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我们现在考虑分析下面的问题，即\begin{equation} \boxed{ \mathcal Q...

Creato il: 21 dicembre 2025

Risposto usando GPT-5.2 Thinking da Chat01

Domanda

我们现在考虑分析下面的问题，即\begin{equation}
\boxed{
\mathcal Q_k=
\frac{1}{N^2}!!\sum_{(p,q,p',q')\notin\mathcal P}!!
\big(\mathbf f_{\modN{k+p-q}}^H\big),
\big(\mathbf f_{\modN{-k-p'+q'}}^H\big),
\exp!\left(-\alpha,\mathbf b^T\mathbf K_2(p,q,p',q')\mathbf b\right).
}
\end{equation}

\begin{equation}\label{eq:Sk_4idx_DFT}
\boxed{
\mathcal Q_k=\mathcal Q_k^{SP}+\mathcal Q_k^{bulk}
}
\end{equation}
\paragraph{Structured hyperplanes (single-pair layers) and matrix forms.}
We partition the non-pairing set into structured hyperplanes and a residual bulk:

(H1) p′=q′,(H2) p=q,(H3) q=q′,(H4) p=p′,bulk: complement.\text{(H1)}\ p'=q',\qquad \text{(H2)}\ p=q,\qquad \text{(H3)}\ q=q',\qquad \text{(H4)}\ p=p',\qquad \text{bulk: complement}.

On each hyperplane, bTK2(p,q,p′,q′)b\mathbf b^T\mathbf K_2(p,q,p',q')\mathbf b reduces (by shift-invariance of Ψ(⋅)\Psi(\cdot)) to a function of a single modular difference d∈\ZNd\in\ZN. We encode this \emph{using K2K_2 itself} via the 1D weight
\begin{equation}\label{eq:kappa_def}
\kappa[d]\triangleq \exp!\left(-\alpha,\mathbf b^T\mathbf K_2(d,0,0,0)\mathbf b\right),\qquad d\in\ZN ,
\end{equation}
which is merely a shorthand (no new model) since it is generated by the original four-argument kernel.

% ===== kappa[d] in closed form and vectorization =====

% Parameters

ω≜2πN,α≜h~2N2,b≜[ 1,−1,−1,1 ]T.\omega \triangleq \frac{2\pi}{N},\qquad \alpha\triangleq \frac{\tilde h^2}{N^2},\qquad \mathbf b \triangleq [\,1,-1,-1,1\,]^T.

% Psi(d)

Ψ(d)≜∑r=1Ndcos⁡(ωrd),Ψ(d)=Ψ(−d).\Psi(d)\triangleq \sum_{r=1}^{N_d}\cos(\omega r d),\qquad \Psi(d)=\Psi(-d).

% Key simplification for the special representative (d,0,0,0)

bTK2(d,0,0,0)b=2(Ψ(0)−Ψ(d))=2(Nd−Ψ(d)).\boxed{ \mathbf b^T \mathbf K_2(d,0,0,0)\mathbf b =2\big(\Psi(0)-\Psi(d)\big) =2\big(N_d-\Psi(d)\big). }

% kappa[d] explicit (sum form)

κ[d]≜exp⁡ ⁣(−α bTK2(d,0,0,0)b)=exp⁡ ⁣(−2α(Nd−Ψ(d)))=exp⁡ ⁣(−2h~2N2(Nd−∑r=1Ndcos⁡(ωrd))).\boxed{ \kappa[d]\triangleq \exp\!\left(-\alpha\,\mathbf b^T\mathbf K_2(d,0,0,0)\mathbf b\right) =\exp\!\left(-2\alpha\big(N_d-\Psi(d)\big)\right) =\exp\!\left( -\frac{2\tilde h^2}{N^2}\Big(N_d-\sum_{r=1}^{N_d}\cos(\omega r d)\Big) \right). }

% ===== kappa as a length-N vector =====

κ≜[κ[0],κ[1],…,κ[N−1]]T∈RN,(κ)d=κ[d],  d∈ZN.\boldsymbol\kappa \triangleq \big[\kappa[0],\kappa[1],\ldots,\kappa[N-1]\big]^T \in \mathbb R^{N}, \qquad (\boldsymbol\kappa)_d = \kappa[d],\ \ d\in\mathbb Z_N.

