FR37 · Conformal bootstrap

Conformal Bootstrap Bound: A Linear Functional Excludes a Gap

Can one linear functional show that no consistent 1D theory has a gap this large? Yes, above a certain gap. Crossing says the operators must add up to cancel the identity term. If one linear functional is non-negative on every operator allowed above the gap and positive on the identity, no consistent spectrum exists. The bound tightens as more derivatives are used and approaches the free fermion, which sits exactly on it.

Pick a gap above the first exchanged operator and search for a linear functional positive on every allowed block. If one exists, crossing rules the gap out.

Trial gap $\Delta_{\mathrm{gap}}=2.4$

Largest gap allowed at this order $\Delta_{\mathrm{gap}}\le 2.153\;(\Lambda=13)$

Analytic limit (free fermion) $2\Delta_\psi+1=2$

Certificate check $\min_\Delta\hat\alpha=-2.72\times10^{-5}$

Rule out a gap

External dimension

0.500

\Delta_\psi

Highest derivative order

13

\Lambda

Trial gap

2.40

\Delta_{\mathrm{gap}}

Drag the gap along the dimension ruler. Above the dashed bound a functional appears.

Model, method and sources

$\sum_{\Delta\ge\Delta_{\mathrm{gap}}}p_\Delta\vec F_\Delta=-\vec F_0,\quad p_\Delta\ge0$

About 125 dimensions are sampled above the gap, plus points added where a candidate functional dips below zero (up to five rounds). A Phase I simplex decides feasibility; its dual vector is the functional. The bound is found by bisection to one part in a thousand. Positivity is checked on a six times finer grid.

Mazáč (2016), Eqs. 2.15–2.21 · Mazáč & Paulos (2018), §4 · Simmons-Duffin, TASI lectures (2016), §10

Derive the functional argument

Model assumptions

Crossing for four identical operators on a line is written as a sum of vectors, one per exchanged operator, built from odd derivatives at the crossing-symmetric point. A Phase I simplex either finds positive weights that add up to the identity term, or returns a dual functional that is non-negative on every sampled dimension above the gap and positive on the identity. Positivity is checked on a finer grid and refined by cutting planes; it is numerical evidence on samples, not a proof over the whole interval. The analytic limit, the first operator of the free fermion, comes from Mazáč (2016) and Mazáč and Paulos (2018).

What the functional does to each operator

More derivatives, tighter bound

More numbers and how they are computed

Components of each crossing vector: $0$
Sampled dimensions after refinement: $0$
Simplex pivots: $0$
Smallest action at, or weights used: $0$

The functional is normalised to one on the identity. The certificate check divides its action by the lengths of the functional and of each crossing vector; a small negative value means the sampled positivity did not fully hold between samples.

Physics tutorial

Ruling out a gap with one functional

BackgroundWrite crossing as a sum of vectors, one per exchanged operator, weighted by squared OPE coefficients that cannot be negative. The identity operator contributes a fixed vector. A consistent theory must add its operator vectors up to cancel the identity.

Why it mattersSuppose every exchanged operator sits above a gap. If some linear functional is non-negative on all the allowed vectors and positive on the identity, no positive sum can cancel the identity, and the gap is impossible.

Start with the essentials

Focus question: Can one linear functional show that no consistent 1D theory has a gap this large?
One-sentence intuition: Above the bound the search returns a separating functional; below it, it returns positive weights that solve the truncated equations. More derivatives move the bound down toward the free fermion, which saturates it.

Core mathematical model

Crossing vectors

\vec F_\Delta=\Bigl(\partial_z^{\,n}\bigl[(1-z)^{2\Delta_\psi}k_\Delta(z)-z^{2\Delta_\psi}k_\Delta(1-z)\bigr]_{z=1/2}\Bigr)_{n=1,3,\dots,\Lambda}

Only odd derivatives survive at the symmetric point. $\Lambda$ sets how many components the search uses.

Exclusion condition

\alpha(\vec F_0)>0,\qquad \alpha(\vec F_\Delta)\ge 0\ \ \text{for all}\ \Delta\ge\Delta_{\mathrm{gap}}

Applying the functional to crossing then gives a positive number equal to zero, a contradiction.

The limit the bounds approach

\Delta_{\mathrm{gap}}\le 2\Delta_\psi+1

Mazáč (2016) and Mazáč and Paulos (2018) proved this bound with analytic functionals; the free fermion has its first operator exactly there.

Common difficulties

Grid positivity is not a proof

Typical misconceptionIf the functional is non-negative at every sampled dimension, the gap is rigorously excluded.

Better mental modelBetween samples the functional can dip below zero. The Lab refines the grid and reports the smallest value it finds; a proof needs positivity on the whole interval.

Run the experiment

01
See the separating line
Start from Gap too large.
What to observe: Every operator vector lies on one side of the violet line; the vector that must be cancelled lies on the other.
02
Lower the gap
Drag the gap handle down past the dashed bound.
What to observe: The line disappears and green arrows add up head to tail to the target.
03
Add derivatives
Raise the derivative order from 3 to 21.
What to observe: The bound curve falls toward the free-fermion line but never crosses it.

Conformal Bootstrap Bound: A Linear Functional Excludes a Gap

Ruling out a gap with one functional

Core mathematical model

Crossing vectors

Exclusion condition

The limit the bounds approach

Common difficulties

Grid positivity is not a proof

Run the experiment

See the separating line

Lower the gap

Add derivatives

Related experiments and articles

From a critical chain to the kink in the 2D bound