Atomic nuclei intimately entangled by a quantum measurement

Oct 17, 2012

Scientists from the Netherlands (Delft University of Technology and the FOM Foundation) and the UK (Element Six) have brought two atomic nuclei in a diamond into a quantum entangled state. This exotic relation was created by subjecting the nuclei to a new type of quantum measurement. These experiments mark an important step towards the realization of a quantum computer. The results were published on 14 October 2012 online in Nature Physics.

is one of the most intriguing phenomena in physics. When two particles are entangled their properties are so strongly connected that they lose their own identity. Measuring both particles yields fully correlated outcomes, even when the particles are very far apart. Einstein famously called this feature 'spooky action at a distance'. It was only after John Bell found an inequality that could prove these weird properties that entanglement was accepted as a fundamental part of nature. Today, quantum entanglement is recognized as a resource for revolutionary new technologies that could provide secure communication and ultra-fast computation.

in synthetic diamond are promising building blocks for a quantum computer. These nuclei behave like a tiny magnet (spin). The two possible orientations of the spin (up or down) can be used to encode . Scientists from Delft University of Technology (Netherlands) reported last year in Nature that they could control and read out individual nuclear spins.  Furthermore, using techniques, the Element Six team (UK) produced synthetic diamond, where due to its exceptional purity the quantum states of the nuclear spins were well protected from their environment. However, interactions between nuclear spins in synthetic diamond are weak, making it challenging to create the entanglement required for . The scientists from Delft, working in partnership with Element Six, have now overcome this challenge by exploiting a special property of quantum measurements.

Quantum measurements not only provide information about a system but also force the system to choose between its possible states. This projective nature of quantum measurements makes them a powerful tool for manipulating quantum systems. The team used an elegant variation on the conventional to generate the desired entanglement. Instead of probing the spin state of each nucleus separately, they measured a joint property of the two nuclei without gaining any knowledge on the individual states. In particular, the measurement forced the nuclei to either assume the same spin orientation or opposite ones, thus imprinting the desired correlations.

The scientists proved that the nuclei were entangled by violating the famous Bell inequality; the first demonstration with spins in a solid. The team now plans to use the entanglement to demonstrate basic quantum algorithms that have no classical counterpart, such as the teleportation of spin states.

Explore further: Physicists design zero-friction quantum engine

More information: Pfaff, W., et al. Demonstration of entanglement-by-measurement of solid-state qubits. Nature Physics. DOI: 10.1038/NPHYS2444

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wrapperband
1 / 5 (1) Oct 17, 2012
Could the entanglement be the result of a third vibration / particle / electromagnetic / Higgs field vibration being created? This would not then be "spooky action at a distance", because the third particle / wave / entanglement particle exists at both end points and in between. Its resulting entanglement or alignment vibration is exhibited at the particles.

This is analogous 2 sounds or water waves which interfere to give a maximum.

There are 2 tests I propose that might intricate this is true.

1. Is there a minimum between the entangled particles?
2. Does the entanglement suddenly stop at a particular distance, where the entanglement particles is stretched beyond its quantum energy state?
holoman
1 / 5 (2) Oct 17, 2012
GOOGLE SEARCH ---- COLOSSAL STORAGE entanglement
antialias_physorg
5 / 5 (2) Oct 17, 2012
Could the entanglement be the result of a third vibration

Nope.

Google for "hidden variable" to find out why not.

Is there a minimum between the entangled particles

A miniumum of what?
Does the entanglement suddenly stop at a particular distance

No it does not (at least to distances of 143 kilometers - which is the current record)
wrapperband
not rated yet Oct 17, 2012
Hi antialias.

If the entanglement doesn't break down at some distance, then isn't every particle in the universe entangled with every other particle ? ...

