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Exact tree-level minimum in the three-field O’Raifeartaigh model

Physical scope

Exact minimization of the classical canonical F-term scalar potential over pointwise field values for the specified rigid N=1 three-chiral-field Wess–Zumino model. It proves nonzero auxiliary F terms at the pointwise minima but, because the adopted model restricts spacetime fields to compact support, does not identify nonzero constant field values as allowed vacuum configurations. It makes no quantum, radiative-stability, or continuum-QFT claim.

Statement

In the classical rigid $N = 1$ Wess–Zumino model with three chiral scalar coordinates $(ϕ_{1}, ϕ_{2}, ϕ_{3}) \in C^{3}$ , canonical Kähler metric, no gauge D-term, and natural units $ℏ = c = 1$ , adopt the holomorphic superpotential

$W (Φ_{1}, Φ_{2}, Φ_{3}) = - k Φ_{1} + m Φ_{2} Φ_{3} + \frac{y}{2} Φ_{1} Φ_{3}^{2},$

where $k, m \in C ∖ {0}$ , $y \in C$ is arbitrary, and $[ϕ_{i}] = [m] = M$ , $[k] = M^{2}$ , and $[y] = 1$ . The classical zero-derivative F-term potential on pointwise field values is

$V (ϕ_{1}, ϕ_{2}, ϕ_{3}) = {∣ - k + \frac{y}{2} ϕ_{3}^{2} ∣}^{2} + ∣ m ϕ_{3} ∣^{2} + ∣ m ϕ_{2} + y ϕ_{1} ϕ_{3} ∣^{2} .$

In the adopted natural-unit convention, $V : C^{3} \to R$ is the classical F-term potential-energy density, with $[V] = M^{4}$ , $[W] = M^{3}$ , and $[F^{i}] = M^{2}$ .

These pointwise field values and eliminated auxiliary values belong to the ordinary complex bosonic sector of the model postulate, obtained by setting fermions and all positive coefficient-algebra degrees to zero; no order on nilpotent formal coefficients is used. Here “tree-level” means this classical potential before quantum-loop corrections. At a field value, F-term supersymmetry is unbroken exactly when all eliminated auxiliary fields $F^{i} = - \overline{\partial_{i} W}$ vanish; otherwise it is F-term broken.

Put $a = ∣ k ∣ > 0$ , $b = ∣ m ∣^{2} > 0$ , and $c = ∣ y ∣$ . If $b \geq a c$ , including $y = 0$ , the global minimum is $V_{\min} = a^{2} = ∣ k ∣^{2}$ , and all minimizing field values are $ϕ_{3} = ϕ_{2} = 0$ with arbitrary $ϕ_{1} \in C$ . If $0 < b < a c$ , then $c > 0$ and

$V_{\min} = a^{2} - \frac{(a c - b)^{2}}{c^{2}} = \frac{b}{c} (2 a - \frac{b}{c}) > 0,$

with all minimizing field values given by arbitrary $ϕ_{1} \in C$ , one of the two distinct square roots $ϕ_{3}$ of

$ϕ_{3}^{2} = \frac{2 k}{y} (1 - \frac{b}{a c}),$

and $ϕ_{2} = - y ϕ_{1} ϕ_{3} / m$ . In both regimes every minimizing field value has at least one nonzero auxiliary F term, so the pointwise classical minimum breaks F-term supersymmetry. Thus the stronger bound $V \geq ∣ k ∣^{2}$ holds for all field values exactly when $∣ m ∣^{2} \geq ∣ k ∣ ∣ y ∣$ ; in the other regime the exact minimum is smaller but remains positive. The set of minima has a flat complex $ϕ_{1}$ direction and is not an isolated point. The adopted field-model postulate restricts spacetime fields to compact support, so this pointwise minimization alone does not establish that nonzero constant minimizing values are allowed homogeneous vacuum configurations. No quantum vacuum, stability against radiative corrections, goldstino conclusion, or continuum-QFT existence is asserted.

Statement references

Stephen P. Martin, A Supersymmetry Primer, §7.3, equations (7.3.1)–(7.3.3).

Adel Bilal, Introduction to Supersymmetry, §6.3.1.

Adopted rigid $N = 1$ Wess–Zumino sigma-model action and field-domain convention: Rigid four-dimensional $N = 1$ Wess–Zumino sigma-model action.

Canonical F-term potential and auxiliary-field equation: Kähler-metric F-term scalar potential in the rigid $N = 1$ model.

Complex arithmetic prerequisite: $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ .

Complex conjugate and modulus definition: Real and imaginary parts, complex conjugation, and modulus.

Complex conjugation and modulus laws: Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive.

Existence of the square roots used to describe the minimizing configurations: Every complex number has a square root, by an explicit Cartesian formula.

The real-part bound and exact phase-alignment equality condition: Ordered-field component bound with its equality case.

The ordered-real-field fact $2 > 0$ used to distinguish the two square roots: The reals form a totally ordered field.

