S-Matrix and T-Matrix Normalization

With covariantly normalized asymptotic states, the S-matrix separates into three layers: the identity contribution describing unchanged particles, a connected interaction operator defined by $S=1+iT$, and a translation-invariant amplitude obtained after stripping one overall momentum-conserving delta distribution. Keeping those layers distinct prevents double counting, undefined squares of delta functions, and mismatched rate formulas.

Required background. In and Out States defines the asymptotic bases whose overlap is $S$.

Helpful background. One-Particle States: Mass, Spin, and Relativistic Normalization derives the sharp-state norm and completeness measure used below.

Identity and transition parts of the S-matrix

For an incoming state $|i,\mathrm{in}\rangle$ and an outgoing state $|f,\mathrm{out}\rangle$,

$$S_{fi}=\langle f,\mathrm{out}|i,\mathrm{in}\rangle.$$

Choose

$$S=1+iT.$$

This is a definition of $T$, including the factor of $i$. It does not imply that $T$ is Hermitian. Unitarity later gives $-i(T-T^\dagger)=T^\dagger T$, not $T=T^\dagger$.

For one scalar species,

$$\langle\mathbf p'|\mathbf p\rangle =(2\pi)^3 2E_{\mathbf p}\, \delta^{(3)}(\mathbf p'-\mathbf p).$$

Multiparticle overlaps contain the appropriate sums over identical-particle permutations, with fermionic signs. That overlap is the matrix element of the 1 in $S$. It is not a connected collision amplitude and should not be inserted into a cross-section formula.

For two identical bosons, for example,

$$\begin{aligned} \langle q_1,q_2|p_1,p_2\rangle ={}&(2\pi)^6(2E_{p_1})(2E_{p_2})\big[ \delta^{(3)}(\mathbf q_1-\mathbf p_1) \delta^{(3)}(\mathbf q_2-\mathbf p_2)\\ &+\delta^{(3)}(\mathbf q_1-\mathbf p_2) \delta^{(3)}(\mathbf q_2-\mathbf p_1) \big], \end{aligned}$$

with the energy factors understood on the delta-function support. Fermions carry a minus between the two permutations. These exchange deltas belong to state normalization, whereas a factor $1/2!$ in an integrated identical-particle final state corrects phase-space overcounting; they are not the same operation.

Translation invariance strips one delta function

Let $P_i$ and $P_f$ be the total incoming and outgoing four-momenta. Because $T$ commutes with translations,

$$(P_f-P_i)^\mu\langle f|T|i\rangle=0.$$

As a distribution, a connected matrix element therefore has support on $P_f=P_i$. Our amplitude convention is

$$\boxed{ \langle f|iT|i\rangle_c =i(2\pi)^4\delta^{(4)}(P_f-P_i)\, \mathcal M_{fi} }$$

or equivalently

$$\langle f|T|i\rangle_c =(2\pi)^4\delta^{(4)}(P_f-P_i)\, \mathcal M_{fi}.$$

Thus

$$\langle f|S|i\rangle =\langle f|i\rangle +i(2\pi)^4\delta^{(4)}(P_f-P_i)\mathcal M_{fi} +\text{disconnected spectator terms}.$$

The final terms matter when only a subset of particles interacts; they carry extra delta functions matching the spectators. The connected subscript means that those factorizable pieces have been removed. Schwartz gives the same $S=1+iT$ and stripped-amplitude convention, including its finite-volume interpretation, in Schwartz 2014, § 5.1, pp. 59–61.

Why the delta distribution is not squared

The transition probability contains the modulus squared of a wave-packet matrix element, not the square of a distribution at a point. A box-and-time regulator gives the familiar shorthand

$$\begin{aligned} (2\pi)^3\delta^{(3)}(\mathbf0)&=V,\\ 2\pi\delta(0)&=T_{\mathrm{obs}},\\ \big[(2\pi)^4\delta^{(4)}(P_f-P_i)\big]^2 &=(2\pi)^4\delta^{(4)}(P_f-P_i)\,VT_{\mathrm{obs}}. \end{aligned}$$

The factors $V$ and $T_{\mathrm{obs}}$ cancel against state normalization, incident density, and the conversion from probability to rate. A packet derivation is conceptually cleaner; the regulated identity is only a compact way to recover the same result. The box derivation and its connected-state restriction are detailed in Weinberg 1995, § 3.4, pp. 134–141.

