research30 min studyLesson 672 in Physics

One-loop scalar self-energy

Evaluate a massive scalar tadpole with an explicit Euclidean cutoff and connect subtraction conditions to propagator poles.

On this page

What you’ll understand

  • Specify the self-energy sign through a dressed propagator.
  • Integrate a cutoff tadpole without confusing its mass shift with a universal observable.
  • Distinguish a momentum-independent mass correction from residue renormalization.

The two-point function reveals how interactions move a particle pole and change its residue. A one-loop tadpole in massive quartic scalar theory is a particularly simple self-energy: it contains an ultraviolet divergence but has no external-momentum dependence. We calculate it with a declared Euclidean cutoff to expose the integration, then impose a mass condition. This cutoff calculation illustrates renormalization bookkeeping; its unrenormalized numerical shift is not a scheme-independent prediction for a measured mass.

Fix the insertion and pole conventions

Retain natural units and Minkowski signature (+−−−)(+---), with free propagator i/(p2−m2+i0)i/(p^2-m^2+i0) and interaction −λϕ4/4!-\lambda\phi^4/4!. Define the one-particle-irreducible insertion as −iΣ(p2)-i\Sigma(p^2). Summing repeated insertions then gives a denominator p2−m2−Σ(p2)+i0p^2-m^2-\Sigma(p^2)+i0. This definition fixes the sign of the mass correction; a source defining its insertion with the opposite sign must change the corresponding denominator. A stable isolated pole is assumed when discussing a real pole mass and residue.

The tadpole’s scalar coefficient is one half times the quartic coupling. No external momentum flows through its loop denominator, so its self-energy is constant at this order. Wick rotation takes the loop into Euclidean momentum with a positive denominator; its contour deformation is justified by the Feynman prescription and absence of obstructing external thresholds for this tadpole. The resulting positive cutoff integral is not a classical self-energy of a localized particle. It is a regulated quantum correction in this particular scalar model.

G(p)=ip2−m2−Σ(p2)+i0,Σtad=λ2∫∣kE∣<Λd4kE(2π)41kE2+m2G(p)=\frac{i}{p^2-m^2-\Sigma(p^2)+i0},\qquad \Sigma_{\mathrm{tad}}=\frac\lambda2\int_{|k_E|<\Lambda}\frac{d^4k_E}{(2\pi)^4}\frac1{k_E^2+m^2}
Sigma has energy-squared units, m and cutoff Lambda are energies, and lambda is dimensionless in four dimensions. This equation uses an explicitly Euclidean spherical cutoff after Wick rotation.↔ Scroll wide equations horizontally.

Perform the radial integral

The four-dimensional unit sphere has area 2π22\pi^2. Combining that angular measure with the denominator’s (2π)4(2\pi)^4 leaves a radial prefactor 1/(8π2)1/(8\pi^2). Substitute u=kE2u=k_E^2, so the radial integral contains u/(u+m2)u/(u+m^2). Splitting it into one minus a simple rational term exposes the quadratic cutoff dependence and the logarithmic mass dependence. The result remains finite for every finite cutoff and positive mass before any subtraction is made.

At large cutoff, the leading term is quadratic and the subleading term contains the mass squared times a logarithm. Other regulators package these contributions differently: dimensional regularization represents the ultraviolet information through meromorphic poles rather than an explicit positive quadratic cutoff term. Comparing bare integral values across regulators is therefore not a comparison of physical masses. Use the same renormalization condition and matched parameters before comparing predictions. A hard momentum cutoff also requires care with momentum shifts and gauge identities in less elementary calculations.

IE(Λ,m)=116π2[Λ2−m2ln⁡(1+Λ2m2)]I_E(\Lambda,m)=\frac1{16\pi^2}\left[\Lambda^2-m^2\ln\left(1+\frac{\Lambda^2}{m^2}\right)\right]
I_E is the cutoff Euclidean integral in the preceding equation and has energy-squared units. The formula assumes m > 0; a massless infrared limit must be assessed before discarding scales or exchanging limits.↔ Scroll wide equations horizontally.