Re.>>>
1. Is there a minimum between the entangled particles? (inter-spacial energy / Higgs frequency?
2. Does the entanglement suddenly stop at a particular distance, where the entanglement particles is stretched beyond its quantum energy state?

vacuum-mechanics
1 / 5 (2) Oct 18, 2012
Quantum entanglement is one of the most intriguing phenomena in physics. .. Measuring both particles yields fully correlated outcomes, even when the particles are very far apart. Einstein famously called this feature 'spooky action at a distance'. It was only after John Bell found an inequality that could prove these weird properties that entanglement was accepted as a fundamental part of nature…


Actually, there is no direct experiment prove which dictate the existence of the entanglement. Bell inequality is just a theoretical prove by using simple algebra which seems not suitable for quantum mechanics, please see detail below
http://www.vacuum...17〈=en
johanfprins
2.3 / 5 (3) Oct 18, 2012
Entanglement occurs when two or more waves MERGE to form a single wave within which the component waves from which it formed TOTALLY lose their separate existences. It is thus NOT a superposition of waves in the normal sense.

Photon-waves MERGE in a laser-cavity to emit a single entangled laser beam.

One can also entangle millions of electron-waves by extracting them into a cavity formed between a suitable cathode substrate (eg. highly doped n-type diamond) and an anode. Once formed, an electron can teleport through this entangled wave from the cathode to the anode or from the anode to the cathode.

Suitable measurements, which change the boundary conditions of an entangled wave, can disentangle this wave into its component parts: eg. a laser beam impinging into a metal disentangles into photon-waves which can be absorbed by the e-waves within the metal. When this absorption increases the mass-energy of an absorbing e-wave to be more than its rest-mass, an electron ejects.
antialias_physorg
5 / 5 (2) Oct 18, 2012
If the entanglement doesn't break down at some distance, then isn't every particle in the universe entangled with every other particle ? ...

No. Why should they?
You can break entanglement (by measurement/interaction). But entanglement doesn't break on its own.
Tausch
1 / 5 (1) Oct 21, 2012
@Johanprins

http://www.mat.un...ides.pdf

Please read.

"...entanglement also exists in classical eld theories..."

You've got support out of Vienna. I like this alternative (optical approach) very, very much. Sidesteps QM dogma.
johanfprins
1 / 5 (2) Oct 22, 2012
@Johanprins

You've got support out of Vienna. I like this alternative (optical approach) very, very much. Sidesteps QM dogma.


Thank you Tausch!

The following postulates unifies Maxwell's equations with all of quantum mechanics; without the need for the Voodoo Copenhagen-interpretation:

1. A moving photon is a COHERENT EM-wave which like any EM-wave is modelled by Maxwell's wave equations.
2. A moving electron is a slower-moving EM-wave which s also modelled by Maxwell's equations.
3. A trapped photon wave MUST form a stationary-wave determined by the boundary consitions that are trapping it. No free photons are possible within a box. The same is valid for electrons. No free electrons within an ideal metal-block!
4. A trapped electron-wave can in turn trap a photon-wave. This requires from the electron wave to instantaneously morph in shape and size. This "quaatum jump" is thus an entanglement of the two waves during which they lose their distinguishabilities.
spinlock
not rated yet Oct 31, 2012
Would our universe be very much different if entanglement didn't exist?
Tausch
1 / 5 (1) Oct 31, 2012
Everything counts. See cardinality in mathematics.
(The difference between existential quantification and universal quantification)
johanfprins
2.3 / 5 (3) Nov 01, 2012
Would our universe be very much different if entanglement didn't exist?

Yes! You would not have had chemical bonding.
Tausch
1 / 5 (1) Nov 01, 2012
Did you take entanglement beyond reification?

http://en.wikiped...llacy%29

and...

Correlations that fit the expected pattern contribute evidence of construct validity.

http://en.wikiped...validity

'Independent' events just became more unreal.
Your answer is consistence.
johanfprins
2.3 / 5 (3) Nov 01, 2012
'Independent' events just became more unreal.
Why?

Two separate entities CANNOT communicate faster than the speed of light. This is a scientific fact that can be experimentally verified.

