Optical Theorem Preview

Unitarity equates the imaginary part of a forward amplitude with the total probability into all accessible final states. In covariant normalization the relation is $2\operatorname{Im}\mathcal M_{ii}=\mathcal F\sigma_{\rm tot}$, with invariant incident flux $\mathcal F$. This page supplies the normalization and channel-counting translation. The theorem and its nonrelativistic partial-wave derivation belong to Optical Theorem.

Required background. Optical Theorem provides the unitarity statement; Invariant Phase Space Preview supplies flux and counting; Relativistic Normalization fixes external states. Helpful background. QFT Bridge: S-Matrix explains multiparticle channels.

Unitarity with a declared amplitude convention

Use $\hbar=c=1$ and write

$$S=I+iT,\qquad \langle f|T|i\rangle=(2\pi)^4\delta^4(P_f-P_i)\mathcal M_{fi}.$$

Here $T$ is the dimensionless transition part of $S$. It is not the potential-scattering $T_{\rm NR}(E)$ in $S=I-2\pi i\delta(E_f-E_i)T_{\rm NR}$. Identifying those symbols without their delta factors and state conventions changes the normalization and sign.

Expanding $S^\dagger S=I$ gives

$$-i(T-T^\dagger)=T^\dagger T.$$

Take the diagonal matrix element in an incoming state with fixed momenta and spin labels, and insert a complete set of physical outgoing states. Use wave packets to control plane-wave delta distributions, or retain a common finite normalization regulator until common factors cancel. After removing the common total-momentum delta, the result is

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

The sum is over all open channels of the same total four-momentum. The phase-space convention is

$$d\Phi_n=(2\pi)^4\delta^4(P_i-P_n) \prod_{a=1}^n\frac{d^3k_a}{(2\pi)^3\,2E_a}.$$

For ordered momenta, $\mathcal S_n=\prod_j n_j!$ counts identical particles by species. If the completeness integral counts each physical state only once, this factor is already included. Fermionic exchange signs affect $\mathcal M$; its squared modulus remains nonnegative.

The disconnected identity part of $S$ is excluded by the definition of $T$. One must not square the unscattered delta as if it were a cross section, or replace a regulated squared delta by an arbitrary finite number.

The invariant-flux form

For a two-particle initial state,

$$\mathcal F=4\sqrt{(p_1\cdot p_2)^2-m_1^2m_2^2}.$$

The right-hand side of the unitarity relation is exactly $\mathcal F\sigma_{\rm tot}$ in the rate convention of the phase-space page. Therefore

$$2\operatorname{Im}\mathcal M_{ii} =\mathcal F\sigma_{\rm tot}.$$

In the center-of-momentum frame, $\mathcal F=4p_i\sqrt s$, so

$$\operatorname{Im}\mathcal M_{ii} =2p_i\sqrt s\,\sigma_{\rm tot}.$$

McGreevy’s section 5.2 gives this invariant-amplitude form. The theorem includes elastic and inelastic outgoing states; the forward quantity on the left remains diagonal in the original incoming channel.

For an unpolarized ensemble, average both sides over the initial spin labels with the same weights. This averages diagonal forward amplitudes on the left and cross sections on the right. It does not replace a sum of squared amplitudes by the square of their average.

For a spinless distinguishable elastic channel, define a center-of-momentum angular amplitude by

$$f_{\rm cm}(\theta)=\frac{\mathcal M(\theta)}{8\pi\sqrt s}.$$

Then $d\sigma_{\rm el}/d\Omega=|f_{\rm cm}|^2$ and the optical theorem reads

$$\sigma_{\rm tot}=\frac{4\pi}{p_i}\operatorname{Im}f_{\rm cm}(0).$$

This resembles the nonrelativistic formula because the angular amplitude has been normalized to make it do so. For unequal incoming and outgoing momenta in an inelastic channel, the additional $p_f/p_i$ rate factor must be retained.

