How do particles become entangled? [Archive]

alexepascual

Nov27-04, 01:51 PM

TheDonk must be frustrated by now. None of the replies really answers his question. I was going to post exactly the same question when I found your post. I'll re-formulate the question to see if this helps and I hope it agrees with the original question's intent.
I would assume we know what entanglement means in a context such as EPR, quantum teleportation, etc. So, this is not the problem. Most descriptions of entanglement describe features of two particles that show that they are indeed entangled, and the more quatitative analyses may describe the degree of entanglement by using the density matrix, but this is not the question we are seeking an answer for either.
Let's say we have a hydrogen molecule. We know from Pauli's exclusion principle that the electrons will have opposite spin. If we separate the atoms, their spins will be entangled. So far so good. But the question is: How did the electrons become entangled when the molecule was created? By what mechanism did the states where both spins are "down" or both are "up" dissapear?.
A similar example could be given in terms of a collision. We heard that every time that two particles interact they become entangled. So if two particles bump into each other, then they must become entangled to a certain degree. Probably the case of a collison is more complex that the one I discussed before (the hydrogen molecule) because in the case of the collision there are more variables involved such as momentum, position, etc.
But in either case, there are combinations of the original tensor product of the separate Hilbert spaces that dissapear. How does this happen? Does entropy for the interacting particles change? How? is there a need to assume some dissipative effect?
I realize a discussion of this phenomenom may become very involved. If anybody here has seen an article on the web where this subject is explained, I would appreciate your pointing us in the right direction. Just remember, we already know what entanglement is, what we want to know is how two particles can become entangled in the first place. We want all the details about what happens when we bring them together. I think an understanding of this is crucial to tackle things such as environment-induced decoherence, the measurement problem, etc.
Once again, If you are knoledgeable about these topics I'll appreciate your guidance.
-ALex Pascual-

TheDonk

Nov27-04, 07:44 PM

alexepascual is right. The answers haven't been exactly what I was asking for. The #4 post, by jvangael was getting closer, but it is more of the way people can do it instead of the general way it happens.

I'm hoping for an answer like:
When two fundamental particles are within x nanometers away.

I know it probably won't involve distance but there must be a single property that two particles can have that will entangle them.

masudr

Nov28-04, 07:48 AM

Two particles must have interacted. If they have, then two measurements represented by operators A and B must behave like [A, B] = ih and then we have entanglement.

blip

Nov28-04, 10:46 AM

So when Bob performs a measurement Alice does not gain any information on her state nor can she tell whether Bob made his measurement. TBut, Bob can conclude in which state Alice's qubit will be in when she measures.

What if they each perform a measurement simultaneously, say one vertical and the other horizontal? Afterwords they send their photon through a wave plate of the type corresponding to the measurement they each made.

Would they pass through their respective plates?

TheDonk

Nov28-04, 07:17 PM

Two particles must have interacted. If they have, then two measurements represented by operators A and B must behave like [A, B] = ih and then we have entanglement.

"Two particales must have interacted."
Can you give me an example of how two particles could interact to become entangled? A simple (if possible) step by step process where two particles start off not entangled and become entangled.

I'm not familiar with this equation and I don't know what i and h are. Can you explain what it means without the equation?

alexepascual

Nov28-04, 08:47 PM

TheDonk:
"i" is the square root of (-1) and "h" is Plank's constant. A and B are what is called "Hermitian Operators" which represent "observables".
If you have little knowledge of quantum mechanics, I suggest you look at some tutorials on the web. Depending on the level at which you want to understand it, the math may become a little intimidating though. In parallel to reading some non-mathematical articles, I suggest you read a book on linear algebra, which is needed for Quantum.
But let me tell you that I think the previous post does not answer your question or mine, and I don't see the conection between what he says and the interaction between two particles.
-Alex-

Eye_in_the_Sky

Nov29-04, 01:33 AM

How do particles become entangled?First of all let's explain the idea of quantum "entanglement".

Suppose that we have two particles, 1 and 2, with corresponding Hilbert spaces H1 and H2. Suppose that particle 1 is in the state |ψ> Є H1, and that particle 2 is in the state |φ> Є H2, and all of this is before the two particles interact. Then, prior to the interaction, the state of the joint system is simply

|ψ>|φ>.

Now, suppose that the interaction between these two particles is such that

|ψ>|φ> → Σk ak|ψk>|φk> ,

where each ak ≠ 0, and there are at least two distinct values for k (and, of course, the |ψk> (|φk>) are linearly independent).

Then, the state of the joint system after the interaction can no longer be written as a simple (tensor) product of one element from H1 with one element from H2 – it must be written as a linear combination of such products. The two particles are now said to be in an "entangled" state.

Next, you ask regarding the interaction itself, referring to it as a sort "bumping" between the two particles:I've heard that it's when two particles bump into each other. How is this "bump" defined? What does it mean for 2 particles to bump? Is it based on distance apart, or something else?It sounds like the type of interaction you have in mind is that of a "collision-like" scenario. So, let's use the example of an "elastic collision". Then, with regards to the "bump" itself, there is nothing really special about it. What is special here is that we are dealing with quantum states.

First let's conceptualize the situation classically. Think of two particles (which repel one another) on a collision course as viewed in their center-of-mass frame. If the particles are directed perfectly "head-on", each one will bounce back in exactly the opposite direction. On the other hand, if their lines of flight are slightly "off-center", each one will be deflected by some angle from its original line of flight, such that:

The smaller the distance between the two lines of flight, the greater the angle of deflection.

However, no matter what the distance between the two lines of flight happens to be, we know that:

The momentum of each particle must be equal and opposite to that of the other.

Clearly, the momenta of the two particles are "correlated" ... and this is due to conservation of momentum.

Classically, we have no difficulty conceptualizing the situation. But, quantum mechanically, we find a bit of a 'twist'.

Suppose that the initial wavefunction for each particle has a very sharp momentum, with particle 1 traveling (to a very good approximation) in the +z direction, and particle 2 traveling (to a very good approximation) in the -z direction. Then, in particular, the wavefunction for each particle's position will show a 'spread' in the xy-plane.

Next, after the particles have gone "bump" and have flown well apart, the wavefunction of the joint system will involve a superposition of the various angles of deflection resulting from each of the possible distances between the two lines of flight consistent with the spread of each particle's initial wavefunction in the xy-plane.

Heuristically, looking at a "reduced" wavefunction |θ>, with θ denoting the angle of the line of flight relative to the z-axis, the above interaction can be summarized as:

|0>|π> → ∫a(θ)|θ>|π+θ> dθ .

Thus, there is nothing really 'special' about the "bump" itself. What is 'special' in all of this is that objects which go "bump" are described by quantum states.

alexepascual

Nov29-04, 12:08 PM

Eye:
Thanks for your detailed answer. I understood most of it, but I still have a lot of questions. My knowledge about the description of multiple-particle systems is kind of defficient. I have a rough grasp of it but there are a lot of things I don't know.
About the original question by TheDonk, he framed the question using the example of a collision between two elementary particles. It may be that he did so because he read some place that particles become entangled after they "bump" into each other. Your analysis of the collision is very detailed and clear. As a matter of fact in your explanation there is a mixture of mathematical expressions with a more intuitive description. I doubt that TheDonk has the mathematical knowledge needed to understand the mathematical description. (I have not asked him how much math or QM he knows), but I am sure he must have learned something from your non-mathematical description, (provided he understands superpositions).
I suspect though, that a collision of two particles may not be the simplest example to study how two particles can become entangled. Wouldn't spin entanglement be easier to understand?
Going back to my difficulties,
I understand that the description of two non-interacting particles that are not entangled whoud consist of a combination of pairs of base states from each separate Hilbert space, and that this is represented by the tensor product of both Hilbert spaces. Now, I have always thought of the tensor product in terms of the combination of base vectors, but I have never thought about what happens to the states of two particles when we describe them in the combined Hilbert space. Assuming non-interaction, what would happen with the complex coefficients? do we just multiply them together for each individual element of the tensor? If so, what about the time dependence?. As I am writing this, I am thinking that it would not make sense to multiply the coefficients because then you might get interferences where there are none. On the other hand, I can see that the probability of finding the composite system in any of the combinations is proportional to the product of the probabilities of each individual base state.
I think what I said a few lines above might be somewhat confising. I said "what happens to the states...". I understand that the states of the individual particles remain unperturbed. What I meant is that I don't understand how to use the coefficients form the individual base states to construct the compound state.
I am now asking you just these simple questions because I think it may be better to tackle one small point at a time.(If I don't understand the simple things I won't be able to undestand the more complex ones). I recall from previous threads that you were very patient with me and I am very grateful for that.

TheDonk:
Eye-in-the_Sky is a very knowledgeable guy and I think he can help us. If you tell us a little more about how much math and quantum mechanics you know, maybe we can give you explanations that not too elementary or too advanced for your level. I consider myself a beginner, but there may be things which I have already understood that I may be able to explain. About myself, although I got a bachellor's degree in physics, which involves two quarters of quantum mechanics and two quarters of classical mechanics, not to mention mathematical methods, I feel that what I studied for school was always in a rush and I never got to understand each of the topics completely. I continue to read books and articles so that I can learn what for one reason or another I didn't learn in school. And I have found this forum to be a great place to learn and to help others.
With respect to the topic of entanglement, I am very optimistic that Eye_in_the_Sky will be able to help us gain a better understanding.

TheDonk

Nov29-04, 04:11 PM

Thanks to both Eye_In_The_Sky and to alexepascual. I have a relatively good understanding of math and bad in QM. I've taken a course on linear algebra, and I understand at least the basics of vector calculus. That's about the extent of my math knowledge. As for QM I've only heard the hype. I've seen the "possibilities" on tv and I've read some stuff. Unfortunately I don't know what superposition or tensors are but I'm going to relook into them because I tried to understand them a couple years ago.

Eye_in_the_Sky, your explanation helped, tho I'm still confused.

I suspect though, that a collision of two particles may not be the simplest example to study how two particles can become entangled. Wouldn't spin entanglement be easier to understand?

So certain properties of two particles become entangled? To properly explain 2 entangled particles, it isn't enough to just say they are entangled but you would need to say which properties are entangled. Is this right? What are all the properties that can be entangled? Is there anything else needed to explain how to particles are entangled? I guess these questions are off topic...
Maybe someone should explain how the entanglement of the spin of two particles would happen unless it's very similar to the collision explanation. What properties are entangled from the collision entanglement?

alexepascual

Nov29-04, 04:57 PM

I think it would be correct to say that some property (observable) of the particle is entangled with that same property or another of the other particle, but maybe Eye can correct me.
In the case of the collision, position and momentum become entangled. In Eye's example, if you find one particle to have certain momentum, the momentum of the other particle has to be per force the opposite. You could also draw conclusions about the other particle's position.
In the case of spins, if you have two hydrogen atoms, each has only one electron, and when you put them in a magnetic field, the spin will take an orientation up or down (with respect to the vertical magnetic field).To be more precise, the spin of each atom will be both up and down at the same time (which is one of the main characteristics of quantum systems). This is called a superposition of states.
If you bring both atoms together, they'll form a molecule. But for those alternatives in which both spins are up or both spins are down, it will be impossible to form a molecule because of Pauli's exclusion principle. So, I guess if you were shooting these atoms towards each other, some would "stick" and some would bounce . About the ones that bounce you can say that their spin is mostly pointing in the same direction while you can be sure that those that stuck to form a molecule have their spins pointing in opposite directions. This relationship between the spins is what entanglement is. Now, you can pull the atoms apart and that relationship (if one is up the other is down) will persist, even if the atoms are taken appart a long distance. When I say "one is up, the other one is down" don't take me wrong. Actually they are both in a superposition of "up" and "down". It is just that when you measure spin on one, at that point you collapse the wave function and one of the two states "up" or "down" becomes reality. If you find one atom to be "up" the other one will be down. In all this discussion we are considering only the spin of the electron in each atom. There are other kinds of spin for the atom but they can be ignored. (at least for a simple treatment?). The nice thing about spin states is that they have only two values (up or down). Momentum and position though are continuous variables, which makes the analysis much more complicated.
I said that a particle can be in different states at the same time which are "superposed". That superposition is represented in a vector space where each dimension represents each of the "base states" that make up the superposition. This vector space is called a Hilbert space. The compuond state is represented by a vector in Hilbert state, and how much of each base state goes into the superposition is represented by the magnitude of the "components" of the vector. (the projection of the vector on each of the base vectors). A Hilbert space can have infinite dimmensions, which is always the case for continuous variables.
I suggest again that you look (maybe google?) for some tutorials on quantum mechanics. You need to learn these things if you want to understand entanglement in some depth. The concept of Hilbert space (also called "state space" is easier to understand when considering spin, because for one electron or proton, it is just a two-dimensional space.
As you are working with vectors, you use matrices to manipulate these vectors, and that's where you need to use linear algebra. You should also try to learn the "Dirac notation". It is not too hard and it is fun. You may be able to find some tutorial on it. Oh! now that I remember, "Wikipedia" has a lot of info on quantum mechanics. Do a search in google for Quantum + Wikipedia and I think you'll find it.

Mike2

Nov29-04, 05:13 PM

If you bring both atoms together, they'll form a molecule. But for those alternatives in which both spins are up or both spins are down, it will be impossible to form a molecule because of Pauli's exclusion principle. So, I guess if you were shooting these atoms towards each other, some would "stick" and some would bounce . About the ones that bounce you can say that their spin is mostly pointing in the same direction while you can be sure that those that stuck to form a molecule have their spins pointing in opposite directions. This relationship between the spins is what entanglement is. Now, you can pull the atoms apart and that relationship (if one is up the other is down) will persist, even if the atoms are taken appart a long distance. When I say "one is up, the other one is down" don't take me wrong. Actually they are both in a superposition of "up" and "down". It is just that when you measure spin on one, at that point you collapse the wave function and one of the two states "up" or "down" becomes reality. If you find one atom to be "up" the other one will be down.
Once they are entangled, can you force the spin of one electron to be up so that the distant electron's spin must be down? That would create instant communication, right?

Eye_in_the_Sky

Nov29-04, 06:53 PM

Concerning my earlier post (#16), two points of clarification are warranted.