Thus two entangled entities cannot be separate entities. "Particles" whatever their definition, are separate entities and can thus never communicate faster than the speed of light. Entanglement is thus a a direct proof that "particles" are not involved: Thus "wave-particle duality" and "complementarity" are crackpot ideas!
Tausch
1 / 5 (1) Nov 01, 2012
Events occurring and 'entities' are synonymous in my world view.
Two different quantum states can represent a single measurably indistinguishable physical state.

see:
http://en.wikiped...Electron

under the subtitle;

Quantum properties

Electrons are identical particles because they cannot be distinguished from each other by their intrinsic physical properties. In quantum mechanics, this means that a pair of interacting electrons must be able to swap positions without an observable change to the state of the system. The wave function of fermions, including electrons, is antisymmetric, meaning that it changes sign when two electrons are swapped; that is, ψ(r1, r2) = −ψ(r2, r1), where the variables r1 and r2 correspond to the first and second electrons, respectively. Since the absolute value is not changed by a sign swap, this corresponds to equal probabilities. Bosons, such as the photon, have symmetric wave functions instead.
johanfprins
1 / 5 (2) Nov 02, 2012
@Tausch
Two different quantum states can represent a single measurably indistinguishable physical state.


This and the rest you have posted I have taught to my students for over 40 years; until I found experimental evdence that this might not be quite correct.

I am of the opinion that two different quantum states can only become a measurable, indistiguishable state when these two quantum states TOTALLY lose their distinguishability as separate entities.

"Identical" separate entities does not mean "indistinguishable" entities, or else a bottle of oxygen molecules whould not have followed Maxwell-Boltzman statistics.

As long as the "identical entities" remain separate entities you can exchange any two of them without requiring a change in the symmetry of the collective entity. Only when these entities merge so that you cannot distinguish them as separate entities anymore, can you argue about symmetric- and antisymmetric macro-wave entities.
johanfprins
1 / 5 (2) Nov 02, 2012
A laser beam is a single entity within which one cannot distinguish separate photons. Each photon is a single distinguisable coherent light-wave, but once they have merged to form the laser beam, they cannot exist as separate distinguishable photon-waves anymore: i.e. there is no "wave-particle" duality involved.

Electrons are a bit more complicated owing to their spins. They can only lose their separate distinguishabilty in pairs: These pairs, however, remain distinguishable, but owing to the opposite spins these new entities are bosons. Nonetheless, it is now possible for these pairs to become indistinguishable to form a single macro-wave.

I have achieved the latter by extracting electrons with an anode from an n-type diamond surface. The macro-wave which forms does not exist of distighuishable electrons, and also not of distinguishable electron-pairs. It is the equivalent to a laser beam, but while a laser beam cannot be stationary, the electron-phase is: And is superconducting.
Deutsch
not rated yet Nov 12, 2012
@johanfprins

Hello, I would like to know more about quantum computers and the technical difficulties. I find them very intriguing and interesting but I do not trust much on Internet documentation or I don´t understand too well. Thank you!
ValeriaT
not rated yet Nov 12, 2012
Here you can see four entangled ions. They still appear like well distinguished particles, not like the single one or even "coherent wave".
johanfprins
1 / 5 (2) Nov 12, 2012
@johanfprins

Hello, I would like to know more about quantum computers and the technical difficulties.
So would I! All the papers are so vague and seems to have been written to bamboozle the funding agencies.
I find them very intriguing and interesting but I do not trust much on Internet documentation or I don´t understand too well. Thank you!
I do not even trust the official versions. I must admit I have not yet had the time to try and study in depth what a qubit is; but I suspect that is like the Higgs boson a hallucination. Most of my time the past ten years have been spent in an effort to teach the quantum-computing physicists (like he late Marshal Stoneham) how a simple capacitor works: So far not one of them could rvrn understand this.!

