# Example of a deterministic model that does not satisfy Bell's inequality.

This is work in progress, help to improve or reject the idea is most welcomed.

There are loop holes like super determinism in Aspects experiment and we will search here for what features such a deterministic model can have in order to bust the magic of Bell's inequality. the magic which is motivated that QM is the result of non determinism. If you are wondering about super determinism this is a fun and informative video on youtoub or see her paper, Hossenfelder. Although the references discusses when the spin is the same in both particles we will discuss the case of two entangled electrons which spin ads up to 0 in order to conserve angular momentum. This system will also brake Bell's inequality.

The picture above shows how we slice for a fixed $$y$$ value e.g. in the measurement the slice will be a disk of possible measurements, The maximal disc is scaled by a factor of

$$\sqrt{1-y^2}.$$

The setup is that we have two measurement devices, $$X,Y$$. Where $$X$$ measures particle 1's spin and $$Y$$ particle 2's spin. E.g we will get a judgment is it is up or down ($$X \in \{1,-1\}$$) and similarly for $$Y$$. We shall consider that the the measurement $$X,Y$$ depends on a hidden variable $$L\in{\lambda,\lambda^*}$$ according to the diagram below (we indicate $$\lambda$$ when $$L=\lambda$$ is selected and $$\lambda^*$$ when $$L=\lambda^*$$ is selected). We will also assume that the true spin is a hidden variables with $$A\in[0,360]$$ particle 1's true spin angle with the $$z$$-component and $$B\in[0,360]$$ particle 2's spin angle with the $$z$$-component.

If we reverse the spin and asks for a spin down, we see that the diagrams above will be the same as they are symmetric under the transform $$A=\beta\to\beta+180$$.

The setup look a bit strange. But the selection of $$\lambda$$ is in relation to the other measurement and as they are related non spherically symmetric, the choice of $$\lambda$$ above do not inflect the spherical symmetry of a single measurement as it is only related to the underlying entanglement.

The target is to measure

$$P(X=1,Y=1|T=\alpha),$$

where $$T$$ is the tilted angle $$T=\alpha$$ of the angle $$\alpha$$ in the $$xz$$-plane (without loss of generality we can assume no tilt in the $$y$$ direction) that measurement device 2 is tilted. Due to the standard entanglement we have that $$B=A+180$$ is always true due to conservation of momentum, hence we can drop $$B$$ in the probabilities below. Furthermore if we reflect along the horizontal axis, we see that particle 1 goes into particle 2 and vice versa and also the angle between them is the same but of opposite angle. Also we will assume that for a spin with angle $$A=\beta$$ we assume we have the probability distribuition

$$P(X=1,L=\lambda,A=\beta \in Q_1) = P(X=1,L=\lambda^*,A=\beta \in Q_2) = C|\sin(\beta)\cos(\beta)|,$$

and note that $$Y$$ is assumed to follow from the condition in this probability density. Also,

$$P(X=1,L=\lambda,A=\beta \in Q_2\cup Q_3 \cup Q_4) = P(X=1,L=\lambda^*,A=\beta \in Q_1\cup Q_3\cup Q_4) = 0.$$

Further more we shall assume that (spherical symmetry in upper half plane)

$$P(X=1,A=\beta) = P(X=1,L=\lambda,A=\beta) \propto \cos(\beta).$$

Now assume that we are measuring a spin up in a quadrant. Then we will have an associated $$L$$ that will decide the corresponding $$L$$ in the second measurement.

Note, the probability measure is essentially the area of the underlying spin vector with the $$xy$$-plane projection of it hence circular symmetric around the $$z$$-axis. Let $$A=\beta$$ be defined as the angle between the $$x$$-axis and $$z$$-axis in the plane intersecting the point $$y$$. Note how the disc at y is scaled by length factor of (we assume without loss of generally a unit sphere),

$$\sqrt{1-y^2}.$$

And hence the Area scale is scaled by

$$1-y^2$$

And that by combining all results for discs with varying $$y$$ we see that the result of the probabilities is invariant of $$\beta$$ below (another constant though) and hence we will just consider one such disc.

As the model is symmetric of the transform $$A \to A+180$$, we will only concentrate ourselves on the region $$A\in[0,180]$$. Also as the transform $$\lambda \to \lambda^*$$ doesn't change the model, we see that if we starting with particle 1 and analyze the result will lead to the same as if we started with particle 2. An obvious symmetry that must holds. Also we see that the system is symmetric to changes in the rotation in the $$xy$$ plane of the setup.

The model defines an underlying hidden state $$A,L$$ selected via a deterministic entanglement and the result how the outcome of the measurement depends on this selected hidden variable. As $$A$$ is the physical spin of the particle, $$L$$ is the only innovation here.

