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Question by Erick Green
What are your thoughts on the higgs boson? The reason for me asking is because the higgs suppose to explain "everything in physics" with the exception of gravity. This sounds like quite far-fetched since quite a bit of "physics" seems to be false. I don't know much about the higgs besides it supposely being the reason mass exist...

Reply
In my opinion, the standard model for particle physics is based on so many false or questionable concepts in physics (e.g. Relativity) that I haven't even tried to understand it in detail. Your question regarding the Higgs boson is therefore a rather academic one to me. There are much more important issues to address (which in all likelhood would show the Higgs boson just to be the product of a phantasy based on flawed concepts)

Question by Svend Ferdinandsen
In descriptions of Bell inequality tests it is said that the expected output from the coincidence detector would vary as cos to the angle between the polarizers. Whereas with hidden variables (no quantum strange effects) it would be a straight line. How do they calculate the straight line, because i can only get a cos dependency in the expected values? I assume the photons are send out pairwise with perpendicular polarization and at any random angle to the polarization detectors which are turned relatively to each other. All photons are detected in any of the detectors pairwise.

Reply
The straight line result applies to spin measurements of particles, not to optical measurements involving light. In the latter case, the hidden variable assumption would also yield a cos2 curve, but with a smaller amplitude (the coincidence rate does not go to 0 at 90o). See the Wikipedia article that discusses both cases. On Mathpages, you can find a more detailed explanation of the linear curve for the spin measurements case.

However, as I have pointed out on my own page, the optical experiments can actually be explained classically if the detection process for the light 'photons' is being considered correctly. I have not looked in detail into the spin-particle case yet, so I don't know whether a valid local realist solution can be found here as well (this would obviously not be possible in terms of classical physics anyway as the spin is a pure quantum concept, but correct modelling of the detection process could still result in a local realist model yielding the observed curve).

Svend Ferdinandsen (2)
Thanks for the answer, it helped a lot. I just wonder why this straight line often is used in combination with optical entanglement.

I have an other question, regarding optical (photon) entanglement:
I have found out that these photons are always produced with a fixed polarization relative to the crystal that produces them. I found it in >this article. That also describes how the entanglement comes alive. Not so special anyway, it is just random if it is vertical or horizontal.I believe that photons with a fixed polarisation (except that it could be "vertical" or "horisontal" but never other directions) could change the way a classical view was calculated. Or it could make the entanglement quantum view stronger. Anyway, the correlation (coincidence counts) must be very dependent on the angles between the polarizers and the crystal and not only to the angle betwen the polarizers as such.

Reply (2)
I have not seen the straight line graph in combination with optical Bell test experiments before. The usual simplistic 'local realist' (hidden variable) theory obtains the function 1/8 + cos2(Θ)/4 in that case (for initially 'unpolarized' radiation). Of course neither of those is what is observed. But this does not disprove that a 'local realist' view is wrong as such, only that the model used here is too simplistic. On my own page I have shown how the correct behaviour is obtained by a much more sophisticated model.

As for your reference: note that here the situation is quite different: the polarizers are defined/replaced here by the beam splitter and are effectively fixed at 90o to each other. It is only the polarization of the light falling onto the beam splitter that is changed. In the first diagram on my own page this corresponds to the red and blue arrows (the polarizers) being both fixed at 45o to the x-axis, so Θ=90o but instead the polarization angle φ of the light being the independent variable. With this, Eq.(7) on this page results then also in the Malus law, but in the form n(φ)=cos2(2φ). Of course, in this case the 'simplistic' realist view would also yield the Malus law, because for orthogonal polarizers there won't be any coincidences if the polarization direction is along one of the polarizers, whereas it is a maximum with the polarization at 45o (so this scenario would not really be an argument against the 'realist' view anyway, not even in its simplistic way).

The crucial point here is that single 'photons' (i.e. individual atomic emissions) are actually not detectable. It takes many photons in order to cause a reaction in the detector, as explained under the above link. This is readily evident if one considers the electromagnetic energy emitted by single decaying atom. According to classical electrodynamics the total power radiated by an oscillator with angular frequency ω and amplitude x is

P = ω4e2x2/3/c3  

where e is the elementary charge and c the speed of light (I am using cgs- units here and in the following). With ω=2.3.1015/sec and x=2.4.10-8 cm (which would be appropriate for an optical light emission at 810 nm) this yields P= 9.8.10-12 erg/sec. For the kind of 'spontaneous parametric down conversion' as discussed in your link, this power will be distributed over an area of a few square millimeters (as is evident from the results in this link). Taking the area as 10 mm2= 0.1 cm2, the radiation flux at the detector from one 'photon' is therefore F= 9.8.10-11 erg/cm2/sec. Now the electric field strength of the wave E is related to the flux F by

F=E2.c

so E=√(F/c) = 5.7.10-11 statvolt/cm. Now in order to transfer the energy h.ν to an electron, a field of strength E would have to act for a time (as shown under this link).

T= 7.10-18.ν /E     [sec]

so T = 2.3 sec here (ω=2πν) However, the temporal length of a photon, as given by the inverse of the atomic decay constant (see this link)

A = 2.π.e2.ω3.x2/3/c3/h   [1/sec]

is in this case 1/A = 5.3.10-8 sec, that is factor 2.3.10-8 too short to be detected. You need thus about 108 photons before the energy required to release a photo electron is transferred to the material. If the light beam is not that spatially concentrated and if the radiation is not fully transmitted by the polarizing device, it would be even correspondingly less

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