research35 min studyLesson 737 in Physics

Unitarity bounds and short representations

Derive dimension restrictions from descendant norms while stating the identity and representation exceptions.

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What you’ll understand

  • Use radial adjoints to compute descendant norms.
  • Apply scalar and symmetric-traceless bounds in d at least three.
  • Interpret null equations without extending the bound to unrelated representations.

A primary dimension cannot be chosen freely in a reflection-positive CFT. Translations generate descendants, and every physical linear combination must have nonnegative norm. Commuting their adjoint special conformal generators back onto the primary turns that requirement into algebraic dimension bounds. Work in Euclidean integer dimension d at least three, with the radial conventions already fixed. The results below cover scalars and symmetric traceless integer-spin tensors; spinors, mixed-symmetry tensors, two-dimensional Virasoro representations and supersymmetric multiplets require their own analysis.

Start with a scalar first-level norm

Normalize a scalar primary state to one. The norm of a derivative in one chosen Cartesian direction is the expectation of a special conformal generator followed by a translation. The primary is killed by the lowering generator, so only their commutator survives. Rotations act trivially on a scalar primary, leaving twice its dimension. A negative dimension would therefore create a negative-norm descendant immediately and is excluded under these assumptions.

This first-level test is necessary but not sufficient. For scalars, the contracted level-two descendant gives a stronger condition. Its norm contains a factor that vanishes at the free-scalar dimension. The separate dimension-zero solution includes the identity, whose translation descendants vanish; it must not be incorrectly ruled out by the positive-dimension scalar bound. In the usual irreducible theory with a unique vacuum, the physical dimension-zero scalar sector is the identity rather than an arbitrary new propagating field.

∥Pμ∣O⟩∥2=2Δ,∥P2∣O⟩∥2=8d Δ ⁣(Δ−d−22)\|P_\mu|\mathcal O\rangle\|^2=2\Delta,\qquad \|P^2|\mathcal O\rangle\|^2=8d\,\Delta\!\left(\Delta-\frac{d-2}{2}\right)
The first formula fixes one μ with no sum; the second uses P² = ΣμPμPμ. The primary is a unit-normalized scalar and P† = K. These are state norms, not Casimir eigenvalues.↔ Scroll wide equations horizontally.

Diagonalize the appropriate spin channels

For a spinning primary, first-level descendants lie in the tensor product of a vector with the primary's rotation representation. The norm matrix is not a single scalar number: rotations mix the tensor components. Decomposing into irreducible rotation sectors identifies its least-positive eigenvalue. For a symmetric traceless tensor of positive integer spin, this gives a lower bound involving spin plus spacetime dimension minus two. Testing only a convenient component can miss the dangerous sector.

At equality, a particular descendant has zero norm. In a unitary Hilbert space it decouples after quotienting null states, leaving a shorter representation. For the symmetric traceless family, that null descendant is the divergence, yielding a conserved current equation at separated insertions. A spin-one conserved current has dimension d minus one, and the stress tensor has spin two and dimension d. Contact terms in Ward identities must still be handled separately from the vanishing separated-point divergence.

Δ=0 (identity),Δ≥d−22 for nonidentity scalars,Δ≥ℓ+d−2 for symmetric traceless ℓ≥1\Delta=0\ \text{(identity)},\qquad \Delta\ge\frac{d-2}{2}\ \text{for nonidentity scalars},\qquad \Delta\ge\ell+d-2\ \text{for symmetric traceless }\ell\ge1
These bounds assume reflection positivity and the stated representation class in d ≥ 3. They are not a common formula for every Lorentz representation or for nonunitary CFTs.↔ Scroll wide equations horizontally.

Read shortening as an operator equation with scope

A scalar saturating its positive-dimension bound has a null Laplacian descendant. In the usual local unitary setting this is a free massless scalar equation, identifying a free scalar sector rather than proving that every other operator in the entire theory is free. An anomalous dimension above the bound restores a nonnull descendant. Additional algebraic symmetries can create other shortening conditions, but those follow from their enlarged representation algebra and cannot be inferred from the scalar calculation alone.

