Bell inequality violation is an illusion.

Top page ( Bohr model including helium )
Leggett   Teleportation   Cheshire cat (14/8/28)
Delayed choice quantum eraser.
Quantum computer is impossible (14/4/13).

Photon is a particle ?

If you have already read the photon page, please proceed to the next section,
Bell inequality violation can be explained by classical electromagnetic waves!.

The photon particle is said to exist. But the interference of a single photon with itself can be explained only by the wave nature of the electromagnetic (E-M) waves, NOT by the photon particle.

(Fig.1) Photon is particle or wave ?

According to QED, a single photon must be a point particle.
But in various real experiments, they often use the classical electromagnetic waves such as left-, right-circularly, vertically, and horizontally polartized lights.

Though they clearly use the property specific to the classical lights, they always try to express them as a photon particle in some papers or something.

As shown in Fig.1, if the light is a point particle, how can it be polarized or interact with other particles ?
So the concept of photon "point" particle is impossible.

"Entangled" photon pair is just classical "polarized" lights.

(Fig.2) Entangled photon pair ?

Entangled photon pair is often created by "parametric down-conversion".
In this process, a photon of wavelength λ enters some crystal, and is split into a pair of photons of wavelength 2λ.
So the total energy is conserved. (= their frequency becomes half ).
( If photon is a particle, it can be split so easily ?? )

One of the produced photon pair is horizontally polarized, and another is vertically polarized.
Though this photon pair is clearly classical electromagnetic waves, they always use the word of a "photon".

If photon is a particle, "interference" is impossible.

(Fig.3) Photon "point" particle can interfere with each other ?

Suprisingly, photon "point" particle can interfere with each other, according to their experiments.
In Fig.3 upper, due to the destructive interference between photons, they vanish !.
Conversely, in Fig.3 lower, due to the constructive interference, the number of photons (= amplitude ) has increased, they insist.
But as shown in Fig.3, these interferences prove that what they call "photon" is clearly classical waves.

Single photon detector just sees photoelectric effect ( NOT a photon ).

(Fig.4) Single photon detector shows photon is particle ?
detector

First, we cannot generate single photon directly.
By weakening coherent laser and seeing their electric current signal at photon detector, we only estimate their existence.
This detector cannot always detect photon ( < 60% ), when it enters, and due to dead time, cannot discriminate between a single and two photons.

Detection "threshold" at detector causes a illusory "photon".

Of course, as we cannot see directly a single photon, this photon's concept is only speculation.
Repeating photoelectric effect, photon detector increases electric current signals from a single photon.
Considering light loss in these processes, it is natural that we think photon detector can show current signal, when the light intensity is above some threshold.

A single photon can be in the "superposition" in two different paths !?

(Fig.5) Mach-Zehnder interferometer.

In the Mach-Zehnder interferometer, when one photon is split into the two different paths at the beam splitter (= BS1 ), we can detect one photon only in one of the two paths at the same time (= Fig.5 ).
( This means that this single photon particle is NOT split into two, according to the quantum mechanical interpretation. )

(Fig.6) Interference of a single photon.

But the interference of electromagnetic waves from these two different paths is actually observed, though a single photon exists only in one of the two paths. It is amazing!
It is said that this strange phenomenon is caused by the superposition of one photon in the two different paths.
But is it really so?
This "superposition" has NO reality, because it is equal to "many fantasy worlds". (= world A and world B in this case. )

Photon = classical lights, which is split at splitter.

(Fig.7) Photon = classical electromagnetic waves !
Mach-Zehnder

Suppose the photon is electromagnetic wave (E-M wave), which is detected as one photon only when the intensity of the E-M wave is above some threshold.
For example, the first light intensity is 10, then it is split by the beam splitter ( BS1 ) into 6 + 4 waves.

We suppose the photon detector can detect it when the light intensity is 6 as a single photon.
So 10 is detected as one photon, not 2 photons (=12).
But the split photon (waves, 6 & 4) can interfere with each other.

And the 4 side is not detected, only 6 side is detected as a photon.
So we can explain the "strange" single photon interference by the combination of the electromagnetic wave and detection threashold !

Detection threshold + classical light = superposition of a photon.

For example, in case of the 50/50 beamsplitter ( BS1 ), if the E-M waves is split into 5 + 5, no photons can be detected. (This case is not recognized at all, so it can be ignored. )
We can recognize the photon existence only when we detect it as a photon.
50/50 beamsplitter divides unpolarized electromagnetic waves into about half on average.
Depending on the polarization and frequency of the light at the beamsplitter, this probability changes.

(Fig.8) Beam splitter 2 ( BS2 ).
BS2

After splitting at the beam splitter 1 (BS1), both the electromagnetic waves are split into almost half also by the beam splitter 2 ( BS2 ).
( For example, 6 → 3 + 3, 4 → 2 + 2 )

In this experiment, only one side reflection is supposed to reverse light phase at the beam splitter 2 (BS2), as shown above.
( The transmitted lights are not changed. )

So at the photon detector 1 (D1), the light amplitude is increased by constructive inteference, and can be recognized as a photon.
Because the light intensity at D1 always reaches "6" by the inteference.
(The light amplitude is a square root of the light intensity.).

Interference + detection threshold causes an illusory single photon.