% (Optional) elementwise definition

(κ)d=exp⁡ ⁣(−2α(Nd−Ψ(d))),d=0,1,…,N−1.\boxed{ (\boldsymbol\kappa)_d =\exp\!\left(-2\alpha\big(N_d-\Psi(d)\big)\right), \qquad d=0,1,\ldots,N-1. }

\emph{Low-rank layers:} the hyperplanes (H1) and (H2) lead to two (Hermitianized) rank-≤2\le2 quadratic forms

Qk,ut(0)≜1N∑d=1N−1κ[d] f−k∗fk+dH,Qk,ut≜12(Qk,ut(0)+Qk,ut(0)H),\mathbf Q_{k,ut}^{(0)}\triangleq \frac{1}{N}\sum_{d=1}^{N-1}\kappa[d]\ \mathbf f_{-k}^{*}\mathbf f_{k+d}^{H}, \qquad \mathbf Q_{k,ut}\triangleq \tfrac12\big(\mathbf Q_{k,ut}^{(0)}+\mathbf Q_{k,ut}^{(0)H}\big), Qk,v0(0)≜1N∑d=1N−1κ[d] fk∗fd−kH,Qk,v0≜12(Qk,v0(0)+Qk,v0(0)H).\mathbf Q_{k,v0}^{(0)}\triangleq \frac{1}{N}\sum_{d=1}^{N-1}\kappa[d]\ \mathbf f_{k}^{*}\mathbf f_{d-k}^{H}, \qquad \mathbf Q_{k,v0}\triangleq \tfrac12\big(\mathbf Q_{k,v0}^{(0)}+\mathbf Q_{k,v0}^{(0)H}\big).

% ===== Normalized DFT convention =====

ω≜2πN,[F]n,d=1Ne−jω(n−1)d,n=1,…,N, d=0,…,N−1,\omega \triangleq \frac{2\pi}{N},\qquad [\mathbf F]_{n,d}=\frac{1}{\sqrt N}e^{-j\omega (n-1)d}, \quad n=1,\ldots,N,\ d=0,\ldots,N-1,

and let fd\mathbf f_d denote the dd-th column of F\mathbf F (indices are taken mod NN).

% ===== Pack kappa into a length-N vector and take one FFT =====

κ~[0]≜0,κ~[d]≜κ[d], d=1,…,N−1,κ~≜[κ~[0],…,κ~[N−1]]T∈RN,\tilde\kappa[0]\triangleq 0,\qquad \tilde\kappa[d]\triangleq \kappa[d],\ d=1,\ldots,N-1, \qquad \tilde{\boldsymbol\kappa}\triangleq[\tilde\kappa[0],\ldots,\tilde\kappa[N-1]]^T\in\mathbb R^N, s≜F κ~=∑d=1N−1κ[d]fd∈CN.\mathbf s \triangleq \mathbf F\,\tilde{\boldsymbol\kappa} =\sum_{d=1}^{N-1}\kappa[d]\mathbf f_d\in\mathbb C^N.

% ===== Key identity: elementwise product of DFT columns =====
For normalized DFT columns, the elementwise products satisfy

fk∗⊙fd=1Nfd−k,fk⊙fd=1Nfd+k,( mod N ).\mathbf f_k^*\odot \mathbf f_d=\frac{1}{\sqrt N}\mathbf f_{d-k}, \qquad \mathbf f_k\odot \mathbf f_d=\frac{1}{\sqrt N}\mathbf f_{d+k}, \qquad (\,\mathrm{mod}\ N\,).

Hence,

∑d=1N−1κ[d]fd+k=N (f−k⊙s),∑d=1N−1κ[d]fd−k=N (fk⊙s).\sum_{d=1}^{N-1}\kappa[d]\mathbf f_{d+k} =\sqrt N\,(\mathbf f_{-k}\odot \mathbf s), \qquad \sum_{d=1}^{N-1}\kappa[d]\mathbf f_{d-k} =\sqrt N\,(\mathbf f_{k}\odot \mathbf s).

