A question about Bell's Inequality and hidden variables

[rede96]

I'd like to start off by saying I'm just a 52 yo interested layman with no back ground in physics so apologize up front for my ignorance!

I understand the basic principle behind Bell's Inequality and how it disproves that when measuring the different spin states of a particle, the particle can't have predetermined properties of spin such as 'spin up' and 'spin left' for example.

But to be honest I never really understood why this ruled out some form of hidden variables that although didn't lead to combinations of spin states being predetermined, it could for example allow for entangled particles to be correlated when measured at the same angle while being random at other angles.

With that in mind I thought it is possible to break bell’s inequality using a classical example if one or more of the variables are random when measured, such as being time or position dependent.

For example, I could ask a group of people three yes/no questions. (a) Were you born in the northern hemisphere? (b) do you like cheese? and (c) are you in America now?

I ask questions (a) and (c) and then at some point later I ask questions (b) and (c). While answer (a) would never change, question (c) may depending on where they were when I asked. So in that way a person could answer yes the first time and no the second, therefore breaking Bell’s inequality.

And although question (c) would appear random, if I knew enough about the person I could maybe predict where they might be and hence predict their answer.

So my question: Is there any reason why this can’t this be applied on a quantum level? Is there any reason why we can’t assume that maybe there are some hidden variables that can cause an entangled pair of particles to be correlated when measured at the same angle but random when measured at a different one?

Sorry if the answer is obvious but I've not been able to think of it.

 

[PeterDonis]

I never really understood why this ruled out some form of hidden variables

It doesn't rule out hidden variables. It rules out local hidden variables. That's a more restrictive concept.

For an example of a theory with non-local hidden variables, take a look at Bohmian mechanics. The hidden variables are the particle positions, which in Bohmian mechanics are always definite. But they are non-local because the particle positions are affected by the quantum potential, which includes influences from outside the past light cone.

 

[rede96]

It rules out local hidden variables.

This is the part that I've still not understood as I saw a difference between local variables that are static (such as an electron having a predetermined state of being spin up and spin left) and local variables that were random depending on other factors that we may not be aware of.

 

[PeterDonis]

I saw a difference between local variables that are static (such as an electron having a predetermined state of being spin up and spin left) and local variables that were random depending on other factors that we may not be aware of.

There's no difference from the standpoint of Bell's Theorem. "Local" doesn't mean "static"; it means, heuristically, "depending only on things in the past light cone". In Bell's mathematical formulation, this shows up as the requirement that the joint quantum state of the two entangled particles, being measured at spacelike separated events, must be factorizable--i.e., it must be the product of individual quantum states for each particle, where each individual quantum state depends only on variables in the past light cone of the corresponding measurement event. That covers both of the possibilities you describe: the variables could be fixed, or they could be randomly determined at each measurement event, as long as, for each individual particle, the variables are restricted to the past light cone.

 

[rede96]

There's no difference from the standpoint of Bell's Theorem. "Local" doesn't mean "static"; it means, heuristically, "depending only on things in the past light cone". In Bell's mathematical formulation, this shows up as the requirement that the joint quantum state of the two entangled particles, being measured at spacelike separated events, must be factorizable--i.e., it must be the product of individual quantum states for each particle, where each individual quantum state depends only on variables in the past light cone of the corresponding measurement event. That covers both of the possibilities you describe: the variables could be fixed, or they could be randomly determined at each measurement event, as long as, for each individual particle, the variables are restricted to the past light cone.

Ah ok. As I'd only looked at Bell Theorem with simple math so I'd not thought of it in those terms. Although I'll be honest I don't fully understand why this is the case and why it still leads to only non local influences.

May I ask does a violation of bell's theorem state that any hidden variables can not be from the past light cone or that they are just not restricted to the past light cone?

 

[PeterDonis]

does a violation of bell's theorem state that any hidden variables can not be from the past light cone or that they are just not restricted to the past light cone?

There is no requirement that a theory that violates the Bell inequalities and reproduces the predictions of QM have hidden variables at all. But if it does, then some of them must be nonlocal. I think that means the latter of your two possibilities is correct, to the extent that they are meaningful.

 

[Justintruth]

"...it could for example allow for entangled particles to be correlated when measured at the same angle while being random at other angles.."

