Bell's inequality for a single measurement within EPR paradox. Vladimir K. Ignatovich Joint Institute for Nuclear Research ICTP-2015 July 03 2015 Moscow

The essence of the EPR paradox The first photon is counted by D1A with probability cos2(φ), or by D2A with probability sin2(φ). But it is counted only by a single detector. If it is registered by D1A , then the photon at Bob is So it will be counted only by D1B, though without Alice’s measurement it could be counted also by D2B with probability sin2(φ) It means that effect of measurement by Alice propagates with the infinite speed the paradox.

The essence of the paradox Either influence of a measurement can propagate with infinite speed (inacceptable), or quantum mechanics is incomplete - there must be some unknown (hidden) parameters in it. We had already introduced a parameter, φ, but it was wrong. We used the notion of two photons, while according to quantum mechanics the source emits not photons but some 2-photon state where x, y are 2 arbitrary orthogonal vectors.

Do photons have individual polarizations? N.D.Mermin. Hidden variables and the two theorems of John Bell. Rev.Mod.Phys, 65 (1993) 803. “It is a fundamental quantum doctrine that a measurement does not, in general, reveals a preexisting value of the measured property. On the contrary, the outcome of a measurement is brought into being by the act of measurement itself. “ Marco Genovese. Research on hidden variable theories: A review of recent progresses Physics Reports 413 (2005) 319–396 “When the two space wave functions, e.g. Gaussians, are widely separated at the moment of measurement, one can therefore realize two wave packets entangled in spin, but well separated in space.” It is impossible to accept these claims for a rational mind like mine

In quantum mechanics there are no photons before any measurement. Only wave function The particle, going to Bob, becomes a photon only after Alice detected her photon. If, for example, D1A made a count, then It looks like a photon with polarization where and it will be counted by D1B After a measurement the photon disappears. Therefore a photon with its polarization exists only after its death

An ideal experiment Alice and Bob arrange their calcite crystals in parallel, mark every photon pair by arriving time, and check, whether their photons are counted only by the same detectors. In practice, experimenters use the detector counts to calculate some quantity, and try to prove that it violates the Bell’s inequality.

Bell’s inequality Let’s ascribe number +1 to a photon, if it is counted by a detector D1, and -1 if it is counted by a detector D2 Consider 4 experiments: Alice arranges her calcite axis oA along unit vectors a and c, and for every option Bob arranges his calcite axis oB along b and d or Every photon measured by Alice is counted as Every photon measured by Bob is counted as or All these numbers can be combined in a mathematical identity: Averaging over all the photons therefore gives the following inequality:

Bell’s inequality So Bell predicts How do experimenters verify this inequality? In a single run of an experiments one measures: Surely, this correlation depends on angle between a and b Instead of The quantity is accepted as the probability for Bob and Alice to count their photons by detectors D1. And so on.

In Quantum Mechanics In quantum mechanics with the wave function One can easily find Experimenters arrange their axes show that and with which violates the Bell’s inequality

But why so complicated? I propose the simplest way I suppose that the source emits two individual photons: A photon polarized along a unit vector φ goes to Alice. Her calcite axis is along a and the second parallel photon goes to Bob, whose calcite axis is along b So, my inequality is

Advantages of my inequality. It is derived for specific hidden parameter: angle of polarization does not contradict to the general one It can be checked in a single measurement for If it is impossible, how reliable are 4 measurements with different axes?

How to recognize the correct state? Let’s count with parallel calcites only those pairs, in which a photon is registered by D1A Then we will see, how many of photons in these pairs are counted by D2B For independent photons one can predict

Spinor particles The similar consideration for spinor particles, where QM predicts Two spinor particles created after decay of a scalar atom can be described by the wave function with or they can be independent oppositely polarized particles with With probability Then so

Scientific censorship The author tries to refute the theory of hidden parameters and proposes his inequality instead of the Universal one There is no sense in this, because the universal inequality, holds for any local classical processes. While his inequality is the model one. It uses a model representation of a couple of flying apart photons that pass polarization analyzers. The only result that can be achieved in the case of refutation of the author’s inequality will be the proof of the inadequacy of his model, which, however, obvious from the outset I do not recommend publication of the paper. Referee of the Russian Journal Uspekhi.

