Quantum Physics Discussion Forum

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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)

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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 90^o). 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.

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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 90^o 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 45^o to the x-axis, so Θ=90^o 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 45^o (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=E^2.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^.e^2.ω^3.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