I think the point is that if a hidden variable were to exist as you say then the degree of correlation is dependent on the angle set in both detectors. This is implied when you say the "same angle" or "other angles" as it is the setting of one detector relative to the setting of the other detector that makes them the "same" or "other". But the detectors can be very far apart and they can be set anytime prior to reception of the entangled photon at the detector and it is not the absolute angle of the detector but the angle relative to the distant detector that determines the degree of correlation. Since the variable you describe does not determine the correlation on its own, but depends on both the local and distant detector angles, then the hidden variable you describe is not local.

Because it is only the correlation that is affected then no information or "signal" can be sent from one detector to the other using this effect. The correlation can be affected but you cannot know what the state of the correlation is without physically comparing the results of the experiments at each detector and so you have to send the result of both experiments somewhere to be compared and that can only be done at light speed or less. You cannot determine information about the correlation from the results of the experiments at either detector alone, so the fact that the correlation changed or did not change cannot be determined locally. However, after you do the comparison you can in fact see that the correlation changed at a time that cannot be explained by some signal traveling from one detector to the other at light speed or less. Just because you cannot determine whether a change in the correlation occurred without comparison does not mean that you will not be able to determine when the change occurred at some latter time after you do the comparison. A detector one light year away from another changes the degree of correlation of entangled pairs as soon as its angle is changed, not one year from the time its angle was changed, but you cannot see that change until you send the information from both detectors somewhere to compare the results and establish what the correlation is. That prevents signaling.

Another thing that has to be remembered is what happens when you have one detector very close to where a stream of entangled pairs are created and another very far. The change of detector angle on the near detector will have no effect on the correlation until the other, opposite, member of the pair is received at the distant end. In other words the correlation is a correlation of one of the pair that was created with the other of the pair created in the creation event. Clearly no correlation exists until both members of the pair are measured and you can't measure them until you receive them so no faster than light signaling.

 

[DrChinese]

Is there any reason why we can’t assume that maybe there are some hidden variables that can cause an entangled pair of particles to be correlated when measured at the same angle but random when measured at a different one?

Sorry if the answer is obvious but I've not been able to think of it.

Sure, you can try to assume that. But any version you come up with that actual characteristic WON'T match the predictions of QM. That is Bell's Theorem, right? After all, QM makes a specific prediction: the correlation for entangled spin 1 particles would be cos^2(theta) where theta is the relative angle between the 2 detector settings. You must match that with your specific hidden variable (HV) model. Usually, everyone forgets that little detail of a step when they are starting where you are starting from. But we know that is not possible, thanks to Bell.

If you actually try to do this, you will quickly see that nothing works out. The easiest (IMHO) one to work with is angles of 0, 120, and 240 degrees. Try for yourself, any pair of those selected should match 25% of the time. Good luck!

 

[rede96]

If you actually try to do this, you will quickly see that nothing works out. The easiest (IMHO) one to work with is angles of 0, 120, and 240 degrees. Try for yourself, any pair of those selected should match 25% of the time. Good luck!

Lol, I think I've been down this route before but it'll be a bit of fun to play around with. Is that 25% of the time including those measured at the same angle or excluding? Thanks

 

[DrChinese]

Lol, I think I've been down this route before but it'll be a bit of fun to play around with. Is that 25% of the time including those measured at the same angle or excluding? Thanks

Right, it's 100% of the time when the 2 angles are the same. And otherwise 25%. If you attempt to put together a set of examples, you will start seeing the issue pretty quickly. You can't get it to work out right unless you select the 2 angles AFTER you know the values.

So suppose A=0 degrees, B=120 degrees and C=240 degrees. Mark down fictional + or - values for all 3 columns and attempt to make them be as close to 25% as possible. You will find that you can get one pair (say AB) to work out, and even a second pair (say BC) to work out. But if you do that, the last pair (AC) will be 50% rather than 25%. No matter how you hand pick the fictional values, the *best* you can do is an average of 33% for any 2 angles. So you need to know the answers in advance in order to agree with QM.

 

[rede96]

suppose A=0 degrees, B=120 degrees and C=240 degrees

May I ask a quick couple of questions. If the angles are set at say 0, 90 and 180 degrees how would I work out the quantum prediction? Also if angle A was 0 and B was 90, Cos 90 = 0. Which is a bit odd as I thought it would be 50% probability?