As far as I can ascertain, with my present limited knowledge, the concept is based on Schroedinger's cat being both alive AND dead. Did you also get that message? And do you think it makes physics sense?
johanfprins
1 / 5 (2) Nov 12, 2012
http://phys.org/n...tml#nRlv you can see four entangled ions. They still appear like well distinguished particles, not like the single one or even "coherent wave".
To quote from above: "
"When two particles are entangled their properties are so strongly connected that they lose their own identity". If they are entangled they do not exist as separate "particles" (etitities), but as a single coherent wave: Such a wave can exist of separate parts: For example, many materials can emit light in opposite directions. In most of these materials these photons are NOT entangled. In the special materials emitting entangled "pkotons" the two parts of the wave being emitted are not separate photons until you make a measurement that disentangles them: That a single coherent wave can consist of two or more parts which are not separate parts, is well-known or else such a wave will not be able to move through two slits simultaneously.
ValeriaT
not rated yet Nov 12, 2012
A laser beam is a single entity within which one cannot distinguish separate photons.
It does apply for squeezed light pulse, not for common laser beam, where the photons aren't entangled and such a beam exhibits a noise in amplitude and spin. Your theory therefore violates the observations (if we neglect the fact, it doesn't provide any testable predictions, so it's useless).
johanfprins
1 / 5 (2) Nov 12, 2012
A laser beam is a single entity within which one cannot distinguish separate photons.
It does apply for squeezed light pulse, not for common laser beam, where the photons aren't entangled and such a beam exhibits a noise in amplitude and spin.
So what? You yourself admit that in this case there is no entanglement.
Your theory therefore violates the observations (if we neglect the fact, it doesn't provide any testable predictions, so it's useless).
How can it violate it if you yourself admit that the experiment you are quoting does not entangle the photons? Another one of your blatant misrepresentations.

Good night!
ValeriaT
1 / 5 (1) Nov 12, 2012
You're just pushing your "everything is a wave, mainstream physics lies" interpretation bellow every article, but this is just an postdiction - not a prediction. If you think, it is so important to consider your view, why don't you present some prediction of it instead?
johanfprins
1 / 5 (2) Nov 13, 2012
If you think, it is so important to consider your view, why don't you present some prediction of it instead?


I can ask the same about your AWT crap! Why do YOU not present some prediction?

My model predicts that a moving photon and a moving electron must diffract because they are both coherent electromagnetic waves and not "particles". Only a complete idiot will believe that a "particle" can diffract.

In the time when we still had sane physicists around it was roundly accepted that Young's double slit experiment proves without any doubt that light consists of waves and NOT of "particles". WHY? Because any person with common sense will tell you that single "particles" being sent through the slits CANNOT form a double slit diffraction pattern. This can also be easily proved experimentally.

Thus when sending entities through double slits and each entity diffracts, each entity MUST be a wave. That is logical physics. Any other explanation is Voodoo!
johanfprins
1 / 5 (2) Nov 13, 2012
@ natello (ValeriaT): Yea! Yea! Yea! The same unscientific nonsense from a demented mind!
johanfprins
1 / 5 (2) Nov 13, 2012
Dare you explain the difference with your theory? Please note, this is not about size of diffraction fringes - but about different level or particle character of photons in each experiment.
Yea! Yea! Yea! Another repeat question by a demented mind: I have already explained this OVER and OVER and OVER again!

johanfprins
1 / 5 (2) Nov 13, 2012
I have already explained this OVER and OVER and OVER again!
Well, at least you're not denying this effect. But you're just repeating, you explained it - but I never saw any explanation of it from your side.
I do not want to correspond with a demented person like you; since you just cannot understand any physics. You do not even realise that the question you asked requires more information: Was the diffraction patterns created by the same number of photons in each case? Did the screen have the same sensitivity for the low and high frequency light?

Obviously the absorbers within the screen for the low frequency light CANNOT be the same as the absorbers for the high frequency light. The lower the frequency of the light the larger the absorber must be and the larger the spot. To conclude that "what is changing here is the PARTICLE CHARACTER of photons" again proves that you know very little about physics and that you argue like a demented idiot! So Yea! Yea Yea!

johanfprins
1 / 5 (2) Nov 13, 2012
Oh and by the way: I have explained OVER and OVER and OVER again, that a material absorbs only light-energy in packets h*nu, since this requires that there must be electron-energy levels spaced an energy (delta)t=h*nu. The light itself does not require to consist of separate qunta since when it frequency resonates with the energy difference of electron0-energy levels (delta)E an amount of light-energy h*nu disentangles from the light wave to be absorbed, since this is the ONLY amount of energy that the levels CAN ABSORB.