Why a real Born amplitude is consistent with unitarity

Suppose a dimensionless real expansion parameter $\lambda$ organizes the amplitudes, $\mathcal M=\lambda\mathcal M^{(1)} +\lambda^2\mathcal M^{(2)}+\cdots$, and there is no lower-order transition probability. Order by order,

$$\begin{aligned} 2\operatorname{Im}\mathcal M_{ii}^{(1)}&=0,\\ 2\operatorname{Im}\mathcal M_{ii}^{(2)} &=\sum_n\frac1{\mathcal S_n}\sum_{\text{spins}} \int d\Phi_n\,|\mathcal M_{ni}^{(1)}|^2. \end{aligned}$$

A leading real amplitude gives a nonzero cross section at second order in $\lambda$. The corresponding imaginary forward part is also second order. Comparing the leading cross section with only the first-order forward imaginary part compares different perturbative orders.

As a factor check, consider the contribution of one distinguishable massless two-particle intermediate channel with constant leading amplitude $\lambda_0$. Its integrated phase space is $1/(8\pi)$, so it contributes

$$\operatorname{Im}\mathcal M_{ii}^{\rm next} =\frac{|\lambda_0|^2}{16\pi}.$$

If the incoming particles are also massless, their flux is $2s$. The corresponding leading partial cross section is $|\lambda_0|^2/(16\pi s)$, and multiplying it by $2s$ reproduces $2\operatorname{Im}\mathcal M_{ii}^{\rm next}$. Other open channels add their own nonnegative contributions. An identical intermediate pair carries the additional $1/2!$ when integrated over ordered full phase space.

What this preview assumes

The relation assumes a unitary scattering description with appropriate asymptotic states and a meaningful forward amplitude and channel sum. It is not a procedure for inserting an arbitrary singular amplitude into an unregulated formula.

In particular, unscreened Coulomb scattering has a divergent total cross section and a singular forward limit. The finite formulas above cannot be tested by substituting $\theta=0$ into the leading Mott expression. Screening, appropriate asymptotic phases, and a controlled limit are needed; Coulomb Scattering owns that qualification.

In theories with massless radiation, unresolved soft emission and the definition of asymptotic charged states also require care. An observable inclusive rate may require summing states that a detector cannot distinguish. This page does not derive that infrared construction or diagrammatic cutting rules. It identifies the unitarity statement those constructions must respect.

Exercises

1. A calculation finds a negative imaginary forward amplitude while all final-state rates are positive. What should be checked first?

Solution

Check the convention for $S-I$, the definition of $\mathcal M$, the complex conjugation in $T^\dagger$, and the forward boundary prescription. With the convention fixed on this page, the right-hand side is nonnegative. A convention change can move a minus sign into the amplitude definition, but not into physical probabilities.

2. Why must a spin average on the left retain diagonal amplitudes for each initial state?

Solution

Unitarity holds for each initial state: $2\operatorname{Im}\mathcal M_{ii}=\sum_n|\mathcal M_{ni}|^2$ with the stated integrals. An incoherent mixture sums these equations with classical weights. Squaring an averaged transition amplitude would introduce interference between differently prepared initial states that is absent in the mixture.

3. Repeat the constant-amplitude two-body contribution for identical massless intermediate particles integrated over the full ordered space.

Solution

The $1/2!$ factor gives $2\operatorname{Im}\mathcal M_{ii}^{\rm next} =|\lambda_0|^2/(16\pi)$, hence $\operatorname{Im}\mathcal M_{ii}^{\rm next} =|\lambda_0|^2/(32\pi)$. The amplitude already includes the appropriate identical-particle exchange structure; the integral factor prevents double counting the same final state.

References

• McGreevy, John. Physics 215B: Particles and Fields. University of California, San Diego, Winter 2019, section 5.2, pp. 136–137. Lecture notes. Covariant optical-theorem normalization.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940. Unitarity and perturbative amplitudes.
• Taylor, John R. Scattering Theory: The Quantum Theory of Nonrelativistic Collisions. Wiley, 1972; Dover reprint, 2006. The potential-scattering optical theorem and its conventions.