 

[DrClaude]

May I ask a quick couple of questions. If the angles are set at say 0, 90 and 180 degrees how would I work out the quantum prediction? Also if angle A was 0 and B was 90, Cos 90 = 0. Which is a bit odd as I thought it would be 50% probability?

The probability of getting anti-correlated results between two detectors set at an angle θ with respect to each other is cos²(θ/2). If the detectors are 90° apart, you do indeed get the 50% probability you expect.

 

[zonde]

May I ask a quick couple of questions. If the angles are set at say 0, 90 and 180 degrees how would I work out the quantum prediction? Also if angle A was 0 and B was 90, Cos 90 = 0. Which is a bit odd as I thought it would be 50% probability?

You have to specify entangled particles and entangled property i.e. is it photon polarization or say electron spin.
For photon polarization it is $\cos^2(\theta)$ or $\sin^2(\theta)$ depending on the type of entanglement. So with orthogonal polarizers you get perfect correlation or perfect anticorrelation.

 

[PeterDonis]

The probability of getting anti-correlated results between two detectors set at an angle θ with respect to each other is cos²(θ/2).

This is for electron spins. As @zonde noted, for photon spins the relevant angle parameter is just $\theta$, since photons are spin-1 while electrons are spin-1/2.

 

[DrChinese]

You have to specify entangled particles and entangled property i.e. is it photon polarization or say electron spin.
For photon polarization it is $\cos^2(\theta)$ or $\sin^2(\theta)$ depending on the type of entanglement. So with orthogonal polarizers you get perfect correlation or perfect anticorrelation.

Yes, I was not clear enough in my example. Entangled photons and entangled electrons have different correlation statistics, and therefore an example could vary for each. I gave representative angles for photons (spin 1) of 0/120/240 degrees, where the are entangled as correlated (as opposed to anti-correlated). For electrons, a similar effect would be achieved at angles of 0/240/480. That happens to work out to be the same as for photons after you rearrange a bit (if I did the arithmetic right).

Also: I always use the "Match" statistics, which is a % and varies from 0 to 1 (and .5 implies complete randomness). True correlation statistics themselves vary from -1 to +1 where 0 is neither correlated or anti-correlated. That is calculated as: Correlation = (Match % - Mismatch %). Most folks just quote the Match statistics because they are simpler, at least they are for me.

 

[rede96]

The probability of getting anti-correlated results between two detectors set at an angle θ with respect to each other is cos2(θ/2). If the detectors are 90° apart, you do indeed get the 50% probability you expect.

You have to specify entangled particles and entangled property i.e. is it photon polarization or say electron spin.
For photon polarization it is cos2(θ)cos2⁡(θ)\cos^2(\theta) or sin2(θ)sin2⁡(θ)\sin^2(\theta) depending on the type of entanglement. So with orthogonal polarizers you get perfect correlation or perfect anticorrelation.

This is for electron spins. As @zonde noted, for photon spins the relevant angle parameter is just θθ\theta, since photons are spin-1 while electrons are spin-1/2.

Thank you :-)

 

[rede96]

I gave representative angles for photons (spin 1) of 0/120/240 degrees, where the are entangled as correlated (as opposed to anti-correlated). For electrons, a similar effect would be achieved at angles of 0/240/480.

Thanks for your help. Just to make sure I'm doing my math correct, could you tell me what the quantum prediction would be for angles a) 0 b) 90 and c) 180
And if you have time maybe a) 10 b) 60 and c) 200. This isn't a test of inequality, I'm just making sure I can calculate the right quantum prediction at different angles.

Also, would I get different results for entangled photons and entangled electrons at these angles?

 

[DrClaude]

This is for electron spins. As @zonde noted, for photon spins the relevant angle parameter is just $\theta$, since photons are spin-1 while electrons are spin-1/2.

Yes, I was assuming spin-1/2, with an initial two-particle state $\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle \right)$. I should have made that clear.

 

[DrChinese]

Thanks for your help. Just to make sure I'm doing my math correct, could you tell me what the quantum prediction would be for angles a) 0 b) 90 and c) 180
And if you have time maybe a) 10 b) 60 and c) 200. This isn't a test of inequality, I'm just making sure I can calculate the right quantum prediction at different angles.

Also, would I get different results for entangled photons and entangled electrons at these angles?