[ One photon "always" detected at D1 by the interference. ]
D1

On the other hand, at the D2 detector, the light amplitudes from the two paths cancel each other, so it can not be detected as a photon.
( The light intensity at D2 side does not reach "6". )

[ At D2, no photon is detected by the interference. ]
D1

Of course, also when the first light "10" is split into 5 + 5 at the beam splitter 1, this result is the same as the above.
In this case the light intensity "5" is split into 2.5 + 2.5, so at D1 side,
halfD1

And at D2 side,
halfD2

In these cases, the total energies are conserved. ( 10 → 10 ).
As a result, we can explain all these phenomena by classical electromagnetic wave model.
So they don't need the concept like "photon particles" at all.
( See also Delayed choice experiment is real ?. )

Delayed choice experiment can be explained by classical lights.

(Fig.9) Delayed choice experiment.
delayed

In the delayed choice experiment, after a single photon has passed through the beam splitter 1 (= BS1 ), that single photon has entered into one of two paths.
Actually, by the photon detector, only one photon is detected ( for example, at D2 in the upper figure ).

What will happen, when we insert the beam splitter 2 (= BS2 ) after the photon has passed BS1 ?
Surprisingly, in this case, the electromagnetic waves suddenly appear in the two different paths, and interfere with themselves ! (= right panel of the upper figure. )

So we can choose the past event (= particle or wave ) from the future (= delayed choice ) !
Of course, in this real world, it is impossible.
So as I said above, the concept of a single photon itself is wrong.
( Unfortunately, ordinary people do not know about these strange things related to "photon" well . )

Photon spin "1" is artificial definition, NO reality.

(Fig.10) Photon spin is really " 1 " ?

One photon is said to have "spin" ±1. What in the world is the "photon spin 1"?
The ordinary textbooks say that the photon spin ±1 correspond to the electromagnetic (E-M) waves with left and right circular polarization.
And the "linear" polarization corresponds to the superposition of these two states.

But this superposition state is very strange and difficult to imagine.
The quantum mechanics often uses a mathematical trick like this.
Then we can take this convenient word "superposition" in every situation arround us!
(e.g. when some things in different states are only mixed, when one thing is "vibrating" very fast, when one thing only keeps still....).
This means that the superposition itself is meaningless.

(Fig.11) Fine-structure = "photon" spin 1 ??
spin

To begin with, how this photon spin 1 was discovered ?
In 1910's, the fine structure of the hydrogen atom meant the relativistic energy difference between the 2S and 2P states of Bohr-Sommerfeld model.
But after the electron spin appeared, the interpretation of this fine structure was changed to the spin-orbital interaction ( = the energy difference between 2P1/2 and 2P3/2 of the hydrogen atom. See this page. )

"Accidentally" this value coincided with that of the Bohr-Sommerfeld model !
So the "unnecessary" transition ( = 2S ---> 1S) needs to be forbidden to explain the spectrum results of the hydrogen with electron spin.
Then the selection rule (2S --x--> 1S) was introduced, and they made the photon spin 1 which probably causes this selection rule.
This means that the photon spin 1 is only an artificial thing.

Photon = point particle ? → Photoelectric effect is impossible.

(Fig.12) Point particle = photoelectric effect ??

It is said that the photoelectric effect needs the concept like "photon particle". But is it really so?
The electrons are always accompanied by the de Broglie's waves around them. And it synchronizes with the electron's momentum. So we must consider the de Broglie's waves rather than the very small point-like electron in case of the photoelectric effect or the Compton effect.
Actually, circularly or linearly polarized electromagnetic waves are emitted from exicited atoms in the experiments.
And Willis Lamb himself, who was great experimentalist dealing with the light, did NOT believe photon particle.

Recent interesting studies indicate that a single photon consists of electromagnetic waves ( in both two slits ) in the weak measurements.
(Science 2011, 332, 1170, S.Kocsis et al., Nature 2011, 474, 188, J.S.Lundeen et al.)

Light frequency = c / λ = wave ( NOT particle ) nature.

(Fig.13) Frequency = photon particle is oscillating ?

The ordinary textbooks often say that the relation between the photoelectric effect and the wave's frequency shows the particle's nature of the photon.
First, the frequency itself is based on the electromagnetic wave nature !
Second, when we heat the air or the water which give rise to "sound wave" or "wave", the frequency of each particle in those waves is increased.
So it is quite natural that the energy is related to the frequency.
( Frequency, amplitude = wave nature ! )

In the Compton effects, if the light is a particle, its velocity would change instead of its frequency, when its momentum changes.
As I said above, the frequency itself is based on wave nature.

Photon "particle" causes "illusory" expanding universe.

(Fig.14) "Convenient" photon can pass through dark energy, matter, Higgs "uninjured"?

As shown on this page, photon particle needs to have strange ability to go through very crowded space "uninjured" for very long time.
Redshift by expanding universe is thought to be caused by the tired light, in which the light loses energy while it is going through the space.

The shining duration from supernova is more elongated from farther stars.
You can imagine this reason from the difference between "marathon" and "100-meter race".
And the concept of photon particle is the origin of many-world interpretation and virtual photon, which violates relativity. ( See this page. )

Bell inequality violation rules out the local realism of the photon particles, but NOT the electromagnetic waves !

(Fig.15) Entanglement of a photon pair ?
entanglement

In 1935, Albert Einstein, Boris Podolsky and Nathan Rosen published a paper about the "EPR Paradox" which criticized the nonlocal quantum mechanical theory [1].
They insisted that the local hidden variables would exist and no actions would be instantaneously (= faster-than light = nonlocal ) transmitted to the places at a distance.

(Fig.15') Nonlocal Schrodinger vs. Hidden variable.

Hidden variable theory means that things are regularly moving obeying some physical principles.
So classical theories such as Bohr model corresponds to this hidden variable theory, and the probability concept of Schrodinger wavefunction does NOT admit it.

