the actual radial algebra remains

$$\{q,q^\dagger\}=D-R-J_3 =D-R_t-J_3+tF.$$

On $|\phi\rangle$, its eigenvalue remains zero. Incorrectly dropping the last term would give $-t$, negative for $t>0$. Choosing a trial R-charge did not change the free scaling dimension or the Hilbert-space norm. The R-generator appearing in the superconformal algebra must be identified, not chosen by convenience among all symmetries of the flat action.

This also explains the meaning of a protected dimension. As long as the same nonzero chiral primary and its shortening persist, its dimension obeys the exact relation to the actual superconformal R-charge. This is stronger than perturbative engineering dimensions. It does not say that an arbitrary trial charge equals a dimension, that the operator survives every deformation, or that its R-charge cannot change between distinct fixed points.

Four-dimensional particle bounds

This section independently uses extended algebras and BPS representations. It concerns massive one-particle states in a four-dimensional Lorentzian $\mathcal N=2$ vacuum with eight real Poincaré supercharges, not local operators in radial quantization.

In a fixed central-charge sector, the two eigenvalues of each rest-frame supercharge block are

$$2(M-|\mathcal Z|),\qquad 2(M+|\mathcal Z|).$$

Here $M$ is a physical mass and $\mathcal Z$ a complex central charge of dimension one. Positivity gives $M\geq|\mathcal Z|$. At positive-mass saturation, four real supercharge combinations annihilate the particle multiplet. This is the massive half-BPS bound. The full diagonalization belongs to the linked lesson. Bilal v1, (3.15)–(3.16) uses a central-charge coefficient twice ours, so its inequality is $2M\geq|Z_{\rm BIL}|$.

A simple family exposes what the bound protects. Suppose a particle exists and stays BPS along a range of vacua labeled by a dimensionless real parameter $t$, with

$$\mathcal Z(t)=\mu(1+it),\qquad \mu>0.$$

Then its exact relation would be

$$M(t)=\mu\sqrt{1+t^2}.$$

It is short at every point of that range, but its mass changes from $\mu$ at $t=0$ to $\sqrt2\mu$ at $t=1$. This is an algebraic example, not an asserted spectrum of a particular model. The algebra does not guarantee that the particle exists or remains stable throughout the range.

The protected statement is the equality to the central charge. It is not a constant numerical mass. Nor can one replace $\mathcal Z$ by an R-charge and interpret this equation as a conformal scaling-dimension formula.

A supersymmetric observable needs its own argument

A line observable is supported on a contour rather than at one point. A supersymmetric line $W[C]$ is annihilated by specified supercharges, possibly after combining the gauge connection with scalar couplings along $C$. Its preserved fraction refers to those charges in the full algebra and to that contour and background. This is not, by itself, a local-primary condition or a massive-particle mass formula.

The distinction can be tested directly for an expectation value. Let a parameter $t$ appear in a Euclidean action $S_t$ and possibly in a bosonic observable $W_t$. Differentiating the normalized path integral gives

$$\frac{d}{dt}\langle W_t\rangle_t =\left\langle\frac{\partial W_t}{\partial t}\right\rangle_t -\left\langle W_t\,\frac{\partial S_t}{\partial t}\right\rangle_{t,c},$$

where the connected expectation subtracts $\langle W_t\rangle_t\langle\partial_t S_t\rangle_t$. The condition $qW_t=0$ does not make either term vanish automatically.

There is an extra sufficient condition. Suppose the charge and integration prescription are fixed, $W$ has no explicit $t$ dependence, and

$$\frac{\partial S_t}{\partial t}=qV,\qquad qW=0,$$

where $V$ is odd. Assume that the measure, action, regulator and integration contour admit the Ward identity $\langle qX\rangle=0$, with no boundary or contact contribution in this use. Then the graded product rule gives

$$q(WV)=WqV,$$

so $\langle WqV\rangle=0$ and $\langle qV\rangle=0$. The derivative vanishes. A deformation of this form is called supersymmetry-exact. General physical couplings need not enter the action in this way. Establishing these hypotheses is part of a protection or localization argument; it does not follow from the word “BPS.”

Even a protected local dimension leaves a normalization question. Multiply the entire free chiral action by a constant $a>0$, while keeping the field symbol $\phi$ fixed. Gaussian inversion gives

$$\langle\phi(x)\phi^\dagger(0)\rangle_a =\frac1{4\pi a|x|}.$$

The dimension stays $1/2$, while the coefficient varies. This particular change is removable by $\phi_c=\sqrt a\,\phi$: it illustrates operator normalization, not a new physical conformal manifold. After fixing a normalization convention, dynamical OPE coefficients or line expectation values still need their own Ward identities or nonrenormalization arguments.

The distinctions can be kept together without conflating their Hilbert spaces:

Object	Algebraic statement	What still needs dynamical input
Massive particle	$M=\lvert\mathcal Z\rvert$ at a stated BPS bound	Existence, stability, and variation of $\mathcal Z$ with the vacuum
Chiral scalar primary in a 3d SCFT	$\Delta=r$ for the superconformal R-generator	Identification of that generator, survival of the operator, and correlator coefficients
Supersymmetric line or other insertion	A specified charge annihilates the insertion	Its expectation value and whether a deformation is supersymmetry-exact

Short multiplets can sometimes recombine into long ones when parameters change. Whether a particular representation can do so is a further representation-theory question, not a consequence of the elementary norm identity alone. Keeping a protected relation, a protected number of states, and a coupling-independent observable separate avoids overclaiming any one of them.

