Background-charge oscillator modes and integral long screenings

Introduce the formal field $J(z)={\sum }_m{\alpha }_m{z}^{-m-1}$. Commuting its nonnegative modes past its negative modes gives the contraction $2p{\sum }_{m>0}m{z}^{-m-1}{w}^{m-1}=2p\mathrm{/}(z-w{)}^2$, expanded in nonnegative powers of $w\mathrm{/}z$. Products of two normally ordered quadratic expressions can be computed by moving each annihilation factor right: a single interchange contributes this scalar contraction; two interchanges contribute its product. There are four single and two double choices for $J(z{)}^{2}J(w{)}^2$. Since the commutators are scalar there are no further terms. This proves the finite contraction rule used in the following steps directly from [F1].

Put $T_0(z)=:J(z{)}^2:\mathrm{/}(4p)$ and $\kappa =(p-1)\mathrm{/}(2p)$. The four single contractions in step 1.1 and the Taylor expansion $J(z)=J(w)+(z-w)\mathrm{\partial }J(w)+\cdots$ give $T_0(z)T_0(w)\sim \frac{1\mathrm{/}2}{(z-w{)}^4}+\frac{2T_0(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }{T}_0(w)}{z-w}\mathrm{.}$ Here $\sim$ records precisely the negative powers of $z-w$ in the indicated formal expansion. One contraction similarly gives $T_0(z)J(w)\sim J(w)\mathrm{/}(z-w{)}^2+\mathrm{\partial }J(w)\mathrm{/}(z-w)$ and $J(z)T_0(w)\sim J(w)\mathrm{/}(z-w{)}^2$. Differentiation in the indicated variables therefore gives $T_0(z)\mathrm{\partial }J(w)+\mathrm{\partial }J(z)T_0(w)\sim \frac{2\mathrm{\partial }J(w)}{(z-w{)}^2}+\frac{{\mathrm{\partial }}^{2}J(w)}{z-w},\mathrm{\partial }J(z)\mathrm{\partial }J(w)\sim \frac{-12p}{(z-w{)}^4}\mathrm{.}$ Thus $T=T_0+\kappa \mathrm{\partial }J={\sum }_n{L}_t(n)z^{-n-2}$ satisfies $T(z)T(w)\sim \frac{c\mathrm{/}2}{(z-w{)}^4}+\frac{2T(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }T(w)}{z-w},c=1-6(p-1{)}^2\mathrm{/}p\mathrm{.}$ All calculations take finite coefficients on a fixed polynomial.

Taking the coefficient commutator of $z^{m+1}T(z)$ and $w^{n+1}T(w)$ uses the difference of the expansions in $w\mathrm{/}z$ and $z\mathrm{/}w$. For a pole $(z-w{)}^{-k-1}$ the inner residue is $\left(\genfrac{}{}{0px}{}{m+1}{k}\right)w^{m+1-k}$. The fourth-order pole consequently gives $c(m^3-m){\delta }_{m+n,0}\mathrm{/}12$. The second- and first-order poles give $2(m+1)L_{m+n}-(m+n+2)L_{m+n}=(m-n)L_{m+n}$. This proves the bracket for all integer indices, since generalized binomial coefficients are exactly the coefficients of the corresponding Laurent expansions. Direct normal ordering at index zero gives $L_t(0)={\sum }_{m>0}m{x}_m{\mathrm{\partial }}_{x_m}+p{t}^2-(p-1)t=D_t$; positive modes kill the vacuum.

Write $E(w)={\mathcal{E}}_{1,t}(w)$, as a map between its source and target charge spaces. The exponential substitution in [F1] gives ${\alpha }_{m}E(w)-E(w){\alpha }_m=2p{w}^{m}E(w)$, including $m=0$ with the different source and target charges. Moving nonnegative $J$ modes right therefore gives $J(z)E(w)\sim 2pE(w)\mathrm{/}(z-w)$. A single contraction of $T_0$ gives $\mathrm{\partial }E(w)\mathrm{/}(z-w)$: explicitly the remaining normal product is $:J(w)E(w):$, equal to $\mathrm{\partial }E(w)$ by differentiating its two exponentials and $w^{2pt}$. Two contractions give $pE(w)\mathrm{/}(z-w{)}^2$. The improvement term gives $-2p\kappa E(w)\mathrm{/}(z-w{)}^2$. Since $p-2p\kappa =1$, the result is $T(z)E(w)\sim \frac{E(w)}{(z-w{)}^2}+\frac{\mathrm{\partial }E(w)}{z-w}\mathrm{.}$ Extracting the $z$ coefficient as in step 3.1 gives $L_{t+1}(n)E(w)-E(w)L_t(n)={\mathrm{\partial }}_w(w^{n+1}E(w))$.

Under $2pt\in \mathbb{Z}$, $E(w)$ has integral powers of $w$ and the residue of a formal derivative is zero: the coefficient of $w^{-1}$ can only arise by differentiating the constant term, which contributes zero. Taking this residue in step 4.1 proves the screening identity. The substitution acts finitely on every input polynomial, and fixed creation coefficients have finite degree, so no analytic limit or convergence is required.