A finite prediction can be expressed through different parameter coordinates
Separate regulation, field normalization, counterterms, and physical renormalization conditions.
On this page
The question for today
What you’ll understand
- Distinguish a regulator from a subtraction convention.
- Rewrite a bare scalar action using renormalized fields and parameters.
- Hold a physical quantity fixed while changing a regulated bare input.
A loop calculation contains several quantities that are often called the mass or coupling. Bare parameters define the regulated action, renormalized parameters define a useful finite coordinate system, and observables are extracted from complete predictions. Renormalization relates these roles without identifying them. We will track the field-normalization factors explicitly and use a cutoff example to show how a changing bare mass can represent the same low-energy mass rather than a changing physical particle.
A regulator and a renormalization condition do different jobs
A momentum cutoff restricts integration momenta, while dimensional regularization analytically continues loop integrals away from four dimensions. Each makes otherwise ill-defined expressions manageable, but neither is by itself a rule for assigning measured values to model parameters. A renormalization condition fixes finite parts by specifying an amplitude, pole, or subtraction prescription. The regulator can then be removed where a continuum limit exists, or retained within an effective theory with a stated cutoff and energy domain.
Use natural units ħ=c=1 and a scalar theory in d=4−ε dimensions. A quartic coupling has mass dimension ε, so a reference energy μ supplies its required dimension while the renormalized λ is dimensionless. This μ is not automatically a physical ultraviolet cutoff. In dimensional regularization it remains after the regulator ε is removed. A divergent bare parameter in that limit is a feature of the chosen parameterization; it does not mean that an observable mass or scattering probability must diverge.
Reexpressing the action fixes the counterterm combinations
Start with a bare kinetic term, mass term, and quartic interaction. After replacing the field, the kinetic coefficient is Zφ, the mass coefficient is ZφZm times m squared, and the quartic coefficient contains Zφ squared times Zλ. Split each into a chosen renormalized term and a counterterm. This algebra explains why a vertex renormalization factor and a coupling renormalization factor need not be the same: changing the field normalization also changes every vertex containing that field.
Counterterms are local interactions constrained by symmetries and the chosen expansion. They are inserted into diagrams at the appropriate perturbative order. They are not independent adjustable corrections for each external momentum or each experiment. Once enough renormalization conditions fix the available parameters, the remaining amplitudes are predictions. In an effective field theory, more local operators appear at higher order in the energy-to-heavy-scale ratio, but a finite set suffices at a specified accuracy when the power counting is controlled.
Holding physics fixed does not hold every input fixed
A sharp-cutoff scalar mass correction can contain a term proportional to the cutoff squared. Its coefficient is model and regulator dependent; dimensional regularization organizes the same matching problem differently and does not require displaying that particular power divergence. The illustrative relation M²=mB²+cλΛ² makes the bookkeeping transparent: if M is fixed while Λ changes, mB² must change. This statement alone does not resolve naturalness questions or establish a UV completion. It only separates an unphysical input from a fitted low-energy output.
The function lab illustrates the cutoff-square contribution. Reset and set both controls to a=0.05 and x=1, predicting y=0.05. Reset and set a=0.05 and x=2, predicting y=0.20. Assign x to a normalized cutoff and y to a normalized loop mass correction. Doubling the cutoff quadruples this illustrative contribution. The compensating bare input must be calculated separately; the lab has no field theory, regulator removal, or measured mass. Each setting is an independent calculation rather than a physical evolution of the sample.
Work it through
Retune a regulated mass while fixing its physical target
In an illustrative cutoff model, M²=mB²+0.01Λ². Express energies in GeV and fix M²=4 GeV². Find mB² at Λ=10 GeV and Λ=30 GeV, then explain the sign of the second bare parameter.
At Λ=10 GeV, the correction is 0.01×100=1 GeV². The required bare input is therefore 4−1=3 GeV².
At Λ=30 GeV, the correction is 0.01×900=9 GeV². The required bare input becomes 4−9=−5 GeV².
Substituting either input returns M²=4 GeV². A negative bare quadratic coefficient in this illustrative regulated relation does not mean the fitted physical excitation has negative mass squared; stability and vacuum structure require the complete regulated action.
The two bare inputs are 3 GeV² and −5 GeV² while the target physical mass remains M=2 GeV in both calculations.
The living notebook
Explore the model
Interactive lab · function
Inspect a cutoff-square contribution
Reset and set BOTH controls to a=0.05 and x=1, predicting y=0.05. Reset and set a=0.05 and x=2, predicting y=0.20. This models an assigned quadratic correction only; compensation and physical renormalization conditions are external to the parabola.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forRegulated contributions can change while a fitted physical quantity stays fixed through parameter retuning.
Pause & reason
Check your understanding
Your turn
Try it yourself
For a finite normalization example take Zφ=1.21, Zm=0.50, and m²=8 GeV². Find mB², the full mass coefficient in the action written through φ, and δmact². Also give φB/φ.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Apply the definitions in order; the action coefficient includes the field factor as well as the mass factor.
Compare with the solution
The bare parameter is mB²=0.50×8=4 GeV². The renormalized-field action coefficient is ZφmB²=1.21×4=4.84 GeV². Hence δmact²=4.84−8=−3.16 GeV². The field ratio is √1.21=1.10. This finite algebraic illustration does not assert these Z values are a controlled small-loop expansion.
Keep the idea
Take it with you
- Regularization and renormalization conditions are separate operations.
- Field normalization contributes to every action coefficient containing that field.
- Bare parameter changes need not imply a change of physical predictions.
Read further
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