You radio antenna does NOT absorb ALL the energy that is being broadcasted from the radio-mast, but ONLY the amount that it can as determined by its size and frequency! Do you NOW get it?

In other words only a part of the energy of the broadcasted radio-wave disentangles to enter your radio-antenna. The radio-mast does not have to broadcast a radio-wave that exists of chunks of energy that is just enough for your antenna to absorb!
johanfprins
1 / 5 (2) Nov 13, 2012
Well, you just told in previous post, you explained this effect many times (OVER and OVER and OVER) - and now you're telling me, this effect cannot be explained without further parameters.

If you were able to understand what I have posted OVER and OVER and OVER, you would not have asked such a stupid question without suppluing the extra information. It ust proves that it does not matter how many times I explain simple physics OVER and OVER and OVER again, it will NEVER penetrate your thick skull!
Tausch
1 / 5 (1) Nov 13, 2012
A ball that rolls will lose orientation.
A ball that expands will not lose orientation.
Touching all paths.

http://en.wikiped...my_group

See Etymology

Comes from playing ball.
johanfprins
1 / 5 (2) Nov 13, 2012
@johanfprins: Does your theory predict the size of photon?
As in the case of ANY electromagnetic wave its size is determined by the boundary conditions. For a freely moving light-wave (thus also for a freely-moving photon-wave) these boundary conditions are determined by the SOURCE of the wave: Have you never done ANY solutions of Maxwell's equations for an EM-wave? Obviously not.

Nowadays we expect first year physics students to be able to do this in South Africa, even though our average mathematical literacy is abyssmal. Where are you from?

Yea! Yea! Yea! You are again proving your total incompetence when it comes to physics!
ValeriaT
not rated yet Nov 13, 2012
So, according to you the size of photon is determined by its source?
johanfprins
1 / 5 (2) Nov 14, 2012
So, according to you the size of photon is determined by its source?


Since it is a relevant question and not another insane argument about AWT, I will answer!

Yes! As well as its history after having been emitted by the source. For example, after it has moved through two slits, its size when reaching the screen is the same as that of the diffraction pattern. It is then absorbed by one of the myriad absorbers in the screen and thus collapses to become the size of the absorber. Thus the size of the spot is not determined by the size of the photon just before it is absorbed, but by the size of the absorber with which it interacts to form the spot.

It has been deduced that photons from faraway stars have cross-sectional areas the size of Australia before they collapse to be detected.
johanfprins
1 / 5 (2) Nov 14, 2012
OK, could you explain after then, why gamma ray photons in spintharioscope tube do appear like tiny bright spots


You can only "observe" the photon-waves by changing their boundary conditions. When you change the boundary conditions of a wave, the wave morphs in shape and size. This is what waves do best.

If the boundary-conditions require the wave to morph to a smaller size, it will morph to a smaller size. If the boundary conditions require the wave to inflate in size and move through two slits, whereafter it further increases in size as it spreads out from the slits, the wave will comply. If you try and "look" through which slit the "particle" has moved, you change the boundary conditions behind the slits so that the wave must again morph: It then collapses to become smaller, and therefore cannot interfere with itself anymore. "Particles" cannot do this even with a Voodoo "guiding wave" attached.
johanfprins
1 / 5 (2) Nov 14, 2012
BTW: You do not directly see the gamma-ray photons, but the light generated by the interaction of the gamma-ray with a flourescent screen. The size of the gamm-ray is determined by the size of the source that emits it: i.e. the radio-active nucleus.

Whe inmpinging into the fluorescent screen, it generates a source for a green photon-wave: The size of this wave is also determined by the size of its source: So you can then see individual photon-waves. It is really quite simple, and meshes perfectly with what we have always known about the behaviour of waves.

Why "particles" were brought into it after Schroedinger proved that electrons are also waves, baffles me. It was probably a political move by Heisenberg and Bohr. What baffles me even more is that physicists who are supposed to be intelligent can cling to this concept even though their own models tell them it cannot be correct.