% ===== Compact forms for Q_{k,ut}^{(0)} and Q_{k,v0}^{(0)} without D_k =====
Using f−k∗=fk\mathbf f_{-k}^*=\mathbf f_k and fk∗=f−k\mathbf f_k^*=\mathbf f_{-k},

Qk,ut(0)=1N fk(∑d=1N−1κ[d]fd+k)H=NN fk (f−k⊙s)H,\boxed{ \mathbf Q_{k,ut}^{(0)} =\frac{1}{N}\,\mathbf f_k\Big(\sum_{d=1}^{N-1}\kappa[d]\mathbf f_{d+k}\Big)^H =\frac{\sqrt N}{N}\,\mathbf f_k\,(\mathbf f_{-k}\odot \mathbf s)^H, } Qk,v0(0)=1N f−k(∑d=1N−1κ[d]fd−k)H=NN f−k (fk⊙s)H,s=Fκ~.\boxed{ \mathbf Q_{k,v0}^{(0)} =\frac{1}{N}\,\mathbf f_{-k}\Big(\sum_{d=1}^{N-1}\kappa[d]\mathbf f_{d-k}\Big)^H =\frac{\sqrt N}{N}\,\mathbf f_{-k}\,(\mathbf f_{k}\odot \mathbf s)^H, } \qquad \mathbf s=\mathbf F\tilde{\boldsymbol\kappa}.

% ===== Hermitian symmetrization =====

Qk,ut≜12(Qk,ut(0)+Qk,ut(0)H),Qk,v0≜12(Qk,v0(0)+Qk,v0(0)H).\mathbf Q_{k,ut}\triangleq \tfrac12\big(\mathbf Q_{k,ut}^{(0)}+\mathbf Q_{k,ut}^{(0)H}\big), \qquad \mathbf Q_{k,v0}\triangleq \tfrac12\big(\mathbf Q_{k,v0}^{(0)}+\mathbf Q_{k,v0}^{(0)H}\big).

% ===== Final low-rank sum (no D_k) =====

Qk,ut+Qk,v0=N2N(fk(f−k⊙s)H+(f−k⊙s)fkH+f−k(fk⊙s)H+(fk⊙s)f−kH),s=Fκ~.\boxed{ \mathbf Q_{k,ut}+\mathbf Q_{k,v0} = \frac{\sqrt N}{2N}\Big( \mathbf f_k(\mathbf f_{-k}\odot \mathbf s)^H + (\mathbf f_{-k}\odot \mathbf s)\mathbf f_k^H + \mathbf f_{-k}(\mathbf f_{k}\odot \mathbf s)^H + (\mathbf f_{k}\odot \mathbf s)\mathbf f_{-k}^H \Big), \quad \mathbf s=\mathbf F\tilde{\boldsymbol\kappa}. }

\emph{Diagonal layer:} the hyperplanes (H3) and (H4) collapse (via the DFT shift/modulation property) to a diagonal matrix

Dd≜diag ⁣(e−jωd(n−1))n=1N,Qdiag≜1N2∑d=1N−1κ[d](Dd+D−d).\mathbf D_d \triangleq \mathrm{diag}\!\big(e^{-j\omega d(n-1)}\big)_{n=1}^N,\qquad \mathbf Q_{\mathrm{diag}}\triangleq \frac{1}{N^2}\sum_{d=1}^{N-1}\kappa[d]\big(\mathbf D_d+\mathbf D_{-d}\big).

% ===== Key idea: represent the diagonal operator by its diagonal vector =====
Since Qdiag\mathbf Q_{\mathrm{diag}} is diagonal, define its diagonal vector

qdiag≜Qdiag1∈RN,\mathbf q_{\mathrm{diag}}\triangleq \mathbf Q_{\mathrm{diag}}\mathbf 1 \in \mathbb R^N,

% ===== Compute q_diag explicitly =====
For each n=1,…,Nn=1,\ldots,N,
\begin{align*}
(\mathbf q_{\mathrm{diag}})n
&=\frac{1}{N^2}\sum{d=1}^{N-1}\kappa[d]\Big(e^{-j\omega d(n-1)}+e^{+j\omega d(n-1)}\Big)\
&=\frac{2}{N^2}\sum_{d=1}^{N-1}\kappa[d]\cos!\big(\omega d(n-1)\big).
\end{align*}

% ===== Pack kappa into a length-N vector and express via DFT columns =====
Define the length-NN vector κ~∈RN\tilde{\boldsymbol\kappa}\in\mathbb R^N by

κ~[0]≜0,κ~[d]≜κ[d], d=1,…,N−1,κ~=[κ~[0],…,κ~[N−1]]T.\tilde\kappa[0]\triangleq 0,\qquad \tilde\kappa[d]\triangleq \kappa[d],\ d=1,\ldots,N-1, \qquad \tilde{\boldsymbol\kappa}=[\tilde\kappa[0],\ldots,\tilde\kappa[N-1]]^T.

