Free fields solve the crossing equation exactly

Correlators built from pairs

A generalized free field has a four-point function that is the sum of the three ways to pair four points, each pair contributing a two-point function. For a fermion one pairing enters with a minus sign. On a line this gives the closed forms below.

Swapping $z$ and $1-z$ only permutes the three pairings, so crossing holds exactly. Nothing about the exchanged operators was put in by hand; they follow from expanding the closed form.

A_\psi(z)=1+\Bigl(\frac{z}{1-z}\Bigr)^{2\Delta_\psi}-z^{2\Delta_\psi},\qquad A_\phi(z)=1+\Bigl(\frac{z}{1-z}\Bigr)^{2\Delta_\phi}+z^{2\Delta_\phi}

These hold for any external dimension. The worked examples use

\Delta=1/2

Source: D. Gaiotto · D. Mazáč

The exchanged operators

Expanding each closed form in blocks gives a ladder of two-particle operators. The fermion exchanges dimensions $2\Delta_\psi+1$ , $2\Delta_\psi+3$ and so on. The boson starts one step lower, at $2\Delta_\phi$ .

Every coefficient $c_n^2$ is positive and falls quickly with $n$ . For $\Delta=1/2$ the fermion values are $1$ , $1/10$ , $1/126$ , $1/1716$ , and the boson values are $2$ , $1/3$ , $1/35$ , $1/462$ .

\Delta_n^{\psi}=2\Delta_\psi+2n+1,\qquad \Delta_n^{\phi}=2\Delta_\phi+2n,\qquad n=0,1,2,\dots

The fermion has no operator below its first rung. That empty interval is what the bound in the next chapter is about.

Source: D. Gaiotto

Worked example: reading coefficients off a Taylor series

Take the boson with $\Delta_\phi=1/2$ . Its closed form minus the identity is $2z+z^2+z^3+\dots$ . The lightest block is $k_1=-\log(1-z)$ . Matching the $z$ term fixes $c_1^2=2$ .

Then $2k_1$ gives $z^2$ with coefficient $1$ , which already matches, so no operator of dimension $2$ is needed. At $z^3$ it gives $2/3$ against the required $1$ . The gap is filled by a block $k_3=z^3+\dots$ with $c_3^2=1/3$ .

The same bookkeeping on the fermion gives $c_2^2=1$ from the $z^2$ term and $c_4^2=1-9/10=1/10$ from the $z^4$ term. The numbers agree with the general formula to every digit printed, which is an independent check of both.

With these coefficients, two boson operators already give $A(0.3)=1.728406$ against the exact $1.728571$ . The full sum matches to six digits.

A_\phi(z)-1=2z+z^2+z^3+\dots,\qquad k_1(z)=-\log(1-z)=z+\tfrac12z^2+\tfrac13z^3+\dots

Matching powers of

z

term by term fixes one coefficient at a time and shows which dimensions are absent.

Source: D. Gaiotto · D. Mazáč

What truncation does

Keep only the first three fermion operators. At $z=1/2$ the crossing residual is exactly zero, because both sides are evaluated at the same point. At $z=0.8$ it is $-4.46\times10^{-3}$ , and at $z=0.9$ it reaches $-3.36\times10^{-2}$ . Ten operators push the residual at $z=0.9$ down to $-5.87\times10^{-6}$ .

This is the honest version of a numerical statement: a finite spectrum satisfies crossing only approximately, and the error depends on where you look. At $z=1/2$ both channels converge equally fast, so numerical studies take derivatives of the crossing equation there.

Source: D. Mazáč · D. Simmons-Duffin

Try it in the Lab

Compare the fermion and the boson

Open the free-boson preset and note the first exchanged operator. Compare it with the free-fermion preset.
Predict how the mismatch curve will look if the first coefficient is wrong, then pick that preset.
With the too-few-operators preset, drag the point toward the right end and watch the residual grow.

Exercises with explained answers

For a fermion with $\Delta_\psi=0.8$ , what is the dimension of the first exchanged operator?
Show the answer
$2\Delta_\psi+1=2.6$ . No operator lies between the identity and this value.
Using $A_\psi-1=z^2+z^3+z^4+\dots$ for $\Delta_\psi=1/2$ and $k_2=z^2(1+z+\tfrac{9}{10}z^2+\dots)$ , show that no operator of dimension $3$ appears.
Show the answer
With $c_2^2=1$ , the block $k_2$ already supplies $z^3$ with coefficient $1$ , exactly what $A_\psi$ needs. A dimension-3 operator would add an extra $z^3$ , so its coefficient must vanish.
Flip the sign of the fermion coefficient $c_4^2$ and keep everything else. Would crossing still hold?
Show the answer
No. The crossing residual at $z=0.3$ becomes $0.114$ . Unitarity would also be violated, because it requires every $c^2\ge 0$ .

Sources and what we recomputed

D. Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, arXiv:1602.07982v1 (2016)
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.
D. Mazáč, Analytic bounds and emergence of AdS2 physics from the conformal bootstrap, arXiv:1611.10060v1; JHEP 04 (2017) 146
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.
D. Gaiotto, D. Mazáč and M. F. Paulos, Bootstrapping the 3d Ising twist defect, arXiv:1310.5078v2; JHEP 03 (2014) 100
Version 2. Section 4.1: one-dimensional generalized free fields. Eqs. (4.17) and (4.20) give the exchanged dimensions and OPE coefficients of the fermionic and bosonic solutions; the course resums them against the closed-form correlators.