If the entangled particles are electrons, they almost always have opposite spins (anti-correlated). The general formula is:

Match%= 1 - (cos^2(theta/2))

So at theta=0 degrees, the result is 0. At 90 degrees, .5. At 180 degrees, 1.

If the entangled particles are photons, they can be correlated or anti-correlated. For anti-correlated, the general form is:

Match%= 1 - (cos^2(theta))

So at theta=0 degrees, the result is 0. At 90 degrees, 1. At 180 degrees, 0.

 

[rede96]

If the entangled particles are electrons, they almost always have opposite spins (anti-correlated). The general formula is:

Match%= 1 - (cos^2(theta/2))

So at theta=0 degrees, the result is 0. At 90 degrees, .5. At 180 degrees, 1.

If the entangled particles are photons, they can be correlated or anti-correlated. For anti-correlated, the general form is:

Match%= 1 - (cos^2(theta))

So at theta=0 degrees, the result is 0. At 90 degrees, 1. At 180 degrees, 0.

Thanks you. The main bit I'm still stuck on is how to calculate the total Match prediction is for a set of three angles.

E.g. For angles 0,120,240 I know it's 0.25 for correlated photons using the formula (cos^2(thetaA - thetaB)) for each of the 3 combinations and then taking the average of the 3 results.

But would this be the same formula for anti correlated electrons at those angles? Or are you saying it's just 1 - (cos^2(theta)) for each of the 3 angles then take the average of those results?

Sorry to ask, but it's just really confusing me :-)

 

[DrChinese]

Thanks you. The main bit I'm still stuck on is how to calculate the total Match prediction is for a set of three angles.

E.g. For angles 0,120,240 I know it's 0.25 for correlated photons using the formula (cos^2(thetaA - thetaB)) for each of the 3 combinations and then taking the average of the 3 results.

But would this be the same formula for anti correlated electrons at those angles? Or are you saying it's just 1 - (cos^2(theta)) for each of the 3 angles then take the average of those results?

Sorry to ask, but it's just really confusing me :-)

No bad questions, are there?

Electron pairs are usually anti-correlated. So the match percentage at any 2 of those 3 angles would be .75 rather than .25. Then the relevant limit you would encounter would be trying to exceed 66% (as compared to 33% for the correlated cases). Hopefully that makes sense. See my post #10.

So as to calculating for 3 angle pairings (AB, BC, AC): Yes, you can just average the 3 expectation values. My example is hand picked so that theta=AB=BC=AC and presumably you would expect that the 3 expectations values would be the same (unless there is some preferred direction). So with my chosen angles, the QM prediction for the anti-correlated case is 75% but no sets of hand values work out to stats better than our 66%.

Again, you have to run through a few specific cases to realize that this is so. An anti-correlated example:

Particle 1, for angles 0,120,240: - + -
Particle 2, for angles 0,120,240: + - +

P1@0 is -, P2 at 120 is -, that's a Match.
P1@0 is -, P2 at 240 is +, that's no Match.
P1@120 is +, P2 at 0 is +, that's a Match.
P1@120 is + P2 at 240 is +, that's a Match.
P1@240 is -, P2 at 0 is +, that's no Match.
P1@240 is -, P2 at 120 is -, that's a Match.

4 matches out of 6 permutations, that's 66%. That's as good as it gets.

 

[rede96]

No bad questions, are there?

Ah but there are questions that won't lead to the answer you are seeking! I tend to ask a lot of those :)

Electron pairs are usually anti-correlated. So the match percentage at any 2 of those 3 angles would be .75 rather than .25

This is the bit I still can't understand. How to calculate the percentage match that would be observed in experiments.

As I understood from your post #19, to calculate the % match for anti-correlated particles, for each angle it would be Match%= 1 - (cos^2(theta/2))

So for angles 0, 120 and 240 that would be:

0 is 1 - (cos^2(0 degrees/2)) = 0
120 is 1 - (cos^2(120 degrees/2) = 0.75
240 is 1 - (cos^2(240 degrees/2)) = 0.75

So the average of those is 0.5 So how does the total match prediction = 0.75? With photons it was (cos^2(angle A - angle B)) So is that not the case here?

This is the one bit I'd really like to understand. How to calculate the total match percentage for any set of 3 angles for both correlated and anti-correlated particles.

