Write the superpotential so that the only combination of X and Φ2 that talks to Φ1 is manifest: W = 1/2 Φ1^2 (h1 X + h2 Φ2) + f X.
Make a linear, unitary change of variables in the X–Φ2 plane:
In this basis, W = 1/2 h S Φ1^2 + f (h1 S − h2 T)/h.
Two immediate consequences follow:
SUSY breaking (tree level): Along the classical pseudomoduli space Φ1 = 0, one has F_S = f h1/h and F_T = f h2/h, so V = |f|^2 and SUSY is broken (since f ≠ 0) with two classically flat complex directions S and T.
What one loop can and cannot do (without computing the Coleman–Weinberg potential explicitly): • Only the multiplet Φ1 has masses that depend on the background pseudomoduli, and those masses depend solely on S through m_f = h S and m_b^2 = |h S|^2 ± |h F_S|. Therefore any one-loop effective potential can depend only on S, not on T. Consequently, the T direction remains exactly flat to all orders in perturbation theory (it decouples from any interaction except for the linear term). • If h1 ≠ 0, then F_S ≠ 0 along the pseudomoduli space, so the Φ1 bosons are split relative to the Φ1 fermion. This necessarily generates an S-dependent Coleman–Weinberg potential (equivalently, an S-dependent one-loop correction to the Kähler metric Z_S(S†S), giving V_eff ≈ |F_S|^2/Z_S). Thus the S direction is lifted at one loop, just as in the standard O’Raifeartaigh model W = κ S + 1/2 λ S φ^2 with κ = f h1/h and λ = h.
Special case:
So, without computing the Coleman–Weinberg potential explicitly, a simple field redefinition shows:
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version: gpt-5-2025-08-07
Status: UQ Validated
Validated: 7 months ago
Status: Needs Human Verification
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