Lemma: Fixed-sector Wick estimates

For a fixed Wick field $B$, input particle number $n$, and integer $s\ge 0$, there exist an integer $r$, a continuous Schwartz seminorm $p$, and a constant $C$ such that

Only finitely many output particle sectors occur.

Proof. Expand the normal product into its finitely many creation–annihilation terms. On a participating set of momenta, each kernel has the form

where $P$ is a fixed polynomial produced by the derivatives. The nonparticipating particles contribute identity kernels.

For the participating variables put

They are nonnegative, and

An output energy weight can therefore be bounded by a polynomial in the input energy and the temporal Fourier variable. Schwartz decay absorbs the latter polynomial.

More explicitly, after inserting a sufficiently large input weight $(1+Q_0{)}^{-r}$, the squared participating kernel is integrable: the radial mass-shell measure of each variable is a constant multiple of $u\mathrm{d}u\mathrm{d}\omega$, the derivative factors have polynomial growth, and a sufficiently high power of $(1+|P_0-Q_0|{)}^{-1}$, together with $(1+Q_0{)}^{-r}$, makes the radial integrals finite. There is no infrared divergence, because $u\mathrm{d}u$ is locally integrable at zero. This gives a Hilbert–Schmidt bound for the weighted participating kernel. If one side has no participating variables, the same calculation uses the rapid decay in the sum of the positive energies.

The finitely many Fock combinatorial factors depend on $n$ and the degree of $B$, and are absorbed in $C$. The spectator energy weight is transferred to the input by the same inequality. Summing the terms proves (eq:wickest).