21 August (2026)

And this one comes from Jacob. What we see here is the distribution of means, each computed from a one-collapse distribution of distances; each collapse had 200 points (he told me) and that's how they compare to the theoretical means for (left) donut, (center) atom = thermal and (right) dipole. As expected, they agree. Now we should see that for a very-high multiphoton observable, where distributions will become fractal, and unique. But their mean could (should) still agree.

18:05 (CET).

---

A first surprise is that the two-atom necessity is far from obvious for four-photon observables (the first to require two atoms). That's what I find, and initially thought there was a mistake in my atomicity requirements, but no, as I discuss later. First I want to see if such a very good agreement remains for higher observables (at some point it should break noticeably).

18:01 (CET).

---

Moment of weakness of the day as I see I've been loosing time looking into this myself instead of asking Claude to solve everything from the start. The basic idea is that the atoms are actually Gauss-Legendre nodes once your regularise the thermal distribution into a flat one, and that allows you to reach high numbers, with which we could produce this following progression of how the thermal ensemble is being populated "canonically" [always choosing the most-covering moments distributions].

There's a more crowded uniform-looking coverage of the full ensemble, striving to touch the extremes (in particular the most-correlated dipole) but failing to. Nothing unexpected or very surprising but a beautiful snapshot of quantum vs ensemble averages.

The derivation goes as follows: the thermal correlator is $C_{\boldsymbol\alpha}=u(1-u)$ with $u$ uniform on $[0,1]$ so that corresponding measure is $$\mu(C_{\boldsymbol\alpha})=\frac{2}{\sqrt{1-4C_{\boldsymbol\alpha}}}\ ,\qquad C_{\boldsymbol\alpha}\in\left[0,\tfrac14\right]\,.$$ Now Claude observes that regularizing that, $$t\equiv1-2u\ ,\qquad C_{\boldsymbol\alpha}=\frac{1-t^{2}}{4}\ ,\qquad t\in[-1,1]$$ then the measure becomes flat: $t$ is uniform with density $\tfrac12$. Then, it claims, the atoms are the Gauss quadrature of a constant weight on $[-1,1]$, which brings it to our polynomial but now in Legendre form: $$\pi_\nu\!\left(C_{\boldsymbol\alpha}\right)\ \propto\ P_{2\nu}\!\left(\sqrt{1-4C_{\boldsymbol\alpha}}\right)$$ where $P_{2\nu}$ is the ordinary Legendre polynomials on $\left[-1,1\right]$, $P_0=1$, $P_1=t$, $P_2=\tfrac12\left(3t^{2}-1\right)$, and so on.

The atoms follow immediately. The roots of $P_{2\nu}$ come in pairs $\pm s_k$, and both members map to the same location, so the $2\nu$ Legendre nodes collapse onto $\nu$ atoms. Their weights are equal by symmetry, and the density $\tfrac12$ converts the Legendre total of $2$ into unity. Hence, with $s_k$ the positive nodes of the $2\nu$-point Gauss-Legendre rule and $\lambda_k$ their standard weights, we get all our atoms easily: $$\boxed{C_k=\frac{1-s_k^{2}}{4}\ ,\qquad w_k=\lambda_k\,.}$$

Nothing is left to solve. No Hankel matrix, no root-finding. The first three cases:

$\nu$	serves $N$	$s_k$	$C_k$	$w_k$
1	1–3	0.5773502692	0.1666666667	1.0000000000
2	4–7	0.3399810436	0.2211032225	0.6521451549
		0.8611363116	0.0646110632	0.3478548451
3	8–11	0.2386191861	0.2357652210	0.4679139346
		0.6612093865	0.1407005368	0.3607615730
		0.9324695142	0.0326251513	0.1713244924
4	12–15	0.1834346425	0.2415879330	0.3626837834
		0.5255324099	0.1809539215	0.3137066459
		0.7966664774	0.0913306309	0.2223810345
		0.9602898565	0.0194608479	0.1012285363
5	16–19	0.1488743390	0.2444591078	0.2955242247
		0.4333953941	0.2030421081	0.2692667193
		0.6794095683	0.1346006596	0.2190863625
		0.8650633667	0.0629163429	0.1494513492
		0.9739065285	0.0128765184	0.0666713443

11:35 (CET).

---

The atoms are basically moment-generating machines. A given number of them can produce more than are needed by a given observable. I got something badly wrong earlier, not appreciating that fully, which is how many atoms we need to cover $N$-photon observables. I thought you need three for $N=6$, but no, you're still good with two atoms. Only, you cannot choose them. You don't have enough moment-freedom to move them around. Let us call "slack" (for margin, or looseness in a rope) the excedent of moments we get.

$$\text{slack}=\left(2\nu-1\right)-M$$

which is either 0 or 1. Whenever it is 0 (no slack), the atoms are uniquely defined. When it is 1, the extra freedom allows to settle for other atoms. So this was the case in 00W0b, there I was giving a family of solutions for $N=4$ and $5$. In the previous post, applying strictly the procedure, which is the one that cuts the slack, I also provide the soutions for $N=6$ and $7$. Those are also valid for $N=4$ and $5$, which, however, have more pairs of atoms.

So the photon-observables served by the same number of atoms go by groups of 4:

• 0-3 photon observables with 1 atom (a 0 observable is no observable but that's the pattern)
• 4-7 photon observables with 2 atoms
• 8-11 photon observables with 3 atoms

etc.

The polynomial for 3 atoms by the way is $$924x^{3}-378x^{2}+42x-1$$ with solutions $$C_k=\frac{3}{22}+\frac{\sqrt{15}}{33}\,\cos\left(\frac{1}{3}\arccos\left(-\frac{\sqrt{15}}{35}\right)-\frac{2\pi k}{3}\right)\ ,\qquad k=0,1,2$$ with corresponding weights $$w_k=\frac{660\,C_k^{2}-160\,C_k+7}{30\left(66\,C_k^{2}-18\,C_k+1\right)}\,.$$

Generalization to higher atoms would maybe look futile, except that Claude found another way that seems to be much more efficient than all the previous material.

10:22 (CET).

---

Now back to particular cases, this time constructive. The procedure as I sketched it has three stages:

• get $\nu$ from $N$.
• solve the Hankel equations for $\pi_\nu$ and take its roots as the locations.
• solve the Vandermonde equations for the weights.

I now do this for $N\le7$, which improves by two additional photons my previous solution and highight and important point to which I come back in the next post, as it had escaped me.

We remind the moments $m_p=\int_0^1u^{\,p}\left(1-u\right)^{p}du=\frac{p!^{2}}{\left(2p+1\right)!}$ so $\left\{\frac{1}{6},\frac{1}{30},\frac{1}{140},\frac{1}{630},\frac{1}{2772}, ...\right\}$

Cases $N=2$ and $N=3$: In this case $M=\lfloor N/2\rfloor=1$ and $\nu=\left\lceil\left(M+1\right)/2\right\rceil=1$. The Hankel equation for $r=0$ only yields $m_0\,a_0=-m_1\ \Longrightarrow\ a_0=-\frac{m_1}{m_0}=-\frac16$. The polynomial is thus of order 1 $\pi_1\left(x\right)=x+a_0=x-\frac16$ with root $C_1=1/6$ which is the value found by Daniel. We don't have to solve the Vandermonde has the weights has to one.

Cases $N\in\{4, 5\}$: $M=2$ and $\nu=\left\lceil3/2\right\rceil=2$ atoms. But this is also for the Cases $N\in\{6,7\}$ in which case $M=3$ but $\nu=\left\lceil4/2\right\rceil$ is still two atoms. This difference is discussed in the next post. The following applies to all four cases.

The Hankel equation goes with $r=0$ and $r=1$ as: $$\begin{pmatrix} m_0 & m_1\\ m_1 & m_2\end{pmatrix}\begin{pmatrix} a_0\\ a_1\end{pmatrix}=-\begin{pmatrix} m_2\\ m_3\end{pmatrix}\ \Longrightarrow\ \begin{pmatrix} 1 & 1/6\\ 1/6 & 1/30\end{pmatrix}\begin{pmatrix} a_0\\ a_1\end{pmatrix}=-\begin{pmatrix} 1/30\\ 1/140\end{pmatrix}$$ which gives the polynomial as $$\pi_2\left(x\right)=x^{2}-\frac27x+\frac{1}{70}\ \Longleftrightarrow\ 70x^{2}-20x+1=0$$ with two locations: $$C_{\pm}=\frac{10\pm\sqrt{30}}{70}$$ i.e., $C_+\simeq0.22$ and $C_-\simeq0.06$. Both lie inside $\left[0,\tfrac14\right].$ The Vandermonde equation reads: $$\begin{pmatrix} 1 & 1\\ C_+ & C_-\end{pmatrix}\begin{pmatrix} w_+\\ w_-\end{pmatrix}=\begin{pmatrix} 1\\ 1/6\end{pmatrix}$$

	$C_j$	$w_j$
$j=+$	$\left(10+\sqrt{30}\right)/70$	$\left(18+\sqrt{30}\right)/36$
$j=-$	$\left(10-\sqrt{30}\right)/70$	$\left(18-\sqrt{30}\right)/36$

This is a particular case and the question of which is the general is more interesting and subtle than I thought.

09:41 (CET).

20 August (2026)

Now for the general case, although still for the phase-invariant case, I provide the equations to solve to get the weights and locations of the atoms.

We had left it at these $M+1$ equations with $2\nu$ unknowns: \begin{equation}\label{eq:00uIJ}\sum_{j=1}^{\nu}w_j\,C_j^{\,p}=m_p\end{equation} for $m_p\equiv\langle C_{\boldsymbol\alpha}^{\,p}\rangle$ and $p=0,1,\dots,M$. Since $M$ is the highest power of $C_{\boldsymbol\alpha}$ needed to describe the state, and in the case of phase-invariance, the unbalanced terms are gone, leaving us with only $p=q$, and since $p+q$ cannot exceed $N$ this caps $p$ at $M=\lfloor N/2\rfloor$. As for $\nu$, this is the number of atoms minimally needed, each coming with two parameters: the weight (mass in the traditional lexicon) and the location (the quantum state for us). We're looking for the $\nu$ pairs $(w_j,C_j)$ for $1\le j\le\nu$. We determine its size below.

We need to solve for $w_j$ the weights and $C_j$ the quantum-state coefficients, or "locations" (of the atoms), whose $p$th powers are weighted by the weight to produce the averages. Each $(w_j,C_j)$ provides an atom for the distribution. There are $M=\lfloor N/2\rfloor.$

These are linear equations in the weights. We solve first for the locations. The proof is constructive, so if we have a solution, it will be enough. Let us thus assume that the locations are algebraic numbers, i.e., that they are solution of a polynomial $$\pi_\nu(x)=x^{\nu}+a_{\nu-1}x^{\nu-1}+\cdots+a_0\ ,\qquad \pi_\nu(C_j)=0\ \text{for all}\ j\,.$$ Multiplying Eq. \eqref{eq:00uIJ} at index $p=s+r$, i.e., $\sum_{j=1}^{\nu}w_j\,C_j^{\,s+r}=m_{s+r}$, by $a_s$ and summing over $s$ from $0$ to $\nu$, $$\sum_{j=1}^{\nu}w_j\sum_{s=0}^{\nu}a_s\,C_j^{\,s+r}=\sum_{s=0}^{\nu}a_s\,m_{s+r}$$ but $\sum_{s=0}^{\nu}a_s\,C_j^{\,s+r}=C_j^{\,r}\sum_{s=0}^{\nu}a_s\,C_j^{\,s}=C_j^{\,r}\,\pi_\nu(C_j)=0$, and therefore the right-hand side vanishes too. Isolating the highest-order term, which is $s=\nu$ with $a_\nu=1$: $$m_{\nu+r}+\sum_{s=0}^{\nu-1}a_s\,m_{s+r}=0\ ,\qquad r=0,1,\dots,\nu-1\,.$$ This is a linear system for the coefficients $a_s$ involving only the moments $m$, and it is in the Hankel form since the entry at position $(r,s)$ is $m_{s+r}$. In matrix form:

$$\mathbf{M}_\nu\,\mathbf{a}=-\mathbf{h}\,,$$ where $\left(\mathbf{M}_\nu\right)_{rs}=m_{r+s},$ with $ r,s=0,\dots,\nu-1,$ $\mathbf{a}\equiv\left(a_0,a_1,\dots,a_{\nu-1}\right)^{\mathsf T}$ and $\mathbf{h}=\left(m_\nu,m_{\nu+1},\dots,m_{2\nu-1}\right)^{\mathsf T}$. Explicitly: $$\begin{pmatrix} m_0 & m_1 & \cdots & m_{\nu-1}\\ m_1 & m_2 & \cdots & m_{\nu}\\ \vdots & \vdots & & \vdots\\ m_{\nu-1} & m_{\nu} & \cdots & m_{2\nu-2}\end{pmatrix}\begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_{\nu-1}\end{pmatrix}=-\begin{pmatrix} m_{\nu}\\ m_{\nu+1}\\ \vdots\\ m_{2\nu-1}\end{pmatrix}\,.$$ And that's where we find the magnitude of $\nu$. The largest value to enter is $m_{2\nu-1}$ in the last entry of the right-hand side $\mathbf{h}$ (and not in $\mathbf{M}_\nu$ where this is $m_{2(\nu-1)}$). We start from $m_0$ and need all the moments in between those two, so $2\nu$ of them, corresponding to the number of unknowns for the atoms. The top indices should match but $2\nu-1$ is odd while $M$ can be anything, therefore we have to settle with the smallest $\nu$ such that $2\nu-1\ge M$, which has solution $\nu=\left\lceil\frac{M+1}{2}\right\rceil$. Substituting $M=\lfloor N/2\rfloor$ gives $\nu=\left\lceil\left(\lfloor N/2\rfloor+1\right)/2\right\rceil$, and the nested floor-ceiling collapses to $$\nu=\left\lceil{N+1\over4}\right\rceil\,.$$

As a Hankel matrix, its antidiagonals are constant, with $\det \mathbf{M}_\nu>0$ so invertible, with also real and distinct roots. Their belonging to the support is unclear to me mathematically but we expect it is the case for semi-classical states and not so for quantum ones. So we solve for two things: first for the coefficients $a_s$ of the polynomial equations, and then for the roots of those equations which provide the locations of the atoms.

Then we have to solve for the weights, but those are from a linear set of equations, so this is, in principle, a formality. This comes out as a Vandermonde system: $$\begin{pmatrix} 1 & 1 & \cdots & 1\\ C_1 & C_2 & \cdots & C_\nu\\ \vdots & \vdots & & \vdots\\ C_1^{\nu-1} & C_2^{\nu-1} & \cdots & C_\nu^{\nu-1}\end{pmatrix}\begin{pmatrix} w_1\\ w_2\\ \vdots\\ w_\nu\end{pmatrix}=\begin{pmatrix} m_0\\ m_1\\ \vdots\\ m_{\nu-1}\end{pmatrix}$$ which is nonsingular if the $C_j$ are non-degenerate, which they should.

A beautiful construction: locations from Hankel, weights from Vandermonde, both guaranteed physical, hence the representation exists with $\left\lceil\left(N+1\right)/4\right\rceil$ atoms.

17:40 (CET).

---

Before I work the general case, the particular cases: for $N=2$ and $N=3$, $C_j=1/6$ indeed (as found by Daniel yesterday) and $w_1=1$ (one atom has all the weight).

The general moment is $m_p=(p!)^2/(2p+1)!$, i.e. $1,\ 1/6,\ 1/30,\ 1/140,\dots$, so for $4\le N\le 5$ (two next cases), then, the possible various minimal solutions are of the type: (minimal because one can take more atoms, mind you, you can still take the full thermal bag): $$C_2=m_1-\frac{\sigma^{2}}{C_1-m_1}\ ,\qquad w_1=\frac{\sigma^{2}}{\sigma^{2}+\left(C_1-m_1\right)^{2}}\ ,\qquad w_2=1-w_1$$

$C1$	$w_1$	$C_2$	$w_2$
$1/5$	$5/6$	$0$	$1/6$
$\left(10+\sqrt{30}\right)/70$	$0.652145$	$\left(10-\sqrt{30}\right)/70$	$0.347855$
$1/4$	$4/9$	$1/10$	$5/9$

where the moment is $1/30$ and $\sigma^2\equiv\mathrm{Var}\left(C_{\boldsymbol\alpha}\right)=m_2-m_1^2=\frac{1}{30}-\frac{1}{36}=\frac{1}{180}\,.$ It's because one atom alone cannot account for the four-to-seven particle fluctuations that you need another one.

One of the extrema as $C_1=1/4$, i.e., the dipole (most classically correlated). But this is contaminated by the $C_2=1/10$, which also has higher probability.

12:10 (CET).

---

In the heterogeneous samples boson-spatial-correlations problem, the one-distribution-does-it-all result is a problem of $\nu$-atom replacements ("atom" in the sense of measure theory). This will bring us to express \begin{equation}\label{eq:00zPI}\overline{\tilde\rho^{(N)}}(\{\theta\})=\sum_{j=1}^{\left\lceil\frac{N+1}{4}\right\rceil}w_j\int_0^{2\pi}\frac{d\phi}{2\pi}\prod_{i=1}^{N}\tilde\rho^{(1)}_{\sqrt{C_j}e^{i\phi}}(\theta_i)\end{equation} where the $\nu\equiv\left\lceil\frac{N+1}{4}\right\rceil$ atoms provide the location $C_j$ and mass $w_j$ as we will show in the following and following post.

From: $$\tilde\rho^{(N)}_{\boldsymbol\alpha}(\theta_1,\dots,\theta_N)=\prod_{i=1}^N\tilde\rho^{(1)}_{\boldsymbol\alpha}(\theta_i)$$

with the one-photon distribution of LG$\pm\ell$ $$\tilde\rho^{(1)}_{\boldsymbol\alpha}(\theta)=\frac{1}{2\pi}\left[1+\gamma e^{2i\ell\theta}+\gamma^*e^{-2i\ell\theta}\right]$$ where $$\gamma\equiv\frac{\alpha_a\alpha_b^*}{\lvert\alpha_a\rvert^2+\lvert\alpha_b\rvert^2}\ ,\qquad \lvert\gamma\rvert^2=C_{\boldsymbol\alpha}\,.$$ The four real parameters of $\boldsymbol\alpha$ enter only through the single complex number $\gamma$, so from here on we index the densities by $\gamma$ and write $\tilde\rho^{(1)}_{\gamma}$ for $\tilde\rho^{(1)}_{\boldsymbol\alpha}$. The structure of this becomes, introducing $z_i\equiv e^{2i\ell\theta_i}$ \begin{equation}\label{eq:00zxQ} (2\pi)^N\,\tilde\rho^{(N)}_{\boldsymbol\alpha}(\theta_1,\dots,\theta_N)=\prod_{i=1}^N\left(1+\gamma z_i+\gamma^*z_i^*\right)=\sum_{p,q\atop p+q\le N}\gamma^p\gamma^{*q}\,S_{pq}(\{\theta\}) \end{equation} where $S_{pq}$ is our long-term combinatoric friend $$S_{pq}(\theta_1,\dots,\theta_N)\equiv\sum_{\substack{P,Q\subseteq\{1,\dots,N\},\ P\cap Q=\varnothing\\ \lvert P\rvert=p,\ \lvert Q\rvert=q}}\exp\left[2i\ell\left(\sum_{i\in P}\theta_i-\sum_{j\in Q}\theta_j\right)\right]$$ which has $\binom{N}{p}\binom{N-p}{q}$ terms (cf. the Distribution of bosons paper).

So far it's only algebra. We did nothing. Now let's turn to quantum states, this will introduce a paradigm shift of retaining a wavefunction, while having performed an average over the quantum states already. For thermal, for instance, $C_{\boldsymbol\alpha}=u(1-u)$ with $u$ uniform so that the thermal measure reads $$\mu(C_{\boldsymbol\alpha})=\frac{2}{\sqrt{1-4C_{\boldsymbol\alpha}}}\ ,\qquad C_{\boldsymbol\alpha}\in\left[0,\tfrac14\right]\,.$$ This average is $\overline{\dots}\equiv\int d\boldsymbol\alpha\,P(\boldsymbol\alpha)$, in which case

$$(2\pi)^N\,\overline{\tilde\rho^{(N)}}=\sum_{p+q\le N}\langle\gamma^p\gamma^{*q}\rangle\;S_{p,q}(\{\theta\})\,,$$

Let's consider our de Finetti representation:

$$\overline{\tilde\rho^{(N)}}(\{\theta\})=\int \prod_{i=1}^{N}\tilde\rho^{(1)}_{\gamma}(\theta_i)d\mu(\gamma)$$ with $\mu$ the distribution of $\gamma$ over the ensemble defined by the quantum state. We now look for a second measure $\mu'$ which gives the same result in terms of atoms. If it exists (and we come to that in a next post), then by assumption: $$\mu'=\sum_{j=1}^{\nu}w_j\,\delta_{\gamma_j}\quad\text{and }\quad\overline{\tilde\rho^{(N)}}=\sum_{j=1}^{\nu}w_j\prod_{i=1}^{N}\tilde\rho^{(1)}_{\gamma_j}(\theta_i)$$ so we have

\begin{equation}\label{eq:00zDS}(2\pi)^N\sum_{j=1}^{\nu}w_j\prod_{i=1}^{N}\tilde\rho^{(1)}_{\gamma_j}(\theta_i)=\sum_{p+q\le N}\langle\gamma^p\gamma^{*q}\rangle\;S_{p,q}(\{\theta\})\,.\end{equation}

At the same time, applying Eq. \eqref{eq:00zxQ} to the case of the atom $\gamma_j$, which is a fixed complex number like any other, we have $$\left(2\pi\right)^{N}\prod_{i=1}^{N}\tilde\rho^{(1)}_{\gamma_j}(\theta_i)=\sum_{p+q\le N}\gamma_j^{\,p}\gamma_j^{*q}\,S_{pq}(\{\theta\})$$

so that after multiplying by $w_j$, summing over $j$ and equating to Eq. \eqref{eq:00zDS}, we find that we must have (because the $S_{pq}$ are linearly independent, which should also be proven but I'll take it as given): \begin{equation}\label{eq:00ECG}\sum_{j=1}^{\nu}w_j\,\gamma_j^{\,p}\gamma_j^{*q}=\langle\gamma^{p}\gamma^{*q}\rangle\ ,\qquad p+q\le N\,.\end{equation}

Let us assume now the particular case of phase-invariance, in which case the atoms become not points but rings (i.e., phase averaged), each $\delta_{\gamma_j}$ being replaced by the uniform measure on the circle of radius $\sqrt{C_j}$—which is the form that was assumed in Eq. \eqref{eq:00zPI} that should eventually be generalized to non-phase-invariant cases—and Eq. \eqref{eq:00ECG} simplifies to $$\sum_{j=1}^{\nu}w_j\,C_j^{\,p}=m_p\ ,\qquad\text{where } m_p\equiv\langle C_{\boldsymbol\alpha}^{\,p}\rangle\ ,\qquad p=0,1,\dots,M$$ with $M$ the largest $p$ allowed, that is $M=\lfloor N/2\rfloor$. This gives us $M+1$ equations with $2\nu$ unknowns. I address the problem of solving this separately. The simplest one gives us the case we discovered yesterday during the meeting. The big result for now is that this depends on the order of the observable, with a step-wise increase that is rank-2 specific, which is Eq. \eqref{eq:00zPI}, that is not yet fully proven, since in addition to providing an explicit construction for $w_j$ and $\gamma_j$ (finding the atoms), we must also justify the $\left\lceil\frac{N+1}{4}\right\rceil$, which I'll do now.

11:05 (CET).

19 August (2026)

I must summarize the past days study of uniformity/homogeneity of CSI violating rank-2 fields led me to this perspective between our case and that of older SPDC literature, which was the question that started this long detour: $$\text{uniform }G^{(1)}\ \wedge\ \text{homogeneous }G^{(2)}\ \wedge\ \text{rank }2\ \Longrightarrow\ \text{CSI never violated}$$ I provided pairwise cases (uniform and violating; homogeneous and violating; something uniform & homogeneous is trivial but non-violating as per the theorem).

12:48 (CET).

18 August (2026)

Now for the 3rd, maybe most ambitious characterization of local violation of CSI: the direction of maximum violation: It is the unit vector which makes the angle $\theta_\mathrm{max}$ (with respect to $\hat{\mathbf{x}}$) where $$\theta_\mathrm{max}={1\over2}\arctan\left({2(g_xg_y-p_xp_y)\over g_x^2-p_x^2-g_y^2+p_y^2}\right)$$ and the corresponding maximum value of this violation is: $$Q_{\max}\equiv\lambda_+(M)=\tfrac12\big[T+\sqrt{T^2+4(g\wedge p)^2}\big]$$ where $\lambda_+$ is the positive-value eigenvalue of $M$ (which is indefinite, so it has one $\ge0$) and $T\equiv|g|^2-|p|^2$. This gives the anisotropy of the landscape.

Proof: To find the direction where $Q(\hat{\boldsymbol\delta})=\hat{\boldsymbol\delta}^{\mathsf T}M\,\hat{\boldsymbol\delta}$ increases the most, we diagonalize its $M=\lambda_+\hat e_+\hat e_+^{\mathsf T}+\lambda_-\hat e_-\hat e_-^{\mathsf T}$, with $\hat e_\pm$ orthonormal. Now, writing $\hat{\boldsymbol\delta}=\cos\psi\,\hat e_++\sin\psi\,\hat e_-$, we have

$$ Q(\hat{\boldsymbol\delta})=\lambda_+\cos^2\psi+\lambda_-\sin^2\psi=\lambda_-+(\lambda_+-\lambda_-)\cos^2\psi $$

which is a convex combination of the two eigenvalues, so $\lambda_-\le Q\le\lambda_+$, with the maximum at $\psi=0$, i.e. $\hat{\boldsymbol\delta}=\hat e_+$. The direction of highest $Q$ (CSI violation) is thus the eigenvector of $\lambda_+$, and $Q_{\max}=\lambda_+$.

So we are left with the eigenproblem of $M$. For a $2\times2$ symmetric matrix the characteristic polynomial is $\lambda^2-T\lambda+D=0$ where $T\equiv\operatorname{tr}M$ and $D\equiv\det M$, so

$$ \lambda_\pm=\frac{T\pm\sqrt{T^2-4D}}{2}\,.$$

We already previously calculated $T=|g|^2-|p|^2$ and $D=-(g\wedge p)^2$, so that $ \lambda_\pm=\frac{1}{2}\left[\big(|g|^2-|p|^2\big)\pm\sqrt{\big(|g|^2-|p|^2\big)^2+4(g\wedge p)^2}\right]$, where $g\wedge p=g_xp_y-g_yp_x$, or explicitly, as the $z$ component of the vector product of the gradients of $\Lambda$ and $\varphi$: $$g\wedge p=\partial_x\Lambda\,\partial_y\varphi-\partial_y\Lambda\,\partial_x\varphi\,.$$

19:44 (CET).

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The map of local violation in the previous post counts directions, weighting a barely-violating one the same as a strongly-violating one. We can quantify the amount of violation by average its amount over all angles. That gives us: $$\boxed{\langle Q\rangle=\tfrac12\big(|g|^2-|p|^2\big)=\tfrac12\big(|\nabla\Lambda|^2-|\nabla\varphi|^2\big)}\,.$$

16:44 (CET).

---

So, now, let me look at ways to see the local violations of arbitrary cases. The first one will be: $$\boxed{f(\mathbf r)=1-\frac{1}{\pi}\arccos\frac{|g|^2-|p|^2}{|g+p|\,|g-p|}\in[0,1]\,.}$$

This quantifies how much around of each point there is a violation. At $\mathbf{r}_1=\mathbf{r}_2$ we always have saturation, of course. But if you step a little bit, then you can, or not, violate. Where? Let's keep introducing new notations until we gag: so now give me $u\equiv g+p$ and $v\equiv g-p$. There is CSI violation iff $$Q=(\hat{\boldsymbol\delta}\cdot u)(\hat{\boldsymbol\delta}\cdot v)>0$$ i.e.,—as already noted—they should have the same sign.

So let's assume $\hat{\boldsymbol\delta}\cdot u>0$. That happens for the full half-circle centered around $u$ (which is a vector, of course, the sum of $g$ and $p$ themselves the gradients of logs of amplitudes and phase yada yada yada). A big of planar geometry now: so we have two vectors with an angle $\Delta$ between them. Two arcs of length $\pi$ whose centres are separated by the angle $\Delta$ between $u$ and $v$ overlap on an arc of length $\pi-\Delta$. The both-negative case is the antipodal copy, and provides another $\pi-\Delta$. So out of the full $2\pi$, we have $f=\frac{2(\pi-\Delta)}{2\pi}=1-\frac{\Delta}{\pi}$ of same-sign quantities and thus CSI violation. This angle $\Delta$ we can get from the law of cosines: $$\cos\Delta=\frac{u\cdot v}{|u||v|}=\frac{(g+p)\cdot(g-p)}{|g+p||g-p|}=\frac{|g|^2-|p|^2}{|g+p||g-p|}\,.$$ Neat, no? So that defines a first useful 2D map of local violations, which is the boxed formula above.

16:31 (CET).

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The differential form of $\cosh\Delta\Lambda+\cos\Delta\varphi>2$ reads $1+(\boldsymbol\delta\cdot\nabla\Lambda/2)^2+1-(\boldsymbol\delta\cdot\nabla\varphi/2)^2>2$, i.e., with $\hat{\boldsymbol\delta}\equiv\boldsymbol\delta/|\boldsymbol\delta|$ the unit vector of distance separation:

\begin{equation}\label{eq:00iBv}(\hat{\boldsymbol\delta}\cdot\nabla\Lambda)^2-(\hat{\boldsymbol\delta}\cdot\nabla\varphi)^2>0\,.\end{equation}

Calling $g\equiv\nabla\Lambda$ the direction of steepest increase of the log amplitude ratio $\Lambda(\mathbf{r})\equiv\ln\big|\phi_b(\mathbf{r})/\phi_a(\mathbf{r})\big|$ and $p\equiv\nabla\varphi$ the direction of steepest increase of the relative phase $\theta_b-\theta_a$, the CSI violation criterion Eq. (\ref{eq:00iBv}) becomes $$Q(\hat{\boldsymbol\delta})\equiv\big(\hat{\boldsymbol\delta}\cdot(g+p)\big)\big(\hat{\boldsymbol\delta}\cdot(g-p)\big)$$ which should be $>0$ for violation, i.e., they should have the same sign. But since, in this form, there exists $M$ such that $$Q(\hat{\boldsymbol\delta})=\hat{\boldsymbol\delta}^{\mathsf T}M\,\hat{\boldsymbol\delta}$$ with $$M\equiv gg^{\mathsf T}-pp^{\mathsf T}=\begin{pmatrix}g_x^2-p_x^2 & g_xg_y-p_xp_y\\ g_xg_y-p_xp_y & g_y^2-p_y^2\end{pmatrix}\,.$$ Since $\det M=-(g\wedge p)^2\le0$ (where $\wedge$ is the $z$-component of the vector product), it is indefinite (in the sense of, it is neither positive definite [making $Q>0$ always], negative definite [$Q<0$] or semi-definite [one sign + zero] but indefinite, i.e., taking both signs). For us, that means it violates CSI in some directions, and does not in others. Except if $p$ and $q$ are collinear, or if either $p$ or $q$ is zero.

The case where $\det M=0$ are interesting, and correspond to different cases, some being the local version of the ones we've been seeing already:

• $\nabla\varphi=0$, i.e., no phase structure: real fields. Then $M=gg^{\mathsf T}$ is positive semi-definite: violation in every direction except along the level curve of $\Lambda$, where $Q=0$. This is local version of the "real fields violate CSI everywhere their amplitude differs".
• $\nabla\Lambda=0$, i.e., amplitude ratio locally flat. Then $M=-pp^{\mathsf T}$ is negative semi-definite: no violation in any direction. The uniform-and-homogeneous no-go, here local but also what happens on the whole plane for LG$_{\pm\ell}$.
• $p$ and $q$ both nonzero but collinear, fully non-violating when $|\nabla\Lambda|<|\nabla\varphi|$, fully non-violating otherwise.

For the rest: no point of a rank-2 2D structured beam can be non-violating in all directions, unless the two gradients are collinear (or zero).

There is a cone of violation.

15:50 (CET).

17 August (2026)

I have given a case of everywhere-violating uniform beam, I will now give the other case of a bimodal homogeneous field violating CSI. From the previous result, this means that this field is non-uniform in space. We'll find that it has the shape of a bathtube with $G^{(1)}=2c\cosh k(x-L/2)$ over $x\in[0,L]$ (translational over $y$). The violation is everywhere on $]0,L]$ iff $k\ge2\pi/L$ (it is only somewhere otherwise).

We work on the domain $[0,L]^2$, and now consider: $$\phi_a(\mathbf{r})\equiv\sqrt c\,e^{-k(x-L/2)/2},\qquad \phi_b(\mathbf{r})\equiv\sqrt c\,e^{+k(x-L/2)/2}\,e^{2\pi ix/L}$$ also only $x$-dependent. Then, we can check that: $$\Lambda=k(x-L/2)\quad\text{and}\quad \varphi=2\pi x/L$$ and also that $$\Psi^{(2)}(\mathbf r_1,\mathbf r_2)=2c\,e^{i\pi(x_1+x_2)/L}\cosh\!\Big[(k+2\pi i/L)\tfrac{x_2-x_1}{2}\Big]$$ so that, from $|\cosh(a+ib)|^2={1\over2}\big(\cos(2b)+\cosh(2a)\big)$: $$G^{(2)}(\mathbf r_1,\mathbf r_2)=|\Psi^{(2)}(\mathbf r_1,\mathbf r_2)|^2=2c^2\big[\cosh(k\delta)+\cos(2\pi\delta/L)\big],$$ depends on $\delta\equiv x_2-x_1$ alone, and thus on $\mathbf{r}_2-\mathbf{r}_1$ alone. This is homogeneous.

Now for the violation: $G^{(2)}(0)=4c^2$ so that we get the violation $G^{(2)}(\boldsymbol\delta)>G^{(2)}(0)$ iif $\cosh(k\delta)+\cos(2\pi\delta/L)>2$. Interestingly, this looks very similar to the iff statement $\cosh\Delta\Lambda+\cos\Delta\varphi>2$. We can choose $k$ so let us define $\kappa\equiv kL/2\pi$ and also $s\equiv 2\pi\delta/L\in(0,2\pi].$ The condition for violation becomes $\cosh(\kappa s)+\cos s>2$ but since $$\cosh s+\cos s=2\sum_{n\ge0}\frac{s^{4n}}{(4n)!}=2+\frac{s^4}{12}+\frac{s^8}{10080}+\cdots$$ the we have violation of CSI for all $s$, i.e., everywhere on $]0,L]$. Reciprocally, if $\kappa<1$, there is a region of no violation, which we can see from the series expansion $\cosh(\kappa s)+\cos s-2=\frac{s^2}{2}\big(\kappa^2-1\big)+\frac{s^4}{24}\big(\kappa^4+1\big)+O(s^6)$. The leading term is strictly negative, so there exists $s_0>0$ with $\cosh(\kappa s_0)+\cos s_0<2$ on $[0,s_0]$, i.e., a whole interval around the diagonal where $G^{(2)}(\delta)<G^{(2)}(0)$, i.e., CSI are satisfied (and exhibiting bunching).

This thus provides quite a flexible homogeneous beam that can violate wholly or partly the CSI.

18:14 (CET).

---

Note that for homogeneous fields, antibunching, i.e., $G^{(2)}(\boldsymbol\delta)>G^{(2)}(0)$, is equivalent to CSI violation $\big[G^{(2)}(\mathbf r_1,\mathbf r_2)\big]^2>G^{(2)}(\mathbf r_1,\mathbf r_1)\,G^{(2)}(\mathbf r_2,\mathbf r_2)$, which is why they insist on homogeneity. Their criterion in this case also yields $\cosh\Delta\Lambda+\cos\Delta\varphi>2$ directly which remains true in our (more) general case.

17:19 (CET).

---

One cannot have CSI violation for a bi-modal field that is both uniform and homogeneous.

This is one or the other.

Homogeneity, i.e., $G^{(2)}(\mathbf r_1,\mathbf r_2)$ depending only on $\mathbf{r}_1-\mathbf{r}_2$ alone, forces $\rho_a(\mathbf{r})\rho_b(\mathbf{r})=c$ a constant where we remind $\phi_{a/b}=\rho_{a/b}(\mathbf{r})e^{i\theta_{a/b}}$. Indeed, since $\Psi^{(2)}(\mathbf r,\mathbf r)=2\phi_a(\mathbf r)\phi_b(\mathbf r)$, then $G^{(2)}(\mathbf r,\mathbf r)=|\Psi^{(2)}(\mathbf r,\mathbf r)|^2=4\rho_a^2\rho_b^2$ and since $G^{(2)}(\mathbf{r}_1,\mathbf{r}_2)=F(\mathbf{r}_1-\mathbf{r}_2)$, then $G^{(2)}(\mathbf{r},\mathbf{r})=F(\mathbf{0})$ pinning the value of $\rho_a\rho_b$ (to the square root of this over 4).

On the other hand, uniformity, pins $\rho_a^2+\rho_b^2=\text{const}$. But sum and product both constant means $\rho_a$ and $\rho_b$ are themselves constant, so uniformity+homogeneity$\implies\Delta\Lambda=0$. And this invalidates the condition for CSI violations. QED.

16:44 (CET).

16 August (2026)

So, now, Claude can help us find an explicit case:

Let us take as our two modes: $$\phi_a(x,y)=\frac{\sqrt2}{\sqrt{1+e^{2\Lambda(x)}}},\qquad \phi_b(x,y)=\phi_a(x,y)\;e^{(1+i)\Lambda(x)}$$ where $\Lambda$, we remind, is $\Lambda(\mathbf{r})\equiv\ln\big|\phi_b(\mathbf{r})/\phi_a(\mathbf{r})\big|$ so here we are making a self-referential definition of sort and need to ensure that the log of the modulus of their ratio spits the $\Lambda$ out, and you can see that it does (the complex exponential goes with the modulus and the log sees $e^\Lambda$). Note that there's only $x$ dependence in the amplitude, but we're looking for a uniform field anyway—and indeed $|\phi_a|^2+|\phi_b|^2=2$ everywhere—so invariance of $y$ shouldn't worry us. Now for $\Lambda$, Claude suggests: $$\Lambda(x)\equiv\operatorname{arcsinh}\!\big((2x-1)\sinh\pi\big)\,.$$ The arcsinh is chosen to ensure orthonormality of the modes, since, using $e^{\Lambda}/(1+e^{2\Lambda})=1/(2\cosh\Lambda)$, $$ \int_0^1\!\!\int_0^1\phi_a^*\phi_b\,dx\,dy=\int_0^1\frac{dx}{\cosh\Lambda(x)}e^{i\Lambda(x)}=\frac{1}{2\sinh\pi}\int_{-\pi}^{\pi}e^{i\Lambda}\,d\Lambda=0\,.$$

Since $\Lambda$ is real, $\theta_a=0$ and $\theta_b=\Lambda$ so that $\Delta\theta=\Lambda$ and that's the 1-Lipshitz criterion of our previous result. Therefore CSI are violated everywhere except on the set $\{\Lambda(\mathbf r_1)=\Lambda(\mathbf r_2)\}$ but this is simply the set where $x_1=x_2$, which is a cube in the hypervolume of dimension 4, and a cube has four-dimensional area zero.

Therefore, we have a two-mode uniform field which violate CSI almost everywhere (in all cases, except where the CS inequalities saturate by construction, but this is both unavoidable and this also has measure zero).

The sum of the two modes is what is uniform. Each varies in $x$ (not in $y$) and they compensate. This is also something true at the plane where the fields are thus defined. In a beam, they will distort, not being eigenmodes of the paraxial propagator.

Note that we don't even need to refer to Lipschitzity in this explicit case, we can also check directly that since both $\Delta\Lambda$ and $\Delta\varphi$ are the same $\Lambda(x_2)-\Lambda(x_1)$, the iff criterion reads

$$ \sinh^2\!\left(\frac{\Delta\Lambda}{2}\right)>\sin^2\!\left(\frac{\Delta\Lambda}{2}\right) $$

which, for $\Delta\Lambda\neq0$, is always true since $\sinh|x|>|x|\ge|\sin x|$.

22:24 (CET).

---

I was just discussing how to get violation everywhere in a homogeneous (constant intensity) beam, providing cases of violations everywhere but on restricted (open) domains. To have it everywhere‒everywhere... follow me:

I remind that violation requires $\sinh^2(\Delta\Lambda/2)>\sin^2(\Delta\varphi/2)$. If $\Delta\Lambda=0$, the lhs is zero and nothing can be violated. So one needs $\Lambda(\mathbf r_1)\neq\Lambda(\mathbf r_2)$ for every pair of distinct points, i.e., $\Lambda$ should be injective on the domain.

But, there is no $\Omega\subset\mathbb R^2\to\mathbb R$ continuous injection.

Therefore rank 2 (bimodal) states in 2D can never violate CSI strictly everywhere.

This sounds like a result but on the diagonal, the CSI are always saturated, so the good criterion is of course "violated everywhere it could".

And this turns out to be easy to achieve.

Since $|\Delta\varphi|<|\Delta\Lambda|$ is a sufficient condition for violation, let us assume a 1-Lipschitz function $g$ of $\Lambda$, such that $\varphi=g(\Lambda)$ with $|g'|\le1$. This ensures, by definition, that $|\Delta\varphi|\le|\Delta\Lambda|$ for all $\boldsymbol{r}$. It's not strict, otherwise we'd be set, but since $x\le y<\pi\implies\sin(x)\le\sin(y)<\sinh^2(y)$ then as long as $|\Delta\Lambda|<\pi$, we have the iff condition for CSI violation.

So now let's turn to the case $|\Delta\Lambda|\ge\pi$: this one is actually even more straightforward as we don't need Lipschitzity, since we always have $\sinh^2(\Delta\Lambda/2)\ge\sinh^2(\pi/2)\approx5.23>1\ge\sin^2(\Delta\varphi/2)$ and therefore CSI are violated again. Therefore:

For $\varphi=g(\Lambda)$ a 1-Lipschitz function, the CSI are violated everywhere except at the set $\{\Lambda(\mathbf r_1)=\Lambda(\mathbf r_2)\}$ which includes at least $\boldsymbol{r}_1=\boldsymbol{r}_2$ and possibly other points but that can be of measure zero almost everywhere.

There is a bit of work to show that the measure is zero:

Let us define $E\equiv\{(\mathbf r_1,\mathbf r_2)\in\Omega\times\Omega:\Lambda(\mathbf r_1)=\Lambda(\mathbf r_2)\}$ the set of CSI saturation. It has measure: $$|E|=\int_\Omega\Big|\{\mathbf r_2:\Lambda(\mathbf r_2)=\Lambda(\mathbf r_1)\}\Big|\,d\mathbf r_1$$ From there, measure theory and calculus establish that $$|E|=\int_\Omega\big|\Lambda^{-1}(\Lambda(\mathbf r_1))\big|\,d\mathbf r_1=\int_\Omega\nu\big(\{\Lambda(\mathbf r_1)\}\big)\,d\mathbf r_1=\int_{\mathbb R}\nu(\{c\})\,d\nu(c)=\sum_{c\ \text{atom}}\nu(\{c\})^2$$ where $\nu$ is the pushforward of area measure by $\Lambda$, i.e., for a set of values $B\subseteq\mathbb R$,

$$ \nu(B)\equiv\big|\Lambda^{-1}(B)\big|=\big|\{\mathbf r\in\Omega:\Lambda(\mathbf r)\in B\}\big| $$

i.e., how much area does a $\Lambda$-value in $B carries in space. An atom of $\nu$ is a single value $c$ with $\nu(\{c\})>0$, i.e., a value which is taken by the function on a set of nonzero measure. The integrand $\nu(\{c\})$ is nonzero only at atoms, and atoms are countable. Therefore, we have established that:

$∣E∣=0$ iff $\nu$ has no atoms, i.e., iff no value of $\Lambda$ is taken on a set of nonzero measure.

Or, equivalently:

$$\boxed{|E|=0\iff \text{no level set of }\Lambda\text{ has positive area}}\,.$$

In such a case, CSI are violated almost everywhere. Now we just need to find a case with uniform intensity.

19:15 (CET).

---

In my ℤ post on structured vs homogeneous (I'm not copying all of them here), __Y_Y__ I was just discussing how to get violation everywhere in a homogeneous (constant intensity) beam, providing cases of violations everywhere but on restricted (open) domains. To have it everywhere‒everywhere... follow me:

I remind that violation requires $\sinh^2(\Delta\Lambda/2)>\sin^2(\Delta\varphi/2)$. If $\Delta\Lambda=0$, the lhs is zero and nothing can be violated. So one needs $\Lambda(\mathbf r_1)\neq\Lambda(\mathbf r_2)$ for every pair of distinct points, i.e., $\Lambda$ should be injective on the domain.

But, there is no $\Omega\subset\mathbb R^2\to\mathbb R$ continuous injection.

Therefore rank 2 (bimodal) states in 2D can never violate CSI strictly everywhere.

This sounds like a result but on the diagonal, the CSI are always saturated, so the good criterion is of course "violated everywhere it could".

And this turns out to be easy to achieve.

Since $|\Delta\varphi|<|\Delta\Lambda|$ is a sufficient condition for violation, let us assume a 1-Lipschitz function $g$ of $\Lambda$, such that $\varphi=g(\Lambda)$ with $|g'|\le1$. This ensures, by definition, that $|\Delta\varphi|\le|\Delta\Lambda|$ for all $\boldsymbol{r}$. It's not strict, otherwise we'd be set, but since $x\le y<\pi\implies\sin(x)\le\sin(y)<\sinh^2(y)$ then as long as $|\Delta\Lambda|<\pi$, we have the iff condition for CSI violation.

So now let's turn to the case $|\Delta\Lambda|\ge\pi$: this one is actually even more straightforward as we don't need Lipschitzity, since we always have $\sinh^2(\Delta\Lambda/2)\ge\sinh^2(\pi/2)\approx5.23>1\ge\sin^2(\Delta\varphi/2)$ and therefore CSI are violated again. Therefore:

For $\varphi=g(\Lambda)$ a 1-Lipschitz function, the CSI are violated everywhere except at the set $\{\Lambda(\mathbf r_1)=\Lambda(\mathbf r_2)\}$ which includes at least $\boldsymbol{r}_1=\boldsymbol{r}_2$ and possibly other points but of measure almost zero.

There is a bit of work to show that the measure is zero:

Let us define $E\equiv\{(\mathbf r_1,\mathbf r_2)\in\Omega\times\Omega:\Lambda(\mathbf r_1)=\Lambda(\mathbf r_2)\}$ the set of CSI saturation. It has measure: $$|E|=\int_\Omega\Big|\{\mathbf r_2:\Lambda(\mathbf r_2)=\Lambda(\mathbf r_1)\}\Big|\,d\mathbf r_1$$ From there, measure theory and calculus establish that $$|E|=\int_\Omega\big|\Lambda^{-1}(\Lambda(\mathbf r_1))\big|\,d\mathbf r_1=\int_\Omega\nu\big(\{\Lambda(\mathbf r_1)\}\big)\,d\mathbf r_1=\int_{\mathbb R}\nu(\{c\})\,d\nu(c)=\sum_{c\ \text{atom}}\nu(\{c\})^2$$ where $\nu$ is the pushforward of area measure by $\Lambda$, i.e., for a set of values $B\subseteq\mathbb R$,

$$ \nu(B)\equiv\big|\Lambda^{-1}(B)\big|=\big|\{\mathbf r\in\Omega:\Lambda(\mathbf r)\in B\}\big| $$

i.e., how much area does a $\Lambda$-value in $B carries in space. An atom of $\nu$ is a single value $c$ with $\nu(\{c\})>0$, i.e., a value which is taken by the function on a set of nonzero measure. The integrand $\nu(\{c\})$ is nonzero only at atoms, and atoms are countable. Therefore, we have established that:

$∣E∣=0$ iff $\nu$ has no atoms, i.e., iff no value of $\Lambda$ is taken on a set of nonzero measure.

Or, equivalently:

$$\boxed{|E|=0\iff \text{no level set of }\Lambda\text{ has positive area}}\,.$$

In such a case, CSI are violated almost everywhere. Now we just need to find a case with uniform intensity.

19:12 (CET).

---

Now at the two-photon level—which for CSI is the one that interests us—the Monken et al. transfer theorem brings the pump profile at the mean transverse coordinate: $$\Psi^{(2)}(\mathbf r_1,\mathbf r_2)\propto\mathcal W\!\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ Z\right)\big(\mathbf H_1\mathbf V_2+\mathbf V_1\mathbf H_2\big)\,.$$

This is also in the limit where $\Phi$ the Fourier transform of the phase-matching function is constant (thin crystals), so we can now get rid of it.

As I just said, we study the full spatial landscape while they study homogeneous beam structures. But they need to get there. Before any homogenization, the link between us and them is simply: $$G^{(2)}(\mathbf r_1,\mathbf r_2)=2\left|\mathcal W\!\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},Z\right)\right|^2,\qquad G^{(2)}(\mathbf r_i,\mathbf r_i)=2\left|\mathcal W(x_i,y_i,Z)\right|^2$$ And we find again the coincidence rate thus giving us the pump profile.

The two-photon wavefront is then processed to be transformed into: $$\Psi^{(2)}(\mathbf r_1,\mathbf r_2)\propto\mathcal W\!\left(\frac{x_1+x_2}{2},\ \frac{y_1-y_2}{2},\ Z\right)\big(\mathbf H_1\mathbf V_2+(-1)^\zeta\mathbf V_1\mathbf H_2\big)$$ where $\zeta=0$ for Caetano and Souto Ribeiro but becomes $\zeta=1$ for Nogueira et al.. In both cases, the authors have flipped the sign of one photon's $y$ coordinate (with a Dove prism and a beam-splitter, respectively). Depending on the parity of $\mathcal{W}$, this makes the function odd (Nogueira) or even (Caetano), which is compensated by the polarization, as the Bose symmetry constrains the amplitude under the combined exchange $(\mathbf r_1,\sigma_1)\leftrightarrow(\mathbf r_2,\sigma_2)$. So the antisymmetry can come from $\mathbb C^d$ and the spatial kernel is then free to be antisymmetric with the state still perfectly bosonic. Polarization can be swept out of the picture with $G^{(2)}=\sum_{\sigma_1\sigma_2}|\Psi^{(2)}_{\sigma_1\sigma_2}|^2$. The homogeneity is only along $y$, so this is only half-spatial: $$G^{(2)}(\mathbf r_1,\mathbf r_2)=2\left|\mathcal W\!\left(\frac{x_1+x_2}{2},\frac{y_1-y_2}{2},Z\right)\right|^2,\qquad G^{(2)}(\mathbf r_i,\mathbf r_i)=2\left|\mathcal W(x_i,0,Z)\right|^2$$

Both now have that at the same (transverse) positions $y_1=y_2$, their functions are identically zero. That's where their CSI violations come from, and how it gets enforced everywhere, with no spatial structure, since cross correlations will not be zero. Nogueira et al. get this from their odd spatial pump profile, obtained by their $\pi$ shift of half of a Gaussian, turning it into basically a fermionic HG mode. Caetano and Souto Ribeiro achieve it by blocking the pump with a wire. The two approaches are, modulo this fairly anecdotal implementation detail—a symmetry-protected zero versus a force-engineered one—fairly similar. This is homogeneous in the sense that they can displace both detectors together, and the result remains the same. In our case, the positions of the detectors determine everything. We could still, get, though, violation of CSI for homogeneous cases too in a restricted sense: violation with a uniform one-photon intensity. The CSI could be violated everywhere on a beam of constant intensity. But our degree of violation has a structure, while theirs remains constant too (in principle, infinite violation).

Concretely, in our case, $\ket{1_{+\ell}1_{-\ell}}$ and $\ket{1_c1_s}$ have the same $G^{(1)}=2R(r)^2$—the same donut structure, indistinguishable to any intensity measurement (that's my "it's the same picture" joke)—and the first never violates anywhere while the second violates over open regions with $\ell$-fold structure. That kills the objection that our violations would just be beam structure.

15:59 (CET).

---

The most important conceptual difference between "us" and "them" (literature of spatial SPDC correlations) is, I now believe, the Schmidt rank gap: we're in the low Schmidt rank case of studying correlations between a few (in fact, 2) modes, while SPDC people not only usually deal with Schmidt number $K\sim\left(\frac{\text{transverse extent of }\mathcal W}{\text{width of }\Phi}\right)^{2}$, typically of order hundreds or thousands, but in the homogeneous case they strive for—which they indeed put at the center of their approach—they are in the $K\to\infty$ limit. They don't have perfectly homogeneous system so if we'd try to recover them from our approach, we'd need the full multimode theory, and we're not there yet. But, the difference is now clear: we have real, spatially structured CSI violations, while they violate it everywhere... and you know what they say: if it's for everybody, then it's for nobody. So I think we can claim we are the real spatial-CSI violating people.

(Here careful about Schmidt rank [number of modes] and Schmidt number [how many modes carry the weight]; Caetano and Souto Ribeiro have infinite Schmidt rank but actually fairly small Schmidts numbers)

For SPDC, the two-photon amplitude factorizes into a pump term at the mean coordinate and a phase-matching term at the relative one:

$$ \Psi^{(2)}(\mathbf r_1,\mathbf r_2)\propto\mathcal W\!\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ Z\right)\Phi(x_1-x_2,\ y_1-y_2;Z) $$

$\mathcal W$ is the pump field propagated to the detection plane $Z$ (although it got transformed into the two-photon beam in the process). A UV filter blocks the original pump beam after the crystal.

Diagonal and trace give different things. On the diagonal, the pump image survives intact:

$$ \Psi^{(2)}(\mathbf r,\mathbf r)\propto\mathcal W(x,y,Z)\,\Phi(0,0;Z) $$

whereas the trace is a convolution:

$$ G^{(1)}(\mathbf r)\propto\int d^2\mathbf v\;\big|\mathcal W(\mathbf r-\mathbf v,Z)\big|^2\,\big|\Phi(2\mathbf v;Z)\big|^2 $$

Let us define $a$ the transverse scale of the structure imprinted on the pump—for Caetano and Souto Ribeiro's imaged 250 µm wire, $a\approx0.5$ mm—and $\sigma$ the width of the pair-correlation factor $|\Phi|^2$ at the detection plane, i.e., how far apart the twins land. From the phase-matching angular spread, $\sigma\sim Z\sqrt{\lambda/L}$, which with $Z=0.75$ m, $\lambda=884$ nm and $L=1$ cm gives $\sigma\approx7$ mm. The convolution then leaves a residual contrast for the wire of order $a/\sigma\approx7\%$, smeared over 7 mm instead of $2a=1$mm. This is the idea of quantum imaging: the object is not visible in one-photon observables but is revealed in two-photon ones.

	rank-2 beam (ours)	SPDC beam (theirs)
pair diagonal $\Psi^{(2)}(\mathbf r,\mathbf r)$	$2\phi_a\phi_b$, dark on a cross of nodes	$\mathcal W(x,y,Z)$, the pump image, sharp
intensity $G^{(1)}(\mathbf r)$	$\lvert\phi_a\rvert^2+\lvert\phi_b\rvert^2$, a doughnut	$\lvert\mathcal W\rvert^2$ convolved with $\lvert\Phi\rvert^2$: flat, cm-wide
link between them	same two functions, exact	convolution over $\sim10^2$ modes

15:08 (CET).

11 August (2026)

I was wondering about bosonization as understood by bosonizers and by exciton people. I asked Nina and Mikhail Glazov about their opinion. Misha replied on the spot, that he was in Karelia and could not attend to the matter in details right away but would try on Thursday or Friday (this was on a Tuesday). And sure enough, first thing in the morning of Thursday came a long and detailed email.

There he explained that «in 1D the classical Fermi-liquid picture for interacting fermions (with repulsive interactions: typically, one band is considered like in metal or doped semiconductor) does not work» and that «it is because fermionic quasiparticles are ill defined», but that «It turned out that the problem of interacting particles can be elegantly solved in 1D by introducing novel bosonic modes (charge density/plasma wave and spin density/spin wave).»

All this is already very relevant, but even more the following:

Note that this bosonization approach is different from “pairing” of electrons and holes to form excitons: In the excitonic case we have (exact) solution of a two-body problem, namely, the “exciton”, and we approximate the manybody state as appropriately (anti)symmetrized products of single “exciton” states. In the case of the Luttinger liquid (bosonized 1D fermions) we consider collective modes right from the start.

That makes the 1D case very specific. He then observes:

Note that in 2D/3D cases the spectrum of Fermi liquid has both single-particle branch (quasielectrons/quasiholes) and manybody excitations (sound/plasmons and spin waves). In 1D case there are no single-particle excitations, at least at small wavevectors.

And back to 1D:

In 1D Luttinger liquid the key problem is somewhat different although it is also related to the commutation relations. Technically, they do the following: The total density of electrons is split into the densities of left- and right-movers, and commutation relations of (the Fourier components of) these densities are shown to have specific approximate form (that can be used to express everything in terms of bosonic operators). The accuracy of fulfilment of these relations is the better, the smaller are the wavevectors [in "Bosonization for beginners — refermionization for experts" (https://arxiv.org/pdf/cond-mat/9805275) it is briefly mentioned in Sec. 5 below Eq. (34) then they introduce the scale a=1/k_F]. The applicability of this procedure is, however, unrelated to the number of bosons, etc. as long as the interactions in the initial fermionic Hamiltonian are of “density-density” form.

Then he also comments on suitability of bosonization for phonons and magnons, which however isn't my point as I worry about the structural impossibility of the concept due to the compositeness of excitons, not on deviations from nonlinearities (which are dealt with correctly through perturbation, in my understanding).

He also makes more—and interesting—comments on regimes where excitonic breakdown would be relevant (small QDs—we actually have a paper on this—Rydberg excitons and polaritons, etc.) and concludes with:

As a historic anecdote, I also remember Robert Suris telling me about his argument with Leonid Keldysh back in 1970s: as you remember, Keldysh took great care of keeping fermionic corrections to the exciton commutation relations when studying exciton condensation [1] while Suris argued that it is enough to consider the exciton-exciton scattering amplitude [2]. In any case, their results agreed both qualitatively and quantitatively… Incidentally, Combescot approach showed that the correct calculation of the two-exciton scattering amplitude (in her technique) yields the same result as other people obtained using exciton language.

So no need of excitons in the case Haldane tackled exactly... but is it because the concept can be bypassed altogether, or because it is ill-defined? If it is bypassed, is it because they do something similar to Combescot, and tackle the full many-body, collective aspect from the start? If it is ill-defined, does it point out in this "simple" case where things can be solved exactly, that excitons are indeed problematic and cannot fit the picture? Or is it because the problems have not been connected and both communities ignore each other? I don't know about Combescot referring to Haldane (though I should check more carefully), but I know that the bosonization experts who visited us at ICMM several times 1/2/3 didn't know about her and her problems with excitons. In this case, what if one takes this exact case, as done by, say, Lee and Eric Yang, write an interacting bosonic Hamiltonian for this and compare the two theories. One remains exact, doesn't it? The other is the standard practice for most of the excitonic/polaritonic community. How do these relate?

11:07 (CET).