Then

s≜Fκ~=∑d=1N−1κ[d]fd,\mathbf s \triangleq \mathbf F\tilde{\boldsymbol\kappa} = \sum_{d=1}^{N-1}\kappa[d]\mathbf f_d,

and

(qdiag)n=1N2(N sn+N sn∗)=2N3/2ℜ{sn}.(\mathbf q_{\mathrm{diag}})_n =\frac{1}{N^2}\Big(\sqrt N\,s_n+\sqrt N\,s_n^*\Big) =\frac{2}{N^{3/2}}\Re\{s_n\}.

Therefore,

qdiag=2N3/2ℜ{Fκ~}=2N3/2ℜ{∑d=1N−1κ[d]fd},\boxed{ \mathbf q_{\mathrm{diag}} =\frac{2}{N^{3/2}}\Re\{\mathbf F\tilde{\boldsymbol\kappa}\} =\frac{2}{N^{3/2}}\Re\left\{\sum_{d=1}^{N-1}\kappa[d]\mathbf f_d\right\}, }

and the diagonal operator can be applied without any diag(⋅)\mathrm{diag}(\cdot) as

Qdiag=(2N3/2Fκ~).\mathbf Q_{\mathrm{diag}}= \left(\frac{2}{N^{3/2}}\mathbf F\tilde{\boldsymbol\kappa}\right).

Thus the structured part is
\begin{equation}\label{eq:Sk_sp_quadratic}
\boxed{
\mathcal S_k^{\mathrm{sp}}(\mathbf g)=\mathbf g^H\mathbf Q_k^{\mathrm{sp}}\mathbf g,\qquad
\mathbf Q_k^{\mathrm{sp}}=\mathbf Q_{\mathrm{diag}}+\mathbf Q_{k,ut}+\mathbf Q_{k,v0}.
}
\end{equation}

% ===== Normalized DFT =====

ω≜2πN,[F]n,d=1Ne−jω(n−1)d,n=1,…,N, d=0,…,N−1,\omega \triangleq \frac{2\pi}{N},\qquad [\mathbf F]_{n,d}=\frac{1}{\sqrt N}e^{-j\omega (n-1)d}, \quad n=1,\ldots,N,\ d=0,\ldots,N-1,

and let fd\mathbf f_d be the dd-th column of F\mathbf F (indices mod NN).

% ===== kappa vector and one FFT =====

κ~[0]≜0,κ~[d]≜κ[d], d=1,…,N−1,κ~≜[κ~[0],…,κ~[N−1]]T,\tilde\kappa[0]\triangleq 0,\qquad \tilde\kappa[d]\triangleq \kappa[d],\ d=1,\ldots,N-1, \qquad \tilde{\boldsymbol\kappa}\triangleq[\tilde\kappa[0],\ldots,\tilde\kappa[N-1]]^T, s≜Fκ~.\mathbf s\triangleq \mathbf F\tilde{\boldsymbol\kappa}.

% ===== s-factor matrix shared by all terms =====

S(s)≜1 sH+s 1T,so [S(s)]n,n=sn∗+sn=2ℜ{sn}.\mathbf S(\mathbf s)\triangleq \mathbf 1\,\mathbf s^H+\mathbf s\,\mathbf 1^T, \qquad \text{so } [\mathbf S(\mathbf s)]_{n,n}=s_n^*+s_n=2\Re\{s_n\}.

% ===== Q_diag without diag(·): diagonal picked by Hadamard with I =====

Qdiag=2N3/2diag⁡(ℜ{s})=1N3/2(I⊙S(s)).\boxed{ \mathbf Q_{\mathrm{diag}} =\frac{2}{N^{3/2}}\operatorname{diag}(\Re\{\mathbf s\}) =\frac{1}{N^{3/2}}\big(\mathbf I\odot \mathbf S(\mathbf s)\big). }

% ===== Low-rank sum expressed with the SAME S(s) =====

Ak≜fkf−kH+f−kfkH,\mathbf A_k\triangleq \mathbf f_k\mathbf f_{-k}^H+\mathbf f_{-k}\mathbf f_k^H, Qk,ut+Qk,v0=N2N(Ak⊙S(s)).\boxed{ \mathbf Q_{k,ut}+\mathbf Q_{k,v0} =\frac{\sqrt N}{2N}\big(\mathbf A_k\odot \mathbf S(\mathbf s)\big). }

% Assume normalized DFT F and its columns f_k.
% Let z = F \tilde{kappa}. To make the Hadamard product well-defined, lift z to an NxN matrix:

z≜Fκ~,S~≜1 zH+z 1T.\mathbf z \triangleq \mathbf F\tilde{\boldsymbol\kappa},\qquad \tilde{\mathbf S}\triangleq \mathbf 1\,\mathbf z^H+\mathbf z\,\mathbf 1^T .