Exercises

1. Annihilation by one charge is not enough

For $h>0$, consider the two-state Hilbert space with orthonormal basis $|0\rangle,|1\rangle$ and

$$q=\begin{pmatrix}0&\sqrt h\\0&0\end{pmatrix}.$$

Compute $q^2$, $\{q,q^\dagger\}$, and the two norms on $|0\rangle$. Is this state at a zero of $\mathcal H_q$? Explain why the scalar-primary argument above can nevertheless use the single condition $q|\mathcal O\rangle=0$.

First hint

The adjoint matrix has its nonzero entry below the diagonal. Act on the state before taking the norm.

Second hint

A superconformal primary is already annihilated by every conformal supercharge, including the radial adjoint of $q$.

Solution

The square is zero and the anticommutator is $hI_2$. On $|0\rangle$, $q|0\rangle=0$ but $q^\dagger|0\rangle=\sqrt h\,|1\rangle$. The two squared norms are $0$ and $h$, so $\mathcal H_q$ has positive eigenvalue $h$. This state is closed under $q$ but does not saturate the zero-eigenvalue bound.

For a superconformal primary, the independent condition $q^\dagger|\mathcal O\rangle=0$ is already available. Imposing $q|\mathcal O\rangle=0$ then removes both positive contributions. The matrix example realizes only this selected nilpotent charge pair, not a full superconformal algebra.

2. Test proposed three-dimensional scalar primaries

With the preparation for the three-dimensional section, test these proposed nonidentity scalar superconformal primary data:

$$(\Delta,r)=\left(\frac25,\frac25\right),\qquad \left(\frac12,\frac34\right),\qquad \left(\frac12,\frac12\right).$$

Compute the two first-level supercharge norms and the norm of the ordinary conformal trace descendant $P^2|\mathcal O\rangle$. Use the previously derived formula $8d\Delta(\Delta-(d-2)/2)$ for a unit-normalized scalar primary. Which tests fail? Does passing these tests alone classify every superconformal multiplet?

First hint

Each diagonal entry of the two supercharge Gram matrices is $\Delta-r$ or $\Delta+r$.

Second hint

In three dimensions the trace-descendant norm is $24\Delta(\Delta-1/2)$. It is independent of the R-charge.

Solution

For $(2/5,2/5)$, the first-level entries are $0$ and $4/5$, but the trace descendant has squared norm

$$24\left(\frac25\right)\left(-\frac1{10}\right) =-\frac{24}{25}.$$

The negative norm excludes it as a physical nonidentity scalar. Chirality cannot evade ordinary conformal unitarity.

For $(1/2,3/4)$, the first-level entries are $-1/4$ and $5/4$, already excluding the proposal. The trace descendant is null, which cannot repair the negative state.

For $(1/2,1/2)$, the entries are $0$ and $1$, and the trace descendant is null. The free chiral supplies an actual representation with these data. The tested inequalities are necessary; additional descendants and the full algebra matter in general. Also, a gauge-charged elementary field by itself is not a gauge-invariant local state to which this Hilbert-space test can be applied.

3. A short particle with a changing mass

With the preparation for the four-dimensional section, use the assumed BPS family $\mathcal Z(t)=\mu(1+it)$. Find $dM/dt$ and the two rest-frame norm eigenvalues. What would happen if one instead kept $M=\mu$ while changing $t$ away from zero? State which conclusion remains conditional on the particle existing.

First hint

The magnitude of the central charge is $\mu\sqrt{1+t^2}$, regardless of its phase.

Second hint

The smaller norm eigenvalue is $2(M-|\mathcal Z|)$. A negative value is incompatible with a positive Hilbert space.

Solution

For a BPS particle,

$$\frac{dM}{dt}=\frac{\mu t}{\sqrt{1+t^2}},\qquad \lambda_{\rm small}=0,\qquad \lambda_{\rm large}=4\mu\sqrt{1+t^2}.$$

Thus the representation stays short while the nonzero scale varies. With $M$ artificially held at $\mu$, the smaller eigenvalue becomes $2\mu(1-\sqrt{1+t^2})<0$ for every $t\ne0$, excluding that mass assignment.

The algebra gives the equality for a state that exists and remains BPS. It does not show that a particle can be followed through every value of $t$ without decay, recombination, or another change in the spectrum.

4. Identify the missing Ward-identity hypothesis

Suppose the Ward identity is valid and $qW_0=0$. First keep the action fixed and define $W_t=e^tW_0$. Next keep $W_0$ fixed but deform the action with $\partial_tS_t=U$, where $U$ is even and is not assumed to be $q$-exact. Derive the response in both cases. What additional statement would make the second response vanish by the argument in the text?

First hint

The first observable has explicit parameter dependence even though it is annihilated by the same charge.

Second hint

For the action deformation, differentiate the normalization of the path integral as well as its numerator.

Solution

With a fixed action,

$$\frac d{dt}\langle W_t\rangle =e^t\langle W_0\rangle.$$

There is no reason for this to vanish. This is a deliberate normalization change, illustrating that charge annihilation alone does not determine a number.

For the second deformation,

$$\frac d{dt}\langle W_0\rangle_t =-\langle W_0U\rangle_t+\langle W_0\rangle_t\langle U\rangle_t =-\langle W_0U\rangle_{t,c}.$$

Knowing that $W_0$ is supersymmetric does not eliminate this connected correlator. A sufficient additional statement is $U=qV$ for an admissible odd $V$, with the fixed charge and all Ward-identity hypotheses remaining valid throughout the deformation. Then both terms vanish by the product-rule calculation above. This is a sufficient mechanism, not a claim that it is the only possible reason a response can vanish.

When reading a protection claim, identify the object, the adjoint or preserved charge, and the quantity fixed by the algebra. The remaining question is then concrete: which charge must be determined, which state must persist, or which Ward identity must be established?

A concrete application is the superconformal bootstrap: combine protected current data with crossing and physical OPE positivity to derive a central-charge constraint.