Thanks Our Editorial Board has considered your paper and come to the conclusion that it cannot be accepted by JETP Letters as not requiring rapid publication. I submitted the paper to JETP on 12 May. Till now referee could not invent a reason for rejection. Thanks

Thanks

Essence of the EPR paradox In QM one can predict that there exist particles with simultaneously precisely defined position and momentum, but UR do not permit such particles. This contradiction between predictions and UR is the essence of the EPR paradox. Resolution of the paradox In QM particles really can have simultaneously precisely defined position and momentum, and UR have nothing to do with it. The paradox is based on wrong definition of such physical quantities as momentum and position of particles

Definitions in the EPR paper For example But position in this state is not defined

Dimensionality cm, and NO normalization! IT IS NOT A PROBABILITY Dimensionality cm, and NO normalization! Eq. (6) is an error, which is a result of the wrong definitions

Ha! We needed to wait 73 years for some one ... would open our eyes to an Einstein's error! Referee report A. Elitzur For students: but if b-a>L? The space is restricted by impenetrable walls! Then and, according to EPR definition, momentum does not exist!

To have probability we must require Correction of the error To have probability we must require No wave packet is an eigen function of momentum operator Therefore momentum does not exist It is wrong Therefore we must redefine Uncertainty relations do not forbid their existence Paradox does not exist Is not an uncertainty, but the size and form of the object

Entangled states do not exist If a wave function of two separated particles is and a measurement of particle at x1 finds it in Then the wave function at x2 is But particle at x2 does not interact with x1, therefore Existed before the measurement But from this logic it follows that before measurement the wave function was the single term And the entangled state is only a non unique list of possible product states. The experiment can only reveal: what was the product

More over the common WF is not a product but the sum Since we have wave packets, then Therefore there are no nonlocality! If measurement of the particle 1 gives The state of the particle 2 remains

QM is an approximation for classical physics Interference in Classical Physics Reflection and transmission on a semitransparent mirror is a bifurcation in a stochastic dynamics

Bell’s inequality The quantity is the probability for Alice to count a photon by detector D2A , and for Bob to count his photon by the detector D1B, and so on.. Note that It is also evident, that Therefore

Quantum Mechanics The photon at Alice can also be registered by D2A with probability 1/2. And the photon at Bob site will be registered by D1B with probability P+- = sin2(β)/2, or by D2B with probability P-- = cos2(β)/2. Again, and we immediately see, that and

Bell’s inequality Let’s ascribe number +1 to a photon, if it is counted by a detector D1, and -1 if it is counted by a detector D2 Consider 4 experiments: Alice arranges her calcite axis oA along unit vectors a and c, and for every option Bob arranges his calcite axis oB along b and d Every photon measured by Alice is counted as or Every photon measured by Bob is counted as or All these numbers can be combined in a mathematical identity: Averaging over all the photons therefore gives the following inequality:

But we cannot average in a single experiment Because we have only 2 axes in every given experiment, we must average consecutively: Let’s take 2 axes a and b and find the average because

Next, we pick axes c and d, and average accordingly is a constant, which will not change with next averaging, for example, over <>c,b Therefore, is equivalent to and in QM it is not violated

We can alternatively pick axes c and d as the second averaging step, then again we obtain a constant, that will not change any more Therefore is equivalent to and in QM it is not violated, again

We can improve derivation. Consider the identity then and therefore, for we get the following inequality: which is not violated in QM because in shown geometry with

Conclusion There are no EPR paradox Uncertainty relations are useful but not important There are no entangled states of separated particles Don’t worry about Bell’s inequalities Don’t believe experiments on Bell’s inequalities Farewell the paper physics of quantum teleportation Farewell the paper physics of quantum cryptography Farewell the paper physics of quantum computing QM is an approximation of classical Mechanics I need a physicist with good mathematical background to prove it.