research30 min studyLesson 678 in Physics

Cutkosky cuts and perturbative unitarity

Match a scalar bubble’s absorptive part to on-shell phase space with explicit state normalization and identical-particle factors.

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What you’ll understand

  • Derive the forward optical theorem from S-matrix unitarity.
  • Track positive-energy delta functions and two-body phase-space normalization.
  • Check the one-loop scalar coefficient against the identical-particle intermediate-state sum.

The imaginary part of a loop diagram is constrained by probability conservation. Perturbative unitarity relates it to lower-order amplitudes summed over real on-shell intermediate states. Cutting rules turn that relation into a diagrammatic calculation, but they do not remove normalization, state-counting, or causal-prescription work. The equal-mass quartic bubble supplies a compact check: the threshold factor obtained from a logarithm must reappear in two-body phase space with exactly the diagram coefficient required by Wick counting.

Fix S and one-particle state conventions

Write S=1+iTS=1+iT and normalize one-particle states by ⟨p∣q⟩=2Ep(2π)3δ3(p−q)\langle\mathbf p|\mathbf q\rangle=2E_{\mathbf p}(2\pi)^3\delta^3(\mathbf p-\mathbf q). Removing the overall four-momentum delta function defines the invariant amplitude. Expanding S†S=1S^\dagger S=1 gives a relation between the anti-Hermitian part of T and T†TT^\dagger T. For a forward matrix element, it becomes twice the imaginary part of the forward amplitude equal to a sum of absolute squares over physical intermediate states.

At a given perturbative order, the right side uses amplitudes of lower orders whose product reaches that order. In quartic scalar theory, tree amplitudes are order lambda, so their squared contribution constrains the order-lambda-squared absorptive part. The intermediate-state sum includes species, polarizations, momentum integrals, and symmetry factors for indistinguishable particles. Counting two identical scalar momenta as ordered without division by two would double the result. This state-counting factor has the same numerical value as the bubble coefficient here, but its origin should still be tracked independently.

2Im⁡Mii=∑X∫dΦX Mi→XMi→X∗2\operatorname{Im}\mathcal M_{ii}=\sum_X\int d\Phi_X\,\mathcal M_{i\to X}\mathcal M_{i\to X}^*
The formula uses the stated relativistic state normalization and consistent momentum delta functions. The intermediate-state measure includes required identical-particle factorials and only physically allowed on-shell states.↔ Scroll wide equations horizontally.

A physical cut imposes positive on-shell energy

The distribution identity for a scalar denominator states that its discontinuity across the pole is minus two pi i times a delta function. For a physical cut, energy flow through the cut selects the positive-energy part. Multiplying by the propagator’s numerator i turns that replacement into a positive two-pi delta factor before the remaining diagram factors and conjugate subamplitudes are assembled. One must not apply a mnemonic replacement while dropping the prescription or reversing the cut orientation silently.

The Lorentz-invariant two-particle phase space integrates two positive-energy measures and one four-momentum delta function. For equal masses and timelike total momentum squared s, its integral is beta divided by eight pi when the two momenta are treated as distinguishable labels. An identical pair requires an additional factorial division in the intermediate-state sum. The phase-space square root vanishes at the same threshold as the bubble logarithm. This equality is a physical consistency test of signs and normalization, not a numerical coincidence.

δ+(q2−m2)=θ(q0)δ(q2−m2),∫dΦ2=β(s)8π\delta_+(q^2-m^2)=\theta(q^0)\delta(q^2-m^2),\qquad \int d\Phi_2=\frac{\beta(s)}{8\pi}
The phase space is defined by (2π)^4 delta^4(P−q1−q2) times the product d³q_j/[(2π)^3 2E_j]. It excludes an identical-particle 1/2! factor, which is supplied separately. The measure is dimensionless in four dimensions.↔ Scroll wide equations horizontally.

