---
title: "How one linear functional rules out a whole spectrum"
description: "Apply one linear functional to the crossing equation. If it is positive on every allowed operator, no spectrum with that gap can exist."
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last_modified: "2026-10-29"
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[Conformal bootstrap](https://sandboxphysics.com/stories/conformal-bootstrap/)

L3 · Chapter 5

# How one linear functional rules out a whole spectrum

Sandbox Physics · 2026-10-29

The crossing equation says a sum of vectors with non-negative weights must add up to zero. That gives a way to prove a spectrum impossible without knowing anything else about it: find one linear functional that is positive on every vector in the sum. This chapter builds such a functional by hand with two derivatives, then follows the bound as more derivatives tighten it toward the free fermion.

## Before you begin

The crossing equation and the free fermion from the previous two chapters, vectors and dot products, and the idea of an inequality between linear expressions. Familiarity with linear programming helps but is not needed.

- Explain why a functional that is positive on every allowed operator rules out a spectrum, and why positivity of the coefficients is essential.
- Construct the two-derivative functional for a one-dimensional fermion and follow the bound toward its analytic limit.

## Crossing as a sum of vectors

Move everything in the crossing equation to one side. The identity contributes a fixed function $F_0$, and each exchanged operator contributes $F_\Delta$ with weight $c_\Delta^2\ge0$. Their sum must vanish for every $z$.

Expand each function in a Taylor series around $z=1/2$. Only odd orders survive, because every $F_\Delta$ changes sign under $z\to1-z$. Keeping orders up to $\Lambda$ turns each function into a vector with $(\Lambda+1)/2$ components.

$$
F_0+\sum_{\Delta\ge\Delta_{\mathrm{gap}}}c_\Delta^2\,F_\Delta=0,\qquad F_\Delta(z)=(1-z)^{2\Delta_\psi}k_\Delta(z)-z^{2\Delta_\psi}k_\Delta(1-z)
$$

The assumption under test is the gap: no exchanged operator lies below $\Delta_{\mathrm{gap}}$.

Source: D. Mazáč · S. Rychkov and A. Vichi

## One inequality instead of infinitely many unknowns

Suppose a linear functional $\alpha$, a fixed set of weights for the Taylor components, is positive on the identity and non-negative on every allowed $F_\Delta$. Apply it to the crossing equation. The left side is a positive number plus non-negative terms, so it cannot be zero.

That contradiction excludes every spectrum with this gap at once, whatever the dimensions and coefficients above it. Searching for such an $\alpha$ is a linear programme. If none exists, the solver instead returns a set of operators with positive weights that satisfies the truncated equations.

$$
\alpha(F_0)>0\quad\text{and}\quad\alpha(F_\Delta)\ge0\ \ \text{for all }\Delta\ge\Delta_{\mathrm{gap}}\ \ \Longrightarrow\ \ \text{no consistent spectrum}
$$

The argument needs $c_\Delta^2\ge0$. Without unitarity a negative weight could cancel the positive terms.

Source: S. Rychkov and A. Vichi · D. Simmons-Duffin

## Worked example: a functional built from two derivatives

Take $\Delta_\psi=1/2$ and keep the first and third derivatives, $\Lambda=3$. In a fixed normalisation the vectors are as below. The second component of $F_\Delta$ is negative for light operators and changes sign at $\Delta=2.4456$.

Set the gap at $2.5$. The functional $\alpha=(-1,\,21.65)$ gives $\alpha(F_0)=1$, zero at $\Delta=2.5$, $12.0$ at $\Delta=3$, $46.8$ at $\Delta=4$ and $123$ at $\Delta=6$. Nothing above the gap can cancel the identity, so the gap is excluded.

Set the gap at $2.3$ instead. Then $F_{2.3}=(0.974,\,-0.111)$, and every normalised functional $(-1,a)$ is negative on it when $a\ge0$, or negative on $F_3$ when $a<0$. The solver finds $0.925\,F_{2.3}+0.091\,F_{3.365}=-F_0$: a positive spectrum, so this gap is allowed at this order.

With two derivatives, the bound is therefore the zero of the third-order component, $2.4456$. The bisection over linear programmes gives $2.4457$, the same number within its tolerance.

$$
F_0=(-1,\,0),\quad F_{2}=(0.888,\,-0.271),\quad F_{3}=(1.074,\,0.604),\quad \alpha=(-1,\,21.65)
$$

First and third Taylor coefficients at the symmetric point, in the normalisation used by the Lab.