Passing all these unitarity tests is still not sufficient to construct a CFT. The proposed primaries must also fit consistent OPE coefficients, crossing, locality and the necessary stress-tensor structure. Conversely, a nonunitary conformal model may violate the positive-norm bounds without an algebraic contradiction because it drops their Hilbert-space assumption. The quadratic lab can illustrate the sign of a simple norm-like function, but it neither computes conformal descendant matrices nor checks a whole representation.

P2∣ϕ⟩=0 ⟺ ∂2ϕ=0at scalar saturation,Pμ1∣Oμ1⋯μℓ⟩=0 ⟺ ∂μ1Oμ1⋯μℓ=0P^2|\phi\rangle=0\ \Longleftrightarrow\ \partial^2\phi=0\quad\text{at scalar saturation},\qquad P^{\mu_1}|\mathcal O_{\mu_1\cdots\mu_\ell}\rangle=0\ \Longleftrightarrow\ \partial^{\mu_1}\mathcal O_{\mu_1\cdots\mu_\ell}=0
The operator statements concern null-state decoupling in separated correlators with the stated locality and unitarity assumptions. Equality of bounds does not eliminate contact terms or replace full CFT consistency.↔ Scroll wide equations horizontally.

A second-level norm detects a forbidden scalar

In d = 3 consider unit-normalized scalar candidates with Δ=0.40\Delta=0.40 and Δ=0.50\Delta=0.50. Compute their fixed-direction level-one norms and contracted level-two norms. Which test distinguishes them?

  1. The first-level norms are 2(0.40)=0.802(0.40)=0.80 and 2(0.50)=1.002(0.50)=1.00. Both pass this limited positive-dimension check.

  2. The stronger factor uses (d−2)/2=0.50(d-2)/2=0.50. The first candidate has norm 8(3)(0.40)(0.40−0.50)=−0.968(3)(0.40)(0.40-0.50)=-0.96, so it contains a negative-norm descendant.

  3. The second candidate has a zero level-two norm. It saturates the scalar bound and requires a null Laplacian descendant, consistent with a free scalar sector. A positive first-level norm alone did not establish this structure.

The result

The 0.40 candidate fails through a −0.96 second-level norm; the 0.50 candidate saturates the scalar bound with a null Laplacian.

Explore the model

Interactive lab

Interactive lab · function

A sign illustration with an explicit limit

Reset and set BOTH Coefficient a = 0.50 and Position x = 1; predict y = 0.50. Reset and set a = −0.50 and x = 1; predict y = −0.50. This quadratic sign exercise is not a descendant norm formula, a conformal representation or a unitarity test.

A sign illustration with an explicit limitThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA positivity argument requires the correct norm matrix, not merely a positive plotted curve.

Check your understanding

Which scalar is an explicit exception to the nonidentity lower bound Δ ≥ (d−2)/2 in the stated d ≥ 3 setting?

Try it yourself

In d = 4, a symmetric traceless spin-one primary is proposed at Δ=2.80\Delta=2.80. Test the relevant bound. What dimension is required if this operator instead belongs to a conserved current multiplet?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the positive-spin symmetric-traceless bound, not the scalar formula. Saturation is the conserved case.

Compare with the solution

The bound is Δ≥1+4−2=3\Delta\ge1+4-2=3. A primary at 2.80 violates it in the stated reflection-positive setting. A conserved spin-one current saturates the bound at Δ=3\Delta=3, with its divergence a null descendant. This conclusion assumes a genuine primary in the specified representation; relabeling an arbitrary derivative as a current would not meet those assumptions. It also does not apply unchanged to a nonunitary theory or another rotation representation.

Take it with you

  • Descendant positivity is stronger than positivity of one primary norm.
  • Scalar bounds have a separate identity solution.
  • Saturation shortens a representation through specific null equations.

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