A renormalization condition selects the mass parameter

Add a mass counterterm to the self-energy and choose its finite part by a stated condition. In an on-shell prescription, require the corrected denominator to vanish at the selected physical mass. At first order, a constant loop shift can be canceled by an equal and opposite constant counterterm. This does not mean interactions have disappeared: other Green functions and higher-loop momentum dependence remain. It means the mass parameter has been defined through that pole measurement.

The residue depends on the derivative of self-energy with respect to momentum squared. Because this tadpole is constant, it gives no residue correction at this order. A momentum-dependent self-energy in another theory or at a higher loop order can affect both pole and residue. The parabola instrument isolates quadratic algebra only. Its normalized height is not the complete tadpole, whose logarithm, regulator choice, and subtraction condition are essential to the result.

M2=m2+ΣR(M2),Zpole=11−ΣR′(M2)M^2=m^2+\Sigma_R(M^2),\qquad Z_{\mathrm{pole}}=\frac1{1-\Sigma_R'(M^2)}
M is the pole mass and Sigma_R includes counterterms in the declared convention. The derivative is with respect to p², so the residue is dimensionless. These expressions assume a simple stable pole and a consistently normalized field.↔ Scroll wide equations horizontally.

A finite cutoff illustration

Let m=E∗m=E_*, Λ=3E∗\Lambda=3E_*, and g=λ/(32π2)=0.01g=\lambda/(32\pi^2)=0.01. Find the unrenormalized tadpole self-energy. What counterterm enforces ΣR=0\Sigma_R=0 at this order?

  1. The dimensionless cutoff bracket is Λ2/m2−ln⁡(1+Λ2/m2)=9−ln⁡10≃6.6974\Lambda^2/m^2-\ln(1+\Lambda^2/m^2)=9-\ln10\simeq6.6974. Keep the logarithm rather than estimating the whole result from its quadratic term.

  2. Multiplying by g and restoring the mass scale gives Σtad≃0.066974E∗2\Sigma_{\mathrm{tad}}\simeq0.066974E_*^2. This is small compared with m2m^2 for the chosen illustrative parameters.

  3. In the convention where the counterterm adds to Sigma, choose δm2=−0.066974E∗2\delta m^2=-0.066974E_*^2. The total first-order self-energy then vanishes under this mass condition, while its derivative was already zero.

The result

The cutoff loop gives +0.066974E_*²; the stated on-shell mass condition assigns an equal negative counterterm.

Explore the model

Interactive lab

Interactive lab · function

Isolate a quadratic term

Reset and set coefficient a=0.5a=0.5, position x=1x=1; predict height 0.5. Repeat from Reset with both controls a=0.5a=0.5, x=2x=2; predict height 2. This is a normalized comparison of the quadratic term alone. The lab does not evaluate the logarithmic tadpole contribution, perform Wick rotation, or choose a mass counterterm.

Isolate a quadratic termThis 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. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

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 forRecognizing a quadratic scaling term does not determine the full regulated mass correction.

Check your understanding

What does the momentum-independent quartic tadpole imply about the first-order pole residue?

Try it yourself

Suppose the loop self-energy is the constant 0.04m20.04m^2 and the chosen pole condition is M=mM=m. Find the mass counterterm and the resulting first-order residue in this convention.

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 pole condition to the total self-energy, then differentiate the constant expression.

Compare with the solution

The pole equation requires ΣR(m2)=0\Sigma_R(m^2)=0. Since the counterterm adds to the loop self-energy, choose δm2=−0.04m2\delta m^2=-0.04m^2. The total self-energy is constant and zero under this subtraction, and its momentum derivative vanishes, so the first-order residue is one. This conclusion is specific to the constant tadpole contribution and does not exclude momentum-dependent residue corrections at higher orders.

Take it with you

  • The propagator denominator fixes the sign convention for the self-energy insertion.
  • Bare cutoff dependence becomes physical only after parameter matching and a subtraction condition.
  • A constant tadpole changes the mass bookkeeping but has no momentum-derivative residue term.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.