First, regarding what I wrote:
... after the particles have gone "bump" and have flown well apart, the wavefunction of the joint system will involve a superposition of the various angles of deflection resulting from each of the possible distances between the two lines of flight consistent with the spread of each particle's initial wavefunction in the xy-plane.This is definitely an oversimplification. In particular, from this simplification, it may appear that the resulting quantum state is merely given by a quantum superposition of the possible classical trajectories. But this is surely not the case! [... However, it is worth noting that for an inverse-square law of force between the two particles, the differential scattering cross section, dσ/dΩ, calculated classically happens to coincide precisely with the quantum-mechanical one (e.g. Rutherford scattering).]

And secondly, the following point must also be clarified:
... with regards to the "bump" itself, there is nothing really special about it.

... there is nothing really 'special' about the "bump" itself.These statements are made only with respect to the phenomenon of "entanglement". On the other hand (as indicated above), the interaction itself is, of course, quantum mechanical – and in that respect, there is certainly something 'special' about the "bump".

alexepascual

Nov29-04, 08:02 PM

Mike2

Nov30-04, 09:35 AM

Mike:
The spins are entangled as far as what you are going to find if you measure them on both ends. Once you force one spin into another direction, you lost your correlations. As far as I know, no communication is possible using entanglement.
Are you saying that forcing the spin into a particular state is the same as measuring that state? Or is there just a greater probability of being in a state with an applied force as opposed to knowing for certain after measuring the spin?

alexepascual

Nov30-04, 10:18 AM

As far as I know, measurement is supposed to tell you something about the state of the particle before you influence it by applying any force which might change the state. If you change the state, then you don't know what it was before. If you are going to measure spin, you make the particle go through a Stern-Gerlach apparatus. The Stern-Gerlach apparatus does apply a force to the particle but this force does not change the spin. It is just that those particles with spin "up" go thorugh one channel and those with spin "down" go thorugh the other channel. Typically, you know that the particle went through a certain channel by having it strike some kind of detector. When the particle hits the detector, the state is destroyed. But at least you know what it was just before detection,(or during detection?) which is what you where looking for.
I think what I have just said makes a little more clear the issue of applying a force to the state to change it. (you don't want to do that). But on the other hand I have to accept that it is a very imprecise description. For instance, when you measure, you don't find what the state really was before measurement, but in reality you collapse the wave function and find the particle to be in one of the possible states in the superposition.
I think it was believed years ago that when you make a measurement of a quantum state you usually get a result that is kind of random because the "observer" is influencing the thing he is measuring. But I think today it is believed that it is not because any kind of influence that you get a random result. There is some built-in randomness in the process of measurement.
I would also like to mention that there is something called "the measurement problem", which in the view of many, has not been satisfactorily solved. So, some of the questions you may ask may not have a definitive answer yet.
I understand this may sound confusing and probably I have not explained it clearly. But I hope it helps a little.

Mike2

Nov30-04, 12:50 PM

As far as I know, measurement is supposed to tell you something about the state of the particle before you influence it by applying any force which might change the state. If you change the state, then you don't know what it was before. If you are going to measure spin, you make the particle go through a Stern-Gerlach apparatus. The Stern-Gerlach apparatus does apply a force to the particle but this force does not change the spin. It is just that those particles with spin "up" go thorugh one channel and those with spin "down" go thorugh the other channel. Typically, you know that the particle went through a certain channel by having it strike some kind of detector. When the particle hits the detector, the state is destroyed. But at least you know what it was just before detection,(or during detection?) which is what you where looking for.
Yes, all honest conversation helps.

I quess what I was getting at is not to actually measure the spin, but only to influence the probabilities with a force - a force not so strong as to make certain the spin after the force is applied, but only strong enough to influence the probability of finding it in a desired state if a measurement were to be done. The communication would be for us to apply a force proportional to the message (e.g. an audio signal) and the other side would actually do the measuring. Is this possible? Or does any interaction at all on our side, whether measured or not, destroy the entanglement? IIRC entangled atoms must be held in isolation in order to be reliable. Entanglement is destroyed when one side interacts with the environment. Is this because the environment is random? Does this still hold if the "environment" is a subtle known force?

selfAdjoint

Nov30-04, 02:52 PM

CharlesP

Nov30-04, 09:26 PM

I am primarily an experimentalist, and although I have been through the formal courses and mathematics of quantum mechanics it sometimes seems more like a ceremony than any aid to conceptualizing. I found the pauli spin matrices most frustrating. To satisfy my conceptualizing one must describe the experiment in great detail. The literature is generally highly abbreviated.
When a hydrogen molecule is formed did a previous wave function collapse to form the paired electrons?
In a laser beam are the photons entangled?
In the Bell experiment he chose angles that were not 0 or 90 degrees thus confusing me beyond help. Isn't there a simpler experiment out now with simpler equations?
I have a real problem with these Alice and Bob experiments because none of the instruments in my lab are named Alice or Bob.
I have found it very hard to corner an expert and ask questions.

Mike2

Nov30-04, 10:38 PM

I don't know what a force strong enough to affect the probability of a spin but too weak to change the spin would be in QM. Spin (around a given axis) only comes in a discrete set of states, so you either leave it in the state it is, that is, you don't interact with it, or less you influence it and then you are operating on the spin state with some operator and the state goes into some set of discrete eigenvalues. It looks to me to be all or nothing.

And therefore yes, if you truly interact with either entangled particle, you collapse the entangled state. Anything that finds out what the spin state of one particle is, disentangles the other particle, if it is still entangled. What you don't know is whether your measured spin of this particle was determined by a prior measurement of the other or not.
The electromagnetic force is governed by interaction with the photon. For a weak enough magnetic field, some electrons will interact with the photons of the magnetic field, and some will not. And for a series of entangles pairs, we would never know on our end which electron spins were effected by the magnetic field photons and which were not. We would only know that on the average more of them are affected with a stronger magnetic field than with a weaker magnetic field. However, if the pairs remain dependent on the electron spins on our side because they are entangle, then a measurement on their side would reveal and average affected by the magnetic field on our side, right?

I guess the question is whether you can have more than two particles involved with entanglement. If the two hydrogen atoms are entangeled, can our electron become entangled with a magnetic field photon (simple because it interacted) without destroying the previous entanglement? How then are many particles entangled if not all at once? I'm assuming entanglement=interaction.

Or is it that we simply cannot change the random nature of the electron spins on the other side? Is it that we can only know what their spin is by examining ours?

Eye_in_the_Sky

Nov30-04, 11:58 PM

I suspect though, that a collision of two particles may not be the simplest example to study how two particles can become entangled. Wouldn't spin entanglement be easier to understand?I don't know if the spin perspective makes it any easier to understand "how" the particles become entangled; but the spin perspective certainly presents some conceptual advantages. It was Bohm, in 1951, who recast the EPR scenario in just such terms. Then, in 1964, with the Bohmian version in mind, Bell discovered his famous inequality.

So, think of two spin-1/2 particles flying off in opposite directions such that the spin state of the joint system is given by

|ψsing> = (1/√2)(|+>|-> – |->|+>) .

Notice that I haven't written the axis with respect to which the spin components |±> are specified. That's because it doesn't matter! It comes out the same no matter what. Go ahead check it out:

|+>z|->z – |->z|+>z = |+>x|->x – |->x|+>x
= |+>y|->y – |->y|+>y .

That's the beauty of the so-called "singlet" state, |ψsing>.

You can immediately see what this means in terms of a spin measurement of just one of the particles relative to any given axis n. Let's say we measure the spin of the first particle, so that the observable for the joint system is

Sn x 1 ,

where "x" denotes "tensor product".

Then, the eigenprojectors corresponding to each of the two possible outcomes are:

P1n,+ = (|+><+|)n x 1 ,

P1n,- = (|-><-|)n x 1 .

Now, project |ψsing> (and then normalize) to get the resulting state of the joint system for each of the two possible results in this measurement of particle 1:

particle 1 is "+" → 'new' joint state is: |+>n|->n ,

particle 1 is "-" → 'new' joint state is: |->n|+>n .

Again, this holds for a spin-component measurement along any given axis n.
______________

Once they are entangled, can you force the spin of one electron to be up so that the distant electron's spin must be down? That would create instant communication, right?Note that in a spin measurement of the above kind, what we have in mind is a Stern-Gerlach device (or the equivalent). For such a physical arrangement, there is no "forcing" of spin to be in any particular direction. However, there is "forcing" of the particle to move either "up" or "down" spatially in relation to the axis n in a manner which correlates with spin. Thus, by detecting the presence of the particle – say particle 1 – in either of these "up" or "down" tracks, a measurement corresponding to the projector P1n,+ or P1n,- (respectively) has been performed.

Note, furthermore, that the interaction of particle 1 with a Stern-Gerlach device has no effect on particle 2. Although particle 1 is deflected from its original line flight, particle 2 will not show any such deflection. Nevertheless, if particle 1 is detected in the upper track, then its state is |+>n, and particle 2 is then necessarily in the state |->n. Similarly, if particle 1 is detected in the lower track, then its state is |->n, and particle 2 is then necessarily in the state |+>n.

Now, let's get to your first question, Mike2.
Once they are entangled, can you force the spin of one electron to be up so that the distant electron's spin must be down?By "forcing", it appears to me that you mean some kind of apparatus which will, for example, leave the |+> as it is, but cause the |-> to become a |+>. Can we do this? Sure! ... why not?

So, let's do this to particle 1, while it is jointly with particle 2 in the |ψsing> state. We then get

(1/√2)(|+>|-> – |->|+>) → (1/√2)|+>(|-> – |+>) .

Do you see what's going on? Here, we are acting directly upon the "spin" of particle 1 to modify it. This is similar to what happens in the Stern-Gerlach scenario, except there we were acting directly upon the "position" of particle 1. In both cases, the property which we 'modify' with regard to particle 1 has no effect whatsoever on the corresponding property of particle 2. To think otherwise is to misconstrue the phenomenon of "entanglement".

Now, you can see this quite clearly by considering the form of the Hamiltonian for the joint system in such a process:

Htotal = (H1 + Hmod) x H2 ,

where Hk is the 'free' Hamiltonian for particle k, and Hmod is the Hamiltonian for the interaction which modifies the status of particle 1. As you can see, the time evolution for the two particles is completely decoupled.

And now, let's get to your second question:
That would create instant communication, right?Indeed, if the "entanglement" phenomenon were such that a direct 'manipulation' of some property particle 1 would induce an (immediate) corresponding change in some property of particle 2, then yes that would be a means for superluminal communication. However, in the above example, we saw that the change "forced" upon particle 1 had no effect upon particle 2.

Nevertheless, one may still wish to contend that somehow, by some as yet unidentified quantum-mechanical effect, perhaps superluminal signaling could be achieved. But no! ... In the late 1970's Eberhard proved that "Quantum nonlocality" does not permit superluminal signaling.
______________

Eye_in_the_Sky, your explanation helped, tho I'm still confused.

So certain properties of two particles become entangled? To properly explain 2 entangled particles, it isn't enough to just say they are entangled but you would need to say which properties are entangled. Is this right? What are all the properties that can be entangled? Is there anything else needed to explain how to particles are entangled?Yes, certain properties (and in general, any dynamical properties) of two particles can become entangled. In a complete description of the entanglement, we must not only specify which properties have become entangled, but also in what manner. If we can write down the quantum state for the joint system, then the description is complete (... least from the quantum-mechanical perspective).

alexepascual

Dec2-04, 12:51 PM

Eye_in_the_sky:
Thanks a lot for your detailed post. I just printed it out and it'll take me a few days to go over it. (I have final exams soon and not much time for this)
As soon as I have read and understood (or failed to understand) everything you write I'll respond to your post. Thanks again.
Mike:
I understand what you want to do is to communicate using entanglement. As far as I know, this has been proven not to be possible. The correlations present in entangled particles are "set" at the time the particles are interacting with each other, before they separate. Once they separate, those correlations persist and they reflect their past condition when they were together. If you make any changes to one, that will not affect the other. Not only that, but you will loose your correlations. If you think that the mere act of destroying the correlations can be used to communicate, I think it could be easily proven that that is not the case. With respect to instantaneous communication, remember that according to special relativity, simultaneity (did I spell it wrong?) is not something that can be defined. Events will look simultaneous in one frame of reference while they will appear at different times in another frame. Not only that, their order in time, if they are space-related, could be reversed depending on the frame of refference.
So I think a causality argument could also be used against the possibility of "instantaneous communication"
About changing the state on one side: let me give you a simple classical example: If we have a set of two cards, one has a 1 written on it and the other one has a 0, and we give them to two different peaple that go to different cities (without knowing what their cards are). We know that when they look at their cards, if the first one has a 0 on it, the other one will have a 1 for sure. That is if nobody changed them. If you have someone changing the zero into a one, then the original correlations don't hold, and, what is more important, the other card didn't magically change to keep the correlation. I know this example is not totally appropriate because is classical (superposition is missing) but except for that, I still think it illustrates this issue very well and the argument can still be used for a quantum system.

TheDonk

Dec2-04, 08:52 PM

alexepascual, I agree with everything you said accept this line.
...I think a causality argument could also be used against the possibility of "instantaneous communication"
Wouldn't instantaneous communication prove that one object isn't causing the other to change at all? Wouldn't it mean that they ARE the same object? It would be like looking at two sides of the same coin. One object (or what we thought was it's own object) is in a different location than the other, but we would now have shown that location has nothing to do with identity.

alexepascual

Dec5-04, 11:35 AM

TheDonk:
I like the fact that you try to see things from a diffetrent angle. I am the same way. But I have some difficulty with your idea.
If we were only talking about two entangled particles of the same type, then your idea might make some sense and it would deserve being explored further. On the other hand, remember that what is entangled is not the particles themselves but properties of the particles. Also consider the things that have entangled properties do not have to be of the same type. How would you apply your reasoning in that case?
On the other hand, about my comment on a causality argument, someone could reject it on the grounds that there is no reason to exclude the possibility of a violation of causality. But it happens that on a macroscopic scale we have a definite direction in which time "runs" (that direction in which entropy increases). So maybe a violation of causality in a macroscopic scale is not possible.
Think of this: You invent this cute machine that allows you to send instantaneous messages. You send your partner experimentalist to a star a good distance from earth. Then you send a message, which in your frame of reference is received "instantaneously". I another person's frame of reference (someone on a spaceship moving at a great speed in a certain direction) the reception of the message happens before it was sent.
Does this violate causality? Well, thinking again about it maybe not, because if the events are "space related" none of them can be considered to be a cause of the other.
Changing the subject: did you have a chance to look at wikipedia?