So in order to analyze the requested probabilities we will first analyze the probability for a tilt $$T$$ in quadrants $$Q_1,Q_2$$ (the other quadrants will follow through the flip symmetry).

We will divide the calculation of the probability through 4 cases.

## Case 1, $$A\in Q_2, L=\lambda$$

Assume we tilt the second device and angle of $$T=\alpha \in Q_1$$. Then we get the diagram above and we note that the only place we have 1 in both particles is if $$A=\beta \in Q_1\cap Q_4^*$$. And this is with probability

$$P(X=1,Y=1,A=\beta\in Q_1,L=\lambda) = C\sin(\beta)\cos(\beta).$$

This means that

$$P(X=1,Y=1,A\in Q_1,L=\lambda) =C\int_0^{\alpha}\sin(\beta))\cos(\beta)\, d\beta = C\sin^2(\alpha)/2.$$

To see this note that

$$\int_0^{\alpha} \sin(x)\cos(x)\,dx = \frac{1}{2} \int_0^{\alpha} \sin(2x)\,dx = \frac{1}{4}(1 - \cos(2x)).$$

This is equal to

$$\frac{1}{4} (1 + (\sin^2(x)-\cos^2(x))\,dx = \frac{1}{4} (1+(2\sin^2(x)-1)\,dx = \frac{1}{2}\sin^2(x).$$

Similarly we see that

$$P(X=1,Y=-1|A \in Q_1,L=\lambda) = C \cos^2(\alpha)/2.$$

## case 2, $$A\in Q_2, L=\lambda$$

For $$\alpha \in Q_2$$ we get the situation in the above figure. We see that the intersecting region is again $$Q_1\cap Q_4^*$$ and we also have

$$\bar \alpha = \alpha^* = 180-\alpha.$$

Then the probability is

$$P(X=1,Y=1,A=\beta\in Q_2,L=\lambda) = C\sin(180-\beta)\cos(180-\beta).$$

So the integral becomes

$$P(X=1,Y=1,A\in Q_2,L=\lambda) = C \int_0^{180-\alpha} \sin(\beta)\cos(\beta)\, d\beta = C\sin^2(180-\alpha)/2.$$

But this is the same as,

$$C\sin^2(\alpha)/2.$$

Similarly we see that

$$P(X=1,Y=-1|A \in Q_2,L=\lambda) = C \cos^2(\alpha)/2.$$

## Case 3, $$A\in Q_1, L=\lambda^*$$

For $$\alpha \in Q_1$$ we get the above situation and we see that the interesting region is $$A=\beta \in Q_2\cap Q_1^*$$ and $$\alpha^*=\alpha$$ and hence $$\bar \alpha = 90-\alpha$$. We read the probability as

$$P(X=1,Y=1,A=\beta\in Q_1,L=\lambda^*) = C\sin(90-\beta))\cos(90-\beta).$$

But

$$\sin(90-\beta)\cos(90-\beta) = \sin(\beta)\cos(\beta).$$

And hence integrating the probability of $$A\in Q_1$$ is

$$P(X=1,Y=1,A\in Q_1,L=\lambda^*) = C\int_{0}^{\alpha}\sin(\beta)\cos(\beta)\,d\beta$$

which we calculated before as,

$$P(X=1,Y=1,A\in Q_1,L=\lambda^*) = C\sin^2(\beta)/2.$$

Similarly we see that

$$P(X=1,Y=-1|A \in Q_1,L=\lambda^*) = C \cos^2(\alpha)/2.$$

## Case 4, $$A\in Q_2, L=\lambda^*$$

The final situation is for the case $$L=\lambda^*,\alpha \in Q_2$$ according to the figure above. Here, for $$A=\beta \in Q_2\cap Q_1^*$$, $$\alpha^* = 180-\alpha$$ and we get the probability,

$$P(X=1,Y=1,A=\beta\in Q_2,L=\lambda^*) = C \sin(180-\beta)\cos(180-\beta).$$

And hence integrating the probability of $$A\in Q_2$$ is

$$P(X=1,Y=1,A\in Q_2,L=\lambda^*) = C\int_0^{180-\alpha}\sin(\beta))\cos(\beta)\,d \beta.$$

