Convention

This singlet diagnostic supports V2's hadronic contact calculations by distinguishing mean charge, charge variance, and invariant states. The matrices below represent field amplitudes on the physical manifold; their sizes introduce no physical spatial dimensions.

Use Hermitian fundamental generators of SU(N)SU(N) with tr⁡(TaTb)=δab/2\operatorname{tr}(T_aT_b)=\delta_{ab}/2. Their completeness relation is

∑a(Ta)ij(Ta)kl=12(δilδjk−1Nδijδkl). \sum_a(T_a)_{ij}(T_a)_{kl} =\frac12\left(\delta_{il}\delta_{jk}-\frac1N\delta_{ij}\delta_{kl}\right).

It is the orthogonal projector onto traceless matrices in this normalization. Contracting with two density matrices gives the purity relation below; contracting the middle indices gives the fundamental Casimir. For a density matrix ρ\rho of trace one, write na=tr⁡(ρTa)n_a=\operatorname{tr}(\rho T_a). Completeness gives

ρ=IN+2∑anaTa,∑ana2=12(tr⁡ρ2−1N). \rho=\frac{I}{N}+2\sum_a n_aT_a, \qquad \sum_a n_a^2=\frac12\left(\operatorname{tr}\rho^2-\frac1N\right).

For a pure fundamental state this norm is (N−1)/(2N)(N-1)/(2N). In contrast,

∑aTa2=CFI,CF=N2−12N. \sum_a T_a^2=C_FI,\qquad C_F=\frac{N^2-1}{2N}.

For SU(3)SU(3) with Ta=λa/2T_a=\lambda_a/2, the pure mean-charge norm squared is 1/31/3, while the quadratic Casimir is 4/34/3. Changing generator normalization changes both consistently; it does not equate these two quantities.

Exact singlet test

On a specified tensor product of particle representations, let Qa=∑iTa(i)Q_a=\sum_iT_a^{(i)}, using −Ta∗-T_a^* for an antifundamental factor. Then

Ctot=∑aQa2≥0,⟨ψ∣Ctot∣ψ⟩=∑a∥Qaψ∥2. C_{\rm tot}=\sum_aQ_a^2\geq0, \qquad \langle\psi|C_{\rm tot}|\psi\rangle =\sum_a\|Q_a\psi\|^2.

Consequently ⟨Ctot⟩=0\langle C_{\rm tot}\rangle=0 exactly when every generator annihilates ψ\psi, equivalently when ψ\psi is an SU(N)SU(N) singlet. For a positive density matrix, zero expectation means that its support lies wholly in the singlet subspace. Merely having ⟨Qa⟩=0\langle Q_a\rangle=0 leaves the sum of variances uncontrolled:

⟨Ctot⟩=∑a⟨Qa⟩2+∑aVar⁡(Qa). \langle C_{\rm tot}\rangle =\sum_a\langle Q_a\rangle^2+\sum_a\operatorname{Var}(Q_a).

Explicit SU(3)SU(3) witnesses are:

State Mean total charge ⟨Ctot⟩\langle C_{\rm tot}\rangle
Single fundamental ρ=I/3\rho=I/3 Zero 4/34/3
Product ∣rrˉ⟩|r\bar r\rangle Zero 22
Meson singlet (∣rrˉ⟩+∣ggˉ⟩+∣bbˉ⟩)/3(|r\bar r\rangle+|g\bar g\rangle+|b\bar b\rangle)/\sqrt3 Zero 00
Ordered product ∣rgb⟩|rgb\rangle Zero 33
Normalized antisymmetric three-quark determinant Zero 00

Here r,g,br,g,b form an orthonormal basis labelled 1,2,31,2,3. The last state is

∣ϵ⟩=16∑i,j,k=13ϵijk∣ijk⟩. |\epsilon\rangle=\frac1{\sqrt6} \sum_{i,j,k=1}^3\epsilon_{ijk}|ijk\rangle.

For the meson, the generator action is proportional to the vectorized matrix Ta−Ta=0T_a-T_a=0. For the determinant, a simultaneous basis change multiplies the state by det⁡U=1\det U=1. Both states are therefore singlets. The ordered product states instead retain nonzero total charge variance.

The distinction agrees with the determinant projection and pair-generator calculation in the hadronic ledger, §§74–76.

Limits of a Casimir penalty

An assumed term κCtot\kappa C_{\rm tot} with κ>0\kappa>0 has zero energy exactly on available singlets. This is a useful conditional selection operator. It does not derive that term from the master kernel, set κ\kappa, or prove spatial confinement, a growing separation energy, or nuclear depth three. If other coloured fields participate, the constraint concerns their total charge too; a valence-only projection requires justification.

interaction_operator_checks.py constructs all eight generators, the two- and three-particle charge operators, and the states in the table. It checks both mean charges and charge variances.