The Weinberg operator, counted from fields

With one Higgs doublet and $f_L$ lepton doublets, the monomial $L^2H^2$ contains

$$N_{L^2H^2}=\frac{f_L(f_L+1)}2$$

independent complex Lorentz-and-gauge invariant polynomials. The answers are $1$, $3$, and $6$ for one, two, and three lepton flavors. This is a useful first calculation in the local-operator workbench: weak-isospin contraction, Weyl spinor contraction, Bose symmetry, and Fermi statistics all matter.

The dimension-five interaction goes back to Weinberg, Baryon- and Lepton-Nonconserving Processes, equation (20). A modern specification of the operator and its uniqueness within the minimal Standard Model field content appears in Grzadkowski–Iskrzyński–Misiak–Rosiek, equation (3.1). The derivation below fixes the conventions needed to reproduce the flavor count and the mass normalization.

Field content and normalization

Use the gauge algebra $\mathfrak{su}(3)_c\oplus\mathfrak{su}(2)_w\oplus\mathfrak u(1)$ and the integer charge $Q=6Y$. The $U(1)$ circle has period $2\pi$; electric charge is $Q_{\mathrm{em}}=T_3+Y$. All leptons in this calculation are left Weyl fields.

Field	Lorentz label	Gauge representation $(SU(3),SU(2))_Q$	Flavors	Canonical dimension
$L_a^{\alpha i}$	$(\tfrac12,0)$	$(\mathbf1,\mathbf2)_{-3}$	$f_L$	$3/2$
$H^i$	$(0,0)$	$(\mathbf1,\mathbf2)_{3}$	$1$	$1$

The indices $\alpha,\beta$ are undotted Lorentz spinor indices; $i,j$ are weak indices; $a,b$ are flavor labels. Fix $\epsilon_{12}=+1$ separately in the spin and weak spaces. The total charge and canonical dimension are

$$2(-3)+2(3)=0, \qquad 2\left(\frac32\right)+2(1)=5.$$

This question involves $L^2H^2$ alone. A Hermitian-conjugate operator has different field rows and belongs to a different complex polynomial sector. Neither field-strength tensors nor derivatives occur.

Write one contraction and determine its flavor symmetry

The composite spinor

$$\chi_a^\alpha=\epsilon_{ij}L_a^{\alpha i}H^j$$

is a gauge singlet: the weak indices are contracted and the abelian charges cancel. It is Grassmann odd because it contains one fermion. A scalar can therefore be written as

$$\mathcal O_{ab} =\epsilon_{\alpha\beta}\chi_a^\alpha\chi_b^\beta =\epsilon_{\alpha\beta}\epsilon_{ij}\epsilon_{kl} L_a^{\alpha i}L_b^{\beta k}H^jH^l.$$

Moving the commuting Higgs factors past the lepton fields introduces no sign. Exchanging $a$ and $b$ gives

$$\begin{aligned} \mathcal O_{ba} &=\epsilon_{\alpha\beta}\chi_b^\alpha\chi_a^\beta\\ &=-\epsilon_{\alpha\beta}\chi_a^\beta\chi_b^\alpha =\epsilon_{\alpha\beta}\chi_a^\alpha\chi_b^\beta =\mathcal O_{ab}. \end{aligned}$$

The Grassmann sign and the antisymmetry of $\epsilon_{\alpha\beta}$ cancel. Thus the independent operators are labeled by unordered flavor pairs $a\le b$. An antisymmetric coefficient matrix multiplies zero, while a symmetric matrix has $f_L(f_L+1)/2$ independent entries. The general spinor exchange rule is given in Dreiner–Haber–Martin, equation (2.60).

This also explains why the one-flavor operator survives. The nilpotence of a Grassmann component does not annihilate $\epsilon_{\alpha\beta}\chi^\alpha\chi^\beta=2\chi^1\chi^2$. There are two independent spin components.

Why there is only one weak contraction after statistics

Four distinguishable weak doublet slots have two singlet channels. Treating the slots as distinguishable would therefore suggest $2f_L^2$ Lorentz-and-gauge contractions. They are not all independent operators when the two Higgs fields and the two leptons are repeated species.

For the Higgs fields,

$$\operatorname{Sym}^2\mathbf2_w=\mathbf3_w.$$

The weak singlet $\epsilon_{ij}H^iH^j$ vanishes because the Higgs components commute. The two-Higgs system must lie in the weak triplet.

