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Vacuum energy and global supersymmetry invariance

Physical scope

Conditional operator-algebra deduction for an existing rest-frame ground-state vector in an adopted positive unitary representation of the four-dimensional extended super-Poincare algebra. The result identifies zero energy with supercharge invariance; it does not establish vacuum existence, construct a QFT, or infer a nonzero vacuum energy from a field-potential calculation.

Statement

Assume the Axiom of Countable Choice and a positive unitary representation of the four-dimensional extended super-Poincare algebra with metric $η = diag (- 1, 1, 1, 1)$ , $σ^{0} = I_{2}$ , natural units $ℏ = c = 1$ , common invariant product domain $D$ , and actual Hilbert adjoints $(Q_{α}^{I})^{*}$ . Let $N \geq 1$ , and let $Ω \in D$ satisfy $∥ Ω ∥ = 1$ , $P^{0} Ω = E_{Ω} Ω$ for a real number $E_{Ω}$ , $P^{j} Ω = 0$ for $j = 1, 2, 3$ , and $⟨ P^{0} ψ, ψ ⟩ \geq E_{Ω} ∥ ψ ∥^{2}$ for every $ψ \in D (P^{0})$ . Say that $Ω$ is supersymmetry-invariant when $Q_{α}^{I} Ω = (Q_{α}^{I})^{*} Ω = 0$ for every $I, α$ , and supersymmetry-noninvariant when at least one of these vectors is nonzero; say it is boundary-neutral when $Z^{I J} Ω = {\overset{ˉ}{Z}}_{I J} Ω = 0$ for every central generator. Then

$4 N E_{Ω} = \sum_{I = 1}^{N} \sum_{α = 1}^{2} (∥ Q_{α}^{I} Ω ∥^{2} + ∥ (Q_{α}^{I})^{*} Ω ∥^{2}) \geq 0.$

Consequently, $E_{Ω} = 0$ if and only if $Ω$ is supersymmetry-invariant, and every supersymmetry-noninvariant such state has $E_{Ω} > 0$ . The identity holds whether or not the state is boundary-neutral. The energy is an eigenvalue, not an energy density. The theorem is conditional on the representation and state data; it proves neither existence of a vacuum state nor existence of a quantum field theory realizing that representation.

Statement references

Stephen P. Martin, A Supersymmetry Primer, §7.1, equations (7.1.1)–(7.1.2).

Adel Bilal, Introduction to Supersymmetry, §6.1.

Four-dimensional real extended super-Poincare algebra, including ${Q_{α}^{I}, (Q_{β}^{J})^{*}} = - 2 σ_{α \dot{β}}^{μ} P_{μ} δ^{I}_{J}$ : Four-dimensional real extended super-Poincare algebra.

Positive unitary representation with a common product domain and actual Hilbert adjoints: Positive unitary super-Poincare representation with spectral-domain covariance.

Global supersymmetry vacuum state, its ground-state condition, and the definitions of boundary-neutrality and supersymmetry invariance: Global supersymmetry vacuum state and invariance.

The Axiom of Countable Choice: The Axiom of Countable Choice ( ${A C}_{ω}$ ).

Facts & Assumptions

Given: The hypotheses in the Statement, in particular $Ω \in D$ , $∥ Ω ∥ = 1$ , $P^{0} Ω = E_{Ω} Ω$ , and the exact operator adjoints on the common invariant product domain.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every sequence of nonempty sets admits a choice function; this is the choice assumption included in the representation supplier.

[F1]

Four-dimensional real extended super-Poincare algebra: ${Q_{α}^{I}, {\overset{ˉ}{Q}}_{\dot{β} J}} = - 2 σ_{α \dot{β}}^{μ} P_{μ} δ^{I}_{J}$ , with $P_{0} = - P^{0}$ , $tr σ^{0} = 2$ , and $tr σ^{j} = 0$ for $j = 1, 2, 3$ .

[F2]

Positive unitary super-Poincare representation with spectral-domain covariance: ${\overset{ˉ}{Q}}_{\dot{α} I} = (Q_{α}^{I})^{*}$ is the actual Hilbert adjoint, and every product in the bracket table is defined on the common invariant domain $D$ .

[F3]

Global supersymmetry vacuum state and invariance: $Ω \in D$ is normalized and satisfies $P^{j} Ω = 0$ , $P^{0} Ω = E_{Ω} Ω$ ; it is supersymmetry-invariant exactly when every $Q_{α}^{I} Ω$ and $(Q_{α}^{I})^{*} Ω$ vanishes.

Proof

Given: [A1], [F1]–[F3], and $N \geq 1$ .

1.1A1F1F2algebra

In the Countable-Choice representation setup [A1, F2], sum the mixed anticommutator in [F1] over the equal spinor labels $α = \dot{β}$ and over $I = 1, \dots, N$ . The Kronecker delta leaves one copy for each flavor, and the Pauli-matrix traces give $\sum_{I = 1}^{N} \sum_{α = 1}^{2} {Q_{α}^{I}, (Q_{α}^{I})^{*}} = - 2 N \sum_{μ = 0}^{3} tr (σ^{μ}) P_{μ} = - 4 N P_{0} = 4 N P^{0}$ . The last equality uses $P_{0} = - P^{0}$ for the stated metric.

2.1F2F3step 1.1algebra

Apply the operator identity from step 1.1 to $Ω$ and take its inner product with $Ω$ . The common product domain in [F2] makes both products defined. Since the inner product is linear in its first variable and $(Q_{α}^{I})^{*}$ is the actual adjoint, $⟨ Q_{α}^{I} (Q_{α}^{I})^{*} Ω, Ω ⟩ = ∥ (Q_{α}^{I})^{*} Ω ∥^{2}$ and $⟨ (Q_{α}^{I})^{*} Q_{α}^{I} Ω, Ω ⟩ = ∥ Q_{α}^{I} Ω ∥^{2}$ . The right side is $4 N ⟨ P^{0} Ω, Ω ⟩ = 4 N E_{Ω}$ by [F3] and normalization. This proves the asserted sum-of-norms identity.

3.1F3step 2.1algebra∎

Every summand in step 2.1 is nonnegative, so $E_{Ω} \geq 0$ . A finite sum of nonnegative real numbers is zero exactly when every summand is zero; hence $E_{Ω} = 0$ exactly when all $Q_{α}^{I} Ω$ and $(Q_{α}^{I})^{*} Ω$ vanish, which is precisely supersymmetry invariance by [F3]. If the state is noninvariant, at least one squared norm is strictly positive, so the identity and $4 N > 0$ imply $E_{Ω} > 0$ . The argument contains no boundary-neutrality assumption.

Vacuum energy and global supersymmetry invariance

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources