August (2026)

2 August (2026)

I don't know. I don't know, I just don't know. I don't even know if I'm thinking about those photons who can't find their place jointly on the rim of a vortex, or if I'm thinking of why a talented, respectable, genius composer would give the best of his art to capture into music how lovers trick each other and get hurt in the process, or if I'm thinking of still other things. It all seems to collapse into the same story. And the more everything becomes one, the less I understand it, the more I find myself outside of the big picture. The universe, and me looking at it from its outside. Time to go back into it. I mean, to go to sleep. I'll have nightmares compete with dreams just as I try to have fermion components fight the bosonic ones.

00:45 (CET).

1 August (2026)

Not going far... $|1+d|=|1-(\frac{\Psi^-_{12}}{\Psi_{12}})^2|=|1-\frac{\Psi^-_{12}}{\Psi_{12}}||1+\frac{\Psi^-_{12}}{\Psi_{12}}|$. If the ratio is real, that becomes $0<(\frac{\Psi^-_{12}}{\Psi_{12}})^2<2$, i.e., $|\frac{\Psi^-_{12}}{\Psi_{12}}|<\sqrt2$ and this is the condition already spotted. I'm increasingly convinced this is not the good reading, however tempting is a ratio of bosonic vs fermionic fluctuations... Also this ratio $R$ might be broken and pathological, and the sign of the difference could be more physical/meaningful.

22:56 (CET).

Remember the "sampling matrix": $$A=\begin{pmatrix}w_1(\mathbf{r}_1)&w_1(\mathbf{r}_2)\\ w_2(\mathbf{r}_1)&w_2(\mathbf{r}_2)\end{pmatrix}\,.$$ This one has everything, it carries both statistics (up to normalization; should I maybe normalized$A$?):

its permanent is the two-boson amplitude $\Psi_{12}=w_1(\mathbf{r}_1)w_2(\mathbf{r}_2)+w_1(\mathbf{r}_2)w_2(\mathbf{r}_1)$.

its determinant the two-fermion amplitude $\Psi^-_{12}\equiv w_1(\mathbf{r}_1)w_2(\mathbf{r}_2)-w_1(\mathbf{r}_2)w_2(\mathbf{r}_1)$.

We want to find $\det M=\Psi_{11}\Psi_{22}-\Psi_{12}^2$ in this and then the CSI violation criterion through $d$ the ratio of $\det A/\det M$ (modulo eigenvalues). Algebra goes like this: $\Psi_{11}\Psi_{22}=4\,w_1(\mathbf{r}_1)w_2(\mathbf{r}_1)w_1(\mathbf{r}_2)w_2(\mathbf{r}_2)$, and calling $x=w_1(\mathbf{r}_1)w_2(\mathbf{r}_2)$ and $y=w_1(\mathbf{r}_2)w_2(\mathbf{r}_1)$, from $(x+y)^2-(x-y)^2=4xy$, we get to $\Psi_{12}^2-(\Psi^-_{12})^2=4\,w_1(\mathbf{r}_1)w_2(\mathbf{r}_2)w_1(\mathbf{r}_2)w_2(\mathbf{r}_1)$ and thus: $$\Psi_{11}\Psi_{22}=\Psi_{12}^2-(\Psi^-_{12})^2$$ so that, finally, $$\det M=-(\Psi^-_{12})^2$$
and so, $d$ is (minus the square of) the ratio of the fermionic over bosonic amplitudes:
$$\boxed{d=-\left(\frac{\Psi^-_{12}}{\Psi_{12}}\right)^{\!2}}$$
and if I link that to my earlier connection between $R$ and $d$, this tells us:

infinite violation ($d=−1$) when $(\Psi^-_{12})^2=\Psi_{12}^2$, i.e. $\Psi^-_{12}=\pm\Psi_{12}$. That sounds deep but actually, since $\Psi_{12}+\Psi^-_{12}=2w_1(\mathbf{r}_1)w_2(\mathbf{r}_2)$ and $\Psi_{12}-\Psi^-{12}=2w_1(\mathbf{r}_2)w_2(\mathbf{r}_1)$, that merely means that one of the cross products vanishes, i.e., one detector is placed at a zero of one of the modes, which is the well-known maximum violation of CSI. So no big deal here... That's also why fermions appear more nonclassical by this standard, by the way, they intrinsically forbid particles to fluctuate locally, because they can't be many present locally in the first place.

strong validation of CSI when $\Psi_{12}\to0$, i.e., the bosonic part vanishes, which seems counterintuitive from the point of view I'm chasing (fermion vs boson), but from $|\Psi_{12}|^2\le|\Psi_{11}\Psi_{22}|$ it just shows it's trivially enforced.

I might be going in circles... the bosonic vs fermionic might be a red herring.

22:12 (CET).

I want to explore the symmetry vs antisymmetry character of states, since that seems to be what rules their CSI violations.
Let us start with the great classic: $$F=\tfrac12(F+PF)+\tfrac12(F-PF)$$ where $P$ is the permutation operator $\mathbf{r}_1\leftrightarrow\mathbf{r}_2$. This splits the Hilbert space of two-particle wavefuntions as $$\mathcal{H}_1\otimes\mathcal{H}_1=\mathrm{Sym}^2\mathcal{H}_1\oplus\Lambda^2\mathcal{H}_1$$ where Sym$^2$ is the symmetric square and $\Lambda^2$ the exterior (or wedge) square. That brings us back to the Takagi decomposition $\Psi=\sum_k\lambda_k\,w_k\otimes w_k$ but now extended to also include fermionic-looking antisymmetric cases.
Sym2 is spanned by $w_j\otimes w_k+w_k\otimes w_j$ for $j<k$, plus the diagonal elements $w_k^{\otimes2}$. It has dimension $K(K+1)/2$. $\Lambda^2$ is spanned by $w_j\wedge w_k\equiv w_j\otimes w_k-w_k\otimes w_j$ for $j<k$. No diagonal, since $w\wedge w=0$ (Pauli principle). It has dimension $K(K-1)/2$. Since $K(K+1)/2+K(K-1)/2=K^2$, we're all accounted for.
Let us work out the general case $\ket{1_a1_b}$ (general in $a$ and $b$, this is still particular about two particles):
For the symmetric (bosonic) part: $$\mathrm{Sym}^2=\mathrm{span}\{a^{\otimes2},\,b^{\otimes2},\,a\otimes b+b\otimes a\}\ \leftrightarrow\ \{|2,0\rangle,|0,2\rangle,|1,1\rangle\}$$
For the antisymmetric (fermionic) part: $$\Lambda^2=\mathrm{span}\{a\wedge b\}\ \leftrightarrow\ \text{the single Slater determinant}$$
The link between the ratio of CSI violation and this distance is important: $$R=\frac{1}{|1+d|^{2}}$$ (this follows from the definition of $d$ such that $1+d={\Psi_{11}\Psi_{22}}/{\Psi_{12}^{2}}$). Thus $R\to\infty$ at the centre $d=-1$ (where $\Psi_{11}\Psi_{22}=0$, the nodal cross) and $R\to0$ as $|d|\to\infty$ (where $\Psi_{12}=0$, the anti-diagonal).
Divergent $R$ always means a diagonal amplitude vanishing at one of the two detectors; vanishing $R$ always means the pair amplitude vanishing between them. This is for any $\ket{1_a,1_b}$.
That doesn't clarify yet the fermionic/bosonic tension, but I wanted $R$ to appear explicitely. Through $d$, it does.

20:33 (CET).

The bowtie boundary for the CSI satisfied/violated for dipoles (HG) is given by the equation: $$\theta_2=\arctan(-(3\mp2\sqrt2)\tan\theta_1)=\arctan(-(\sqrt2\mp1)^2\tan\theta_1)$$ I leave below the Mathematica code to plot the relevant quantities for this case.

This plots the densityplot of CSI violations:

ClearAll[viol, safe, cfun];
viol[t1_, t2_] := Sin[t1 + t2]^2/Abs[Sin[2 t1] Sin[2 t2]];

With[{cap = 10., gam = 0.45}, 
 safe[x_] := 
  If[TrueQ[Element[x, Reals]] && NumericQ[x], Clip[x, {1/cap, cap}], 
   cap];
 cfun[r_] := 
  Module[{f}, 
   f = Sign[#] Abs[#]^gam &@Clip[Log[r]/Log[cap], {-1., 1.}];
   If[f < 0, Blend[{GrayLevel[0.05], White}, 1 + f], 
    Blend[{White, RGBColor[0.11, 0.69, 0.48]}, f]]];
 With[{n = 700}, 
  Module[{g}, 
   g = Table[
     safe@N@viol[t1, t2], {t2, Pi/(2 n), Pi - Pi/(2 n), Pi/n}, {t1, 
      Pi/(2 n), Pi - Pi/(2 n), Pi/n}];
   Show[ListDensityPlot[g, DataRange -> {{0, Pi}, {0, Pi}}, 
     InterpolationOrder -> 0, Mesh -> None, Frame -> True, 
     ColorFunctionScaling -> False, ColorFunction -> cfun, 
     PlotLegends -> 
      BarLegend[{cfun, {1/cap, cap}}, ScalingFunctions -> "Log", 
       Ticks -> {0.1, 0.3, 1, 3, 10}, LegendLabel -> "R", 
       LegendMarkerSize -> 260], 
     FrameLabel -> {Subscript[\[Theta], 1], Subscript[\[Theta], 2]}, 
     FrameTicks -> {{{{0, "0"}, {Pi/2, "\[Pi]/2"}, {Pi, "\[Pi]"}}, 
        None}, {{{0, "0"}, {Pi/2, "\[Pi]/2"}, {Pi, "\[Pi]"}}, None}}],
     ContourPlot[viol[t1, t2] == 1, {t1, 0, Pi}, {t2, 0, Pi}, 
     PlotPoints -> 300, 
     ContourStyle -> Directive[Red, Dotted, Thick]], 
    Graphics[{Red, Dotted, Thick, Line[{{0, 0}, {Pi, Pi}}]}], 
    PlotRange -> {{0, Pi}, {0, Pi}}]]]]

This shows a cut in the density plot: wild variations (to infinity)

Manipulate[
 LogPlot[viol[t1, t2], {t1, 0, Pi}, 
  PlotRange -> {{0, Pi}, {0.05, 50}}, PlotPoints -> 400, 
  Exclusions -> {Sin[2 t1] == 0, Sin[t1 + t2] == 0}, 
  PlotStyle -> Directive[RGBColor[0.11, 0.69, 0.48], Thick], 
  GridLines -> {{{t2, Directive[Red, Thick]}, {Mod[Pi - t2, Pi], 
      Directive[Gray, Dashed]}}, {{1, Directive[Black, Thick]}}}, 
  Frame -> True, Axes -> False, 
  FrameLabel -> {Subscript[\[Theta], 1], "R"}, 
  FrameTicks -> {{Automatic, 
     None}, {{{0, "0"}, {Pi/2, "\[Pi]/2"}, {Pi, "\[Pi]"}}, None}}, 
  ImageSize -> 460], {{t2, 0.35 Pi, Subscript[\[Theta], 2]}, 0.001, 
  Pi - 0.001, Appearance -> "Labeled"}, SaveDefinitions -> True]

17:39 (CET).

The first thing is when is $\sin2\theta_1\sin2\theta_2>0$? Both must be positive or both negative, so that's the quadrant 1 and 3 of $\begin{smallmatrix} 2&1\\3&4 \end{smallmatrix}$. So, there are trivial violations of CSI along the diagonal, where photons are detected together. Interesting! In the 2 and 4 quadrants is where we have restricted violations:
The condition to retain is:
$$\boxed{\;\text{CSI violated}\iff\lvert\sin(\theta_1-\theta_2)\rvert<\sqrt2\,\lvert\sin(\theta_1+\theta_2)\rvert\;}$$ with saturation at $\sin(\theta_1-\theta_2)=0$, i.e., $\theta_1=\theta_2\mod\pi$ and infinite violation at $\sin(\theta_1+\theta_2)=0$, i.e., $\theta_1=\theta_2+\pi/2\mod\pi$.
This compare the difference (fermionic like) to the sum (bosonic like) with the conclusion that the fermionic amplitude must not exceed the bosonic one by more than $\sqrt{2}$. This number is probably not random and its physical meaning would pop up more clearly at the multiphoton level. There's some bosonic fuckery going on there.
We thus satisfy CSI (and saturate them) when bosons are aligned, although this is impossible for dipoles, and we violate them maximally when they are perpendicular, where they want to be found.
This is the region of CSI violations for two bosonic Fock dipoles $\ket{1_u,1_v}$ in the $(\theta_1,\theta_2)$ plane:

It seems to agree with what Daniel showed me the other day, except it was on $[0,2\pi[$ (hence four times this) and not showing the degree of violation, which is important as in fact it's weakly violated in most places and gets wild when it becomes impossible; also it was with his own color code as opposed to the one we used with Ref. _^[1]. Red here is saturation of the inequalities.

15:02 (CET).

The trick is the one I used previously, to separate the cases of the modulus, i.e.,

$\sin2\theta_1\,\sin2\theta_2>0$ and we drop the absolute value

$\sin2\theta_1\,\sin2\theta_2<0$ and the numerator gets a minus (to drop $|\cdot|$)

and, depending on the cases, we have $|\sin2\theta_1\sin2\theta_2|=\pm\big(\sin^2(\theta_1+\theta_2)-\sin^2(\theta_1-\theta_2)\big)$ and, therefore:
$$\lvert d+1\rvert=\pm\left(1-\frac{\sin^2(\theta_1-\theta_2)}{\sin^2(\theta_1+\theta_2)}\right)$$ and this is to be compared to unity: smaller or larger?
The $+$ is a trivial violation everywhere unless saturated, as $|d+1|<1\iff 0<\frac{\sin^2(\theta_1-\theta_2)}{\sin^2(\theta_1+\theta_2)}$ which is true except if $\sin^2(\theta_1-\theta_2)=0$ where the CSI is saturated, which happens when $\theta_1=\theta_2\mod\pi$.
The $-$ is the case I already touched upon with the golden ratio for $t=p/q$. In this case $\lvert d+1\rvert<1$ becomes
$$\sin^2(\theta_1-\theta_2)<2\sin^2(\theta_1+\theta_2)$$
If I compare with Daniel's result, which states violations iff $$\sin^2(2\theta_1)\sin^2(2\theta_2)<\sin^4(\theta_1+\theta_2)$$ one can check numerically that it fulfills the same conditions—the two are equivalent—so no issues there and it's certainly not a big deal to show cleanly (not entirely immediate to me in five minutes, $\sin2\theta_1\sin2\theta_2=\sin^2(\theta_1+\theta_2)-\sin^2(\theta_1-\theta_2)$ and there has many things to look at; not my business for now). Now for the geometric consequences of that, it's what interests me: where does this happen?

13:50 (CET).

Carrying on with which areas violate CSI spatially, in the concrete case of $u=h_{10}\propto x\,e^{-r^2/w^2}$ and $v=h_{01}\propto y\,e^{-r^2/w^2}$, and recalling (it's been a long break) that $p\equiv u(\mathbf r_1)v(\mathbf r_2)$ and $q\equiv u(\mathbf r_2)v(\mathbf r_1)$, while $d\equiv-(p-q)^2/(p+q)^2$ and we're interested in the region $|d+1|<1$, then, since $p=x_1y_2 e^{-(\mathbf{r}_1+\mathbf{r}_2)/w^2}$ and $q=x_2y_1 e^{-(\mathbf{r}_1+\mathbf{r}_2)/w^2}$, calling $G\equiv e^{-(\mathbf{r}_1+\mathbf{r}_2)/w^2}$ the part which factors out, we have $p-q=(x_1y_2-x_2y_1)G= \left|\begin{smallmatrix} x_1 & y_1\\ x_2 & y_2 \end{smallmatrix}\right|G $ and $p+q=(x_1y_2+x_2y_1)G= \left\Vert\begin{smallmatrix} x_1 & y_1\\ x_2 & y_2 \end{smallmatrix}\right\Vert G$ so that $G$ neatly cancels and from trigonometric identities such as $\cos\theta_1\sin\theta_2-\cos\theta_2\sin\theta_1=\sin(\theta_1-\theta_2)$, we get, again with cancellation of $r$, $p-q\propto\sin(\theta_2-\theta_1)$ and $p+q\propto\sin(\theta_1+\theta_2)$, and thus $d=-\frac{\sin^2(\theta_2-\theta_1)}{\sin^2(\theta_1+\theta_2)}$ so that, finally:
$$\qquad \lvert d+1\rvert=\frac{\lvert\sin2\theta_1\,\sin2\theta_2\rvert}{\sin^2(\theta_1+\theta_2)}\,.$$
The problem is then, when is this strictly less than 1?

12:39 (CET).

This month will be in the company of Falla

It'll be a long one, a lone one, a none one, a noon one too. A full dive into El Amor Brujo has long been overdue. Even L'Amour Monstre was not so urgent or compelling.

11:02 (CET).

A science friend asked me about my plans for the 12th of August, and I asked what was going on on the 12th of August. Way to lose a science friend! She just replied, rightly «The World Cup is one thing and a total solar eclipse is another!» Ouch. I would have had given my utmost attention to this sort of things still some little time ago. I had seen that ICMM and BP (the oil company) provide equipment for "safe viewing", and I'm scared to tamper more with my eyesight than I already do by closing my eyes very little (not sleeping much) and focusing them on either books or screens, and when not one or the other, usually one and the other. So I guess I had kind of filed it under "avoid; keep distance from attractive things, this hasn't turned out too well for you recently anyway". Maybe I should still give the moon meeting the sun the attention it deserves. This is the sort of impossible union that has been occupying me a lot and indeed not turned out too good. But what can be done about it? All those stuff are cosmic events.

Where to see the 12th August eclipse.
01:17 (CET).

↑ Violation of classical inequalities by photon frequency filtering . C. Sánchez Muñoz, E. del Valle, C. Tejedor and F.P. Laussy in Phys. Rev. A 90:052111 (2014).

[cite-sanchezmunoz14b-1] Violation of classical inequalities by photon frequency filtering . C. Sánchez Muñoz, E. del Valle, C. Tejedor and F.P. Laussy in Phys. Rev. A 90:052111 (2014).

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