Source: D. Mazáč · S. Rychkov and A. Vichi

## More derivatives, a sharper bound

Each extra pair of derivatives gives the functional more freedom, so the bound can only fall. For $\Delta_\psi=1/2$ it is $2.4457$ at $\Lambda=3$, $2.2316$ at $9$, $2.1533$ at $13$ and $2.0893$ at $21$.

The free fermion has its first operator at $2\Delta_\psi+1=2$ and satisfies crossing exactly, so no valid bound can go below $2$. Mazáč and Paulos constructed functionals in closed form showing that $2\Delta_\psi+1$ is the optimal bound. The numbers above approach it from above, slowly.

Source: D. Mazáč · D. Mazáč and M. F. Paulos

## Try it in the Lab

### [Exclude a gap, then let it through](https://sandboxphysics.com/labs/linear-functional-exclusion/)

1. Open the Lab at its default gap and check that the functional curve stays above zero everywhere above the gap.
2. Choose the three-derivative preset. Predict the bound, then compare it with the readout.
3. Lower the trial gap below the bound and look for the operators the solver places instead.

## Exercises with explained answers

1. Why would the exclusion argument fail if one coefficient $c_\Delta^2$ were allowed to be negative?

   Show the answer

   Then $c_\Delta^2\,\alpha(F_\Delta)$ could be negative and cancel $\alpha(F_0)>0$. The sum could vanish, and the functional would prove nothing.

2. At $\Lambda=3$, is the gap $2.4$ excluded? Use the sign change of the third component.

   Show the answer

   No. $2.4$ lies below the zero at $2.4456$, so the operator at the gap has a negative third component, and as in the worked example no normalised functional can be non-negative on it and on $F_3$ together.

3. A claimed one-dimensional theory has $\Delta_\psi=1/2$ and first exchanged operator at $2.1$. Which of the bounds in this chapter rules it out?

   Show the answer

   The bound at $\Lambda=21$, $2.0893$, is below $2.1$, so that order excludes it. At $\Lambda=13$ the bound $2.1533$ still allows it.

[Previous: Free fields solve the crossing equation exactly](https://sandboxphysics.com/stories/conformal-bootstrap/free-fields-as-exact-solutions/) [Next: What a numerical bootstrap bound proves, and what it does not](https://sandboxphysics.com/stories/conformal-bootstrap/what-a-bound-proves/)

## Sources and what we recomputed

1. [D. Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, arXiv:1602.07982v1 (2016)](https://arxiv.org/abs/1602.07982v1)

   Version 1. Eqs. (157)-(158): the four-point function as a sum over conformal multiplets, one conformal block each. Eqs. (164)-(168): the Casimir equation for blocks and their closed form in two dimensions, built from the one-variable function k. Eqs. (194)-(199), Sections 10.5-10.7: the crossing equation for identical scalars as a sum of vectors, the derivative functionals and the assumptions behind the scalar-gap bound. Fig. 26: 2D crossing vectors. Fig. 27 and Table 1: 2D bounds for derivative cutoffs 6 to 28 with the kink at the Ising point. Fig. 28: the 3D bound with its kink.

2. [D. Mazáč, Analytic bounds and emergence of AdS2 physics from the conformal bootstrap, arXiv:1611.10060v1; JHEP 04 (2017) 146](https://arxiv.org/abs/1611.10060v1)

   Version 1. Eqs. (2.12)-(2.17): the four-point function of identical operators on a line, the cross-ratio, the one-variable block and the crossing equation. Eq. (2.18) and Fig. 2: the numerical gap bound, which the text says converges to twice the external dimension plus one as derivatives are added. Eqs. (2.19)-(2.23): the generalized free fermion that saturates it and the zeros of the extremal functional. Fig. 1: the action of a typical extremal functional.

3. [S. Rychkov and A. Vichi, Universal constraints on conformal operator dimensions, arXiv:0905.2211v2; Phys. Rev. D 80, 045006 (2009)](https://arxiv.org/abs/0905.2211v2)

   Version 2. Eq. (2.8): the crossing sum rule written with conformal blocks. Section 3, Eqs. (3.1)-(3.2): the positivity argument, a linear functional of derivatives at the symmetric point that rules out a spectrum. Table 2: the two-dimensional bounds of that paper.

4. [D. Mazáč and M. F. Paulos, The analytic functional bootstrap. Part II, arXiv:1803.10233v2; JHEP 05 (2019) 088](https://arxiv.org/abs/1803.10233v2)

   Version 2. Section 4: extremal functionals in one dimension and the optimal gap bound equal to twice the external dimension plus one, saturated by the generalized free fermion.