Eye_in_the_Sky

Dec5-04, 11:57 PM

There is a problem with some of the things I wrote in an earlier post of this thread.

Back in post #29 (http://www.physicsforums.com/showthread.php?p=387740#post387740), I wrote:
By "forcing", it appears to me that you mean some kind of apparatus which will, for example, leave the |+> as it is, but cause the |-> to become a |+>. Can we do this? Sure! ... why not?The said process, as seen from within the Hilbert space of that particle, is clearly non-unitary. Nevertheless, there must be some physical arrangement which can bring it about ... but I am not precisely sure what it is, nor how to represent it mathematically.

Furthermore, the claim I made that in such a process we would get
(1/√2)(|+>|-> – |->|+>) → (1/√2)|+>(|-> – |+>)... is definitely wrong! (... because the final state shows a preferred direction which can be identified on the "particle-2 end".)

Moreover, to say that such process is describable by a Hamiltonian such as
Htotal = (H1 + Hmod) x H2 ,

where Hk is the 'free' Hamiltonian for particle k, and Hmod is the Hamiltonian for the interaction which modifies the status of particle 1.... is also wrong! (... because such a Hamiltonian implies that the said process, as seen from within the Hilbert space of particle 1, is unitary.)

Nonetheless, the 'spirit' of my answer still stands. It is just as alexepascual puts it at the end of post #30 (http://www.physicsforums.com/showthread.php?p=389249#post389249) (note, I have inserted boldface for emphasis):

About changing the state on one side: let me give you a simple classical example: If we have a set of two cards, one has a 1 written on it and the other one has a 0, and we give them to two different peaple that go to different cities (without knowing what their cards are). We know that when they look at their cards, if the first one has a 0 on it, the other one will have a 1 for sure. That is if nobody changed them. If you have someone changing the zero into a one, then the original correlations don't hold, and, what is more important, the other card didn't magically change to keep the correlation. I know this example is not totally appropriate because is classical (superposition is missing) but except for that, I still think it illustrates this issue very well and the argument can still be used for a quantum system.

TheDonk

Dec6-04, 12:31 AM

If we were only talking about two entangled particles of the same type, then your idea might make some sense and it would deserve being explored further. On the other hand, remember that what is entangled is not the particles themselves but properties of the particles. Also consider the things that have entangled properties do not have to be of the same type. How would you apply your reasoning in that case?
I agree that this reasoning would point towards the entangled objects not being the same object but who's to say that the properties are not objects of some type? I think it could still work either way... maybe not tho, because of the next thing you said:

Think of this: You invent this cute machine that allows you to send instantaneous messages. You send your partner experimentalist to a star a good distance from earth. Then you send a message, which in your frame of reference is received "instantaneously". I another person's frame of reference (someone on a spaceship moving at a great speed in a certain direction) the reception of the message happens before it was sent.
Does this violate causality? Well, thinking again about it maybe not, because if the events are "space related" none of them can be considered to be a cause of the other.
Let's say we "connect" two objects for instantaneous communication. Then one person flies really fast away towards another star. Unless the people come back to the same frame of reference, there is no agreement on who is older than who. I may be wrong but each person would be able to say they are older than the other person. So if one person starts communicating when will the other person get the info? I don't think they could decide on what instantaneous is, could they? I think we need someone who knows special relativity well to answer this. But if I'm right, this will only enforce the point I quoted you on above.

Back to the topic of what causes entanglement. I've seen some explanations but am having big problems learning a few things that I need to. What is the bare minimum concepts I need to know to understand an explanation of the cause of entanglement? Superposition? Wave functions? Do I need to understand the quantum operators? Better understand Hilbert space? All of these I only have a vague idea in. I thought I knew everything to know about Hilbert space, but now I'm not sure.

Thanks for the info so far. :smile:

alexepascual

Dec6-04, 10:21 PM

The Donk:
You definitely need to understand about superposition of states. Also look into the EPR experiment but in the version where they use the spin of particles (not position and momentum).
It may also help to learn a little more about the special theory of relativity. The twin paradox is not related to the topic of our discussion. But you may want to look into instanteneity and get a good understanding of the Lorentz transformations. Different frames of reference refer to observers moving at different velocities.
Again: take a look at Wikipedia
-Alex-

Eye:
Your posts require a lot more thinking. I'll reply in about three weeks. (after classes are over). Thanks again.

Eye_in_the_Sky

Dec8-04, 10:55 PM

What is the bare minimum concepts I need to know to understand ... entanglement? Superposition? Wave functions? Do I need to understand the quantum operators?As a bare minimum for a good understanding of "entanglement", I think you will need the following:

1) a "2-dimension vector space" with "inner product";

2) an "orthonormal basis" on that space;

3) a "tensor product" of two such spaces;

4) the "singlet state" in the resulting tensor-product space;

5) "linear operators" on the resulting tensor-product space;

6) the "Projection Postulate" of Quantum Mechanics.

Eye_in_the_Sky

Dec14-04, 01:06 AM

When a hydrogen molecule is formed did a previous wave function collapse to form the paired electrons?If we imagine looking at the Schrödinger time-evolution for a pair of hydrogen atoms coming together, we would find that the resulting state shows a superposition of two general scenarios:

(i) the two atoms 'bounced off one another' and no molecule was formed;

(ii) the two atoms 'joined together' and formed an H2-molecule.

A subsequent "measurement" of some kind is then required to establish which scenario prevails, and (in an established terminology) that measurement then 'collapses' the wavefunction.

This is analogous to what takes place when a particle is incident upon a step potential. In general, an incident wave packet will split up into two wave packets, one corresponding to reflection and another to transmission. A subsequent measurement is then required to 'decide' which scenario prevails.
______________In a laser beam are the photons entangled?Here, it sounds like you are alluding to the property of "coherence", which in this context means an extremely large number of photons can (to a very good approximation) be described collectively by the same (single-particle) quantum state.

yanniru

Dec14-04, 08:42 PM

Well, after reading all these posts, I conclude that no one understands entanglement.

I have read elsewhere of two explanations that make some sense. Particles in superfluids are entangled because their wavefunctions overlap and are in quantum coherence, like photons in a laser beam. Essentially they act like a single particle.

And the second is like unto it: the EPR paradox is explained by overlapping wavefunctions with the conceptual addition of the extra dimensions of string theory.
The entangled particles were made in ten dimensions at a point in space and time. Six of these dimensions compactify and continue to keep the two particles in contact as they separate in space by a thread of the compactified dimension. That is, their wave functions overlap in the compactified dimensions.

Of course I am not sure if any of that is true. What I do understand is that when physicists cannot explain some phenomenon, they give it a name and go on with life. So far entanglement is one of the things that they still cannot explain. Thay can predict, but they cannot explain.

Richard

TheDonk

Dec15-04, 12:00 AM

I thought wave functions had no cut-off value at a certain distance and spanned the whole universe... If so, when are they combined? Unless there is degrees of entanglement and everything is slightly entangled. I'm quite sure that isn't the case.
Maybe I'm wrong about wave functions... if not, could you explain?

alexepascual

Dec15-04, 11:26 AM

Richard:
Your version of entanglement using the compactified dimensions of string theory is interesting. Could you give us your source?
I hope this is something someone has worked out to the point of making the idea feaseable.
Thanks,
-Alex-

yanniru

Dec21-04, 07:09 AM

Louis Cypher

Dec22-04, 09:47 AM

:eek:

--------------------------------------------------------------------------------

Posted this on another forum and they just deleted it and wouldn't respond to my questions so I'm assuming it's nonsense, what I wanted too know though is why it's nonsense, to help me learn a bit more about qm; after reading an article on qbits and how they can use entanglement to check parity in a different way so that q information doesnt become lost by observation I had the thought if we entangle a photon and then observe the two photons entangled state a computer could take this information via observing another entangle photon, and do it computationally to check the quantum or just use error bit from then on any information that computer and the other chose to send quantumly would instantaneously be transfereed wireless without broadband information exchange at insatantaneous speed; the applications of the idea are endles, why is this wrong, please answer?!? I'm dying to know.

thanks

-------------------------------------------------------------------------------- :rolleyes:
Last edited by Louis Cypher : Today at 03:45 PM.

Louis Cypher

Dec22-04, 10:38 AM

finding a way to keep the photon hanging around might be a little tricky but then, you could send off for a new photon every once in a wile which could equally be sent by being entangled with the original; obviously a feed back loop from quantum to computer would need to recode information to be transmitted to re entangle the photon after this, but qbit parity check would maintain integrity? Any responses appreciated, even if you do tell me I'm an idiot?

alexepascual

Dec22-04, 10:59 AM

Sorry. I am the source. Although when I posted it once before on this forum, someone replied that he had also thought of the same thing.
I am not trained to make the idea feasible. Perhaps you are. I take that explanation as evidence that the extra dimensions exist. Not sure why no bono fide string theorist has pursued it.
Richard:
So far I have found the "many worlds" interpretation of quantum mechanics as the most acceptable and when I think about quantum phenomena I think in terms of this interpretation. I have only read popularization articles and books about string theory. Since I read these articles, I have wondered if there is any connection between the extra dimmensions of string theory and the branches of the many worlds interpretation.
I thought you had read some article explaining entanglement in terms of the string theory. I think if you believe your scheme might work, the best for you is to continue learning about QM and string theory until you can prove yourself how these extra dimmensions account for entanglement or that they don't.

instantaneously be transfereed wireless without broadband information exchange at insatantaneous speed; the applications of the idea are endles, why is this wrong, please answer?!? I'm dying to know.
Louis:
Most physicists agree that it is not possible to send information "instantaneously" using entanglement. Of course the applications would be many if you could do this. But it appears you can't. If you think most physicists are wrong, then, I think you should study the subject deep enough so that you can put your ideas in mathematical form.
If you just want to know why your idea is wrong, you can do a Google search for: entanglement + causality or "teleportation".
Other place where you can search about these topics is in Wikipedia. You can actually learn a lot about QM there.
If you wonder why someone can't just give you a detailed explanation. I think it is because the effort it woud take to explain it to you is much, much, greater than the effort you put in formulating your hypothesis in the first place.

Eye_in_the_sky:
I'll be reading your posts in the comming days (done with finals now) . I promise I'll give you a response soon.

TheDonk:
Where are you?

Louis Cypher

Dec22-04, 02:32 PM

Any ideas guys?
see previous post

Kea

Dec22-04, 04:01 PM

Elsewhere I posted this:

A categorical semantics of quantum protocols
S. Abramsky and B.Coecke
http://users.ox.ac.uk/~mert1596/QUOXIC/talks/samson.pdf

Y. is no doubt correct in saying that noone understands it, but
this is a pretty good start.

TheDonk

Dec22-04, 07:36 PM

I looked into superposition and the EPR experiment and now understand what both of them are. (I already had an idea what they were)

But back to my original thread topic question. How do particles become entangled? Let's use spin for the example. When do two electrons that aren't entangled go together and become entangled? What needs to happen for the electrons to enangle? Shouldn't this be easy to answer? It seems like it would be simple to answer this in laymens terms and without math, tho any answer would be appreciated.

Could it be that almost everything entangles particles? Maybe most measurements entangle the particle and the particle measuring it? Can two different particle types (ie electron and photon) be entangled?

Could every single particle be entangled with a pair and the only way to untangle it is to entangle it with something else? But what would happen to the original? Maybe it entangles with the other left over particle. I obviously just made this up and have no evidence at all but it sounds cool. Would this be hard to disprove?

Louis Cypher:
Your idea makes sense, I think, but only if you can send info instantaneously, which we don't know how to do. What you're talking about is a little bit off topic. (tho I'm bad for going off topic too)

Mike2

Dec22-04, 07:50 PM

I looked into superposition and the EPR experiment and now understand what both of them are. (I already had an idea what they were)

But back to my original thread topic question. How do particles become entangled? Let's use spin for the example. When do two electrons that aren't entangled go together and become entangled? What needs to happen for the electrons to enangle? Shouldn't this be easy to answer? It seems like it would be simple to answer this in laymens terms and without math, tho any answer would be appreciated.

Could it be that almost everything entangles particles? Maybe most measurements entangle the particle and the particle measuring it? Can two different particle types (ie electron and photon) be entangled?

Could every single particle be entangled with a pair and the only way to untangle it is to entangle it with something else? But what would happen to the original? Maybe it entangles with the other left over particle. I obviously just made this up and have no evidence at all but it sounds cool. Would this be hard to disprove?
Obviously if particles were not entangled and then become entangled, then they become entangled when they start to effect one another. This means that at the point that a photon goes from one to the other, then information from one goes to the other so that electron spins must line up accordingly. I suppose that even the most distant particles would eventually transfer a photon between them and become entangled. But distant atoms could also remain hidden from one another for a long time.

TheDonk

Dec22-04, 09:04 PM

Obviously if particles were not entangled and then become entangled, then they become entangled when they start to effect one another. This means that at the point that a photon goes from one to the other, then information from one goes to the other so that electron spins must line up accordingly. I suppose that even the most distant particles would eventually transfer a photon between them and become entangled. But distant atoms could also remain hidden from one another for a long time.
Thanks for the response Mike. I'm not too familiar with QM yet so I still don't know some of these obvious things. You said they get entangled when they effect each other. In what ways can this happen. Is sending a photon the only way? Does sending a photon always entangle the electron that absorbs it? Can you give me an example of how a different property other than spin can become entangled?

In every atom that has an even amount of electrons, are all the electron pairs entangled? Id that what defines them as a pair in the first place?

Louis Cypher

Dec23-04, 10:10 AM

Thanks that's the sort of answer I was looking for, the have used photon entanglement to exchange info though haven't they or am I mistaken?

Quantum encription bob and alice thingy, I'm not talking about teleprtation there just firing the light along a wire or at another computer as in the Bob and Alice experiment.

Just starting a couple of A level style maths courses before I go on to the Degree, it's a pain it means I have to do eight years of study to go with the foundation year I've allready done, but then no one ever claimed physics was easy. look at my post as a reflection of quantum encryption not telportation then; ok if entanglement doesnt teleport it at least qould mean we could us light instead of electricity to transmit signals, and in turn a device could interpret this much like a modem and encode back to an electronic signal, dial up users would be chuffed, and we would no longer have need of dial up, everyone could use broadband and light is better than electricity for sending signals plus it would be very safe from hackers compared to the internet now.

