Equation 4 · Part 5 · How Particle Physics Beyond the Standard Model Actually Works

Symbol Lambda^2

\mathcal{L}_{\text{eff}} = \mathcal{L}_{\text{SM}} + \sum_i \frac{c_i}{\Lambda^{2}}\,\mathcal{O}_i

\Lambda^{2}

What this part means

the square of Lambda; the mass scale of whatever new physics has been integrated out.

Its job in the formula

Lambd $a^2$ occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

Full expressionSymbol Lambda^2Article meaning

Where the article explains it

Here $\mathcal{O}_i$ are Standard-Model-field operators of dimension six, $c_i$ are unknown coefficients, and $\Lambda$ is the mass scale of whatever new physics has been integrated out.

The passage around this formula

If a new particle is too heavy to produce directly, its effects can still appear as small deviations in the interactions of known particles, encoded as additional terms in an effective Lagrangian: $\mathcal{L}_{\text{eff}} = \mathcal{L}_{\text{SM}} + \sum_i \frac{c_i}{\Lambda^{2}}\,\mathcal{O}_i$ . Here $\mathcal{O}_i$ are Standard-Model-field operators of dimension six, $c_i$ are unknown coefficients, and $\Lambda$ is the mass scale of whatever new physics has been integrated out. This is the Standard Model Effective Field Theory (SMEFT) framework used across current LHC and dark-matter searches. Its value is precise and limited: it lets an experiment quote a model-independent bound on $c_i$ / $\Lambda^2$ from a measured deviation (or non-deviation) without committing to what…

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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