Classical lights + detection threshold causes "illusory" entanglement.

(Fig.16) Bell inequality violation is impossible in this real world.

In 1982, Alain Aspect showed that the Bell inequality (CHSH) was violated using entangled photons ([3], [4]).
These facts seem to show that the quantum mechanical predictions are correct.
( So they insist that this experiment succeeded in removing "reality" from the quantum mechanics. )

But in fact, Aspect's experiment of the Bell inequality violation using the photon pair denied the local realism of only particles.
It did NOT deny the local realism of the classical ( electromagnetic ) waves !

If we suppose electromagnetic waves can be detected as a single "photon particle" above some threshold, we can naturally explain this Bell inequality violation as a "local" and "realistic" thoery.

Pass probability of a photon (= light ) at filter ?

(Fig.17) Pass probability of a photon.
intensity

When the angle between the polarization axis of the E-M wave and the polarizing filter is θ, the pass light intensity is I cos2θ.
So they insist the pass probability of a single photon is cos2θ, and the reflection is sin2θ.
This is an "artificial" definition. ( See Fig.18. )
( But I think "rigid" photon particle can NOT satisfy this Malus law correctly. )

Probabilities of photon's pass or reflect at filter ?

(Fig.18) Pass or reflect probability of a single photon at filter.
intensity

Suppose two pair photons with the opposite (= parallel ) polarization axes are flying in the opposite directions.
The photon A bumps into the polarizing filter A, while photon B bumps into the filter B. (Fig.19.)

Classical case -- Probability = probability × probability.

(Fig.19) The probability of both photons passing their filters ?
both

When the angle between the photon A and the filter A is α, the probability that photon A passes through the filter A is cos2 α ( ← (E cos α)2 ). (The reflection is sin2 α).
When the angle between the photon B and the filter B is β, the pass is cos2 β, and the reflection is sin2 β.

In this case, for example, the probability that both the two photons pass their filters is ( cos2 α × cos2 β ).
It's very natural, because it's probability × probability, which is "generally accepted concept".
But this case satisfies the Bell inequality and is different from the experimental results !

Superluminal "spooky" link ?

The experimental results shows that the instant the photon A has passed through the filter A and detected, the polarization axis of the photon B becomes parallel to the filter A, which is very far away from photon B ( see " change ! " in Fig.20. )

When the photon A reflects from the filter A, the polarization axis of the photon B becomes perpendicular to the filter A. ( See also Fig.21. )

(Fig.20) Experimental results. ( = entanglement ! )
violation

So the probability that both the two photons pass each filter is cos2 (β - α) ( not cos2 α × cos2 β ) !
( To put it accurately, the average probability of both photons passing is 1/2 cos2 (β - α). )
Here, θ = (β - α) means the angle between the filters A and B.
This case violates the Bell inequality in some angle θ.

The instant one photon passes ( or reflect ) its filter, another photon's polarization changes !?

(Fig.21) Entanglement is caused by faster-than-light "spooky" link !
filter

Do you think that the existence of a filter is treated too specially ?
Though a filter is only one of many things in the universe ....
( This is a very convenient way to interpret it. )

Photon "rigid" particle is the cause of "unreal" spooky link.

(Fig.21') Photon particle is not divided → Wave is divided.

If the photon is a wave, the passing photon A can be detected only when the polarized angle of the photon A is near the angle of the filter A due to the detecting threshold.
For example, when α is about π/4 (= middle point), the photon A is divided into pass and reflect sides equally at the filter and detected at neither pass nor reflect sides.
Because both pass and reflect light intensities I' become weaker than the detection threshold (= Fig.21'').
( For example, we can suppose the detector can detect it above threshold of 0.7 × I. )
When we cannot detect both two photons at the same time, this case is NOT counted as a entanglement result.

When classical lights are divided into weak light → cannot reach detection threshold.

(Fig.21'') Wave is equally divided = not reach detection threshold.

As a result, when we detect two photons A,B at the same time, it is possible that they originally have the polarized angle near the filters,due to the detection threshold
This means that the "wave nature" of the light causes the illusion that the entanglement is really occurring.

The source of the two entangled photons in this page is the "excited" atom, in which the directions of the two photons are paralell to each other.
If we separate the light into the vertical and horizontal directions, change cos2 θ into sin2 θ, which meanings are the same as far as they satisfy Malus's law.

Classical lights can explain Bell inequality violation.

(Fig.22) Directions of Filter A and B are the same. ( θ = 0 degrees )
0angle

This figure shows the case in which the angles ( = the directions of the arrows) of the polarizing filters A and B are the same ( = 0 degree).
As I said, the photon A and B have the opposite ( = parallel ) polarization axes.

So in this case, when photon A passes through filter A, photon B always passes through filter B, ( = cos2 0 = 1 ), in entanglement state.
Here we try to explain this strange phenomenon using the classical electromagnetic waves.

Detection "threshold" at photodetector is a key player.

When the polarization axis of the photons A ( this axis is the same as the photon B, because they are parallel ) points toward ( ++ ) of Fig.22, both the photons A and B pass each filter, and are detected as "photon particles", because these light can pass these filters enough to be detected (= Fig.23 ).

The pass light intensity is I cos2α ( or I cos2β ). When the light polarization points to (++), this angle α (or β ) is near 0 degrees, so the pass intensity is strong enough.

(Fig.23) Both waves pass their filters ( ++ ) enough to be detected.