Like I say just thinking

ideas are the cornerstone of science, thanks for replying, and for refraining from calling me an idiot :smile:

reilly

Dec23-04, 12:55 PM

Entanglement occurs in boh classical and quantum systems. Entanglement is a fancy word for correlation. The primary source of entangelment/correlations beween FREE particles -- EPR, Stern-Gehrlach (well almost free) is CONSERVATION LAWS, which are at their most striking for two-body systems. Note that without entanglement, the neutrino might well have remained undiscovered.

In two photon or two electron systems, the drill is to work with a s-wave source. Thus total angular momentum is zero. This means, measuring along the quantization axis, that if particle one is in a +1 Jz state, then immediately you know the quantization axis eigenvalue of the other particle -- just go to your already prepared table lookup. You knew, before the experiment, that if one gives spin up, the other necessarily must be spin down. QM allows only two state here; spin up/spin down or vica versa. The rest is Clebsch-Gordon coefficients, and appropriate rotation matricies.
There's no superluminal communication, nor any magic involved; the experimenter knows all the conditionals and probabilities prior to the experiment.

So, what am I missing? What other than conservation laws creates entanglement?

Regards,
Reilly Atkinson

Louis Cypher

Dec23-04, 03:38 PM

Thanks for that; I do now the postions of entanglement all I was saying is why can't we use a computer to take this Q information and code it into electrons? Or use the computer to take the quantum, change it and parity check it and then use this to transfer information via a modem from electrical to QM and vice versa, what's the problem here? If we can use QM to transfer info why cant we use a computer to decrypt encode and the recrypt and then send; a new internet without band width problems we use light as a medium of internet transfer, not electrons.

Am I being deranged?

I know we couldnt do it now, but if we could set up a network of cables to transfer photons not electrons could we not use it as an internet connection instead of broad band, using q encryption alice and bob stuff to transfer info much more safely than the internet using encryption, I feel I'm missing something here, cos everyones posting facts, which I appreciate alot; why couldnt we in "theory" if we had the technology use the quantum to exchange information via an electrical medium ie a computer?

Am I being dense again someone explain?

:confused:

alexepascual

Dec23-04, 07:05 PM

Louis,
Most of internet communications are in fact being transmitted by optical fibers. Up until now these fibers reach some local switching stations which convert the optical signal to electrical. The technology has been evolving so that more and more "conversations" can be sent over the same fiber at the same time. There are discussions about having these optical fibers reach the home, but it appears this is not economically feaseable at the time. If this were done, then you would have a system similar to what you propose except for the entanglement part. There has been a lot of discussion about the possibility of using entanglement to send secure messages over an optical channel. This would not necessarily increase the bandwidth but it would allow for an ultra-secure way to transfer information. Also, of course this system would not allow for transfer of information faster than the speed of light. If you want to learn more about these things, there is plenty of information on the web. Google for: Quantum cryptography, quantum encryption, etc. Every time you come up with an idea, it is better to first check to see if it hasn't been invented yet. Most of the time, whatever idea you (or I) come up with, someone else has thought about it before and even published it. If they haven't, there is a good chance the idea is wrong.

alexepascual

Dec23-04, 07:16 PM

Reilly,
I am not sure I understand your observation, but it appears to me that you are putting classical correlations and quantum entanglement on the same footing. In quantum entanglement you have a superposition of different possible "realities' all existing at the same time. I think this has implications that are quite troubling (or exciting) depending on how you look at it.
One of the features of entanglement that makes it unique is "non-locality".
I think Einstein was very well versed in classical mechanics, but even so, he found non-locality quite troubling. So I don't think there is anything trivial about this phenomenon. It is true that some of the people posting here, who don't have enough formal training in QM have a wrong idea about what is entanglement, but that doesn't change the fact that there are aspects of entanglement that are quite puzzling.

Wave's_Hand_Particle

Dec23-04, 08:43 PM

How do particles become entangled? I've heard that it's when two particles bump into each other. How is this "bump" defined? What does it mean for 2 particles to bump? Is it based on distance apart, or something else?

How do particles become entangled? I've heard that it's when two particles bump into each other. How is this "bump" defined? What does it mean for 2 particles to bump? Is it based on distance apart, or something else?

Entanglement is when 'ONE' particle, 'EXISTS' at two different locations, in the same instant.

One can make many speculations about the devision of a quantity?..if one cuts an whole orange in half, there does not appear two whole oranges, in two seperate frames of reference, there exists 'ONE' orange at two locations, but their difference is not Time-Dependant.

It is a wonder that nature provides many aspects of observation, a "cause will have an effect".

Entanglement is like an opposite to the Pauli Exclusion Principle, there are other seemingly paradox's, resulting in definate Observation constraints, some may be Conscious derived, for instance:Consciousness is the Effect of the surrounding Spacetime, with Memory the Path Integral of Entangled 'event' States?

Just as you cannot remember all events 100% ( to do so would invoke a present-time event ), because you cannot be at two TIME locations in the same instant, to do this is to nullify the Uncertainty Principle, with Measurment, in fact some have speculatated that Entanglement is really just giving a Product, Two of the same identifyable quantities, or in simplistic terms, a precise measurement of Location/position, and disregarding the Momentum Measurment? :wink:

The 'cause' of entanglement (effect), can now be attributted to the QM 'splicing' experimental setups, or forgeting about the value of one property of Measure?

Louis Cypher

Dec24-04, 03:38 AM

thanks guys, yeah lets go optical, and have a safe internet connection without all the annoying hackers and idiots ruining our fun, people may have concieved of going optical but what I suggested was going entangly optical, a leap I know but just an idea, just wanted sum feedback, if we could use the optical 1bit thing as a bus in a computer it would still travle at the speed of light but we would have a 1 a 0 and a 1 and 0 this would let us calculate faster; or would it as you say; anyway marry existing technology with the quantum is an idea, so how come no ones even trying as such, quantum modems, telephone lines the applications are endless, k its not possible now but why dont people at least try it, if they are theyre doing small scale experiments, thinking small, lets push things forward and try to use the quantum we can use entanglement as a modem, transfer optically, then decode using entanglement as a modem then convert to electricity? I know we cant but it does no harm to think about it, if we could use the quantum then why not use it and marry it to current stuff, instead of trying to ditch the old for the new?

reilly

Dec24-04, 11:14 AM

[QUOTE=alexepascual]Reilly,
I am not sure I understand your observation, but it appears to me that you are putting classical correlations and quantum entanglement on the same footing.

Indeed, I am.

(ap)In quantum entanglement you have a superposition of different possible "realities' all existing at the same time. (Why? In what sense do they exist?))

Arguable. If true in QM, then why not, say, in using Kalman Filtering to deduce aircraft location in airport radar's? Just because it might be does not mean that it is -- still a valid take on the world.

Happy Holidays to all,
Regards,
Reilly Atkinson

alexepascual

Dec26-04, 05:28 PM

(ap)In quantum entanglement you have a superposition of different possible "realities' all existing at the same time. (Why? In what sense do they exist?))

Reilly,
I think we are getting into interpretation territory here. Heisenberg would have talked about "potentialities" instead of "realities". If a photon has gone through a double slit, I guess you could say that it went through both (in a way). But as you can't say that one quantum went through each slit, I guess you would have to think of these split personalities of the photon I guess as less that full "realities". But again I think this may be a matter of interpretation.
The second part of your post I don't understand and I suspect it may be a little off topic.
But I still think that quantum correlations and classical correlations are fundamentally different. Once again, I think a discussion about this would deserve its own thread and is not essential for our discussion on how two particles become entangled.
-Alex-

Eye_in_the_Sky

Dec27-04, 03:21 AM

Reilly,
I am not sure I understand your observation, but it appears to me that you are putting classical correlations and quantum entanglement on the same footing.

Indeed, I am.Classical correlations can be construed in terms of a local "reality", BUT quantum correlations cannot! This is the essential difference between the two.

That quantum correlations imply nonlocality can be seen from the following argument:

1) Suppose that the value obtained in a measurement is 'determined' at the time of the measurement and not before.

Then, clearly, for a pair of spin-entangled particles far apart in space, the joint 'collapse' which ensues on account of a spin-component measurement of only one of the particles implies nonlocality. //

2) On the other hand, suppose that the value obtained in a measurement is 'determined' before the time of the measurement. That is, this value – prior to measurement – is "definite but unknown".

Then, assume locality and derive a "Bell inequality". Such an inequality, however, contradicts the QM formalism. //

Putting 1) and 2) together leads to the inescapable conclusion that quantum correlations imply nonlocality.

... HOWEVER, there is another separate point which Reilly seems to be making, namely:

A "knowledge-based approach" to the 'collapse' phenomenon is consistent with the QM formalism.

The above statement is true!

Nevertheless, it should be noted that anyone who subscribes to such a "knowledge-based approach" is necessarily also subscribing to a nonlocal hidden-variable interpretation of QM.

yanniru

Dec28-04, 07:24 AM

Non-local hidden variable interpretation- why that's Bohmian mechanics. Is that a dirty word that cannot be said?

alexepascual

Jan4-05, 10:11 PM

Eye_in_the_Sky:
Going back to your post #16:
Now, suppose that the interaction between these two particles is such that
|ψ>|φ> → Σk ak|ψk>|φk> ,
where each ak ≠ 0, and there are at least two distinct values for k (and, of course, the |ψk> (|φk>) are linearly independent).
I would like to look at this in more detail. How do we go from |ψ>|φ> to Σk ak|ψk>|φk> ?
The k seems a little confusing to me. What basis are you using? Shouldn't we define a new basis in the product Hilbert space? If k refers to that basis, shouldn't we write the state as Σk ak|ψ>|φ>k?
If H1 is n-dimmensional and H2 is m-dimmensional, k would run from 1 to nxm right? Would the ak be determined by the particular interaction? What about time evolution? Can we arbitrarily use either the Schodringer picture or the Heisenberg picture?

I have read somewhere else that the state |ψ>|φ> is "not the most general one". But I don't understand why or what they mean by "general". If you would like to look at the source, I can post a link or copy the paragraph here.

In my post #17 I expressed my puzzlement at how a composite state of two non-interacting particles is described. I understand you choose some basis for each of the particles and then make a tensor product of these single-particle base states. But what do you do with the complex coefficients? If you were to multiply them together, wouldn't you loose information about each individual particle?. Maybe my questions don't make sense, but in any case, I think you might be able to give me some orientation. Thanks in advance.

alexepascual

Jan5-05, 02:18 PM

Today I went to the library and tried to find information about my questions in the previous post. I looked at Sakurai and Shankar but I didn't find any info on composite Hilbert spaces, interacting and non-interacting particles, etc. Maybe I didn't look in the right chapter.
Anyway, I have been doing some thinking. If we start with the state vectors for the two (non-interacting) particles (in some basis) at a particular time, and do the tensor product of them, we get a complex amplitude for each element in the tensor. It makes sense that the probability to find a particular produc-state will be proportional to the products of the probabilities for each of the components in the product state. So I guess it makes sense to multiply the amplitudes of the individual-particle state vectors.
But I realize that the states in H1 might have different energy than those in H2, therefore their phases would change at different speeds. If we had the time dependence of the phase encoded in the state vectors, then we would not have a single number for the amplitude of each of the product states but a pair of complex functions of time.
I guess another way of doing this would be to leave the state vectors alone and to plug the time dependence into the operators. (Heisenberg picture).
But all this is just my speculation. I would like to see a good explanation of these topics in a book or article or get it from someone who knows it very well. I can imagine that this topic is very elementary for those who understand quantum mechanics well. For me it is still confusing.
I'll appreciate any help.

DrChinese

Jan5-05, 04:28 PM

1) Suppose that the value obtained in a measurement is 'determined' at the time of the measurement and not before.

Then, clearly, for a pair of spin-entangled particles far apart in space, the joint 'collapse' which ensues on account of a spin-component measurement of only one of the particles implies nonlocality. //

2) On the other hand, suppose that the value obtained in a measurement is 'determined' before the time of the measurement. That is, this value – prior to measurement – is "definite but unknown".

Then, assume locality and derive a "Bell inequality". Such an inequality, however, contradicts the QM formalism. //

Putting 1) and 2) together leads to the inescapable conclusion that quantum correlations imply nonlocality.

This argument is not correct in the context of Bell. In 1) above you essentially assume that which you want to prove.

In Bell, the argument is that locality and reality cannot both be true and still yield results consistent with QM. Locality can hold if "realistic" assumptions are abandoned. By failing to acknowledge this possibility in your argument, you naturally conclude it is locality which fails. That doesn't fly.

Caroline Thompson

Jan5-05, 05:15 PM

I am primarily an experimentalist ... In the Bell experiment he chose angles that were not 0 or 90 degrees thus confusing me beyond help. Isn't there a simpler experiment out now with simpler equations?
I have a real problem with these Alice and Bob experiments because none of the instruments in my lab are named Alice or Bob.
I have found it very hard to corner an expert and ask questions.

If you look at the real experiments you will find that most have been done using light and assuming it to come in "photons". If you want to see the logic really intuitively, though, it's easiest to look at a model of the Bohm-type idea that uses the spins of two particles. The "local realist" logic, and in particular that relevant to the "detection loophole", is covered by my Chaotic Ball model (see http://arxiv.org/abs/quant-ph/0210150 ). You might look at this first then try and follow what I'm on about in my discussions of real optical experiments and their various other loopholes in http://arxiv.org/abs/quant-ph/9903066. Alternatively, you might start by looking at the pages I contributed last summer to wikipedia (hoping someone else has not edited them out of recognition!). The key page is http://en.wikipedia.org/wiki/Bell%27s_Theorem, from which you can follow links to pages on the actual experiments and on the loopholes.

As you may have gathered, I'm a local realist and am firmly convinced that, once you allow for the actual experimental conditions, all the results can be explained by local realist models. An important feature of these models that you will not find in popular books on EPR is that the hidden variables set at the source do not completely determine (in conjunction with the detector settings) the outcomes. They determine only their probabilities.

