Spectral weights in oscillator response

Summary: Weighted Hockey-Stick source strengths have a sharp response-stiffness convergence threshold, and a four-normal-direction affine spectrum retains shared-mode-dependent cutoff growth that a common additive constant cannot remove.

Keep the source map and the response spectrum together

The oscillator source marginal integrates a specified linear scalar source over $q$ ordinary normal spatial directions. At shared oscillator degree $k$ and total cutoff $N$, its paired normal occupations have weights

$$a=\frac q2,\qquad M=\left\lfloor\frac{N-k}{2}\right\rfloor, \qquad w_q(r)=\frac{(a)_r}{r!},\qquad \sum_{r=0}^M w_q(r)=\binom{M+a}{M}. \tag{1}$$

Here $(a)_r=a(a+1)\cdots(a+r-1)$ and $(a)_0=1$. The final equality is the weighted Hockey-Stick identity. It follows by multiplying the shell generating function $(1-t)^{-a}$ by $(1-t)^{-1}$. All spatial coordinates retain their ordinary physical meaning; this flat oscillator comparison introduces no compactification.

Let $R$ be that full complex linear source map and let $L$ be a specified positive response operator on modes of the same field. If its eigenvalue at total degree $k+2r$ is $\ell_{k+2r}>0$, then

$$G_{N,k}=(RL^{-1}R^\dagger)_k =c_0^{2q}\sum_{r=0}^{M}\frac{w_q(r)}{\ell_{k+2r}}, \qquad c_0=\sqrt2\,\pi^{1/4}w^{-1/4}. \tag{2}$$

The oscillator parameter $w$ uses the convention in the source-marginal note. Dividing by $c_0^{2q}$ expresses the sum in ground-integral units; it does not perform physical kinetic normalization.

For example, the real energy

$$E[u,v]=E_0[u]+\frac12\langle v,Lv\rangle -g\operatorname{Re}\langle u,Rv\rangle \tag{3}$$

has stationary response $v=gL^{-1}R^\dagger u$ and correction $-g^2\langle u,Gu\rangle/2$. Both variations use the same complex map and its adjoint, as in common field response. The positivity assumption belongs to this response example; it is not an assertion that the full signed Dirac Hamiltonian is positive.

The exact threshold for removing the mode cutoff

Fix $k$, and suppose at large $r$

$$\ell_{k+2r}\sim\lambda r^s,\qquad \lambda>0. \tag{4}$$

The gamma-ratio asymptotic gives $w_q(r)=\Gamma(r+a)/(\Gamma(a)\Gamma(r+1)) \sim r^{a-1}/\Gamma(a)$. Thus the positive summand in (2) is asymptotic to $r^{a-s-1}/(\lambda\Gamma(a))$. Comparison with a power integral proves

$$\boxed{G_{\infty,k}<\infty\quad\Longleftrightarrow\quad s>q/2.} \tag{5}$$

More precisely, writing $\widetilde G=G/c_0^{2q}$,

$$\begin{aligned} \widetilde G_{N,k}&\sim \frac{M^{a-s}}{\lambda\Gamma(a)(a-s)},&&s<a,\\ \widetilde G_{N,k}&\sim\frac{\log M}{\lambda\Gamma(a)},&&s=a,\\ \widetilde G_{\infty,k}-\widetilde G_{N,k}&\sim \frac{M^{a-s}}{\lambda\Gamma(a)(s-a)},&&s>a. \end{aligned} \tag{6}$$

Finite lower levels affect finite constants, not this classification. The result concerns a fixed source map and its stated spectrum as a regulator is removed. If $N$ instead labels a physically selected finite branch, the finite sum remains valid without taking this limit. Those distinct uses of a band are also separated in consistent band contacts.

An affine spectrum has explicit borderline cases

Take the comparison spectrum

$$\ell_{k+2r}=\kappa+\nu(k+2r)=2\nu(r+b_k), \qquad b_k=\frac{\kappa+\nu k}{2\nu}>0, \qquad \kappa,\nu>0. \tag{7}$$

It has $s=1$. Define $S_M(a,b)=\sum_{r=0}^M(a)_r/[r!(r+b)]$, so $G_{N,k}=c_0^{2q}S_M(a,b_k)/(2\nu)$. For $0<a<1$, positivity permits termwise integration:

$$\begin{aligned} S_\infty(a,b) &=\int_0^1 t^{b-1}\sum_{r=0}^\infty\frac{(a)_r}{r!}t^r\,dt\\ &=\int_0^1t^{b-1}(1-t)^{-a}dt =\frac{\Gamma(b)\Gamma(1-a)}{\Gamma(b+1-a)}. \end{aligned} \tag{8}$$

The last equality is Euler's beta integral. Among positive integer codimensions, this convergent case is $q=1$. For $a\ge1$ the endpoint integral and the positive sum both diverge; analytically continuing the gamma expression does not assign their ordinary positive response a finite value.

