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The ideal two-shower event fraction is $B_e^2$, approximately $4\%$ in simple channel counting or $3.18\%$ using the measured tau lepton electronic branching fraction. The selected particle decay chain isThe electron and Positron each make a contained electromagnetic shower; muons penetrate, and generic hadrons do not deposit all their energy as a contained electromagnetic shower.

To estimate $B_e$, use lepton universality and neglect final masses and Quantum chromodynamics corrections. The electronic and muonic decay widths each carry weight one. The accessible inclusive hadronic channels have weights $3|V_{ud}{|}^2$ and $3|V_{us}{|}^2$, where three is the color multiplicity in a decay width. The charm quark is inaccessible in physical tau lepton hadronic decays. With $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$, the total weight is $1+1+3=5$, givingThe 2021 measured value $B_e=0.1782$ instead gives $B_e^2=0.0317552$.^[ref] The difference reflects effects omitted by the simple free-channel estimate, especially the hadronic contribution to the total decay width.

There are two tau leptons per produced pair and two per selected pair, so the fraction of all produced tau leptons belonging to these two-shower events is also $B_e^2$. The unrestricted fraction of individual tau leptons decaying electronically is $B_e$; these are different selections. A real high-energy shower cut gives an additional acceptance factor: $N_{\mathrm{selected}}/N_{\mathrm{pairs}}=B_e^2{\epsilon }_{ee}$, which cannot be computed without the cut and detector response. No factor of two is needed for the unique $e^{+}{e}^{-}$ assignment.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive Quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before Quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.