But when the polarization axis of the photon A( = photon B ) points toward the blue part of Fig.22, the transmitted (or reflected) light intensities are below detection threshold, because these lights are divided almost equally into pass and reflect sides (= Fig.24 ).

For example, the pass light intensity is I cos2α and the reflect intensity is I sin2α.
When this angle α is near 45 degrees (= π/4 = blue part of Fig.22 ), both pass and reflect intensities become 1/2 × I. ( 1/2 = cos2 π/4 = sin2 π/4. See also Eq.21''. )

When the filter's directions are the same, both photons (= lights ) always give the same results (= pass or reflect ).

(Fig.24) Both waves are split almost equally = not detected.

In this case, both pass and reflect light intensities become weaker.
So neither photon A nor B can be detected as a "photon particle" by the detector (this case is not used even in calculating the detection efficiency).

There is NO spooky link. It's caused by detection "threshold" + classical lights.

(Fig.24') Both waves reflect ( - - ) enough to be detected.

And when the polarization axis of the photon points toward the ( -- ) of Fig.22, both the photons A and B are reflected by the filters enough to be detected as "photon particles".
For example, the reflect light intensity is I sin2α. when this angle α is near 90 degrees (= (--) part of Fig.22 ), reflect intensity becomes strong enough.
So in this case, the "illusion" of the entanglement of these pair photons are occurring.
(In Fig.22, the results of the photon A and B become the same !)

There is NO "classical" photon.

(Fig.I) Classical photon pair ?

As shown in Fig.I, classical photon pair behaves like usual classical mechanics.
( But I think classical photon itself does not exist. )
So, when photon A passes the filter A, the photon B passes or reflects.
So the results of photon A and photon B are NOT always the same, they insist.
( There are cases other than pass - pass, reflect - reflect, too. )

Entangled photons = "Split" classical light + detection threshold.

(Fig.II) Entangled photon = classical waves.

In entangled photon pair, when the photon A passes the filter A, the photon B always passes the filter B due to spooky link.
Instead of accepting this unrealistic idea, it is more natural that we think photon is classical electromagnetic waves.
As I said, due to the detection threshold, these waves always show the same detection results.

When the angle between two filters is 90 degree...

(Fig.25) Angle between Filter A and B is θ = 90 degrees.
90angle

Next, this figure shows the case in which the angle between the filters A and B is 90 degrees.
In this case, when the photon A passes through the filter A, the photon B always is reflected by the filter B ( = cos2 90 = 0 ) according to the entanglement.

When the polarization axis of the photon A ( = photon B) points toward (+ -), the photon A passes the filter A, and the photon B is reflected by the filter B (and can be detected as a "photon particle").

Two photons always give different results (= one is "pass", another is "reflect" )

(Fig.25') Photon A passes, photon B reflects ( + - ).

But the polarization axis points to the blue part of this figure, neither photon A nor B can be detected as a "photon particle" due to the detection threshold.
(Because they are "almost equally" divided into pass and reflect directions. So both pass and reflect light intensities become weaker.)

When the angles between filters is 45 degree...

(Fig.26) Angle between Filter A and B is θ = 45 degrees.
45angle

This figure shows the case in which the angle between the filter A and B is 45 degrees.
In this case , when the photon A passes through the filter A and is detected, the probability that the photon B passes the filter B is 50 percent ( = cos2 45 = 1/2 ) according to the entanglement.

When the polarization axis of the photon A ( = photon B) points toward ( ++ ), both the photons A and B can pass each filter and be detected as "photon particles".
In the direction of ( +- ), the photon A pass the filter A and the photon B is reflected by the filter B and can be detected as "photons".
So the probability that both the photons A and B pass each filter becomes 50 % as I said.

Detection "threshold" of a single photon is the key point.

(Fig.26') Photon A pass, reflect, not detected regions.

But in the directions of the blue parts, only one of the two photons A and B can be detected as a photon, because one of them has weaker transmitted (or reflected ) light intensity than the detection threshold.

If only one of them can be detected, this case is not used as the correct experimental result.
Because only when we detect both the two photons at the same time, we consider these photons to be in the entangled states.
(In the actual Bell test experiment using the photon pairs, the coincidence ( = entangled photon pairs ) detection rate is very low ( < 20% ). So the loophole-free experiments are almost impossible. )

This example violates the Bell inequality due to the cos2 θ probability. But of course, the strange phenomenon such as the entaglement is not occurring here.

Here, let's calculate the Bell inequality (CHSH type).
We add one more case in which the angle between the filter A and filter B is 30 degrees.

As light's polarization is closer to filter's direction, it can pass.

(Fig.27) Angle between Filter A and B is θ = 30 degrees.
30angle
Red parts (+ +) --- Both photon A and B pass the filters and detected.
Purple parts (+ -) --- Photon A passes and photon B is reflected by the filters and both are detected as photons
Shaded parts (light blue) --- Only A (pass) or B (reflect) can be detected as a photon. (= This case is not used as a correct experimental result.)

As shown in this figure, when the photon A passes through the filter A and is detected, the probability that the photon B passes the filter B is 3/4. ( = 45/(45+15) = cos2 30 = 3/4 )
(In the case of 60 degrees, this probability becomes 1/4).

Here we explain about the Bell inequality simply.

Considering two different angles in each filter.

(Fig.28) Photon A → filters A1 or A2. Photon B → filters B1 or B2.
four filters
Suppose, photon A bumps into two kinds of filters A1 (0 degree) and A2 (60 degrees).
And photon B bumps into two filters B1 (30 degrees) and B2 (90 degrees).
Fig.28 shows the relations among these four filters.
( Caution: The filters A1 and A2 are at the side of photon A, and the filters B1 and B2 are at photon B side.)