This kind of model is described in Clauser and Horne's 1974 paper (Physical Review D, 10, 526-35 (1974)), which is much less widely read than it deserves. It presents versions of Bell's inequality that apply to the real conditions, in which detectors are not anywhere near 100% efficient. You will find here that they state quite categorically that the usual CHSH test cannot be used unless you know the number of pairs emitted and use this as denominator in the estimate of the quantum correlation. Though they do not, iirc, use the term "fair sampling", what they mean is what I and others have realised: that you cannot assume it. To do so will almost inevitably bias the test.

But please read some of my work and get back to me if this does not make sense!

If what you want is to understand quantum theory then you've come to the wrong place, since my studies have indicated that the QT formula is in fact wrong! The logic has to be the local realist one. That this is so could, I believe, be shown experimentally if, instead of publishing just one "Bell test statistic" each time, experimenters were to publish actual counts for a range of settings of parameters such as the efficiencies of the detectors.

Caroline

Eye_in_the_Sky

Jan6-05, 03:54 AM

Non-local hidden variable interpretation- why that's Bohmian mechanics. Is that a dirty word that cannot be said?... Not in the least!

In using the expression "nonlocal hidden-variable interpretation of QM", I was only trying to be as general as possible.

Eye_in_the_Sky

Jan6-05, 04:14 AM

... I would like to look at this in more detail. How do we go from |ψ>|φ> to Σk ak|ψk>|φk> ?By means of Schrödinger-evolution of the joint interacting system.
__________Would the ak be determined by the particular interaction?Yes.
__________The k seems a little confusing to me. What basis are you using? Shouldn't we define a new basis in the product Hilbert space? If k refers to that basis, shouldn't we write the state as Σk ak|ψ>|φ>k?In the above, the vectors |ψk> and |φk> do not refer to any particular basis. Indeed, in saying what I said back in post #16,
Now, suppose that the interaction between these two particles is such that

|ψ>|φ> → Σk ak|ψk>|φk> ,

where each ak ≠ 0, and there are at least two distinct values for k (and, of course, the |ψk> (|φk>) are linearly independent).... my point was only to make clear that, on account of the interaction, the joint system can no longer be written as a "simple (tensor) product" of one element from H1 with one element from H2.
__________I have read somewhere else that the state |ψ>|φ> is "not the most general one".I think the intention of such a remark is along the lines of what I just said above:

An arbitrary vector in the tensor-product space cannot in general be written as a "simple (tensor) product" of one vector from H1 with one vector from H2.

For example, consider a Hilbert space H spanned by {|1>,|2>} and construct the tensor-product space H' of H with itself. Consider the vector

|1>|2> + |2>|1> Є H' .

Can you find vectors |ψ> and |φ> in H such that

|1>|2> + |2>|1> = |ψ>|φ> ?

... Good luck!
__________What about time evolution? Can we arbitrarily use either the Schodringer picture or the Heisenberg picture?Either picture can be used. In the above, where I wrote

|ψ>|φ> → Σk ak|ψk>|φk> ,

I was working in the Schrödinger picture. The left-hand-side is at some initial time to, and the right-hand-side is at later time t.
__________In my post #17 I expressed my puzzlement at how a composite state of two non-interacting particles is described. I understand you choose some basis for each of the particles and then make a tensor product of these single-particle base states. But what do you do with the complex coefficients? If you were to multiply them together, wouldn't you loose information about each individual particle?No, you don't lose any information. It sounds to me like you are trying to "read off" the amplitude for measuring particle 1 in some eigenstate in a measurement where "you don't care" what happens to particle 2. But you can't just do that. You have to combine all the cases where particle 1 is in the prescribed eigenstate and particle 2 can be in "anything".

On the other hand, if you wish to measure each particle in some eigenstate, then you can merely "read off" the amplitude (... if you get what I mean). This part, it appears to me, you have subsequently understood, because, as you write:
Anyway, I have been doing some thinking. If we start with the state vectors for the two (non-interacting) particles (in some basis) at a particular time, and do the tensor product of them, we get a complex amplitude for each element in the tensor. It makes sense that the probability to find a particular produc-state will be proportional to the products of the probabilities for each of the components in the product state. So I guess it makes sense to multiply the amplitudes of the individual-particle state vectors.Correct. ... But, then, you go back to the earlier confusion:
But I realize that the states in H1 might have different energy than those in H2, therefore their phases would change at different speeds. If we had the time dependence of the phase encoded in the state vectors, then we would not have a single number for the amplitude of each of the product states but a pair of complex functions of time.There is no problem here.

Write the state of the joint, non-interacting, non-entangled system as

|ζ(t)> = |ψ(t)>|φ(t)> ,

and say that

|ψ(t)> = ∑i ai(t)|ψi>

and

|φ(t)> = ∑j bj(t)|φj>

(where, of course, {|ψi>} and {|φj>} are each orthonormal sets).

Thus,

|ζ(t)> = |ψ(t)>|φ(t)>

= ∑ij ai(t)bj(t) |ψi>|φj> .

Now ... let's find the probability that particle 1 is in the state |ψn>. The associated projection operator is just

|ψn><ψn| (x) 1 ,

and therefore the required probability is given by

<ζ(t)| |ψn><ψn| (x) 1 |ζ(t)>

= <ψ(t)|ψn><ψn|ψ(t)> ∙ <φ(t)|φ(t)>

= |an(t)|2 .

... Does this clear up the confusion?

Eye_in_the_Sky

Jan6-05, 04:35 AM

Back in post #59 (http://www.physicsforums.com/showpost.php?p=412828&postcount=59) of this thread, I wrote (among other things):
That quantum correlations imply nonlocality can be seen from the following argument:

1) Suppose that the value obtained in a measurement is 'determined' at the time of the measurement and not before.

Then, clearly, for a pair of spin-entangled particles far apart in space, the joint 'collapse' which ensues on account of a spin-component measurement of only one of the particles implies nonlocality. //

2) On the other hand, suppose that the value obtained in a measurement is 'determined' before the time of the measurement. That is, this value – prior to measurement – is "definite but unknown".

Then, assume locality and derive a "Bell inequality". Such an inequality, however, contradicts the QM formalism. //

Putting 1) and 2) together leads to the inescapable conclusion that quantum correlations imply nonlocality.However, DrChinese has suggested some point(s) of error on my part in the above. Specifically,

This argument is not correct in the context of Bell. In 1) above you essentially assume that which you want to prove.

In Bell, the argument is that locality and reality cannot both be true and still yield results consistent with QM. Locality can hold if "realistic" assumptions are abandoned. By failing to acknowledge this possibility in your argument, you naturally conclude it is locality which fails. That doesn't fly.Let me try to understand what you are saying here. In (a hopefully understandable) shorthand, my argument above can be summarized as:

---------
1) (QM & not 'pre-determined') → nonlocality ;

2) (QM & 'pre-determined') → nonlocality ;

therefore: QM → nonlocality .
---------

Now, let's see if I have understood you correctly.

With regard to 1), basically you are saying that the argument is "trivial".
... If so, I agree.

With regard to 2), you are saying that it omits the mention of an implicit assumption, namely, that of "realistic-ness". If we include this assumption explicitly in 2), then that argument becomes:

2') ("realistic-ness" & QM & 'pre-determined') → nonlocality .

... DrChinese, have I understood you correctly?

DrChinese

Jan6-05, 10:26 AM

However, DrChinese has suggested some point(s) of error on my part in the above. Specifically,
Let me try to understand what you are saying here. In (a hopefully understandable) shorthand, my argument above can be summarized as:

---------
1) (QM & not 'pre-determined') → nonlocality ;

2) (QM & 'pre-determined') → nonlocality ;

therefore: QM → nonlocality .
---------

... DrChinese, have I understood you correctly?

Eye,

I am not saying your arguments are trivial, and I don't think we disagree on the basic science. I do think that your argument embodies a reformulated version of Bell which ends up losing a little something in the process.

The real alternatives from Bell are:

1) QM & Locality Fails (which does imply non-locality); this is more or less the case you call "determined at time of measurement" and therefore "not predetermined". (Locality=Bell's condition that the result of a measurement at one place does not affect the result of a measurement at another.)

2) QM & Reality Fails (which does NOT imply non-locality) which you can see does not quite match your 2) "pre-determined" alternative. (Reality=Bell's condition that the chance of an outcome is within the range of 0 to 1.)

It always comes back to your view of what is reality. If you think the measurement "caused" the photon spins to take on a definite value, then the question becomes: which measurement? The one on the "left" causing the one on the "right" (arbitrary assignments of left and right)? Or vice versa? Or perhaps, you might say this is an example of the future influencing the past.

The ultimate problem is that our concepts of reality do not readily correspond to the mathematical formalism. All you can conclude from the Bell side is that the relative angle between the non-local polarizers acts AS IF it is the fundamentally real variable being measured. This is true of many experimental setups in QM, such as demonstrations of photon self-interference. You have been tempted to conclude that non-locality is demonstrated. But this is not correct because no causal effect propagated at a speed faster than light (after all, we don't even know the direction of the travel, much less what the actual effect is).

Unless, of course, you define it that way. QED. There is still a light cone which limits everything that has happened, which is just as strong as a counter-argument to your definition of non-locality.

alexepascual

Jan6-05, 12:27 PM

Eye_in_the_Sky:
Again you have given me a lot of material to chew on. I hope this time I won't take as long to respond as I have more time available.

Caroline:
Thanks for your contribution to the thread. I guess many of us here are just trying to gain a better understanding of QM as it has been formulated and generally accepted. I understand the value of questioning the accepted dogma, but at least in my case I think I'll benefit more from gaining full understanting of the subject as presently understood and not to venture into concepts that deffy the commonly accepted theory. (at least not yet)
From all I know, and as Dr.Chinese clearly states, it appears that the tests (which you question) of Bell's inequalities sucessfully prove that locality or reality must give way. Your position is at odds with the accepted theory, and that must be the reason you didn't get a response. You might get better results by initiating a thread just on your proposal. Good luck.

CarstenDierks

Jan6-05, 03:44 PM

"Two particales must have interacted."
Can you give me an example of how two particles could interact to become entangled? A simple (if possible) step by step process where two particles start off not entangled and become entangled.
Hi Donk,
Jurgen was not so bad with his example in post #4. Alain Aspects experiment shows how photons become entangled: The 2 photons are sent off in opposite directions with opposite spins.

Maybe someone should ask the physicists at Innsbruck University, Austria, how they produce entangled particles. Their research is very advanced and they should be able to say how their particles become entangled.

http://www.sciam.com/article.cfm?articleID=00014CBD-7633-1C76-9B81809EC588EF21

Or does anybody here know how they did it?

Carsten

Eye_in_the_Sky

Jan7-05, 03:33 AM

I do think that your argument embodies a reformulated version of Bell which ends up losing a little something in the process.I agree that my argument does not follow the structure of that given by Bell. (In my argument, I used as the "starting point": 'pre-determined' or not 'pre-determined'. Bell's "starting point" is simply 'locality' (according to Einstein's definition). Originally, I had specific reasons for choosing the "starting point" as I did, but now I am not so sure that that was entirely necessary. Moreover, I am now suspecting that the first part of my argument (labeled "1)", i.e. not 'pre-determined') is less trivial than I thought it to be.)

Let us therefore consider Bell's argument exactly as he originally presented it. You are saying that in that formulation there is a "reality", or a "realistic-ness", assumption of some kind being made. You have also indicated one implication of such an assumption:

Reality=Bell's condition that the chance of an outcome is within the range of 0 to 1.On the face of it, to negate such a proposition is absurd ... isn't it?

Tell me, is this the only assumption of Bell which physicists consider negating on account of a not-"reality" proposal?

DrChinese

Jan7-05, 10:05 AM

Let us therefore consider Bell's argument exactly as he originally presented it. You are saying that in that formulation there is a "reality", or a "realistic-ness", assumption of some kind being made. You have also indicated one implication of such an assumption:

Reality=Bell's condition that the chance of an outcome is within the range of 0 to 1.

On the face of it, to negate such a proposition is absurd ... isn't it?

Tell me, is this the only assumption of Bell which physicists consider negating on account of a not-"reality" proposal?

I agree that on the face of it, the proposition is absurd. But further investigation shows it is far from absurd, and that is part of what makes QM so powerful (as you know).

If you look at the outcomes, you will see that certain combinations are actually predicted (by QM) to have negative probabilities, flying in the face of the "obviously reasonable" counter-position of reality shown above. QM makes plenty of similar counter-intuitive predictions, all of which have so far been pretty well verified. Aspect verified this particular one in his famous experiments.

The "loophole" from Bell's Theorem was that a non-local theory could also account for the observed behavior. So there is your option, and you are free to choose it if you prefer.

But there are other tests of QM which show similar negative probabilities. For example, take the reflection of light off a mirror from a source to an intensity detector. There are many paths to the detector that provide negative intensity, in violation of common sense. If you prevent the light from taking those paths (such as via etching the mirror in appropriate spots), the detected intensity increases. This is because the "negative" probability cases are being excluded, which cause the sum of the various path intensities to the detector to increase.

So ultimately, what is "reasonable" to one person may not be reasonable to another. Is non-locaity more reasonable than negative probabilities?

You can also map the hidden variable concept to the negative probabilities in the Bell paper. It is the manipulation of the local hidden variable (your "predetermined") outcomes that directly leads to the negative probabilities in the first place. Another way to say it is that certain combinations are suppressed. The difficulties lie in the words used to describe it more than the actual underlying formalism.

Caroline Thompson

Jan7-05, 11:39 AM

If you look at the outcomes, you will see that certain combinations are actually predicted (by QM) to have negative probabilities, flying in the face of the "obviously reasonable" counter-position of reality shown above. QM makes plenty of similar counter-intuitive predictions, all of which have so far been pretty well verified. Aspect verified this particular one in his famous experiments.
"Pretty well" is, in this instance, surely not good enough! We're talking about a claim that something happens that is not merely (in my view) "counterintuitive" but actually "impossible". If such a thing is to be believed we need incontrovertible evidence, not experiments whose interpretation only backs the belief if you also accept a number of assumptions that are, to a local realist, simply not reasonable.

The "loophole" from Bell's Theorem was that a non-local theory could also account for the observed behavior. So there is your option, and you are free to choose it if you prefer.
Quite! [Perhaps I should have read a little further before reacting.]

But there are other tests of QM which show similar negative probabilities. For example, take the reflection of light off a mirror from a source to an intensity detector. There are many paths to the detector that provide negative intensity, in violation of common sense. If you prevent the light from taking those paths (such as via etching the mirror in appropriate spots), the detected intensity increases. This is because the "negative" probability cases are being excluded, which cause the sum of the various path intensities to the detector to increase.