Facts & Assumptions

Given: The canonical three-field model and its classical pointwise zero-derivative potential in the Statement; all field values range over $C^{3}$ , and $k, m \neq 0$ .

[A1]

Rigid four-dimensional $N = 1$ Wess–Zumino sigma-model action: the adopted rigid $N = 1$ model has a holomorphic superpotential, a canonical Kähler special case, engineering dimensions $[ϕ_{i}] = M$ , $[W] = M^{3}$ , smooth compactly supported spacetime component fields, and the displayed algebraic auxiliary Lagrangian. Its explicit component conversion gives the complex bosonic auxiliary Lagrangian with $F_{r a w} = 2 F$ , physical D bracket $- I_{D} / 4$ , and physical F bracket $I_{F} / 2$ . Its full formal even auxiliary coefficient is unrestricted before taking this ordinary complex sector. This proof uses the resulting complex scalar potential, without ordering nilpotent coefficients or inferring a quantum measure.

[F1]

Kähler-metric F-term scalar potential in the rigid $N = 1$ model: for the canonical Kähler metric, $V_{F} = \sum_{i} ∣ W_{i} ∣^{2}$ , and the algebraic auxiliary fields satisfy $F^{i} = - \overline{W_{i}}$ .

[F2]

$C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ : $C$ is a field and every element has unique coordinates $a + b i$ ; in particular it has no zero divisors and every nonzero element is invertible.

[F3]

Real and imaginary parts, complex conjugation, and modulus and Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive: $∣ z ∣^{2} = z \overset{ˉ}{z}$ , $∣ z ∣ \geq 0$ , $∣ z ∣ = 0$ exactly when $z = 0$ , and $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ .

[F4]

Every complex number has a square root, by an explicit Cartesian formula: every complex number has a square root.

[F5]

Ordered-field component bound with its equality case: if $x, u, r \in R$ , $r \geq 0$ , and $r^{2} = x^{2} + u^{2}$ , then $x \leq r$ , with equality exactly when $u = 0$ and $x \geq 0$ .

[F6]

The reals form a totally ordered field: $2 = 1 + 1 > 0$ in $R$ , so $2 \neq 0$ in $C$ under its real-coordinate embedding.

Proof

Given: [A1], [F1]–[F6], and the parameter hypotheses in the Statement.

1.1F1algebra

Differentiate the stated polynomial with respect to its three chiral coordinates: $W_{1} = - k + (y / 2) ϕ_{3}^{2}$ , $W_{2} = m ϕ_{3}$ , and $W_{3} = m ϕ_{2} + y ϕ_{1} ϕ_{3}$ . Substituting these derivatives into [F1] gives exactly the displayed potential.

2.1F2F3F5step 1.1algebra

Set $a = ∣ k ∣$ , $b = ∣ m ∣^{2}$ , $c = ∣ y ∣$ , and $t = ∣ ϕ_{3} ∣^{2} \geq 0$ . Write $\overset{ˉ}{k} y ϕ_{3}^{2} = x + i u$ , where $x, u \in R$ , and put $r = ∣ \overset{ˉ}{k} y ϕ_{3}^{2} ∣$ . By [F2] and [F3], $r \geq 0$ , $r^{2} = x^{2} + u^{2}$ , and modulus multiplicativity gives $r = a c t$ . Applying [F5] gives $x \leq a c t$ , with equality exactly when $u = 0$ and $x \geq 0$ . Expanding the first square gives ${∣ k - \frac{y}{2} ϕ_{3}^{2} ∣}^{2} = a^{2} + \frac{c^{2} t^{2}}{4} - x \geq a^{2} + \frac{c^{2} t^{2}}{4} - a c t$ . Adding the second term $b t$ and the nonnegative third square yields $V \geq q (t) : = a^{2} + (b - a c) t + \frac{c^{2} t^{2}}{4}$ . Equality in this lower bound requires the third square to vanish and, when $c t > 0$ , requires $Im (\overset{ˉ}{k} y ϕ_{3}^{2}) = 0$ and $Re (\overset{ˉ}{k} y ϕ_{3}^{2}) \geq 0$ , equivalently that $\overset{ˉ}{k} y ϕ_{3}^{2}$ is nonnegative real.

3.1F2F3step 2.1algebra

Suppose $b \geq a c$ . If $c > 0$ , then $(b - a c) t \geq 0$ and $c^{2} t^{2} / 4 \geq 0$ , with their sum zero only at $t = 0$ . If $c = 0$ , then $q (t) = a^{2} + b t$ and $b > 0$ , so again its unique minimum occurs at $t = 0$ . Thus $V \geq a^{2}$ . At $ϕ_{3} = ϕ_{2} = 0$ , every term except $∣ k ∣^{2}$ vanishes, so the lower bound is attained for every $ϕ_{1}$ . Conversely, any minimizing value must have $t = 0$ ; then the third square is $∣ m ϕ_{2} ∣^{2}$ , which vanishes only for $ϕ_{2} = 0$ since $m \neq 0$ . These are exactly all the minimizers in this regime.

