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

2. 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.

3. 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.

4. 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

5. 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

6. 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.

7. 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:

8. 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.

9. 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

10. 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

11. 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?

12. 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

13. 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

14. 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.

15. 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

16. Thanks

17. 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

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

19. 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

20. 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!

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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:

28. 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

29. 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

30. 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

31. 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

32. 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.