I guess you've been reading Feynman's QED? But all the above is only a matter of interpretation, conducted by someone who is determined to believe that light consists of photons. Allow it to be pure waves and take account of the way interference works and these apparently negative intensities simply cease to exist. I wonder if you've read Mach's "Principles of Physical Optics"? It was first published around 1926 but has recently been re-published. It is written totally without any reference to photons or quantum theory and describes some quite amazing phenomena, all of which it explains using wave theory.

Caroline

reilly

Jan7-05, 01:40 PM

[QUOTE=DrChinese]I agree that on the face of it, the proposition is absurd. But further investigation shows it is far from absurd, and that is part of what makes QM so powerful (as you know).

If you look at the outcomes, you will see that certain combinations are actually predicted (by QM) to have negative probabilities, flying in the face of the "obviously reasonable" counter-position of reality shown above. QM makes plenty of similar counter-intuitive predictions, all of which have so far been pretty well verified. Aspect verified this particular one in his famous experiments.

*************************************

I have a problem with negative probabilities. So, I'd be most grateful if you could provide details and chapter and verse on this idea. Thank you.

Regards,
Reilly Atkinson

Caroline Thompson

Jan8-05, 04:09 AM

Caroline:
Thanks for your contribution to the thread. I guess many of us here are just trying to gain a better understanding of QM as it has been formulated and generally accepted. I understand the value of questioning the accepted dogma, but at least in my case I think I'll benefit more from gaining full understanting of the subject as presently understood and not to venture into concepts that deffy the commonly accepted theory. (at least not yet)
From all I know, and as Dr.Chinese clearly states, it appears that the tests (which you question) of Bell's inequalities sucessfully prove that locality or reality must give way. Your position is at odds with the accepted theory, and that must be the reason you didn't get a response. You might get better results by initiating a thread just on your proposal. Good luck.

Thanks. I may do some day (I'm not sure I'm allowed to at present) but for the time being will just carry on trying to make sure that people know what they are doing. It is entirely possible that there is no future in "entanglement" and hence no benefit (other than passing exams and/or impressing your friends) in trying to gain a "better understanding" of it.

I maintain that the future of physics lies in a return to the experimental evidence and intuitive approaches. This is as true of Einstein's theories as it is of quantum theory. I think the era in which we are content with mathematical algorithms that give us the right answer but leave us unsatisfied as regards understanding is drawing to a close. We can carry on using the algorithms but we need a new fundamental theory. So far as the experiments are concerned, there is no reason that this should not be entirely intuitive. It need not necessarily produce quantatitive predictions. Its sole purpose is to give us understanding of our universe, with local causes for everything.

Of two things I'm quite certain: the "true" physics of the future will model light as pure waves, with no "photons", and will not involve entanglement. All actual experiments can be explained using ordinary classical correlations, once you allow for the imperfections and the use of inappropriate versions of the Bell test.

Incidentally, unless allowance is made for the fact that real experiments have hidden variables that are "stochastic", only accounting for the probabilities of outcomes and not the outcomes themselves, rational interpretation of the results is impossible. Surely we all agree that when it comes to interpreting real experiments we need to use a model that covers what is actually done, not just an ideal situation dreamt up by theorists? Do have a look at my wikipedia pages (which DrChinese thinks should be somehow made to vanish!). Start with http://en.wikipedia.org/wiki/Bell%27s_Theorem and follow links to pages on the actual experiments and their loopholes.

Caroline
http://freespace.virgin.net/ch.thompson1/

DrChinese

Jan8-05, 12:10 PM

Do have a look at my wikipedia pages (which DrChinese thinks should be somehow made to vanish!). Start with http://en.wikipedia.org/wiki/Bell%27s_Theorem and follow links to pages on the actual experiments and their loopholes.

Caroline
http://freespace.virgin.net/ch.thompson1/

Not vanish, just move to a page called Caroline's Theorem! :)

alexepascual

Jan8-05, 12:51 PM

Caroline:
I can understand that DrChinese and you don't get along too well. On the other hand, from reading the posts it appears you might find some compatibility with Reilly's ideas. He also has expressed the opinion that quantum correlations are no different than classical correlations.
For the moment I'll accept the standard theory and try to find my own intuitive framework to make sense of it. With respect to the need for intuitive understanding, I see some compatibility between your position and mine. I think that quantum theory still needs to be modified or complemented until it becomes a better explanatory system. (I don't accept the Copenhagen interpretation) But given its success in predicting results (which you may question) , I would not question the validity of the equations.
But when it comes to finding a particular intuitive structure that makes the theory easier to make sense of, there is where you and I take different routes. I understand you are a local realist, which makes you reject anything that departs from classical thinking. I would say you are a conservative person in this respect, although being a conservative may appear as being a rebel nowadays when most people have accepted the revolutionary ideas and these don't appear revolutionary anymore.
I, on the other hand feel more confortable with an explanation that seems absurd to many people, and it is the "many worlds" interpretation. There is an apparent problem with this interpretation (there may be other problems too) and it is the proliferation of worlds. But I think perhaps a new version of this interpretation could turn out to eliminate that proliferation.
Paradoxically, I think in the end the "many worlds" interpretation is also a realist interpretation. Except that "reality" is not constrined to a 4-dimensional space-time but includes all possible outcomes. In this sense, I would say the many worlds interpretation is "super-realist" (that's my own idea).
But I can understand that such interpretation would be repulsive to you, because you are a positivist and would not accept the existence of other "worlds" which cannot be detected or measured directly.
I think Mach's ideas had an important role in the philosophy of science but are not always beneficial when taken to extremes. Remember that Mach didn't believe in the existence of atoms.
I think that some times it is advantageous to consider elements which are not directly measurable but which serve to organize the results of measurement in a more elegant way. Being the fact that nature appears to (most of the time) turn out to be pretty elegant, these non-observable constructs stand a chance of eventually becoming observable or at least so pervasive and necessary that could be considered as part of reality.
On the other hand, I think the original purpose of this thread was to explore the process of entanglement (whose existence you denny. For that reason, I think we continue to debate this controversial topic in the wrong place. You say you are not sure you can create a thread. As far as I know anybody who is a subscriber to the forum can create threads. You should have a button visible in your screen (when you are looking at the list of threads) that says "create new thread" or something similar.

DrChinese

Jan8-05, 12:54 PM

I have a problem with negative probabilities. So, I'd be most grateful if you could provide details and chapter and verse on this idea. Thank you.

Regards,
Reilly Atkinson

It will probably take me a day or two :) to present it fully, but here is the starting point so you can see where I will be going:

a. 2 single channel detectors I will call Left and Right. The Left is set at angle A=67.5 degrees. The Right alternates between B=22.5 degrees and C=0 degrees. The selection of the angles is done to allow some sleight of hand with the math later. I will call it + if there is a detection, and a - if there is no detection. Efficiencies and actual experimental requirements are ignored as I am just trying to show the concepts involved.

b. In the Realistic case (the assumption of Realism, that the variables exist and have values independent of their observation), you could imagine that both B and C exist at the same time. Therefore, there are 8 combinations that must total to 100%. They are:

[1] A+ B+ C+
[2] A+ B+ C-
[3] A+ B- C+
[4] A+ B- C-
[5] A- B+ C+
[6] A- B+ C-
[7] A- B- C+
[8] A- B- C-

c. In the quantum world, 2 of the above cases are suppressed: [3] and [6]. The reason is that they don't actually exist as possibilities. B is the angle between A and C in my example, and B must always yield the same +/- value as either A or C. In these two cases, B is opposite to A and C. So in the Realistic scenario, [3]>=0 and [6]>=0 (and the sum of all 8 cases=1).

d. So what I have to do is to demonstrate that the quantum mechanical predictions for these 2 cases is actually less than zero. If I can do this, it will demonstrate a big conflict. The difficulty is that I have to do it using terms involving measurements of B or C, but not both at the same time. Can I do it? To be continued...

-DrC

Caroline Thompson

Jan8-05, 04:58 PM

[1] A+ B+ C+
[2] A+ B+ C-
[3] A+ B- C+
[4] A+ B- C-
[5] A- B+ C+
[6] A- B+ C-
[7] A- B- C+
[8] A- B- C-

This setup looks very similar to the one used in http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality
The page covers my interpretation of the Wigner-d'Espagnat inequality, which is alternatively covered by a quantum theorist at http://en.wikipedia.org/wiki/Wigner-d%27Espagnat_inequality . Your idea seems subtly different, but perhaps it would make your task easier to switch to this better-known case? Since the inequality violates local realism it would be expected to also produce negative probabilities if analysed in the manner suggested by Feynman in
Feynman, R P, Simulating Physics with Computers, International Journal of Theoretical Physics 21, 467-488 (1982).
[I read this a very long time ago and don't guarantee anything!]

Caroline

Caroline Thompson

Jan8-05, 05:13 PM

Not vanish, just move to a page called Caroline's Theorem! :)

Now that's not fair and you know it! :wink:

I discuss Bell's theorem in the spirit originally intended, though, not in the manner to which people have become accustomed. Bell was originally a realist. When experiments were found to (apparently) back quantum theory he was saddened. He said:

"So, for me, it is a pity that Einstein's idea does not work. The reasonable thing just does not work." [P Feyerabend "Historical Comments on Realism", p 194 of A Van der Merwe, F Selleri and G Tarozzi, “Bell's Theorem and the foundations of modern physics” (World Scientific, Singapore, 1992)]

Caroline
http://freespace.virgin.net/ch.thompson1/

Caroline Thompson

Jan8-05, 05:19 PM

Caroline:
I can understand that DrChinese and you don't get along too well. [Actually we understand each other quite well, I think.]

On the other hand, from reading the posts it appears you might find some compatibility with Reilly's ideas. He also has expressed the opinion that quantum correlations are no different than classical correlations. Thanks, I'll look him up.

Re your multiworlds ideas, you are quite right: I have no place for them in my world!

On the other hand, I think the original purpose of this thread was to explore the process of entanglement (whose existence you deny. For that reason, I think we continue to debate this controversial topic in the wrong place. You say you are not sure you can create a thread. As far as I know anybody who is a subscriber to the forum can create threads. You should have a button visible in your screen (when you are looking at the list of threads) that says "create new thread" or something similar.

Thanks for the advice, but I can't find that button. I have found, though, a comforting little note that says I have the right to create new threads. Some day I'm sure I'll find out how to do so!

Caroline

DrChinese

Jan8-05, 05:34 PM

Now that's not fair and you know it! :wink:

I discuss Bell's theorem in the spirit originally intended, though, not in the manner to which people have become accustomed. Bell was originally a realist. When experiments were found to (apparently) back quantum theory he was saddened. He said:

"So, for me, it is a pity that Einstein's idea does not work. The reasonable thing just does not work." [P Feyerabend "Historical Comments on Realism", p 194 of A Van der Merwe, F Selleri and G Tarozzi, “Bell's Theorem and the foundations of modern physics” (World Scientific, Singapore, 1992)]

Caroline
http://freespace.virgin.net/ch.thompson1/

Sorry, but I really don't think it is all that important whether Bell believed in local realism at various points in his life or not. His theorem holds great significance and I only wish that Einstein could have lived to see it.

As to the spirit of your Wikipedia presentation of Bell's work: I think you would be disingenuous if you said you were not aware of your presentation bias. Anyone familiar with his work will recognize the hand of a highly biased author. As I have said before, your anti-establishment bias is a good thing when it is properly disclosed. It is not fair to the less educated reader to mis-characterize your stand as commonly accepted science when it is not. That statement remains true even if you are eventually shown to be correct, because then your position would itself become mainstream. That is the way of science.

freep2

Jan8-05, 06:03 PM

There is no space or matter in the univeres. Just Energy. Light=Energy. Thought= Energy.

DrChinese

Jan9-05, 11:42 AM

[QUOTE=DrChinese]

I have a problem with negative probabilities. So, I'd be most grateful if you could provide details and chapter and verse on this idea. Thank you.

Regards,
Reilly Atkinson

I have pretty well completed my post in response, Reilly. I will be creating a separate thread to discuss this, no later than tomorrow. I think the result should be as straightforward as is possible, and the math is such that anyone can easily follow. I hope I don't bungle it :)

-DrC

Caroline Thompson

Jan9-05, 01:39 PM

Sorry, but I really don't think it is all that important whether Bell believed in local realism at various points in his life or not. His theorem holds great significance and I only wish that Einstein could have lived to see it.

As to the spirit of your Wikipedia presentation of Bell's work: I think you would be disingenuous if you said you were not aware of your presentation bias. Anyone familiar with his work will recognize the hand of a highly biased author. As I have said before, your anti-establishment bias is a good thing when it is properly disclosed. It is not fair to the less educated reader to mis-characterize your stand as commonly accepted science when it is not. That statement remains true even if you are eventually shown to be correct, because then your position would itself become mainstream. That is the way of science.

See my response to your entry http://en.wikipedia.org/wiki/Talk:Bell%27s_theorem#Caroline_Thompson.27s_POV_an d_Self_Promotion_in_this_topic.

I have never disguised my bias. In my opinion, as I think you know, it is unfair on the public to be told that quantum entanglement has been confirmed experimentally when this is not the case. Why, if it had been satisfactorily proven, would people still be trying to find "loophole-free" tests?

caribou

Jan9-05, 05:09 PM

DrChinese

Jan9-05, 06:10 PM

1. I have never disguised my bias.

2. In my opinion, as I think you know, it is unfair on the public to be told that quantum entanglement has been confirmed experimentally when this is not the case.

3. Why, if it had been satisfactorily proven, would people still be trying to find "loophole-free" tests?

1. Your participation in public forums usually omits reference to your unorthodox views, so I would say you disguise your bias by omission.

2. My opinion, shared by most scientists, is that it has been shown. Unambiguously.

3. The "closing of loopholes" is one of the most important elements of modern science, and in no way negates existing experiements or lessens their significance. That is why ever more elaborate tests of General Relativity are being performed today - even though GR is here to stay regardless of the outcome. Refinements and improvements to theory are good, even if minor.