Note that diag(S~)=2ℜ{z}=2ℜ{Fκ~}\mathrm{diag}(\tilde{\mathbf S})=2\Re\{\mathbf z\}=2\Re\{\mathbf F\tilde{\boldsymbol\kappa}\}.

Define

Ak≜fkf−kH+f−kfkH,Qksp=(1N3/2I+N2NAk)⊙S~.\mathbf A_k\triangleq \mathbf f_k\mathbf f_{-k}^H+\mathbf f_{-k}\mathbf f_k^H, \qquad \mathbf Q_k^{\mathrm{sp}} =\Big(\tfrac{1}{N^{3/2}}\mathbf I+\tfrac{\sqrt N}{2N}\mathbf A_k\Big)\odot \tilde{\mathbf S}.

If Qksp\mathbf Q_k^{\mathrm{sp}} is Hermitian PSD, then

Sksp(g)=gHQkspg=∥(Qksp)1/2g∥22=∥((1N3/2I+N2NAk)⊙S~)1/2g∥22.\boxed{ \mathcal S_k^{\mathrm{sp}}(\mathbf g) =\mathbf g^H\mathbf Q_k^{\mathrm{sp}}\mathbf g =\big\|(\mathbf Q_k^{\mathrm{sp}})^{1/2}\mathbf g\big\|_2^2 = \left\| \left( \Big(\tfrac{1}{N^{3/2}}\mathbf I+\tfrac{\sqrt N}{2N}\mathbf A_k\Big)\odot \tilde{\mathbf S} \right)^{1/2}\mathbf g \right\|_2^2 . }

我们现在在上面的推导思路下，我们要推导Q_bulk的解析解，用DFT矩阵的形式和索引矩阵的方法来表达

Pensiero

Deriving Q_bulk in DFT matrix form

User wants help deriving the Q_bulk expression in DFT matrix form, starting with the structured and bulk parts. It looks like bulk involves a sum over non-pairing tuples and hyperplanes, which needs to be represented in matrix form. I'll explore Kronecker products or convolutions and DFT vector manipulation to express this.

Interpreting Q_k and matrix product

Q_k is intended for a quadratic form: g^H Q_k g. I’m thinking that in the original sum, the f vectors likely form rank-1 matrices via an outer product with conjugates. Q_k might involve row vectors, but the multiplication of two row vectors doesn't fit. Maybe the sum involves an entrywise product or Hadamard product?

Analyzing Q_k bulk part and kernel K2

Maybe they left out transposes: f_{...} f_{...}^H makes sense. In the low-rank case, f_k (sum ...)^H was used. Q_k is an NxN matrix that results in a scalar after applying to g. For the bulk part, I'll relate it to the remaining tuples, potentially involving a 2D convolution with kernel K2 and modular differences.

Reconstructing K2 matrix and analyzing variance

I know that for the representative (d,0,0,0), b^T K2 b = 2(N_d - Psi(d)). For a general four-index case, K2 might involve index differences with covariance Psi(p-q). It seems like b^T K2 b corresponds to the variance of a linear combination of x_p, x_q, and so on. Let's check that.

Finalizing Q_bulk in matrix form

The expression for b^T K b checks out and matches. The weight exp(-α b^T K b) depends on modular differences, and due to shift invariance, we only need three independent differences. For Q_bulk, I need to express it with DFT and index matrices, potentially involving Hadamard products or outer products between DFT columns.

Deriving Q_bulk matrix form

Alright, the Q_bulk matrix needs to match the structure of previous terms. We define it as a sum over the bulk indices, weighted by an exponential function. I think the DFT shift properties help with defining it correctly. The expression uses modulation matrices and permutation matrices to get those frequency shifts.

Reparameterizing indices for bulk weight

So, to handle the bulk weight, I’ve set up a reparameterization of the indices. By shifting the base to t=q, I’m changing the differences (e.g., a, b, c) to depend on shifts rather than specific indices. This simplifies things, especially with shift invariance. Now, there's a consistent dependence on just differences.

