Efimov Universality from Decomposition Inconsistency

Corrected Derivation — Honest Status of All Claims

Preamble: What This Document Does and Does Not Establish

The previous version claimed to have independently derived the scalar values $\varepsilon_0 = -1/4$ and $W_0 = -s_0^2/2$. That claim was wrong. This document corrects it with precision: the operator structure of the 3×3 decomposition is fully derivable; the scalar split of $\lambda_\text{sym}$ into $\varepsilon_0 + 2W_0$ is convention-dependent and not independently fixed by the Bethe–Peierls condition alone.

The Conceptual Chain (Fully Established)

The following sequence is well-founded and unambiguous:

1. Three pair-spectator charts $

   1\rangle,

   2\rangle,

   3\rangle$

2. Zero-range Bethe–Peierls matching couples those charts
3. Permutation symmetry selects the fully symmetric collective mode

4. Scale-independence in the window $r_0 \ll \rho \ll

   a

   $ forces $\mathcal{H}_\Omega \propto \rho^{-2}$

5. The symmetric eigenvalue $\lambda_\text{sym} = -s_0^2 - 1/4$ produces an attractive supercritical channel
6. Log-periodic solutions give the Efimov geometric spectrum

No part of this chain is disputed. The issue is in the intermediate step of separately identifying $\varepsilon_0$ and $W_0$.

Step 1: The Faddeev Ansatz and the Bethe–Peierls Condition

Write $\Phi = \phi_{12} + \phi_{13} + \phi_{23}$, each component singular only in its own pair. The Bethe–Peierls condition at $r_{12} \to 0$ is:

At unitarity $a \to \infty$:

This is the coupling equation. The key structural fact, which holds exactly at unitarity, follows immediately:

$\phi_{12}$ does not source itself. The zero-range condition with $a = \infty$ has no regular part, so $V_{12}\phi_{12} = 0$ at unitarity. Only $\phi_{13}$ and $\phi_{23}$ source $\phi_{12}$.

In operator language, the Faddeev equation for the amplitude $t_{12}(p)$ (the Fourier-space residue conjugate to $\phi_{12}$) is:

with $K_\text{self} = 0$ exactly at unitarity. This gives the operator structure of the 3×3 matrix:

Here $K_\text{cross}$ is an integral operator (not yet a scalar), and the matrix entries are operator-valued, not numbers.

Step 2: The Cross-Channel Coupling Kernel — Derived Explicitly

The cross-channel coupling $K_\text{cross}$ is computed by expressing $r_{13}$ in the pair-(12) Jacobi frame and projecting to s-wave. For equal-mass bosons, this gives the momentum-space STM kernel:

The denominator $p^2 + pp’ + p’^2$ is the exact kinematic result of expressing $r_{13}^2$ in terms of pair-(12) Jacobi momenta after s-wave angular projection for equal masses:

This kernel is independently derived from the kinematic recoupling, with no use of the known Efimov result as input.

Step 3: Self-Consistency → The Transcendental Equation

In the symmetric sector $t_{12} = t_{13} = t_{23} \equiv t$, each component receives contributions from two cross-channels:

This is a fixed-point equation (not a standard eigenvalue equation at this stage). Substituting the power-law ansatz $t(p) = p^{is_0 - 1}$, the integral becomes a Mellin transform:

Self-consistency — that $p^{is_0-1}$ is indeed the fixed-point — reduces to:

with unique solution $s_0 \approx 1.00624$. This derivation is complete and uses no Efimov result as input.

Step 4: The Hyperradial Channel — Derived from the Scaling Exponent

The power-law $t(p) \sim p^{is_0-1}$ in momentum space corresponds to a position-space wavefunction $\phi \sim r^{-is_0}$ at threshold. Translating to the hyperradial function $f_0(\rho)$ via the standard reduction of the 6D kinetic energy (with the variable change $F(\rho) = f_0(\rho)/\rho^{5/2}$ to remove the first-derivative term), the hyperradial equation becomes:

where $\nu = -2 + is_0$ is the hyperradial scaling exponent. With $\nu + 2 = is_0$:

Therefore:

Numerical check: $\nu = -2 + i(1.00624)$, $(\nu+2)^2 - 1/4 = -(1.00624)^2 - 0.25 = -1.2625$ ✓

This step is fully derived. The $-1/4$ here is the Langer correction from the variable change $F \to f_0/\rho^{5/2}$; it is kinematic, not interaction-generated.

Step 5: Where the Scalar Split Stands — Honest Assessment

The symmetric eigenvalue is:

The democratic matrix structure gives:

What is convention-independent and fully derived:

What is convention-dependent:

The scalar split $(\varepsilon_0, W_0)$ exists in (at least) two consistent forms:

Form A — integral equation (momentum space):

$\varepsilon_0^\text{A} = 0$ is exact at unitarity ($K_\text{self} = 0$). The $-1/4$ is absorbed into $W_0$.

Form B — hyperspherical (position space, after Langer reduction):

$\varepsilon_0^\text{B} = -1/4$ is the Langer correction from $F \to f_0/\rho^{5/2}$; $W_0^\text{B}$ is the remainder.

Both satisfy $\varepsilon_0 + 2W_0 = -s_0^2 - 1/4$. Neither is more fundamental than the other — they are representations of the same operator in different functional spaces (momentum amplitudes vs. hyperangular functions after a specific variable change).

What the Framework Adds

Even without fixing the scalar split independently, the framework contribution is genuine:

The standard treatment identifies the Efimov effect via a hyperangular eigenvalue. The decomposition framework identifies it as the failure of three pair charts to be simultaneously self-consistent under zero-range matching.

The precise statement, now fully supported, is:

At unitarity, the zero-range Bethe–Peierls condition on $\phi_{12}$ at $r_{12} = 0$ is sourced entirely by $\phi_{13}$ and $\phi_{23}$ (the self-coupling $K_\text{self}$ vanishes exactly). By identical-boson symmetry, the same holds for each chart. The only mode in which all three boundary conditions can be simultaneously satisfied is the symmetric collective mode $(t,t,t)$. The self-consistency condition for this mode — that the STM kernel acting on $t$ returns $t$ — produces the transcendental equation for $s_0$, and the resulting hyperradial potential is $U_\text{eff} = -(s_0^2 + 1/4)/\rho^2$.

This is a complete and accurate statement of what the decomposition picture establishes.

Summary Table

One-Line Summary

The Bethe–Peierls condition forces $K_\text{self} = 0$ at unitarity, so three-body consistency is entirely cross-chart; the symmetric fixed point of the STM kernel produces the transcendental equation for $s_0$; the effective potential $-(s_0^2+1/4)/\rho^2$ follows from the Langer reduction; and the scalar split $\varepsilon_0 + 2W_0 = -s_0^2 - 1/4$ is a consistent parametrization whose individual terms are convention-dependent.