Now separate the lepton spin space $S=\mathbb C^2_\ell$ from its weak-and-flavor space $W\otimes F$. The Lorentz-singlet part of

$$\Lambda^2\bigl(S\otimes(W\otimes F)\bigr)$$

uses $\Lambda^2S=\mathbf1_\ell$, and therefore carries $\operatorname{Sym}^2(W\otimes F)$. Expanding this remaining symmetric power gives

$$\operatorname{Sym}^2(W\otimes F) =(\operatorname{Sym}^2W\otimes\operatorname{Sym}^2F) \oplus(\Lambda^2W\otimes\Lambda^2F).$$

Only the weak-triplet term $\operatorname{Sym}^2W\otimes\operatorname{Sym}^2F$ can combine with the Higgs triplet to give a singlet. The triplet square contains that singlet once, leaving precisely $\operatorname{Sym}^2F$ as the multiplicity space. This establishes both the count and the absence of an independent antisymmetric-flavor operator, without choosing Clebsch–Gordan coefficients.

Lepton flavors	Distinguishable-slot count	Invariant polynomial count	Flavor space
$1$	$2$	$1$	$\operatorname{Sym}^2\mathbb C$
$2$	$8$	$3$	$\operatorname{Sym}^2\mathbb C^2$
$3$	$18$	$6$	$\operatorname{Sym}^2\mathbb C^3$

Flavors here are ungauged degeneracies. Projecting onto a flavor-group singlet would be a different problem. The symmetry of the Weinberg coefficient provides a standard example of this distinction in Fonseca, Section 1.

Electroweak components and the mass term

Write

$$L_a=\begin{pmatrix}\nu_a\\e_a\end{pmatrix}, \qquad H=\begin{pmatrix}H^+\\H^0\end{pmatrix}.$$

Then $\chi_a=\nu_aH^0-e_aH^+$. Suppressing the two-component spin contraction, the one-flavor operator is

$$\mathcal O =(\nu\nu)(H^0)^2 -2(\nu e)H^0H^+ +(ee)(H^+)^2.$$

The relative factor two follows because $(\nu e)=(e\nu)$ for anticommuting Weyl spinors. The three component terms form one gauge-invariant operator. Counting them as separate interactions before imposing gauge symmetry would overcount.

Choose the following coefficient convention, with a sum over all ordered flavor pairs and a symmetric matrix $C=C^T$:

$$\mathcal L_5=-\frac{1}{2\Lambda} \sum_{a,b}C_{ab}\mathcal O_{ab}+\mathrm{h.c.}$$

For $\langle H\rangle=(0,v/\sqrt2)^T$, the operator becomes

$$\mathcal O_{ab}\longrightarrow \frac{v^2}{2}\,\epsilon_{\alpha\beta} \nu_a^\alpha\nu_b^\beta.$$

Comparison with

$$\mathcal L_{\mathrm{mass}} =-\frac12\sum_{a,b}(m_\nu)_{ab} \epsilon_{\alpha\beta}\nu_a^\alpha\nu_b^\beta +\mathrm{h.c.}$$

therefore gives

$$m_\nu=\frac{v^2}{2\Lambda}\,C.$$

The prefactor depends on how $C$ is defined; the displayed Lagrangians fix it. Copying a mass formula from a convention with a different overall factor in $\mathcal L_5$ would produce an apparent factor-of-two discrepancy. Diagonalizing this symmetric mass matrix is a separate flavor-basis operation; see Dreiner–Haber–Martin, Section 3.2 and Appendix D.2.

If lepton number is assigned as $L(L)=1$ and $L(H)=0$, the operator carries two units of lepton number. Gauge invariance permits it; imposing an additional exact lepton-number symmetry would forbid it. The workbench does not silently impose that additional symmetry or infer a mechanism that generates the Wilson coefficient.

Global form and reproducibility

Both fields descend to the standard $\mathbb Z_6$ quotient: the descent expression $2(p+2q)+3n+Q$ is $0$ for $L$ and $6$ for $H$. They consequently descend to its order-two and order-three subgroups as well. The invariant count is the same for each compatible quotient. This agreement is a property of these local representations, not an equivalence of all global gauge theories; see the Standard Model global group.

To reproduce the calculation, select one of the Weinberg presets in the workbench, retain one Higgs flavor, and choose lepton flavors $1$, $2$, or $3$. Keep powers $m_L=m_H=2$ and derivative order zero. The exported physical setup records the integer charge convention, the global quotient, all fields, and the monomial.

The count is six complex coefficients at three flavors before flavor-basis identification. Adding the conjugate monomial gives a conjugate six-dimensional space; in a Hermitian Lagrangian its coefficients are fixed by $C^*$ and do not constitute six additional independent complex parameters. The full minimal-SM statement that this is the unique dimension-five operator class uses the complete field-content analysis in the cited literature. A calculation of this one monomial verifies its invariant space, rather than enumerating every possible dimension-five monomial.