3.2F2F3step 2.1algebra

Suppose $0 < b < a c$ , so $a, c > 0$ . Completing the square gives $q (t) = a^{2} - \frac{(a c - b)^{2}}{c^{2}} + \frac{c^{2}}{4} {(t - \frac{2 (a c - b)}{c^{2}})}^{2}$ . The unique minimizing value of $t \geq 0$ is therefore $t_{*} = 2 (a c - b) / c^{2} > 0$ , and the lower bound is $q (t_{*}) = a^{2} - (a c - b)^{2} / c^{2} = (b / c) (2 a - b / c)$ . Since $0 < b / c < a$ , we have $(b / c) (2 a - b / c) > 0$ , while $a^{2} - q (t_{*}) = (a - b / c)^{2} > 0$ . Thus this lower bound is positive and strictly below $a^{2}$ .

4.1F1step 3.1algebra

At each minimizer in step 3.1, $W_{1} = - k \neq 0$ . The auxiliary equation in [F1] gives $F^{1} = - {\overset{ˉ}{W}}_{1} \neq 0$ ; hence the classical F-term order parameter is nonzero at every such pointwise minimum.

4.2F2F3F4F5step 3.2algebra

Put $ρ = (2 k / y) (1 - b / (a c))$ . It is nonzero because $k, y \neq 0$ and $0 < b < a c$ . By [F4] at least one square root exists; let $ϕ_{3}$ be any root, so $ϕ_{3}^{2} = ρ$ . Since $c = ∣ y ∣ > 0$ , [F3] and $∣ y y^{- 1} ∣ = ∣ 1 ∣ = 1$ give $∣ 1 / y ∣ = 1 / c$ . The real factor $1 - b / (a c) = (a c - b) / (a c) > 0$ has modulus equal to itself, and $∣ 2 ∣ = 2$ , so $∣ ρ ∣ = \frac{2 a}{c} \frac{a c - b}{a c} = \frac{2 (a c - b)}{c^{2}} = t_{*}$ . Multiplicativity in [F3] now gives $∣ ϕ_{3} ∣^{2} = ∣ ϕ_{3}^{2} ∣ = ∣ ρ ∣ = t_{*}$ . Also $\overset{ˉ}{k} y ϕ_{3}^{2} = \overset{ˉ}{k} y ρ = \frac{2 ∣ k ∣^{2} (a c - b)}{a c} = \frac{2 a (a c - b)}{c} = a c t_{*} > 0$ . For arbitrary $ϕ_{1}$ , set $ϕ_{2} = - y ϕ_{1} ϕ_{3} / m$ , which is defined because $m \neq 0$ . The third square vanishes, and the first-square bound is an equality by [F5], so every such field value attains $q (t_{*})$ .

5.1F2F3F5F6step 2.1step 4.2algebra

Conversely, a global minimizer must have $t = t_{*}$ , since $V \geq q (t) \geq q (t_{*})$ and the value $q (t_{*})$ is attained in step 4.2. Equality in step 2.1 then forces $m ϕ_{2} + y ϕ_{1} ϕ_{3} = 0$ and $\overset{ˉ}{k} y ϕ_{3}^{2} = a c t_{*} = 2 a (a c - b) / c$ by [F5]. Using $∣ k ∣^{2} = a^{2}$ , the latter equation is equivalent to $ϕ_{3}^{2} = \frac{2 a (a c - b)}{c \overset{ˉ}{k} y} = \frac{2 k}{y} (1 - \frac{b}{a c})$ . The third-square equation gives $ϕ_{2} = - y ϕ_{1} ϕ_{3} / m$ . Thus the list in the Statement contains every minimizer. To count the roots, let $u$ be one root from step 4.2, so $u^{2} = ρ \neq 0$ . Any other root $w$ satisfies $(w - u) (w + u) = w^{2} - u^{2} = 0$ ; because $C$ is a field, $w = u$ or $w = - u$ . They are distinct: equality would give $2 u = 0$ , but [F6] gives $2 \neq 0$ , and $u \neq 0$ . Hence there are exactly two distinct roots.

6.1A1F1F3step 3.1step 3.2step 4.2step 5.1algebra∎

At the minima of steps 4.2–5.1, $W_{1} = - k + \frac{y}{2} ϕ_{3}^{2} = - \frac{k b}{a c}$ , so $∣ W_{1} ∣ = \frac{b}{c} > 0$ by [F3]. Hence [F1] gives $F^{1} = - {\overset{ˉ}{W}}_{1} \neq 0$ , proving F-term breaking also in this regime. Step 3.1 gives the stronger bound and equality set when $b \geq a c$ ; the value $q (t_{*})$ computed in step 3.2 is attained in step 4.2 and is strictly below $a^{2}$ , disproving that bound for $b < a c$ . The displayed minimizers in each case leave $ϕ_{1}$ arbitrary, so the minimum is not isolated. All conclusions concern the pointwise potential; [A1] restricts spacetime field maps to compact support, and the proof does not treat nonzero constant backgrounds as admissible fields.

Exact tree-level minimum in the three-field O’Raifeartaigh model

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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Sources