Which integrated means,

$$P(X=1,Y=1,A\in Q_2,L=\lambda^*) = C\sin^2(180-\alpha)/2 = C\sin^2(\alpha)/2.$$

Similarly we see that

$$P(X=1,Y=-1|A \in Q_2,L=\lambda^*) = C \cos^2(\alpha)/2.$$

This implies that indeed we reproduce the quantum mechanical results and by noting that combining both $$\lambda,\lambda^*$$ we get,

$$P(X=1,Y=1|T=\alpha) = P(X=1,Y=1,L=\lambda|T=\alpha)+P(X=1,Y=1,L=\lambda^*|T=\alpha),$$

is the same ,

$$P(X=1,Y=1|T=\alpha,A\in{Q1\cup Q_2}) = C \sin^2(\alpha).$$

Hence, symmetry gives,

$$P(X=1,Y=1|T=\alpha) = 2C \sin^2(\alpha) = \sin^2(\alpha)/2,$$
$$P(X=1,Y=-1,T=\alpha) = \cos^2(\alpha)/2.$$

We see that this is the same as in the Aspects Experiment using QM.

Therefore, the correlation is,

$$\text{Cor}(\alpha) = E(XY|T=\alpha) = -\cos^2(\alpha) + \sin^2(\alpha) = - \cos(2\alpha).$$

We know that for $$\alpha=45/2$$, we get essentially the same situation as in the example in Bell's, But with P=Cor.

So if you viewed the YouTube video linked in the beginning section, you will see that super determinism or a violation of the statistical independence condition meaning,

$$P(A,L|XY^*) \neq P(A,L)$$

Now for $$\beta,\alpha \in Q_1$$ we know that we can only select $$\lambda=1$$ if we know that $$X==Y$$. This means,

$$P(A = \beta \in Q_1,L=\lambda^*|XY=1) = 0$$

But

$$P(A \in Q_1,L=\lambda^*)=P(A \in Q_1,L=\lambda^*|XY=-1) > 0.$$

So the statistical independence is broken and hence the model we describe, as it produces the same results as we get in the Aspect experiment, we know it must be super deterministic as the model defined here is deterministic.

## Discussion

The following section is still under development and a bit too much handwaving involved, but it shows the direction of developing this blog post. It most certainly is sketchy and shaky.

Now you may argue that the model can't be physical, and there is a point to be made about this. So we need to find a more traditional model that can show some light abut how come we have this statistic. The best way going forward is to try define a model in terms of classical physics that is not too extreme. One of Einsteins ideas was that if the intrinsic spin has a property that when applied to a magnetic field, it will precess at a certain angle. This can explain Stern Gerlash result and is therefore an attractive assumption. So somehow we must introduce a torque into the particle that turns it around. One of the simplest ways to do this, and to me feels natural and not to complex, is to assume loops with center in the origin and having the spin's uniform in a half spherical shell. This idea is inspired by Randell Mill's GUTCP model and actually the main inspiration behind this super deterministic model. He also sugges and calculates the angle of precession and frequency. But bare in mind, as his book has bad reputation, that in this discussion it is just enough to consider an ensamble of intrinsic spin's in stead of a single one and assume the intrinsic torque will behaves similarly as for loops just as it is assumed in the traditional Stern Gerlash measuring device.

the torque for a spinning particle is

$$\tau \propto B\times S \propto \cos(\theta)$$

We can schematically describe this in teh following diagram,

$$<center> ![spin](images/pma.svg) </center>$$

This mean that we would end up with a proper spin up or a proper spin down depedning on the sign of this expression. and we would not be able to reproduce the prerequsits of the super deterministic example we had above.

The question now is what more do we need to add in order to have better luck. The obvious next step in the analysis are that the incoming particles is precessing when the magnetic fields is "turned on". If the precession vector is $$L$$, also in the plane, then the influence of that onto the spinn in the direction of the old spin direction is,

$$S_+ \propto \sin(\gamma).$$

Hence the torque from this direction given that the stedy point (overall spinn) is angled $$\beta$$ is,

$$\tau \propto \pm \cos(\beta)\sin(\beta+\gamma).$$

integrating the torque in the half sphere, then lead t0,

$$\tau \propto \pm \cos(\beta)\sin(\beta).$$

This leave us with the following differnt signs,

We see that in the spin up case the attractive regions are up/down, and the spin down system has attractive regions in the left/right direction. Now if we assume that $$\lambda,\lambda^*$$ both will be measured as a down particles when refelcted, which is needed, to be physical, we see that a $$\lambda$$ particle need to reflect to a particle that attracts to the right considered as down, and the $$\lambda^*$$ particles attracts to the left and also for this particle the left is measured as down in all producing the same logic as the diagram below.

This reproduce the phenomena in the introduction. Also if we assume that the density is related to the torque, we finally can produce the model we analyzed in the beginning. Also if we assume that the probabilities is reflected through the magnetude of the torque we used, then we basically have, albeight not yes set in stone argument (may contain bugs) for what kind of physcis is needed to produce the quantum mechanical analysis of aspects experiment.