S-Matrix and T-Matrix Normalization

With covariantly normalized asymptotic states, $S=1+iT$ separates the identity contribution from the transition operator $T$. Its connected matrix elements define invariant amplitudes after one overall momentum-conserving delta distribution is stripped. The full $T$ also contains disconnected collision clusters. Keeping the operator, its connected part and the stripped amplitude 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

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

The final terms include untouched spectators and products of two or more nontrivial collisions. They carry additional momentum delta functions for the separate clusters. For example, two independent scattering systems have

$$\begin{aligned} S&=S_A\otimes S_B,\\ T&=T_A\otimes1+1\otimes T_B+iT_A\otimes T_B. \end{aligned}$$

The last term describes two nontrivial transitions and has no untouched spectator. The connected subscript removes all these factorizable pieces. Schwartz gives the same $S=1+iT$ and stripped-amplitude convention in Schwartz 2014, § 5.1, pp. 59–61, and discusses products of disconnected collision clusters in Schwartz 2014, § 7.3, p. 96.

Why the delta distribution is not squared

The transition probability contains the modulus squared of a wave-packet matrix element. To identify the normalization behind the usual large-time rate formula, replace the energy-conservation factor by the integral over a finite observation interval:

$$\begin{aligned} I_T(\omega)&\equiv\int_{-T/2}^{T/2}\mathrm dt\,e^{i\omega t} =\frac{2\sin(\omega T/2)}{\omega},\\ I_T(0)&=T,\qquad T=T_{\mathrm{obs}},\quad\omega=E_f-E_i. \end{aligned}$$

At finite $T$, $|I_T(\omega)|^2$ is an ordinary nonnegative function and is generally not $T I_T(\omega)$. The correct statement is the weak limit

$$\int\frac{\mathrm d\omega}{2\pi}\, \frac{|I_T(\omega)|^2}{T}f(\omega) \longrightarrow f(0), \qquad T\to\infty,$$

for a smooth, rapidly decreasing test function $f$. To see this, Fourier norm preservation for the interval indicator gives

$$\int\frac{\mathrm d\omega}{2\pi}\, \frac{|I_T(\omega)|^2}{T}=1.$$

For any fixed $\delta>0$, the weight outside $|\omega|<\delta$ is at most $4/(\pi T\delta)$, since $|I_T(\omega)|\le2/|\omega|$. The weight therefore concentrates at zero, and continuity of $f$ gives the limit. Equivalently, $|I_T|^2/T\to2\pi\delta(\omega)$ as a distribution. This is the energy-kernel step in the large-time rate argument; it is not an exact factorization theorem for arbitrary finite-time interacting dynamics.

Spatial normalization can instead be kept discrete. In a periodic box of volume $V=L^3$, with allowed momenta $\mathbf p=2\pi\mathbf n/L$,

$$W_V(\mathbf p-\mathbf p') \equiv\int_V\mathrm d^3\mathbf x\, e^{i(\mathbf p-\mathbf p')\cdot\mathbf x} =V\delta_{\mathbf n,\mathbf n'}, \qquad |W_V|^2=V W_V.$$

This last identity is exact on the discrete momentum set because a Kronecker delta squares to itself. It supplies the volume factor; the weak time limit supplies $T_{\mathrm{obs}}$. State normalization, incident density and conversion of probability to rate then cancel the corresponding factors. The familiar informal replacement of a squared momentum delta by one delta times $VT_{\mathrm{obs}}$ summarizes these operations under integrals; it does not define a square of a Dirac distribution. Weinberg explicitly presents the box argument as a mnemonic, defines the finite-time integral, and restricts the rate discussion to connected transitions Weinberg 1995, § 3.4, pp. 134–136.

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.” That product is undefined. Square the regulated kernel or packet amplitude first; only then take the weak large-time and continuum limits appropriate to the observable.

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 $S$: $T_{\mathrm{src}}=iT$. For the connected part, if the source writes $\langle f|T_{\mathrm{src}}|i\rangle_c=(2\pi)^4\delta^{(4)}\mathcal A$, then $\mathcal A=i\mathcal M$ in this page’s convention. Disconnected products and spectator overlaps must be separated before making this amplitude identification.

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.