They will rather invent "many-worlds", retro-action in time, renormalization of infinities! LOL!
johanfprins
1 / 5 (2) Nov 14, 2012
In your (or J.A.Wheeler's) theory the electron is a EM wave
Up to this point yes!
confined by its own mass
Not in my theory!

It's size would be limited with the space-time curvature,
No the photon's size is not limited by space-time curvature: Initially its size is limited by the source that emits it but its shape and size can change when the boundary conditions change: For example, when it is absorbed its size must adapt to be the size that the absorber can absorb.

Between being emitted and being absorbed, any light wave (also a photon) can change its shape and size, and even be sliced into subparts, when the boundary conditions change so that it must do this; for example, to move through separate slits.

Only when a light-wave enters a gravitatiional field around the rest-mass of matter, does it slow down owing to refraction. But this is altogether another aspect than "limiting" the "size of a photon".
johanfprins
1 / 5 (2) Nov 14, 2012
You told us, that the EM wave in electron is moving "very slowly" because of space-time curvature existing here.
You are distorting what I have said: It moves slowly because it has rest-mass. Only within the inertial reference frame within which it is NOT moving, must there be curvature in space to ensure that the EM-wave is trapped within a cavity so that it is a stationary wave. Within the other reference frames the electron-wave is a coherently moving EM-wave, and such a wave does not have a cavity within which it is stationary.

This space-time curvature you derived from mass of electron by E=mc^2 formula.
Nope! I did not just use m*c^2 fob the electron but also for the photon. The only difference is that part of the mass-energy of the electron is rest-mass energy; while this is not the case for the photon. This is why the photon cannot be stationary within any IRF.
johanfprins
2.3 / 5 (3) Nov 14, 2012
If you trap a photon within an artifial cavity, so that it becomes a "standing" stationary wave, its energy becomes rest-mass energy, This happens when photons are generated within the walls of a black-body cavity and injected into the cavity. Each photon forms an extended stationary wave, or entangles with a stationary wave that is already present.

Within the cavity the are not photons moving around like "particles" since all the photons injected can only enter the cavity if they can resonate with the dimensions of the cavity, in order to form stationary waves; each of which fills the whole cavity. As I have stated time and again, when the boundary conditions change the wave morphs in shape and size!

The oscilators within the walls can generate photons of any energy; while we know that only certain energies are allowed to enter the cavity: This is so since each photon MUST have the ability to morph into an allowed stationary wave filling the whole cavity.
johanfprins
1 / 5 (2) Nov 14, 2012
If the photons are slices into two parts with their passing trough double slits, how do you explain, they still form a pin-point marks at the target? The size of photons should correspond the mutual distance of both slits instead.
I have explained OVER and OVER and OVER again, that the wave morphs in shape and size when the boundary conditions change. The two slices collapse into forming a wave with the correct shape and size which can be accomadated by the detector in the screen with which it resonates.

This also happens for electron-waves: For example a p-wave consists of two parts. If it emits a photon so that it must become an s-wave, the two lobes collapse into forming an s-wave. This morphing of a wave has been assumed, by the Copenhagenists, to be a "quantum jump" of a "particle". As I have stated OVER and OVER and OVER again, "particles" are not required at all to explain quantum wave-mecahnics.
johanfprins
1 / 5 (2) Nov 14, 2012
It moves slowly because it has rest-mass.

EM wave has never rest-mass. In your theory therefore no particle formed with "coherent EM wave" can have the rest mass.


I have derived it THREE times on this forum that when you start from the relativity equation for an electron with rest-mass, from which Dirac also started, and use the same substitutions WITHOUT fudging the mathematics as Dirac had done, you get the Maxwell's differential equation for an EM-wave that moves at a speed lower than c. I am not going to derive it again just because you are too stupid to follow mathematics.
johanfprins
1 / 5 (2) Nov 14, 2012
This is just the problem: if the mass of electron is the cause of lower speed of EM wave in it, then the photon has an energy E = h/f = m/c^2, which corresponds some mass too - so its wave should move slower than the speed of light too.
Oh my GOD!! I just knew that I should not have been so kind to try and explain simple physics to you! ONLY when part of the mass-energy is REST-MASS energy can an entity NOT move with the speed of light!!!!!!!!!