Expressing Q_bulk in terms of DFT matrices

The bulk weight term depends on indices a, b, and c, with a focus on differences like b-c. By reparameterizing, I aim to express Q_bulk in DFT form, factoring out certain components. Summing across indices with the correct constraints, we multiply by a normalization factor N.

Representing Q_bulk in the frequency domain

I can express the sum over a and e=b-c by creating a coefficient matrix C. This reduces the 3D sum to a 2D matrix. I then represent Q_bulk as a sum of modified DFT matrices, leading to a compact expression. It simplifies to a circulant operator model in the Fourier domain, with clear coefficient structure.

Deriving Q_bulk with DFT matrices

I need to express Q_bulk by shifting the frequency indices and using a coefficient matrix, C. After transforming into the Fourier domain, terms may correspond to sparse matrix entries. The coefficient matrix can be updated via 3D sums, and Q_bulk eventually relates to the Kronecker product for efficient computation. Let's finalize the analytic expression based on this approach.

Defining Q_bulk with DFT matrix and constraints

The expression for Q_bulk can be parameterized with differences like a, d, and t. These represent shifts within pairs, simplifying the dependence on DFT indices. Constraints exclude certain pairings to avoid overlaps and simplify the coefficient matrix. The final expression uses a 2D matrix, reducing the sum over four indices to three, allowing for efficient computation in Terms of DFT matrices.

Examining scaling and coefficient expression for Q_bulk

I’m working through scaling for Q_bulk, considering it as a sum of coefficients and outer products. I’ll define the mapping from (a,d) to (u,v) using permutation matrices. This lets me simplify expressions for the coefficients, ultimately giving a DFT-based form. Also trying various exponentials that factor into Bessel sums—still figuring out the optimal way to express coefficients efficiently!

Expressing Q_bulk with DFT and shift matrices

I’m defining Q_bulk using index permutation and shift matrices. Incorporating a Hermitian structure, I can represent it through a combination of DFT techniques and selections, along with permutation matrices. Then, express the coefficient matrix CkC_k using frequency domain operations. I’m working toward deriving a formula involving Bessel functions and Kronecker products for more efficiency. Let's refine step-by-step!

Thought for 3m 34s

Risposta

下面给出在你现有“结构超平面 + 余项 bulk”的推导框架下，把 QbulkQ_{\text{bulk}} 写成 “DFT矩阵形式 + 索引(置换)矩阵形式” 的一套干净表达。核心思路是：对 bulk 区域做移不变重参数化，把四重和化为“三个模差”的求和；再把 kk 的作用抽离成 频域索引置换。

1) 先把四指标核的指数写成“模差”形式

在你前面同样的平稳/移不变假设下（即 Ψ\Psi 只依赖差分），对 b=[1,−1,−1,1]T\mathbf b=[1,-1,-1,1]^T 与四点协方差核 K2(p,q,p′,q′)\mathbf K_2(p,q,p',q')，可写成差分组合（与之前 (d,0,0,0)(d,0,0,0) 一致）：

bTK2(p,q,p′,q′)b=4Ψ(0)−2Ψ(p−q)−2Ψ(p−p′)+2Ψ(p−q′)+2Ψ(p′−q)−2Ψ(q−q′)−2Ψ(p′−q′)\boxed{ \mathbf b^T\mathbf K_2(p,q,p',q')\mathbf b = 4\Psi(0) -2\Psi(p-q)-2\Psi(p-p')+2\Psi(p-q') +2\Psi(p'-q)-2\Psi(q-q')-2\Psi(p'-q') }

其中 Ψ(0)=Nd\Psi(0)=N_d，Ψ(d)=∑r=1Ndcos⁡(ωrd)\Psi(d)=\sum_{r=1}^{N_d}\cos(\omega r d)。

2) bulk 的关键重参数化：(q,a,t,d)(q,a,t,d)

为了让 DFT 列索引只落在**两条“对内差分”**上，推荐用下面三差分参数：

a≜p−q,t≜q′−q,d≜p′−q′(均在 ZN)a \triangleq p-q,\qquad t \triangleq q'-q,\qquad d \triangleq p'-q' \qquad(\text{均在 }\mathbb Z_N)

则

p=q+a,q′=q+t,p′=q+t+d.p=q+a,\quad q'=q+t,\quad p'=q+t+d .

2.1 DFT 列索引的简化（这是最关键的一步）

k+p−q=k+a,−k−p′+q′=−k−(q+t+d)+(q+t)=−k−d.k+p-q = k+a, \qquad -k-p'+q' = -k-(q+t+d)+(q+t)= -k-d.

