QFT Bridge: Optical Theorem and Unitarity

The optical theorem is a consequence of unitarity. In nonrelativistic scattering, it relates the imaginary part of the forward scattering amplitude to the total cross section. In QFT, the same principle relates the imaginary part of a forward invariant amplitude to a sum over all allowed on-shell intermediate states. Unitarity derives the channel constraints; here we translate the state sum into relativistic phase space.

The formulas look different because the normalization of states and the phase-space measure are different. The principle is the same:

$$S^\dagger S=I.$$

The optical theorem in potential scattering

With the potential-scattering amplitude convention

$$\psi(\mathbf r) \sim e^{ikz} + f(\theta) \frac{e^{ikr}}{r},$$

the optical theorem is

$$\sigma_{\mathrm{tot}} = \frac{4\pi}{k} \operatorname{Im}f(0).$$

This equation says that the total probability removed from the incident beam is encoded in the forward amplitude.

The Transition Operator in the S-Matrix

Define the dimensionless transition part $\mathbb T$ by

$$S=I+i\mathbb T,$$

then unitarity gives

$$S^\dagger S=I \quad \Rightarrow \quad i(\mathbb T^\dagger-\mathbb T)=\mathbb T^\dagger\mathbb T.$$

This $\mathbb T$ is not the energy-valued Lippmann–Schwinger operator $T(E)$; energy delta functions relate their matrix elements. Taking a forward matrix element produces a relation between the imaginary part of the forward amplitude and a sum over final states.

This is the algebraic heart of every optical-theorem formula. The rest is normalization.

Relativistic Phase Space and Total Cross Section

Set $\hbar=c=1$, use $\langle\mathbf p'|\mathbf p\rangle=2E_{\mathbf p}(2\pi)^3\delta^3(\mathbf p'-\mathbf p)$, and define $\mathcal M$ after factoring $i(2\pi)^4\delta^4(P_f-P_i)$ from $S-I$. Then for a specified initial state,

$$2\operatorname{Im}\mathcal M_{ii} =\sum_n\frac1{\mathcal S_n}\int d\Phi_n\,|\mathcal M_{ni}|^2,$$

where $\mathcal S_n$ avoids overcounting identical final particles and

$$d\Phi_n=(2\pi)^4\delta^4\!\left(P_i-\sum_jp_j\right) \prod_j\frac{d^3p_j}{(2\pi)^3\,2E_j}.$$

Sum final spin/internal labels as well. For two incident stable particles, divide the right side by the invariant incident flux

$$F=4\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2} =4k_{\mathrm{cm}}\sqrt s$$

to obtain the total cross section. Hence $2\operatorname{Im}\mathcal M_{ii}=F\sigma_{\mathrm{tot}}$. Initial spin averages, if desired, must be applied consistently to both sides. This assumes an ordinary scattering problem with finite cross sections; long-range Coulomb or massless infrared singularities need further treatment.

Cuts and Imaginary Parts

In perturbative QFT, imaginary parts of amplitudes arise when intermediate states can go on shell. Diagrammatic cutting rules make this precise: cutting internal lines places them on shell and relates the discontinuity of a diagram to products of lower-order amplitudes.

This is not a new principle beyond the optical theorem. It is unitarity implemented in relativistic perturbation theory.

Resonance Connection

Resonances have complex poles and widths because probability can flow into open channels. In a simple one-channel model, the width appears in a denominator such as

$$E-E_R+i\Gamma/2.$$

In QFT, widths and imaginary self-energies similarly reflect allowed decay channels. The optical-theorem perspective explains why an imaginary part is not optional: it is required by unitarity once channels open.

Why a Real Tree Amplitude Is Consistent

For a real scalar interaction $\mathcal L_{\mathrm{int}}=-\lambda\phi^4/4!$, the tree $2\to2$ amplitude is $-\lambda$. It is real, while the cross section begins at $\lambda^2$. Unitarity therefore constrains the imaginary part at one loop, also order $\lambda^2$, rather than demanding an imaginary part in the order-$\lambda$ answer. For $s>4m^2$,

$$2\operatorname{Im}\mathcal M_{ii}^{\mathrm{one\ loop}} =\frac12\lambda^2\int d\Phi_2 =\frac{\lambda^2}{16\pi}\sqrt{1-\frac{4m^2}{s}}.$$

The $1/2$ counts identical final scalars. This is the same order-counting lesson as the first/second Born relation in the nonrelativistic optical theorem.

Exercises

1. For the scalar example, find the leading $\operatorname{Im}\mathcal M_{ii}$ at $s=16m^2$.

Solution

The two-body velocity factor is $\sqrt{1-4/16}=\sqrt3/2$. Divide the displayed unitarity equation by two to obtain $\operatorname{Im}\mathcal M_{ii}=\lambda^2\sqrt3/(64\pi)$. The result vanishes at threshold, where two-body phase space closes.

2. Why does the QFT optical theorem involve a phase-space integral?

Solution

QFT final states can contain several particles with continuous momenta. Summing over all final states therefore requires the Lorentz-invariant phase-space measure, including momentum conservation and the on-shell measures for each particle. This replaces the simpler angular integration of elementary two-body potential scattering.

References

• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• S. Weinberg, The Quantum Theory of Fields, Vol. I: Foundations, Cambridge University Press, 1995.