Bell inequality ( CHSH type ).

In this case, there are 16 patterns (= 24) in which photons A and B pass (+) or are reflected (-) by the filters ( A1,A2,B1,B2 ), as follows,

bell

where the sum of each probability is 1 ( P1 + P2 + P3 + .....+ P15 + P16 = 1 ).
Here we define as follows,

---------------------------------------------------------------------
< A1, B1 > = ( the probability that the signs of A1 and B1 are the same, + + and - - ) - (the probability that the signs of A1 and B1 are different, + - and - + ) = (P1+P2+P5+P6+P11+P12+P15+P16) - (P3+P4+P7+P8+P9+P10+P13+P14)

In the same way,
< A2, B1 > = ( the probability that the signs of A2 and B1 are the same, + + and - - ) - (the probability that the signs of A2 and B1 are different, + - and - +) = (P1+P2+P7+P8+P9+P10+P15+P16) - (P3+P4+P5+P6+P11+P12+P13+P14)

< A2, B2 > = ( the probability that the signs of A2 and B2 are the same, + + and - - ) - (the probability that the signs of A2 and B2 are different, + - and - +) = (P1+P3+P6+P8+P9+P11+P14+P16) - (P2+P4+P5+P7+P10+P12+P13+P15)

< A1, B2 > = ( the probability that the signs of A1 and B2 are the same, + + and - - ) - (the probability that the signs of A1 and B2 are different, + - and - +) = (P1+P3+P5+P7+P10+P12+P14+P16) - (P2+P4+P6+P8+P9+P11+P13+P15)
---------------------------------------------------------------------

So we arrive at the following equation,

< A1, B1 > + < A2, B1 > + < A2, B2 > - < A1, B2 > = 2 ( K - L )
K = P1+P2+P6+P8+P9+P11+P15+P16
L = P3+P4+P5+P7+P10+P12+P13+P14

Using the relation, (K + L) = Σ Pi = 1,
-1 = - (K + L) ≦ (K - L) ≦ (K + L) = 1

So the following inequality (Bell inequality, CHSH type) must be satisfied,
| < A1, B1 > + < A2, B1 > + < A2, B2 > - < A1, B2 > | = 2 | K - L | ≦ 2

Bell inequality violation = beyond classical limit "2" ?

For example, the probability that the results of the photons A (→ A1) and B (→ B1) are the same ( ++ and -- ) is cos2 (A1-B1),
where (A1-B1) is the angle between filters A1 and B1 (= θ of Fig.20 ).
The probability that both photon A and B reflect ( - - ) from filters is also related to cos2 (A1-B1).
When the photon A reflects from filter A1, the polarizarion of the photon B beomes perpendicular to the filter A1.
(This means that the angle between photon B and filter B becomes π/2 - (A1-B1).)

So the probability that the photon B reflects from filter B is sin2 π/2 -(A1-B1) = cos2 (A1-B1).
( To be precise, the sum of the cases of (+ + ) and (- -) becomes cos2 (A1-B1). So each (+ +) or (- -) is half of it.)

And the probability that the results are different ( + - and - + ) is sin2 (A1-B1).
So, < A1, B1 > is
(Eq.1)
minus
where θA1 - B1 means the angle of (A1 -B1).

In the case of Fig.28, each cosine value becomes
(Eq.2)
cos
and
cos2

From Eq.2, the above Bell inequality (CHSH type) is,
(Eq.3)
cos
This violates Bell inequality !

Detection "threshold" at photodetector causes Bell inequatlity violation.

In conclution, if the photon is a indivisible particle, the strange nonlocal phenomena such as the entanglement are actually occurring, and we must believe this faster-than-light transmission.
In other words, if the photon is a dividable electromagnetic wave, we need not consider this fantasy.
You can easily judge which case (photon is a particle or wave) is more natural.

For example, in the case of beryllium ion (Be+) ( or ytterbium ion ( Yb+ ) ) trap qubits, the ion trap is continuously used to trap the ion pair correctly.
It is said that this case is the entanglement state and the detection efficiency is higher than the photon case.

But this entanglement state of the two Be+ is much more artificial than that of the photons, because the continuous influence of the external electronic and magnetic force by the trap may cause the illusion of the entanglement.
The distance between Be+ ions is very short ( several µm ), so locality loophole exists.
So even if the Bell inequality is violated, it is very difficult to say the entanglement is "naturally" occurring.
( See also this page. )

Local realistic loophole ( using CHSH ) is NOT closed.

[ It's about time that some top journal should change basic idea into "reality". ]

Because, they are obstructing development of "real" science under their present stance .

First, all researchers and news writers should make efforts to explain to ordinary people to make them understand easily, and should NOT confuse them ONLY by fascinating words "entanglement".
It cannot be helped that some top journal had spreaded illusory fields such as entanglement and quantum computer due to the time gap related to three-body helium. ( I understand their feelings well. )
But aiming at illusory ideas such as closing loopholes forever, means we admit only unreality and "Shut up and calculate !" forever, which obstructs the development of science.