Caroline Thompson

Jan10-05, 03:04 AM

Here's a quote from Roland Omnes' book Understanding Quantum Theory:

An entangled state is a quantum superposition of two distinct physical systems. This is a very frequent situation because any composite system whose wave function is not simply a product of the wave functions of its components is entangled. The existence of these states is proclaimed by the Pauli principle, and in that sense, it is responsible for a host of physical properties from the hardness of a stone to the laser.

I consider the hardness of stones and the existence of lasers to be experimentally confirmed. :wink:

OK, so that was Omnes opinion and you are happy to go along with it. You may not be surprised to know that I do not and that I have my own ideas as to why stones are hard and how a laser works. See my web site: http://freespace.virgin.net/ch.thompson1/

Caroline

Caroline Thompson

Jan10-05, 03:13 AM

1. Your participation in public forums usually omits reference to your unorthodox views, so I would say you disguise your bias by omission.

2. My opinion, shared by most scientists, is that it has been shown. Unambiguously.

3. The "closing of loopholes" is one of the most important elements of modern science, and in no way negates existing experiements or lessens their significance. That is why ever more elaborate tests of General Relativity are being performed today - even though GR is here to stay regardless of the outcome. Refinements and improvements to theory are good, even if minor.

Surely nobody is quite so naive as to think that local realists don't exist? Surely anyone can quickly deduce that I am one?

Re (2), on what is your opinion based? Perhaps it does not depend on the Bell tests, but really there is no other known way to test for entanglement of separated particles and, as you know, there has been no loophole-free experiment. Not only that but it is possible to find perfectly straightforward local causal explanations for all experiment to date if you take account of the actual variant of Bell test used and actual conditions.

You statement (3) is misleading. The kind of revision of theory that will be needed if entanglement turns out to be not a fact is a fundamental one. It may not involve many changes in the equations that are used, but it involves a huge change in outlook. It would mark the beginning of a new era (or is it a return to an old one?) in which magic (i.e. anything not due to local realist causes) was banished from science.

Caroline

Eye_in_the_Sky

Jan10-05, 05:57 AM

I still do not quite see it. I am unable to convince myself that it is wrong to say "quantum correlations imply nonlocality". I acknowledge that Bell's argument as he originally presented it is not enough to establish that claim, and moreover, that my original argument way back at post #59 (http://www.physicsforums.com/showpost.php?p=412828&postcount=59) is inadequate in this regard. However, if I shift the terms of that argument ever so slightly, my 2) becomes:

2') hidden variables "exist" → nonlocality .

And that is fine.

What then would my 1) become? It would become:

1') hidden variables do not "exist" → state-vector description is "complete" → state-vector description gives a full account of the "real factual situation" → under appropriate conditions, the "real factual situation" in one place will change on account of an action performed at another place arbitrarily far away → nonlocality .

I suppose that the main problem with my 1') has to do with this construct of a "real factual situation". Perhaps it is not necessarily a valid one. Is that so? If we (tentatively) grant validity to that construct, can 1') be invalidated on other grounds?

caribou

Jan10-05, 08:18 AM

OK, so that was Omnes opinion and you are happy to go along with it. You may not be surprised to know that I do not and that I have my own ideas as to why stones are hard and how a laser works.

I think you'll find it's the opinion of Omnes and most physicists who either founded or developed quantum theory.

Caroline Thompson

Jan10-05, 12:08 PM

I think you'll find it's the opinion of Omnes and most physicists who either founded or developed quantum theory.

Well yes, the majority of physicists may well believe that without QT we could not explain the hardness of stone of how a laser works. Clearly those who developed the theory fondly hoped that they had now explained everything (and that despite, incidentally, the fact that Bohr and Von Neumann did not think lasers possible until one had been demonstrated to them! See http://freespace.virgin.net/ch.thompson1/People/CarverMead.htm)

Several of the Founding Fathers, though, recognised later in life that they could have been wrong -- that better theories might be possible. And I rather like this quotation:

The scientific community ... is like a pack of hounds ... where the louder-voiced bring many to follow them nearly as often on a wrong path as a right one, where the entire pack even has been known to move off bodily on a false scent.
Samuel Pierpont Langley, 1889

I first encountered it in the report on one of the Solvay lectures from the 1920's, though I have not been able to trace the exact reference. It was one of the older generation of physicists who said it.

Caroline

DrChinese

Jan10-05, 02:11 PM

Well yes, the majority of physicists may well believe that without QT we could not explain the hardness of stone of how a laser works. Clearly those who developed the theory fondly hoped that they had now explained everything (and that despite, incidentally, the fact that Bohr and Von Neumann did not think lasers possible until one had been demonstrated to them!

Your point that even famous scientists change their minds as new information becomes available has been discussed ad nauseum in the past. It is well recognized that current theory will be adjusted in the future as we learn more. So what? You use that to justify your own otherwise meritless position by trying to convince us that "maybe" you will turn out to be right in the end. Well, maybe pigs will fly tomorrow as well. In the meantime, I am waiting for you to produce something more than an elaborate conspiracy to hide the "truth" according to Caroline.

Science is about creating useful theory. QM is useful. What you are espousing is essentially that no one can ever know anything - because no level of proof is sufficient for you. Your viewpoint lacks utility.

caribou

Jan10-05, 02:58 PM

Caroline Thompson

Jan10-05, 05:15 PM

Your point that even famous scientists change their minds as new information becomes available has been discussed ad nauseum in the past. It is well recognized that current theory will be adjusted in the future as we learn more. So what? You use that to justify your own otherwise meritless position by trying to convince us that "maybe" you will turn out to be right in the end. Well, maybe pigs will fly tomorrow as well.

I'm not talking about pigs flying, DrChinese, but about common sense. It has been out of fashion ever since Einstein's 1905 relativity paper (well actually rather later than that: since he rose to fame after the 1919 eclipse data was claimed to support his GR theory) to believe that we can hope to use our intuition to arrive at valid models of the real world. All I am saying is that the actual data in no case forces this fashion on us. The reason we accept counterintuitive theories is that the mathematics behind them is manageable. They enable us to build formulae that appear to work.

The Bell tests, though, could be shown to be the exception if only a wider range of parameters were used in each experiment. The published results suport QM, but what about other unpublished ones or results that would have been obtained under slightly different conditions? For these conditions, the mathematics of QM becomes intractable. Nobody even attempts it unless it is absolutely forced upon them, and when this happens the services of an expert are required. With my "local realist" models, on the other hand, once you understand the basics of how each piece of apparatus works, anyone can see how the model should be modified. The realist model would, in the hands of someone knowledgeable about the actual apparatus, be more flexible and have greater predictive power than QM

In the meantime, I am waiting for you to produce something more than an elaborate conspiracy to hide the "truth" according to Caroline.
I am not talking of conspiracy, merely of ignorance, together with the consequences of living with a theory to which you cannot apply your intuition. It can lead to absurdities and you have no way of judging when this has happened. It happens, I maintain, in just about every "quantum optics" experiment, not just in the Bell tests. They can all benefit by switching to a realist model.

Science is about creating useful theory. QM is useful.

QM in general may be useful. Quantum optics (the area that includes the Bell test experiments) is redundant. The whole area is already covered by what is essentially classical optics, with just a few extensions to cover things that were not known 100 years ago.

What you are espousing is essentially that no one can ever know anything - because no level of proof is sufficient for you. Your viewpoint lacks utility.

I trust you have read sufficient of my work to know what you are talking about here. I am not talking in general, only about a certain area of physics in which previously accepted standards of scientific explanation have been abandoned on insufficient evidence.

Though I am more concerned with understanding than with utility, in the long term I maintain that applications of "entanglement" will benefit from recognition of the fact that this it not any magical quantum-theoretical phenomenon but merely a result of ordinary correlations of ordinary variables such as polarisation direction. The act of looking at "coincidences" produces results that look strange until you stop to work out (as I did in my Chaotic Ball model) what is really happening. Surely it is unwise to just look at them and say "How strange!" and not investigate further?

Sorry to gabble on, but if it's just my voice against the masses I feel it to be necessary.

Caroline
http://freespace.virgin.net/ch.thompson1/

reilly

Jan10-05, 05:53 PM

Caroline Thompson writes:I'm not talking about pigs flying, DrChinese, but about common sense. It has been out of fashion ever since Einstein's 1905 relativity paper (well actually rather later than that: since he rose to fame after the 1919 eclipse data was claimed to support his GR theory) to believe that we can hope to use our intuition to arrive at valid models of the real world. All I am saying is that the actual data in no case forces this fashion on us. The reason we accept counterintuitive theories is that the mathematics behind them is manageable. They enable us to build formulae that appear to work.

****
With what time i have for this forum, I should be doing my due diligence on Dr. Chinese's ideas on negative probabilities. But, where do you ever get the idea that common sense and intuition no longer play a role in physics? (If you don't use these, what can you do?)

Very few, if any professional physicisits would agree. Rather, many could cite many occasions in their careers in which common sense/and intuition played a huge role. And, the literature is full of such examples. Your notion that we accept what you term "counterintuitive theories: because of mathematical nicities is, with all due respect, quite wrong. Einstein, Bohr were highly intuitive thinkers, Cooper's approach to explaining superconductivity was highly intuitive -- which, in fact, allowed him, to simplify the math of an interacting electron gas in an appropriate lattice. The WKB approximation is highly intuitive, the Fermi-Thomas approach to heavy atoms is again highly inuitive. Much of basic nuclear physics, thermal neutron sattering for example -- giving the idea of a scattering length, a very useful "intuitive" tool.

There's much more to say, but I've not the time now. I'll say that the history and literature of 20th century physics does not support your contention.

Regards, Reilly Atkinson

Hans de Vries

Jan11-05, 03:21 AM

Quantum theory is the most successful physical theory of all and entanglement is a very important part of it. Many people may not understand the true meaning of the superposition of states of distant particles in entanglement but that is very much their problem and not the theory's.

There are no non-local "action on a distance" terms in the Standard Model
which describes the Electro Magnetic, Weak and Strong Interactions.
It is a local Quantum Field Theory.

Non-local theories are in serious conflict with both Special Relativity
and Quantum Mechanics.

Non-local theorist claim that instantaneous action on a distance doesn't
violate Special Relativity with arguments that seem more like a word play.

Quantum teleportation is supposed to work because it "Circumvents"
Heisenberg's uncertainty principle according to Zeilinger.

Regards, Hans

Caroline Thompson

Jan11-05, 03:32 AM

... where do you ever get the idea that common sense and intuition no longer play a role in physics? (If you don't use these, what can you do?)

Perhaps I did not choose my wording sufficiently carefully. What I said was "It has been out of fashion [recently] to believe that we can hope to use our intuition to arrive at valid models of the real world."

What I meant was simply that it has become acceptable to use models that are not intuitive when this seems to be necessary. This may not matter in some contexts, but if the idea is extended to mean that no intuitive model is ever going to be possible we need to think again. Have we been sufficiently careful in deciding whether or not the observed facts really do imply this?

I maintain that Dirac et al's mathematics forced the idea of entanglement of separated particles on us and our reluctance to relax the assumption of the particle nature of light has blinded us to alternative, purely intuitive, explanations of the observations.

Einstein set the scene for this in two ways:
a) by inventing the "photon" (though this was not consistently used as meaning an indivisible particle: at one stage he referred to it as "needle radiation", which has no such implications) and
(b) by postulating the constancy of the speed of light, declaring the notion of an underlying wave medium -- the aether -- to be superflous.
Heisenberg, when he first introduced his Uncertainty Principle, actually referred to Einstein's counterintuitive idea of the constancy of the speed of light as a precedent. See page P117 of:
Hendry, John, “The Creation of Quantum Mechanics and the Bohr-Pauli Dialogue”, D Reidel Publishing Company 1984

... Your notion that we accept what you term "counterintuitive theories: because of mathematical nicities is, with all due respect, quite wrong. Einstein, Bohr were highly intuitive thinkers, Cooper's approach to explaining superconductivity was highly intuitive -- which, in fact, allowed him, to simplify the math of an interacting electron gas in an appropriate lattice. The WKB approximation is highly intuitive, the Fermi-Thomas approach to heavy atoms is again highly inuitive. Much of basic nuclear physics, thermal neutron sattering for example -- giving the idea of a scattering length, a very useful "intuitive" tool.

There's much more to say, but I've not the time now. I'll say that the history and literature of 20th century physics does not support your contention.

Right, so the initial maths tends to be based on intuition and experience, but what happens next is what concerns me. The maths gets extrapolated. It is assumed to apply in situations broader than its original purpose. Perhaps I should not generalise, but this is what has, I maintain, happen in the case of quantum entanglement. It was an unwanted side effect of trying to apply mathematics intended for single particles to pairs of them. A decision was taken to assume that these could be modelled by one, "nonseparable", wave function. This assumption led to a nice neat formula, one of whose consequences was that Malus' Law would apply to coincidence counts between two separate polarisers just as it was (so long as you assume detection rates proportional to input intensities) known to apply to the output from crossed polarisers placed "in series".

But this extrapolation was not based on any physical model and is, under local realist theories, false. Experiments only approximately back it up, and only in a limited set of conditions, sometimes only after data adjustments that themselves rely for their justification on further quantum-theoretical mathematics. [See Thompson. C H: “Subtraction of ``accidentals'' and the validity of Bell tests”, Galilean Electrodynamics 14 (3), 43-50 (May 2003).
http://arXiv.org/abs/quant-ph/9903066]

Incidentally, if I am right then there is no need to delve into the question of negative probabilities. These have never been shown to happen, which is scarcely surprising since the concept is mathematically impossible.

Caroline
http://freespace.virgin.net/ch.thompson1/

vanesch

Jan11-05, 04:11 AM

Quantum optics (the area that includes the Bell test experiments) is redundant. The whole area is already covered by what is essentially classical optics, with just a few extensions to cover things that were not known 100 years ago.

I defy you to explain the results of:

Am. J. Phys. Vol 72, No 9, September 2004

with classical optics.

cheers,
Patrick.

caribou

Jan11-05, 06:51 AM

There are no non-local "action on a distance" terms in the Standard Model
which describes the Electro Magnetic, Weak and Strong Interactions.
It is a local Quantum Field Theory.

Non-local theories are in serious conflict with both Special Relativity
and Quantum Mechanics.

Non-local theorist claim that instantaneous action on a distance doesn't
violate Special Relativity with arguments that seem more like a word play.

Quantum teleportation is supposed to work because it "Circumvents"
Heisenberg's uncertainty principle according to Zeilinger.