Put $H_M(b)=\sum_{r=0}^M(r+b)^{-1} =\psi(M+b+1)-\psi(b)$, where $\psi=\Gamma'/\Gamma$ is the digamma function. The equality follows by telescoping its recurrence. Then

$$\boxed{S_M(1,b)=H_M(b),\qquad S_M(2,b)=M+1+(1-b)H_M(b).} \tag{9}$$

The second formula uses $(2)_r/r!=r+1$ and $(r+1)/(r+b)=1+(1-b)/(r+b)$.

These are scalar-source response sums in the stated units, not particle charges or experimentally measured coupling factors.

Four normal directions retain a shared-mode-dependent divergence

Codimension four is relevant to a ten-to-six flat spatial comparison, although the occupied particle modes and their physical kernel have not been identified with this oscillator example. For fixed shared degrees $k\ne l$, use the same total cutoff $N$ in both responses. Their paired cutoffs differ only by a bounded integer. Equation (9) therefore gives

$$\boxed{G_{N,k}-G_{N,l} =\frac{c_0^8}{2\nu}(b_l-b_k)\log(N/2)+O(1) =\frac{c_0^8(l-k)}{4\nu}\log(N/2)+O(1).} \tag{10}$$

The common linear growth cancels, but the displayed logarithm does not. Subtracting one cutoff-dependent constant times the identity changes neither side's difference, so it cannot make every shared-mode response finite. This is an additive-subtraction statement, not a prohibition of all effective theories. Further direct terms or a different source and spectrum would need to follow from the same action and normalization.

The divergence is not automatically a short-distance singularity. Here the cylindrical source dual is constant in the normal coordinates and is not in their untruncated $L^2$ space. High oscillator levels also explore increasing spatial extent. The mode cutoff does not by itself separate large-distance and short-distance contributions.

Field norm and effective response have different thresholds

For one retained source coefficient $u_k$, the stationary field in (3) has canonical squared norm

$$\|v\|^2=g^2|u_k|^2c_0^{2q} \sum_{r=0}^M\frac{w_q(r)}{\ell_{k+2r}^2}. \tag{11}$$

Under (4), this positive sum converges precisely when $2s>q/2$, or $s>q/4$. Its energy norm instead obeys $\langle v,Lv\rangle=g^2|u_k|^2G_{N,k}$ and needs the stronger condition (5). For an affine spectrum, two or three normal directions therefore permit a finite canonical field norm while the effective response still diverges. At four normal directions even the canonical norm has logarithmic growth. A finite norm alone does not establish a finite source energy or a finite induced coupling.

What additional source structure would have to do

Suppose an actual projection supplies a shell amplitude $f_r$ as well as the marginal. The positive response then contains $w_q(r)|f_r|^2/\ell_{k+2r}$. If $|f_r|^2\sim C r^{-t}$ with $C>0$, the same proof gives the convergence threshold $s+t>q/2$; an upper bound of that order gives sufficiency. This specifies a requirement on a derived source profile, not a new adjustable suppression assigned to a chosen particle depth.

Finally, the linear scalar-source map must not be applied to a wave and then squared in place of its physical current. A bilinear current $\Psi^\dagger\Gamma\Psi$ has its own coherent coefficient contractions before projection. Its normal density marginal can be finite even when the unrestricted linear-source response in (2) diverges. The source-marginal derivation gives that distinction explicitly.

Verification

oscillator_response_tails.py evaluates the finite weighted response and its closed borderline forms. oscillator_response_tail_checks.py runs 105 checks. It independently enumerates normal occupations using exact central-binomial weights, integrates the convergent beta case, checks growing and decaying tail blocks, and compares shared degrees at a common total cutoff. Full complex source matrices also reproduce the reciprocal stationary response, its canonical norm and its Schur energy correction. The all-cutoff conclusions follow from (4)–(10), rather than from a finite numerical scan.