因此 bulk 中每一项的“两个频率列”只由 (a,d)(a,d) 决定，与 q,tq,t 无关。

2.2 指数权重只依赖 (a,t,d)(a,t,d)

把上面的差分代回第1节的公式，可得

ϕ(a,t,d)≜bTK2(p,q,p′,q′)b=4Ψ(0)−2Ψ(a)−2Ψ(a−t−d)+2Ψ(a−t)+2Ψ(t+d)−2Ψ(t)−2Ψ(d)\boxed{ \phi(a,t,d) \triangleq \mathbf b^T\mathbf K_2(p,q,p',q')\mathbf b = 4\Psi(0) -2\Psi(a) -2\Psi(a-t-d) +2\Psi(a-t) +2\Psi(t+d) -2\Psi(t) -2\Psi(d) }

于是权重

w(a,t,d)=exp⁡ ⁣(−α ϕ(a,t,d)).w(a,t,d)=\exp\!\big(-\alpha\,\phi(a,t,d)\big).

3) bulk 的集合约束用“模δ”写成一个掩码

定义模δ函数

δN(x)={1,x≡0 (mod N)0,else\delta_N(x)=\begin{cases} 1,& x\equiv 0\ (\mathrm{mod}\ N)\\ 0,&\text{else} \end{cases}

你定义的四个超平面在 (a,t,d)(a,t,d) 下变成：

  • (H2) p=q  ⟺  a=0p=q \iff a=0
  • (H1) p′=q′  ⟺  d=0p'=q' \iff d=0
  • (H3) q=q′  ⟺  t=0q=q' \iff t=0
  • (H4) p=p′  ⟺  q+a=q+t+d  ⟺  t=a−dp=p' \iff q+a=q+t+d \iff t=a-d

另外，“pairing set P\mathcal P” 里常见的交换配对 (p,q)=(q′,p′)(p,q)=(q',p') 在这里等价于

p=q′  ⟺  a=t,q=p′  ⟺  t+d=0,p=q' \iff a=t,\qquad q=p'\iff t+d=0,

即 (t=a, d=−a)(t=a,\ d=-a)（如果你的 P\mathcal P 定义包含这类配对，需要一并剔除）。

因此可以用一个 bulk 掩码写成（给出一个“包含你列出的四超平面 + 常见swap pairing”的版本）：

χbulk(a,t,d)=(1−δN(a)) (1−δN(d)) (1−δN(t)) (1−δN(t−a+d)) (1−δN(t−a) δN(d+a))\boxed{ \chi_{\text{bulk}}(a,t,d) = (1-\delta_N(a))\, (1-\delta_N(d))\, (1-\delta_N(t))\, (1-\delta_N(t-a+d))\, \Big(1-\delta_N(t-a)\,\delta_N(d+a)\Big) }

若你的 P\mathcal P 还有别的配对类型，就在 χbulk\chi_{\text{bulk}} 里继续乘上对应的 (1−⋯ )(1-\cdots) 因子即可；后面的矩阵形式完全不变。

4) 把四重和化成“(a,d)(a,d) 的系数矩阵 + DFT外积基”

先把原式中对 qq 的求和用移不变性消掉：对每个固定 (a,t,d)(a,t,d)，qq 有 NN 个取值且权重/DFT索引都不变，所以四重和变成一个 NN 的因子，从而：

Qkbulk=1N∑a∈ZN∑d∈ZNΓbulk[a,d]  fk+a∗ f−k−dH\boxed{ \mathbf Q_k^{\text{bulk}} = \frac{1}{N} \sum_{a\in\mathbb Z_N} \sum_{d\in\mathbb Z_N} \Gamma_{\text{bulk}}[a,d]\; \mathbf f_{k+a}^{*}\,\mathbf f_{-k-d}^{H} }

其中 bulk 的二维系数矩阵（独立于 kk）定义为

Γbulk[a,d]≜1N∑t∈ZNχbulk(a,t,d)  exp⁡ ⁣(−α ϕ(a,t,d))\boxed{ \Gamma_{\text{bulk}}[a,d] \triangleq \frac{1}{N} \sum_{t\in\mathbb Z_N} \chi_{\text{bulk}}(a,t,d)\; \exp\!\big(-\alpha\,\phi(a,t,d)\big) }