Also I am assuming that the total match prediction is a percentage of the total number of tests - any tests where the angles where the same.

Then the relevant limit you would encounter would be trying to exceed 66% (as compared to 33% for the correlated cases).

I'm not questioning Bell theorem, so don't disagree. But I am exploring different scenarios just for a bit of fun. Hence why I wanted to know how to calculate the total Match % for any set of 3 angles.

 

[zonde]

This is the one bit I'd really like to understand. How to calculate the total match percentage for any set of 3 angles for both correlated and anti-correlated particles.

I don't see the point of calculating average match of different measurement settings. You never do that in experiments. Measurement settings are very well known and recorded variables. Why pretend that they are hidden variables? That's the reason why I don't like Mermin's version of Bell inequality.

If you have not seen this version of simple proof of Bell's inequality you can give it a try. IMO it is easier to understand.

 

[rede96]

I don't see the point of calculating average match of different measurement settings. You never do that in experiments

Just to be clear, all I am asking is for a given set of 3 angles how do I calculate how many matches I'd expect to measure in total if I ran the experiment on those 3 angles. (For both correlated and anti correlated entanglement) And for comparing the match % do I ignore any results when the angles are the same.

 

[rede96]

If you have not seen this version of simple proof of Bell's inequality you can give it a try. IMO it is easier to understand.

Thanks for that.

 

[zonde]

And for comparing the match % do I ignore any results when the angles are the same.

In another recent thread Lord Jestocost gave a link to http://www.theory.caltech.edu/classes/ph125a/istmt.pdf. So you can read it yourself. The answer is at the very end of page 10. Mermin calculates average from all three angles. 1/3*1 + 2/3*(1/4)=1/2

 

[zonde]

Just to be clear, all I am asking is for a given set of 3 angles how do I calculate how many matches I'd expect to measure in total if I ran the experiment on those 3 angles. (For both correlated and anti correlated entanglement)

But the answer was already given. So where do you have a problem?
Correlated photons: match % $= \cos^2(\theta)$
Anticorrelated photons: match % $= \sin^2(\theta)$
For all three angles if the fractions of different settings are equal you just take average. (But you don't do such a thing in real experiments.)

 

[wle]

But would this be the same formula for anti correlated electrons at those angles? Or are you saying it's just 1 - (cos^2(theta)) for each of the 3 angles then take the average of those results?

If you have two spin-1/2 particles entangled in the singlet state ($\lvert \psi_{-} \rangle = \bigl( \lvert \uparrow \downarrow \rangle - \lvert \downarrow \uparrow \rangle \bigr) / \sqrt{2}$) then the probability of getting the same result (a "match") is $\frac{1 - \cos(\theta)}{2}$ (or $1 - \cos(\theta/2)^{2}$, which is the same). In case this is the point you weren't clear on: $\theta$ means the angular difference between the orientation of the measurements on both particles.

If you represent the orientations of the measurements on the two particles with angles $\theta_{1}$ and $\theta_{2}$ on some plane then this means you should use $\theta = \lvert \theta_{1} - \theta_{2} \rvert$, but this only works if the measurements are restricted to a 2D plane (e.g. you never measure the spin along an axis that has a component in the $y$ direction), since one angle is not enough to represent all the possible orientations in 3D space.

Photon polarisation works like spin-1/2 (they are both 2-level/"qubit" systems described by the same mathematics in quantum mechanics) except with vertical and horizontal polarisation (which are 90 degrees apart) corresponding to the spin up and spin down states (which are 180 degrees apart). So the results for polarisation measurements are the same as the results for spin-1/2 measurements except with the angular differences multiplied by an extra factor of 2.

For both correlated and anti correlated entanglement

There isn't really such a thing in general. There is a spin-1/2 entangled state (the singlet state) where you always get anticorrelated results if you do the same measurement on both particles. But there isn't any spin-1/2 entangled state that always produces correlated results when you do the same measurement on both particles. The best you can do is produce correlated results for measurements restricted to some given 2D plane. For example, with the state $\lvert \phi_{+} \rangle = \bigl( \lvert \uparrow \uparrow \rangle + \lvert \downarrow \downarrow \rangle \bigr) / \sqrt{2}$ you only get correlated results if you do the same measurement in the $x-z$ plane on both particles and you get anticorrelated results if you measure the spin of both particles along the $y$ axis.