So Yea! Yea! Yeah! You ARE a moron and will DIE a moron. I hope VERY soon!
ValeriaT
1 / 5 (1) Nov 14, 2012
when part of the mass-energy is REST-MASS energy can an entity NOT move with the speed of light
Well, this is just what the circular reasoning is called. You're just assuming, that photon has no rest mass - so it cannot move with speed of light. But I don't see any reason, why not to apply the same logic for photon, like for electron - which is already localized well.

Regarding your heartily expressed hope, it appears somewhat undiplomatic for me under situation, I'm by nearly fifty years younger and I've apparently way more stable blood pressure than you...;-)
johanfprins
1 / 5 (2) Nov 15, 2012
Well, this is just what the circular reasoning is called.
Einstein's Special Theory of Relativity is NOT circular reasoning! It is based on the FACT that a light-has the SAME speed c within ANY and ALL inertial reference frames. Thus, light CANNOT be stationary within an inertial reference frame. From Galileo's inertia and Newton's laws, it is known that only an object with rest mass can be stationary within an inertial reference frame. A photon is a light-wave: THUS it cannot have rest-mass. For God's sake stop being so stupid!

I'm by nearly fifty years younger
This explaisn why you know nearly nothing about physics
and I've apparently way more stable blood pressure than you...;-)
Then you should be ashamed of chasing up my blood pressure by posting the same inane and insane ideas time and again, after enough proof has been given on this forum that your arguments are based on hallucinations.
pancake
not rated yet Nov 17, 2012
If you trap a photon within an artifial cavity, so that it becomes a "standing" stationary wave, its energy becomes rest-mass energy, This happens when photons are generated within the walls of a black-body cavity and injected into the cavity. Each photon forms an extended stationary wave, or entangles with a stationary wave that is already present. ...



A laser. Yes , I follow. And standing waves build up enough energy to escape through the brewster window of the laser cavity. That much i understand. Having tuned them.

this is considered entanglement? cool. carry on.
johanfprins
1 / 5 (2) Nov 17, 2012


A laser. Yes , I follow. And standing waves build up enough energy to escape through the brewster window of the laser cavity. That much i understand. Having tuned them.

this is considered entanglement? cool. carry on.


Thank you. Since the laser beam is a single holistic wave, it can split and move through two slits while still remaining a single wave. The two parts moving through the slits thus remain entangled and thus coherent. For this reason they can interfere on the other side to form diffracted wave-fronts. When these wave-fronts reach rhe diffraction screen, they encounter many detectors of atomic size each of which can only detect a single photon at a time: Therefore, they disentangle into photons which are then detected by the atomic-sized detectors to form many spots simultaneously.

johanfprins
1 / 5 (2) Nov 17, 2012
A photon is the smallest coherent light wave that can be emitted and detected. It is a mini laser-beam that can move through both slits so that its two parts remain entangled.

Therefore a single photon also forms a diffracted wave-front which arrives at the detection screen: This wavefront cannot disentangle into more photons, and can thus only be detected by ONE of the available atomic-sized detectors. Thus it leaves a single spot on the screen.

For nearly 80 years it has been concluded that this single spot proves that a photon is a "particle". But since the outstanding nature of a particle is that it cannot diffract, the Voodoo of "wave-particle duality" and "probability-waves" have been invented to overcome this apparent paradox, which has NEVER been a paradox at all; since the contorted reasoning based on "wave-particle duality" is obviously not required.
Noumenon
3 / 5 (4) Nov 18, 2012
The light itself does not require to consist of separate qunta since when it frequency resonates with the energy difference of electron0-energy levels (delta)E an amount of light-energy h*nu disentangles from the light wave to be absorbed, since this is the ONLY amount of energy that the levels CAN ABSORB.


But, there is no such ΔE, between levels, in a free electron , and so no "boundary conditions" for the light wave to "disentangle" to match.

A free electron can have any arbitrary energy, and upon Compton scattering, the particle nature of light is evident. The electron gains momentum proportional to hυ of the photon and not to the intensity of the light. And the photon increases in wavelength showing that is loses momentum, a characteristic of a particle.

