Koide with AI

Well, first koide formula find by an AI agent afair:

The crucial discovery: the **»dual Koide»** — the Koide ratio of inverse masses `Q(1/m_d, 1/m_s, 1/m_b) = 0.665`, deviating from 2/3 by only **0.22%**. The triple (d, s, b) is nearly «self-dual» under the Seiberg seesaw.

Here the prompt:

Write and run a Python script /tmp/seiberg_bloom.py to investigate how the Seiberg effective superpotential for N_f = N_c = 3 SQCD produces mass spectra related to Koide triples.

## Background

In Seiberg's SQCD with N_f = N_c = 3, the low-energy theory has meson superfields M^i_j and baryons B, B?, with constraint det(M) - BB? = ??. Adding tree-level masses W_tree = Tr(m? M), the vacuum has B = B? = 0 and:

M^j_j = -?? / (m_j · det(M)/M^j_j·M^j_j)  ... actually let me be precise.

For diagonal M and m? = diag(m_1, m_2, m_3), the F-term equations give:

m_j + X · (det M) · (M^{-1})_j = 0

For diagonal M = diag(M_1, M_2, M_3):

m_j + X · M_1·M_2·M_3 / M_j = 0

This gives M_j = -X · (?_k M_k) / m_j, so m_j · M_j = -X · det(M) for all j.

Since det(M) = ??: M_j = ??/(m_j · product ratios)

More precisely: m_j · M_j = constant C for all j, and ? M_j = ??.

So M_j = C/m_j, and ?(C/m_j) = ??, giving C³ = ?? · m_1·m_2·m_3.

Therefore C = ?² · (m_1·m_2·m_3)^{1/3}.

And M_j = ?² · (m_1·m_2·m_3)^{1/3} / m_j.

## Tasks

1. **Compute the SQCD meson VEVs** as functions of quark masses (m_1, m_2, m_3) and ?_QCD.

   Use m_u=2.16, m_d=4.67, m_s=93.4 MeV, and ?_QCD ? 330 MeV.

   For 3-flavor SQCD: compute M_j = ?² (m_1 m_2 m_3)^{1/3} / m_j for (u,d,s).

2. **Physical meson masses from SQCD**: In the SUSY limit, the scalar component of M^i_j has mass equal to the fermionic component. When SUSY is broken (F ? 0), the scalar (meson) picks up an additional F-term contribution. The fermionic mass matrix of the Seiberg theory is:

   W_{ij} = ?²W/?M_i?M_j

   For the effective superpotential W = ? m_j M_j + X(? M_j - ??), compute the Hessian ?²W/?M_i?M_j at the vacuum.

3. **Seed limit**: Take m_1 ? 0 (the "chiral limit" for one flavor). What happens to the meson VEVs and the mass matrix? Show that one F-term is necessarily nonzero, giving spontaneous SUSY breaking.

4. **Test the Koide condition**: For the 3-flavor case (u,d,s), the meson masses M_uu, M_ds, etc. should be related to physical meson masses (??, K, ?, etc.). The question is whether the effective mass spectrum of the fermionic components (leptons in the sBootstrap) satisfies any Koide-like relation.

   Compute the eigenvalues of the fermionic mass matrix W_{ij} = ?²W/?M_i?M_j for the (u,d,s) sector and check Q = ??/(???)² including all sign choices.

5. **5-flavor extension**: For N_f = 5 (u,d,s,c,b), the constraint becomes det(M^{5×5}) = ?^{2·3} = ?^6 (wait — for N_f > N_c, the Seiberg duality changes). Actually for N_f = 5, N_c = 3, we're in the conformal window (N_c < N_f < 3N_c). The magnetic dual has N_f - N_c = 2 colors. Compute the relevant quantities in this regime if tractable, otherwise just note the complications.

Print all results clearly with physical interpretations.