 

[rede96]

In another recent thread Lord Jestocost gave a link to http://www.theory.caltech.edu/classes/ph125a/istmt.pdf. So you can read it yourself. The answer is at the very end of page 10. Mermin calculates average from all three angles. 1/3*1 + 2/3*(1/4)=1/2

Thanks for that. I'll need to read and digest properly as it's still not obvious to me how I'd calculate the average for other angles that are not equally distributed.

But the answer was already given. So where do you have a problem?

What I'm trying to do is make a simple simulation so I can play around with different angles and see what results I get. But to do this I need to know what results I'd expect experimentally at different angles so I see the differncres between my simulator and what would happen though experiment.

The problem is I'm just an interested layman with no back ground in physics at all and to be honest, after reading al the post still don't understand how to calculate the expected results. For example what would be the expected match results for angles 10, 60 and 200 for say photons?

Is it just a case of adding up the expected results for each pair of angles tested?

What I understood was it would be cos^2(thetaA - thetaB) for each angle, which gives me the expected match % at those two angles, then multiply that by the number of tests done on those angles.

E.g assuming I do 100 tests at each angle (I ignore where the angles are the same), then:

60 - 10 = cos^2 (50) = .72 x 100 = 72.3 matches (on average)

Using the same math...
200 - 10 = 99.9 matches
200 - 60 = 45.6 matches

So over the total 300 tests I'd expect to see 217.9 matches on average.

But if I took each angle individually then I'd get a different set of results e.g.

cos^2(10) x 100 = 99.9
cos^2(60) x 100 = 45.6
cos^2(200) x 100 = 92.8

So the total is 238.4

So which is correct?

And what's the right formula to use for electrons that are anti correlated?

Thanks for your help

 

[zonde]

60 - 10 = cos^2 (50) = .72 x 100 = 72.3 matches (on average)

Using the same math...
200 - 10 = 99.9 matches
200 - 60 = 45.6 matches

Where do you get those numbers?
My calculator gives:
cos^2(50 deg.)=0.413
cos^2(190 deg.)=0.970
cos^2(140 deg.)=0.587

But if I took each angle individually then I'd get a different set of results e.g.

You take relative angles. You don't take each angle individually.

 

[rede96]

Where do you get those numbers?
My calculator gives:
cos^2(50 deg.)=0.413
cos^2(190 deg.)=0.970
cos^2(140 deg.)=0.587

Opps! Sorry, was in a rush as I'm at work and didn't bracket my formula correctly in excel.

You take relative angles. You don't take each angle individually.

Ok great, thanks. So for electrons which are anti correlated then do I use cos^2((angleA - AngleB)/2) or is it cos^2((angleA/2)-(angleB/2))

 

[zonde]

Ok great, thanks. So for electrons which are anti correlated then do I use cos^2((angleA - AngleB)/2) or is it cos^2((angleA/2)-(angleB/2))

Well, the two formulas are the same.
But it's sine not cosine: sin^2((angleA - angleB)/2) or alternatively 1-cos^2((angleA - angleB)/2) (it's the same). Try to calculate sin^2(0), it's 0 as it should be for anticorrelation. And sin^2(180/2)=1.

 

[rede96]

But it's sine not cosine: sin^2((angleA - angleB)/2) or alternatively 1-cos^2((angleA - angleB)/2) (it's the same). Try to calculate sin^2(0), it's 0 as it should be for anticorrelation. And sin^2(180/2)=1.

Great, thanks very much for your help.

Could I just confirm one other thing. When I compare the total number of matches for a number of tests, I only comparable them against the tests where the angles were not the same. E.g if I do 90 tests on all permutations of the three angles, I'll have 30 on average where the angle A = Angle B. So if I got 15 matches in total, I'm comparing that figure as a percentage against 60 tests. (0.25) and not the full 90. Is that correct?

 

[zonde]

Could I just confirm one other thing. When I compare the total number of matches for a number of tests, I only comparable them against the tests where the angles were not the same. E.g if I do 90 tests on all permutations of the three angles, I'll have 30 on average where the angle A = Angle B. So if I got 15 matches in total, I'm comparing that figure as a percentage against 60 tests. (0.25) and not the full 90. Is that correct?

It seems correct.

 

[rede96]

It seems correct.

Thank you.