A scalar contact check

Take

$$\mathcal L_{\mathrm{int}}=-\frac{\lambda}{4!}\phi^4.$$

The connected tree-level four-point vertex is $-i\lambda$. Since our definition assigns the connected S-matrix element the factor $i\mathcal M$, comparison gives

$$i\mathcal M=-i\lambda, \qquad \mathcal M=-\lambda.$$

The overall sign disappears from $|\mathcal M|^2$ for this isolated diagram but remains important in interference. This one-line check fixes the relation between a Feynman-rule result called “$i\mathcal M$” and the $\mathcal M$ used in rates.

Dimensional and normalization checks

In four dimensions, the covariant sharp ket has mass dimension $-1$, because its norm has dimension $-2$. For a connected amplitude with $N$ external particles,

$$[\mathcal M_N]=4-N.$$

For $2\to2$, $\mathcal M$ is dimensionless. Combined with a flux of dimension $2$ and dimensionless two-body phase space, it produces a cross section of dimension $-2$, as required.

If instead sharp states are normalized as $\delta^{(3)}(\mathbf p'-\mathbf p)$, each external ket differs by $[(2\pi)^3 2E_{\mathbf p}]^{-1/2}$. The displayed amplitude formula then acquires corresponding external factors. One may use either convention; one may not take the amplitude from one and the phase-space formula from the other.

Common pitfalls

“The identity term is the zero-coupling limit of a connected amplitude.” It is a separate distributional overlap, with its own permutation delta functions. Connected scattering begins with $iT$.

“$T$ is Hermitian because $S$ is unitary.” With $S=1+iT$, unitarity relates the anti-Hermitian part of $T$ to $T^\dagger T$. The unitarity chapter develops the full nonlinear identity.

“Momentum conservation is part of $\mathcal M$.” In this convention $\mathcal M$ is stripped of exactly one overall $\delta^{(4)}(P_f-P_i)$. Spectator deltas signal disconnected pieces and must be classified separately.

“$\delta^{(4)}(P)^2$ is an ordinary function.” It is not defined pointwise. Packets or a controlled box-and-time limit convert it into one delta distribution times the observation volume and time.

Check your understanding

1. A source defines $S=1+T_{\mathrm{src}}$ with no factor of $i$. What is the map to this page’s convention?

   Answer

   Match the same operator $S$: $T_{\mathrm{src}}=iT$. If the source writes $\langle f|T_{\mathrm{src}}|i\rangle=(2\pi)^4\delta^{(4)}\mathcal A$, then $\mathcal A=i\mathcal M$ in this page’s convention.

2. Why does a disconnected process with one untouched spectator carry more than one delta function?

   Answer

   The spectator overlap contributes its own on-shell momentum delta distribution, while the interacting sub-process contributes a conservation delta for its participants. A connected amplitude has only the single overall conservation delta.

3. In the scalar contact example, what changes if one quotes the Feynman-rule result itself as the “amplitude”?

   Answer

   The vertex is $-i\lambda=i\mathcal M$, so a convention calling the vertex factor the amplitude is quoting $i\mathcal M$, not this chapter’s stripped $\mathcal M=-\lambda$. Rate formulas use $|\mathcal M|^2$ in the stated normalization; interference phases require translating the factor of $i$ consistently across all diagrams.

Continue

Relativistic Scattering Kinematics encodes the support of the overall delta function in invariant variables. LSZ Reduction explains how $\mathcal M$ is extracted from correlator poles. For the nonlinear consequence of $S^\dagger S=1$, continue to S-Matrix Unitarity.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014. DOI.
• Weinberg, Steven. The Quantum Theory of Fields. Volume I: Foundations. Cambridge: Cambridge University Press, 1995. DOI.