[ Bell inequality violation is "still" an illusion. ]

In some website, we see the news about closing all local realistic ( fair sampling ) loophole.
But unfortunately, they did NOT close all loopholes by usual CHSH-type bell inequality.
As far as I see, different-type inequality allows lower-efficiency and some photon's loss.
And I cannot find all single (= not coincident ) events. ( They seem to use some limited single events, in inequality calculation. Is this really allowed ? )

Different from usual CHSH-type, this inequality relies ONLY on incomplete and Not enough informations = about a half of usual CHSH-type.
Anyway, this result cannot elimilate electromagnetic wave's local realistic model, as I explain above.
This different-type inequality experiment uses some light rotator, and NOT detect reflection from splitter.
In this method, our realistic electromagnetic wave model can give much more than 100 % efficiency (= their definition, coincident / single events ).

So this result cannot eliminate electromagnetic wave model.
( "Light rotator" uses phase change of a part of a photon, which means photon particle is just classical waves. )

[ All three loopholes have NOT been closed "simultaneously". ]

You may read some articles such as all loopholes have been closed, so Einstein is wrong.
But this experiment closed only one loophole of "fair-sampling".
( The fair-sampling loophole is the only one left that hasn't been closed until now. )

So all three important loopholes of "detection", "locality" and "fair-sampling" loopholes have NOT been closed simultaneously.
This means spooky link has NOT been proved to be correct in any experiment.
I confirmed this fact in some site, but all webnews should explain this fact more clearly to ordinary people, I think.

"Freedom-of-choice" is important, because we can "artificially" choose the special situation which can violate Bell inequality. At least, they should violate usual CHSH inequality, not only in very special inequality.

[ Time gap causes various serious illusions and paradoxes. ]

As I say, very complicated three-body helium couldn't be solved without computers.
Due to this time gap ( from 1920s to 1990s ), various illusory fields such as entanglement and quantum computers need to be created to fill this gap in physical world.
And serious paradoxes of special relativity have been hidden and eliminated from usual textbooks.

[ Entangled "photons" at different times ? ]

(Fig.T-1) "1" and "4" photons can be really entangled at different times ?

They claimed that recent experiment showed entangled "photons" that never coexisted in time.
But as I said many times, the trick of these kinds of experiments is they always use the word of "photon", NOT electromagnetic wave.

In Fig,T-1, entangled photon pair (= 1, 2 ) is generated at first BBO.
( One of them is horizontally polarized (= H ), and another is vertically polarized (= V). )
After detecting photon 1 state, another entangled pair of (= 3, 4 ) photons are generated.
Then, mixing photons 2 and 3, and they detect 2 and 3 photon states (= H or V polarizations. )
Of course, we cannot know which photon enters upper or lower detector, because they are mixed.

(Fig.T-2) Entangled photon pair is generated at BBO ?

Entangled photon pair is often created by "parametric down-conversion" at BBO.
In this process, a photon of wavelength λ enters some crystal, and is split into a pair of photons of wavelength 2λ.
So the total energy is conserved. (= their frequency becomes half ).
But as I said many times, if photon is a particle, it cannot be split so easily.

And "horizontally" or "vertically" polarized states mean they are just classical electromagnetic waves ( NOT a particle ).
How can a point particle of photon be polarized ?? It's impossible.

(Fig.T-3) Pattern I   photons ( 2, 3 ) = ( V, V ) → ( 1, 4 ) = ( H, H ).

When we detect both photons "2" and "3" as vertically polarized (= V ) photons, both photons 1 and 4 are automatically horizontally polarized photons (= H ).
It is quite natural, because they generated photon pair, one of which is H, and another is V at BBO.

So they insist photons 1 and 4 are in the mysterious "entangled" states.
But as you feel, these results are natural things from the usual classical light's viewpoint.
Then why they insist these states are faster-than-lihgt entanglement ?

(Fig.T-4) They suppose "virtual" classical "photon" pair,

I don't think ordinary people understand why they consider these results as mysterious entanglement.
As I said in above section, these kinds of experiments assume, what we call, "classical" photons.
If classical photons really exists, they cannot be devided at beam spilitter, so they cannot always show the same polarization results at detector, even if their polarizations are in the same direction.

For example, in Fig,T-4, both filters A and B point in the same directions.
( And both photons A and B have the same polarization. )
In the entangled photons, when one of them passes the filter, another photon always passes the filter by spooky link.
But in classical photons, even when photon A passes the filter A, photon B does NOT always pass filter B, as shown in Fig,T-4 (= it reflects at the probability of sin2 α ).
If you have not read the sections above in this page, read them first.

(Fig.T-5) Both waves pass their filters ( ++ ) enough to be detected.

If we think classical "photon" as illusion, and use the concept of classical "electromagnetic waves", we can easily explain these kinds of experiments from the local realistic viewpoint.

First, when we can detect both "2" and "3" photons at detectors, it means both "2" and "3" photons, which have passed the beam splitter, have strong light intensities enough to be detected as a "photon".
( If their intensities are weak, we cannot detect them as a single photon below threshold, and they are neglected. )

So in this case, both "1" and "4" also have strong light intensities, which are polarized in the specific directions, and we can detect them as "entangled" states, at the single photon detector (= H and H in this case ), because they are more than detection threshold.

For example, if you can detect photon "2" as "V", it means photon 2 almost points towards "V" before it enters beam splitter.
Because if photon 2 points between "V" and "H" (= 45 degree ) in advance, it is divided equally into pass and reflect directions ( cos2 45 = sin2 45 = 1/2 ), and cannot reach detection threshold.

(Fig.T-6) Pattern II.

When we detect photons 2 and 3, as one is "V" and another is "H", it means one is H, another is V in 1 and 4 photons.
Again it is quite natural from the usual classical viewpoint.