This is all that is meant by particle,... no ontological statements are being made in QED, while your ideas must presuppose them.
johanfprins
1 / 5 (2) Nov 18, 2012
But, there is no such ΔE, between levels, in a free electron , and so no "boundary conditions" for the light wave to "disentangle" to match.
Oh MY GOD!!!! Do you have a metal with infinite size! If you have PLEASE show it to me. When you have stationary waves the energy-spacing between tyem is determined by the dimensions which cause them to be stationary waves. The smaller the dimensions, the LARGER the eneergy spacing! Even my grandchildren in grade 5 know this!!

johanfprins
1 / 5 (2) Nov 18, 2012
A free electron can have any arbitrary energy,
Correct this is why such an electron is modelled by Maxwell's wave equation for light which moves at a speed v less than c!

and upon Compton scattering, the particle nature of light is evident.
When and where has Compton scattering been measured from freely moving electrons?

This is all that is meant by particle,... no ontological statements are being made in QED, while your ideas must presuppose them.
You are the one who presupposes that a photon-wave must be a "particle"; while it is totally unnecesary to presuppase this. A "particle" CANNOT DIFFRACT, Goofball! All experiments prove that a photon CANNOT be a "particle" since a photon CAN diffract! Goofball!

johanfprins
2.3 / 5 (3) Nov 19, 2012
@ Noumenon,

I apologise for calling you a "goofball". I am more frustrated with myself that I have been the "goofball" who for years taught my students "wave-particle duality" and "complementarity" as if these econcepts have been the greatest breakthroughs in physics ever; while, in fact, they have been the greatest delusions in physics ever.

How did this come about? When Einstein modelled the photo-electric-effect, The accepted model for electron transport within a metal was that of Peter Drude, which is based on the assumption that there are FREE electrons within a metal. Therefore it was assumed that the photon and the electron collide like two classical particles; whereas this is wrong.

In fact, Einstein's own Special Theory of Relativity immediately proved that this "classical collision" is not possible and must be wrong, since a free electron must have energy that is larger than its rest-mass energy.
johanfprins
2.3 / 5 (3) Nov 19, 2012
The rest-mass energy of an electron is the energy that an electron will have when it is far away from the metal and stationary relative to the metal. Within the metal its energy is less than its rest-mass energy by at least the workfunction of the metal: Thus, such an electron IS NOT a free particle within the metal that can experience a classical collision with a "photon-particle". If it could have been a free-electron, there would not be any chemical bonding of the atoms; since the workfunction would be zero.

Thus to have assumed classical collisions between the impinging light and the valence-electron states, which are chemically binding the atoms together, is utter nonsense: In mitigation,the physicists did not know in 1905 how valence electrons bond atoms together.

By 1927, they did know about chemical bonding, but still decided to interpret the photo-electric effect and Compton-scattering as being caused by classical collisions between "particles". How utterly stupid of us all
Widdekind
1 / 5 (1) Dec 17, 2012
According to the DOI article, the scientists seemingly entangled to separate, and non-interacting, nuclei (within the diamond crystal lattice), by interacting both of them, simultaneously, with another "ancilla" quantum particle, namely an electron. The electron interacted with both nuclei, thereby becoming like a medium between them, through which "bridge" they could influence each other, and so becoming entangled. Measurements of the electron then revealed said entanglement.
johanfprins
1 / 5 (1) Dec 17, 2012
According to the DOI article, the scientists seemingly entangled to separate, and non-interacting, nuclei (within the diamond crystal lattice), by interacting both of them, simultaneously, with another "ancilla" quantum particle, namely an electron. The electron interacted with both nuclei, thereby becoming like a medium between them, through which "bridge" they could influence each other, and so becoming entangled. Measurements of the electron then revealed said entanglement.


A covalent bond is an entanglement of two electron-waves to form a single holistic entity; which does not consist of two separate electron-waves. Since these two electron-waves lost their separate identities they are in immediate contact with one another: Time does not play a role within the volume of the covalent bond.