这已经是一个“解析表达”：bulk 的复杂性全部被压缩进 Γbulk\Gamma_{\text{bulk}} 这个 N×NN\times N 的系数矩阵（由一个长度 NN 的 tt-和给出）。

5) 用“索引置换矩阵”把 kk 抽离出来（频域矩阵形式）

令 F\mathbf F 为你定义的归一化 DFT 矩阵，fℓ\mathbf f_\ell 为其第 ℓ\ell 列，且 fℓ=Feℓ\mathbf f_\ell=\mathbf F\mathbf e_\ell。

再定义两个索引矩阵：

  • 循环移位置换 Πs\mathbf\Pi_s：Πsen=emod(n+s)\mathbf\Pi_s\mathbf e_n=\mathbf e_{\mathrm{mod}(n+s)}
  • 取负号置换 J\mathbf J：Jen=e−n\mathbf J\mathbf e_n=\mathbf e_{-n}

并记

Pk≜Π−kJ\boxed{ \mathbf P_k \triangleq \mathbf \Pi_{-k}\mathbf J }

则

e−(k+a)=Pk ea,e−k−d=Pk ed,\mathbf e_{-(k+a)}=\mathbf P_k\,\mathbf e_a,\qquad \mathbf e_{-k-d}=\mathbf P_k\,\mathbf e_d,

从而

fk+a∗=f−(k+a)=F Pk ea,f−k−d=F Pk ed.\mathbf f_{k+a}^*=\mathbf f_{-(k+a)}=\mathbf F\,\mathbf P_k\,\mathbf e_a, \quad \mathbf f_{-k-d}=\mathbf F\,\mathbf P_k\,\mathbf e_d.

代回第4节的和式，得到一个非常干净的“DFT 对角化/频域表示”：

Qkbulk=F  Pk  Γbulk  PkT  FH,[Γbulk]a,d=Γbulk[a,d].\boxed{ \mathbf Q_k^{\text{bulk}} = \mathbf F\;\mathbf P_k\; \boldsymbol\Gamma_{\text{bulk}}\; \mathbf P_k^{T}\;\mathbf F^{H}, \qquad [\boldsymbol\Gamma_{\text{bulk}}]_{a,d}=\Gamma_{\text{bulk}}[a,d]. }

这就是你要的 “DFT矩阵形式 + 索引矩阵(置换矩阵)形式”：
bulk 的全部数值结构在 Γbulk\boldsymbol\Gamma_{\text{bulk}}（与 kk 无关），而 kk 只通过 Pk\mathbf P_k 做行列置换进入。

6) 另一种等价的“时域调制矩阵”表达（可选）

若你也想保留你前面用过的对角调制矩阵 Dℓ=diag(e−jωℓ(n−1))\mathbf D_\ell=\mathrm{diag}(e^{-j\omega \ell(n-1)})，利用 fk+a=Dafk\mathbf f_{k+a}=\mathbf D_a\mathbf f_k、fk+a∗=D−af−k\mathbf f_{k+a}^*=\mathbf D_{-a}\mathbf f_{-k}，可得

Qkbulk=1N∑a,dΓbulk[a,d]  D−a (f−kf−kH) Dd\boxed{ \mathbf Q_k^{\text{bulk}} = \frac{1}{N} \sum_{a,d} \Gamma_{\text{bulk}}[a,d]\; \mathbf D_{-a}\, (\mathbf f_{-k}\mathbf f_{-k}^H)\, \mathbf D_{d} }

这也是“索引矩阵/结构矩阵”的一个版本（用 D±a\mathbf D_{\pm a} 编码索引）。

7) Hermitian 化（与你 structured 部分一致）

bulk 求和集合若不是天然对称，建议同样做

Qkbulk←12(Qkbulk+QkbulkH)\boxed{ \mathbf Q_{k}^{\text{bulk}} \leftarrow \frac12\Big(\mathbf Q_{k}^{\text{bulk}}+\mathbf Q_{k}^{\text{bulk}H}\Big) }

以保证最终二次型实值/PSD分析的便利。

如果你把 P\mathcal P 的精确定义（到底包含哪些 pairing）用一句话写出来，我可以把第3节的 χbulk(a,t,d)\chi_{\text{bulk}}(a,t,d) 精确到“完全等价于 (p,q,p′,q′)∉P(p,q,p',q')\notin\mathcal P 且不在 H1–H4”的最简 δ-乘积形式；但即使不补充，第4–5节的 DFT+索引矩阵主结构已经是通用且可直接写进论文的解析表达。

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