"Bell state measurement" needs to detect both photons.
So if we can detect only one of 2 and 3 photons, their results are NOT used as experimental results. (= trick ! )
I'm sorry for experimentalists of entanglement, but if you spread "supernatural" phenomena such as faster-than-light entanglement, you should explain to ordinary people in more detail for them to understand your procedure well, I think.

Entanglement of all particles (= atoms, spin1/2, hole ) are illusions.

[ Why various top journals continue "Shut up and calculate !" about spin "forever" ? ]

Even in 2013 of 21th century, various top journals continue "Shut up and calculate !" about "spin".
( For example, the role of spin in organic photovoltaics, or quantum many-body spin system in an optical lattice clock even in 2013. )
As shown on this page, it is impossible that very weak spin magnetic moment can cause the energy difference between singlet and triplet states.

And S states (= angular momentum zero ) in each energy levels are impossible, too.
Because outer electron cannot penetrate many inner electrons and nucleus.

[ Doublet = spin or circular motion ? ]

If the two electrons are entangled to the spin 0 state (spin up + spin down = spin 0), it is said that the magnetic moments are canceled out and become zero.
But this explanation is wrong. Because the two electrons are apart from each other by the repulsive Coulomb force. (If the two electrons having the opposite magnetic moments are completely sticking to each other, the sum of their magnetic moments becomes zero. But it is impossible.)

[ Electron spin is real ? ]
electron-spin

And the spinning fermions don't go back to their original configurations by the 2π rotation. (By the 4π rotation, they return.)
And if the electron spin is caused by the real spinning of the electron, the electron rotation speed becomes more than 100 times faster than the speed of light.

So the spin is only the "mathematical" thing, not a "real" physical object. (See also this page.)
So it is unnatural that we use this "mathematical" spinor for the "real" entanglement experiment (for example, like cosθ/2).
The original spin state may be changed by each experimental environment having some external magnetic fields. It is "artificial" manipulation

Bell inequality of spin 1/2 particles often appears in usual textbooks, though actual experiments treat other things such as photons.
Because this experiment with spin 1/2 particles is very difficult.
The important point is that the Bell inequality itself of spin 1/2 particle is "unreal" thing, which is a little different from that of photons.
We explain this reason later.

[ Two neutral atoms really entangle each other ? ]

[ Rydberg blockage. ]

It is said that they succeeded in entanglement of neutral atom-atom.
But as far as I see, these kinds of entanglement use imaginary (two) photons to detect them.
So we can explain this phenomenon by usual local classical mechanics ( NOT fantasied nonlocal ), as I said in above section.

When two Rb atoms are trapped and very close to each other (= several micrometers ), the energy to excite both two atoms is a little bigger due to dipole-dipole interactions.

[ Faster-than-light entanglement ? Really ?? ]

So, they can excite only one of two atoms, and only this excited atom goes down to the lower hyperfine state (= red arrow ) by additional irradiation.
As a result, one of the two atoms is in the original upper hyperfine state, and another is the lower hyperfine state. (= 5s, upper and lower arrows. )
This state is an entanglement, they insist. But is it really so ?

To confirm these two hyperfine states, they use, what we call, (two) photons.
So at this moment, atom-atom entanglement is equal to photon- photon entanglement.
Furthermore, they just artificially make two atoms upper and lower hyperfine states, which is NOT related to entanglement.

[ Electron hole entanglement is an illusion. ]

[ Faster-than-light entanglement ? Really ?? - part II. ]

Basically, electron and hole, which are generated from each other, are correlated.
They often use very vague concept of "spin", this spin state can be measured by light polarization.
At this moment, spin-hole entanglement can be explained by classical lights, as I said above.
And spin has the same magnetic moment (= Bohr magneton ) as Bohr's orbit.

Or they insist the entanglement can be confirmed by measuring vague noise of thin insulator between them.
But as you see, the distance between electron and hole is so short, that we cannot confirm they are really faster-than-light.

Actually, these entanglement things are NEVER useful for our daily lives, though these kinds of experiments have been performed extremely many times.
These facts clearly proved the entanglement itself is just an illusion.

[ Spin 1/2 entanglement is only a mathematical and imaginary thing. ]

First we define the next unit vector using the polar coordinates as follows,
(Eq.4)

When the spin points to the direction of this unit vector, the spin operator can be expressed as
(Eq.5)

This is an inner product of vectors n and S.

So the spinor of the eigenfunction, which points to the direction of n is
(Eq.6)

And the spinor of the eigenfunction, which points in the opposite direction from n is
(Eq.7)

Suppose two spin 1/2 particles of the opposite spins are flying in the opposite directions.
The spin of the particle A is measured by A, while the spin of particle B is measured by B.

First, we suppose the direction of measurement A points to the unit vector n of Eq.4.
If the spin of the particle A is +1/2 in the n direction at the measurement A, the particle B instantly becomes - 1/2 in the n direction. (= entanglement.)
This means that the instatnt the particle A becomes α of Eq.6, the particle B becomes β of Eq.7.

Next we suppose the direction of measurement B is z direction.
(So the angle betweem measurements A and B is θ.)
According to Eq.7, the probability that the spin of the particle B becomes -1/2 in the z direction is
(Eq.8)

When the particle A is +1/2 at the measurement A and the particle B is -1/2 at the measurement B, this case is named "( + + )".
Because they are usually opposite from each other.

And the probability that the spin of the particle B becomes +1/2 in the z direction is
(Eq.9)

When the particle A is +1/2 at the measurement A and the particle B is +1/2 at the measurement B, this case is named "( + - )".