Compare the bubble with the physical state sum

The scalar s-channel loop contributes λ2F(s)/(32π2)\lambda^2F(s)/(32\pi^2) to the finite amplitude. Its imaginary part is therefore lambda squared times beta divided by thirty-two pi. Twice that quantity equals one half times the two-body phase space times the squared tree amplitude. In physical forward equal-mass scattering, the t and u channels do not acquire this same s-channel two-particle cut. A real ultraviolet counterterm also has no absorptive contribution to this check.

A cut constrains a discontinuity; it does not determine every real subtraction constant of an amplitude. Dispersion relations need appropriate subtraction information to reconstruct the real part. Gauge theories add the requirement that only physical states survive the complete unitarity sum, even when an intermediate gauge-fixed calculation uses additional fields. The Bloch lab demonstrates norm conservation for one closed pure two-state model. It neither sums multiparticle phase space nor implements Cutkosky rules or an interacting scattering matrix.

2Im⁡Ms(1)=λ2β16π=12!λ2∫dΦ22\operatorname{Im}\mathcal M_s^{(1)}=\frac{\lambda^2\beta}{16\pi}=\frac1{2!}\lambda^2\int d\Phi_2
This one-loop equality is for one real scalar species with tree amplitude −lambda and a two-identical-scalar intermediate state. It uses the same F definition and positive beta as the preceding threshold lesson.↔ Scroll wide equations horizontally.

Check a two-scalar cut

Take λ=1\lambda=1 and s=16m2/3s=16m^2/3, so beta is one half. Compute the integrated two-body phase space, the identical-state unitarity sum, and Im of the one-loop s-channel amplitude.

  1. The labeled two-body phase space is (1/2)/(8π)=1/(16π)(1/2)/(8\pi)=1/(16\pi). This expression has no mass units in four-dimensional natural units.

  2. The squared tree amplitude is one. Divide by 2!2! for identical intermediate scalars, obtaining the unitarity right side 1/(32π)1/(32\pi). This equals twice the required imaginary part.

  3. Thus Im⁡Ms(1)=1/(64π)\operatorname{Im}\mathcal M_s^{(1)}=1/(64\pi). Independently, the bubble coefficient times Im⁡F=π/2\operatorname{Im}F=\pi/2 gives (1/32π2)(π/2)(1/32\pi^2)(\pi/2), the same result.

The result

The identical-scalar state sum is 1/(32π), matching twice a loop imaginary part of 1/(64π).

Explore the model

Interactive lab

Interactive lab · quantum

A limited norm-conservation illustration

Reset and set mixing angle 60∘60^\circ, phase 0∘0^\circ: predict populations three quarters and one quarter. Play or Step and confirm their sum stays one during phase precession. Repeat from Reset with both controls 60∘60^\circ, 180∘180^\circ; the same normalization holds. This pure two-state instrument is not an S-matrix or multiparticle phase-space calculator.

A limited norm-conservation illustrationA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. 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 forProbability conservation motivates unitarity, but a physical optical-theorem check additionally needs amplitudes and complete on-shell states.

Check your understanding

Where does the extra factor one half enter for two identical intermediate scalars?

Try it yourself

For a hypothetical distinguishable two-particle intermediate channel with the same phase-space value and constant tree amplitude as the worked example, remove the identical-particle factor. What are the unitarity sum and the corresponding required imaginary part?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The labeled phase space is unchanged; only the state-counting factor is different in this comparison.

Compare with the solution

Without the factorial division, the right side is 1/(16π)1/(16\pi) rather than 1/(32π)1/(32\pi). The optical theorem therefore requires imaginary part 1/(32π)1/(32\pi) for this hypothetical channel contribution. Both values are twice their identical-pair counterparts. This is a comparison of normalized state sums with specified equal amplitudes, not a claim that changing particle species leaves every diagram coefficient or interaction in a complete theory unchanged.

Take it with you

  • The optical theorem follows from S-matrix unitarity with declared normalization.
  • Physical cuts impose positive-energy on-shell conditions and complete intermediate-state counting.
  • Discontinuities test absorptive coefficients while real subtraction information remains separate.

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