In the same way, when the particle A is -1/2 at the measurement A, the particle A means β of Eq.7, and the particle B instantly becomes α of Eq.6.
So the probability that the spin of the particle B becomes +1/2 in the z direction is
(Eq.10)

When the particle A is -1/2 at the measurement A and the particle B is +1/2 at the measurement B, this case is named "( - - )".

And the probability that the spin of the particle B becomes -1/2 in the z direction is
(Eq.11)

When the particle A is -1/2 at the measurement A and the particle B is -1/2 at the measurement B, this case is named "( - + )".

So like the above photon case, the probability of ( + + ) and ( - - ) minus the probability of ( + - ) and ( - + ) is
(Eq.12)

where θA1-B1 means the angle between the measurements A1 and B1.

Here we do the following four measurement (each particle has two). (Fig.29).

(Fig.29) Particle A → measurement A1 or A2. Particle B → B1 or B2.
four measurement

So each cosine value can be calculated as (like Eq.12),
(Eq.13)

and

So the Bell inequality of the spin 1/2 particle is
(Eq.14)

This violates Bell inequality !

But we go back to Eq.6 and Eq.7 again.
According to Eq.6 and Eq.7, the spinor can't return by one rotation !
(It needs two rotations (= 720 degrees) for returning to its original form.)
So this spinor is NOT a real thing, either.
This means that Bell inequality itself of Eq.14 is wrong in the case of spin 1/2 particles.

As a result, it is very difficult to say that the "real" entanglement of spin 1/2 "real" particles is occurring.
( The illusions of Entanglements are caused by "unreal" particles such as photons and spin 1/2 particles.)

All interpretations of QM such as MWI, Copenhagen and BM are meaningless ?

The many worlds interpretation(MWI) was proposed first by Hugh Everett [5].
And now the MWI is considered as the best interpretation of QM.
But many people including I think it is only fantasy.

In many-worlds, the collapse of the wavefunctions in superposition states is explained by the mechanism of quantum decoherence.
By decoherence, many-worlds claims to resolve all of the correlation paradoxes of quantum theory, such as the EPR paradox and the Schrödinger's cat.
(But I think "decoherence" is very intentional and a very convenient way to interpret.)
In MWI, there is no strange observation-triggered wavefunction collapse which the Copenhagen interpretation proposes.
(Surely, it is unnatural that we treat only "observer" as a "special" thing.)

(Fig.30) All interpretations are unnatural ? → "Shut up and Calculate !"
interpretation

The most "real" interpretation of the quantum mechanics is the Bohmian mechanics (BM) ( De Broglie - Bohm theory or pilot wave ).
But strange to say, the BM electron doesn't have the charge and spin 1/2.
Because if the BM electron has the charge, it radiate energy as it is moving.
And the BM electron may be faster than the speed of light. See also this page.

On the other hand, in the Bohr model, when the orbital length is a integer times the de Broglie's wavelength, the electron's motion becomes stable not radiating the energy.
The BM obeys the probability density of the Schroedinger equation, so the electron's movement is much more complex than the Bohr's model.
And strange to say, the probability density near the point at infinity of the ground state hydrogen electron is not zero.

More important problem is that all these interpretations are inconsistent with the relativistic quantum field theory(QFT).
All interpretations of QM are based on the Schroedinger equation(S.E.) which is acausal and "probabilistic".
But the nonrelativistic S.E. can't explain the relativistic effects including the fine structure (spin-orbital interactions) .

As the atoms become heavier (which means the atomic nucleus charge becomes larger), the experimental results become more different from that of S.E., because the electon's speed becomes faster. In S.E., irrespective of the nucleus charge, the electrons of any atoms are all static as electron clouds obeying the probability density.
So the Schroedinger equation is incomplete in explaining the experimental results..

On the other hand, the Dirac equation (D.E.) which satisfies the (relativistic) causality can explain this relativistic effects.
Of course, D.E. is not "probabilistic".(= D.E. doesn't show the probability density.) So the causal Dirac equation can be said to be only a "mathematical" model.
So the relativistic QFT is inconsistent with all interpretations and the nonlocal phenomena such as the entanglement.
This means the QM contains self-inconsistency.

And according to "superposition" of the QM, the Benzene's electrons are in the many-world like superposition state !
From the classical viewpoint, the Benzene's electrons are only oscillating and moving.
So the strange "superposition" state can be replaced by the classical movement "naturally".

How about the Bohr-Sommerfeld model ?
Sommerfeld developed the original Bohr model to the model which can explain the relativistic effect completely, (which result accidentally coincided with that of the Dirac hydrogen model).
Of course, the Bohr-Sommerfeld model dosn't contain self-inconsistency at all.

References :

[1] A. Einstein, B. Podolsky, N. Rosen: "Can quantum-mechanical description of physical reality be considered complete?" Phys. Rev. 41, 777 (1935).

[2] J. Bell: "On the problem of hidden variables in quantum mechanics" Reviews of Modern Physics 38 #3, 447 (1966).

[3] A. Aspect, Dalibard, Roger: "Experimental test of Bell's inequalities using time-varying analyzers" Phys. Rev. Letters 49 #25, 1804 (1982).

[4] A. Aspect, P. Grangier, G. Roger: "Experimental realization of Einstein-Podolsky-Rosen-Bohm gedanken experiment; a new violation of Bell's inequalities" Phys. Rev. Letters 49 #2, 91 (1982).

[5] Hugh Everett: "Relative State Formulation of Quantum Mechanics" Reviews of Modern Physics 29 454 (1957).

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2010/3/8 updated. Feel free to link to this site.