Regards, Hans

Actually, the "true meaning" I was referring to is that apparently people have been mistaken when they think there are "spooky action-at-a-distance" and faster-than-light effects in quantum theory.

If I understand experts like Zurek and Omnes, things like the EPR experiment apparently involve a superposition of measurement outcomes and no faster-than-light effects.

Hans de Vries

Jan11-05, 07:17 AM

If I understand experts like Zurek and Omnes, things like the EPR experiment apparently involve a superposition of measurement outcomes and no faster-than-light effects.

There is definitely faster then light correlation in the teleporting theories.

The twisted argument is that it doesn't count because it can not
be checked immediately if an event is correlated to another distant
one. The check would have to go with the speed of light.

With some real bad luck I could get killed instantaneously by some
astronomical catastrophe some 100 lightyears away via entanglement.
It would however take 100 years to check that my dead was correlated
to the distant catastrophe.

So, am I not dead then as long as the correlation is not checked?
Doesn't it count?

Regards, Hans

vanesch

Jan11-05, 07:51 AM

If I understand experts like Zurek and Omnes, things like the EPR experiment apparently involve a superposition of measurement outcomes and no faster-than-light effects.

Yes ! That's also what's my opinion. It is the observer who considers the correlation that does "the collapse thing", and not the two distant "observers" who just entangle their message with the part of the state of the received subsystem.
(but the price to pay for that opinion is a kind of solipsism...)

alexepascual

Jan11-05, 12:44 PM

Caroline:
To make new thread, while looking at the posts on this thread, look at the top of the page. There should be a box that says:
Physics > Quantum Physics > How do particles become entangled ?
Click on "Quantum Physics", which will take you to the list of threads under this topic.
On the top left of that screen, just above the first thread, you'll see a button that says: "New Thread". Good luck!

Eye_in_the_sky:
Things came up and I am still reading your post. Some of it I understand, some of it I have questions. But I'll read it again and do a little more thinking before I respond.

Caroline Thompson

Jan12-05, 04:22 AM

Yes ! That's also what's my opinion. It is the observer who considers the correlation that does "the collapse thing", and not the two distant "observers" who just entangle their message with the part of the state of the received subsystem.
(but the price to pay for that opinion is a kind of solipsism...)

Never mind the interpretation! What Bell showed was (as EPR had suspected), that the QM theory of entangled particles involves something in conflict with local realism. It is this that bothers me -- the failure of basic assumptions such as the ablity to predict probabilities of joint outcomes of independent events by multiplying the individual probabilities.

Before some people jump on me and say that the two sides are not independent, do look at the actual local realist calculation given in Appendix C of http://arXiv.org/abs/quant-ph/9903066! You will see then just what role the hidden variable, lambda, plays and at what stage the multiplication takes place. It is before integration over lambda.

Bell discusses the necessary assumptions in pp 36-37 of
Bell, John S, The Speakable and Unspeakable in Quantum Mechanics, Cambridge University Press 1987

Caroline

vanesch

Jan12-05, 06:31 AM

Never mind the interpretation! What Bell showed was (as EPR had suspected), that the QM theory of entangled particles involves something in conflict with local realism.

That's true if you think that the result of a measurement is classical and "graved in stone". However, if you consider that the "measurement" just transmits the entanglement, and that the result also comes in a superposed state, you just get, when you look at the correlations of the results from both distant photodetectors, a local interference of entangled systems. So this saves the "local" part. I'm not sure it saves the "realist" part :tongue2:
Imagine sending out Bob to the left photodetector, and Alice to the right photodetector. They come back in a superposed state, with all different possible results in a superposition, and it is when you look at both of them that the actual von Neuman measurement takes place, for you, as an observer. It's all in the mind :smile:

But ok, if you don't buy into photons, you won't buy into this. Although it is 100% in agreement with all of existing physics :smile:

alexepascual

Jan14-05, 12:10 PM

Eye_in_the_Sky:
I figured that to get a good understanding of your posts I would have to study more about composite systems, tensor products, etc.
With respect to your expression:
|ψ>|φ> → Σk ak|ψk>|φk>
Now I understand what your intent was when you wrote that.
Even if that expression wasn't totally correct, it could be used to convey your idea. But I am not totally satisffied with that, because it appears that your expression is actually correct, and I suspect there is more to it than meets the eye.
I found that expression in an article on the web, but I don't know if the context is exactly the same. I could copy from that article but I'll try to make it shorter by just briefly telling you what it says.
The article says that it is always possible to find bases for H1 and H2 such that the composite state is represented by the right side of the equation above. It also says that the procedure to find such basis is known as "Schmidt decomposition" or "biorthogonal or polar expansion"
I looked up these on the web and there were many entries that came up which are related to entanglement.
The article also says that the dimmension of the index k is that of the smallest Hilbert space. In one of my previous responses I was assuming you meant that k was running from 1 to nxm, which would not be correct in the context of a "polar expansion".
So, I would like to know if a "polar expansion" is what you had in mind when you wrote that post.
Thanks Eye,

chrismuktar

Jan14-05, 01:29 PM

I know that generally speaking entanglement of photons is done using 'parametric down conversion', where photons are collided in a material exhibiting non linear refractive indices. This process renders two entangled photons.

For the purpose of quantum cryptography, people are trying to do this process inside an optical fibre. The process is called four wave mixing (FWM).

Eye_in_the_Sky

Jan17-05, 09:21 AM

I figured that to get a good understanding ... I would have to study more about composite systems, tensor products, etc.In that context, it would also be a good idea to look at "density operators" and the operation of "partial tracing" to obtain "reduced density operators".

With respect to your expression:
|ψ>|φ> → Σk ak|ψk>|φk>
Now I understand what your intent was when you wrote that.
Even if that expression wasn't totally correct, it could be used to convey your idea. But I am not totally satisffied with that, because it appears that your expression is actually correct, and I suspect there is more to it than meets the eye.As I said, my intent was merely to indicate that, on account of the interaction, the initial joint product-state evolves into a final state which cannot be written as simple product-state.

The condition I gave was:

|ζ> = ∑k ak|ψk>|φk> ,

where each ak ≠ 0, and there are at least two distinct values for k, and {|ψk>} and {|φk>} are each sets of linearly-independent vectors.

This condition is necessary and sufficient for the joint state |ζ> to have no expression as a simple product-state.

(Is there more to this than meets the eye? ... Well, maybe.

The sort of "interaction" I like to keep in mind as an example for an 'entanglement' scenario (although in this context, the word 'correlation' has a more appropriate connotation) is based upon the following type of unitary transformation. Consider two sets {|ψk>} and {|φk>} in H1 and H2 respectively, where the vectors in each of the sets are orthonormal (and not just linearly independent). Let |φ> be a specific state in H2. Give a partial definition of a unitary operator U as follows:

U(|ψk>|φ>) = |ψk>|φk> .

Then, for any |ψ> ≡ ∑k ak|ψk>, we have by linearity

U(|ψ>|φ>) = ∑k ak|ψk>|φk> .

A unitary operator with this sort property is what is usually considered in elementary discussions of the so-called "Measurement Problem".)

... "Schmidt decomposition" or "biorthogonal or polar expansion" ... So, I would like to know if a "polar expansion" is what you had in mind when you wrote that post.No, that is not what I had in mind.

"Schmidt Decomposition" goes like this:

For a given |ζ> Є H1(x)H2, there exist orthonormal bases {|αi>} and {|βj>} of H1 and H2 respectively, and nonnegative real numbers pk, such that

|ζ> = ∑k √pk|αk>|βk>

and

∑k pk = 1 ,

where the summations are over k = 1, ..., min(dimH1,dimH2).

This condition is much stronger than mine. To know that two systems are entangled, you don't have to go to a Schmidt Decomposition. My weaker condition is enough.

alexepascual

Jan25-05, 10:24 PM

Eye_in_the_Sky:
I am a little confused with respect to the process of obtaining an entangled state by applying a unitary operator.
Considering the case of two non-identical particles in which we look at spin in the z-direction, the composite Hilbert space would be 4-dimmensional. That is, in general we would have 4 components, each represented by a product state. Let's say we have two particles in a superposition of spin up and spin down. Let's also assume that, as you proposed in a previous post, the particles interact in a way that the final result is a state that contains |+>|-> and |->|+>. So the states |+>|+> and |->|-> now have a zero coefficient.
As I am writing this I realize that we could think of this transformation as a rotation of the state vector (originally spanning all four base vectors) into the subspace spanned by the resulting vectors. The thing is, I have seen something like this done with a projector instead of a unitary transformation. Actually, if we wanted to do it with a rotation in Hilbert space, the angle of rotation would depend on the original angle, which would preclude the use of the operator to represent such a process for an arbitrary vector.
Now, even if that unitary transformation was possible and correct, as far as I know, interactions are described in terms of Hamiltonian operators right?
Well, I don't see how I could use a Hamiltonian to get that entanglement. Would I have to apply the Hamiltonian for a very precise time interval?
I think it would perhaps be asking too much if I asked you to write the Hamiltonian Matrix, as this would be a 4x4 matrix, but maybe you can give me a few more hints.
Something that has me puzzled is that the process of entanglement would entail a reduction of the realm of possibilities, a "compression" of the Hilbert space, which has for me more the taste of a non-unitary process with a change in entropy.
I'll appreciate your comments Eye,
Alex

Eye_in_the_Sky

Jan28-05, 12:55 AM

I am a little confused with respect to the process of obtaining an entangled state by applying a unitary operator.
Considering the case of two non-identical particles in which we look at spin in the z-direction, the composite Hilbert space would be 4-dimmensional. That is, in general we would have 4 components, each represented by a product state. Let's say we have to particles in a superposition of spin up and spin down. Let's also assume that, as you proposed in a previous post, the particles interact in a way that the final result is a state that contains |+>|-> and |->|+>. So the states |+>|+> and |->|-> now have a zero coefficient.Actually, when I spoke of an interaction leading to entanglement, I did not have in mind a spin-spin system. (The spin-spin entangled "singlet" state which I mentioned earlier is something which, so to speak, just 'pops out' on account of conservation of angular momentum, where the original system was already in a state of zero total angular momentum.)

The example you are suggesting is not so straight forward. Presumably, the proposed interaction would respect conservation of angular momentum. But, as you can see, in your example the total spin for the joint system will not be conserved. So, we would then need to consider something more than just the joint spin-space of the two-particle system. Perhaps then, we should take into account the position parts of the joint system as well, and then think about if (and how) we can get a 'trade-off' between orbital angular momentum and the spins. Or perhaps we should have to consider a third entity which interacts with these two, which can thereby 'carry off' or 'put in' some angular momentum. ... This is too difficult.

... However, what you are really looking for is a clear, simple, and physically real example of a process which leads to "entanglement". So, here is one.

Consider a single electron with corresponding Hilbert space spanned by the "spin-position" basis {|z+>|r >, |z->|r > | r Є R3}. Suppose that the electron is initially in the state

(a|z+> + b|z->)|φo> ,

and it serves as input to a Stern-Gerlach device which is set to measure the spin component along the z-axis. Let the initial spatial wavefunction
φo(r) be a packet centered at the origin and moving in the +x direction towards the Stern-Gerlach device.

What will you get when you solve the Schrödinger equation for this situation? ... You will find that the initial spatial wave packet splits into two parts, |φ+> and |φ–>, one part correlated with |z+>, and the other with |z->. That is,

(a|z+> + b|z->)|φo> → a|z+>|φ+> + b|z->|φ–> ,

where the "spin-up" packet φ+(r) will be centered at a point with a positive value of z, and the "spin-down" packet φ–(r) will be centered at a point with a negative value of z.

In this example, it is with regard to the state vector for a single electron that the spin and position parts become "entangled" (or (perhaps, more appropriately) "correlated") with one another. This is accomplished by means of the unitary (Schrödinger) evolution of an electron in the magnetic field of a Stern-Gerlach device.

alexepascual

Jan30-05, 08:45 PM

The example you are suggesting is not so straight forward. Presumably, the proposed interaction would respect conservation of angular momentum. But, as you can see, in your example the total spin for the joint system will not be conserved.
I can't see how momentum is not conserved. It was zero to start and is still zero after the interaction.
With respect to your example, I understand how that works physically, and I thank you for trying to find an example that is experimentally feasible. I think I understand your description. But the part where the wave packet splits is a little too complex for me. What I mean is that even I can understand that this is exactly what would happen in a Stern-Gerlach apparatus, working out the math in detail might be a little over my head. I am also affraid that we might need to consider the coupling between the field and the electron, and I don't know anything about quantum field theory.
So, while your example helps, I am still looking for a simpler case. Maybe that's not possible and things are more complex than I would like them to be.
Also, what I was originally looking for was a case where two separate particles get entangled. I wonder if, after sacrificing the advantage of a physically reallistic example, we could find a mathematical description of the entanglement process for two two-state systems.
Let's say that we have these two two-state systems, and that we ignore the physical nature of the states. These systems are not entangled. Then we come up with some matrix U, expressed in the same basis as the states, which transforms the compound system into an entangled system. I wonder if this is possible. I understand that it migh be hard to come up with a physical process that will achieve this, but I was wondering if at least in the abstract it is possible (using some unitary matrix).

Eye_in_the_Sky

Feb2-05, 02:54 PM

I can't see how momentum is not conserved.I now retract my earlier statement in which I wrote:
... in your example the total spin for the joint system will not be conserved.The most general unitary transformation consistent with conservation of total spin for the composite system is given by:

|+>|+> → eiα|+>|+>

|->|-> → eiα|->|->

|+>|-> → [(eiα + eiβ)/2]|+>|-> + [(eiα - eiβ)/2]|->|+>

|->|+> → [(eiα - eiβ)/2]|+>|-> + [(eiα + eiβ)/2]|->|+>

where α and β are arbitrary real parameters.

[Sorry about my earlier confusion.]

Thus, for example, we can take α = 0 and β = π/2 ... and this would give:

|+>|+> → |+>|+>

|->|-> → |->|->

|+>|-> → a|+>|-> + a*|->|+>

|->|+> → a*|+>|-> + a|->|+>

where a = (1+i)/2 .

----------------------------
There is, however, one important point to note here in relation to what you said about the total spin:

It was zero to start and is still zero after the interaction.Note that in a {|J,mJ> | J = 0,1 ; mJ = -J,...,J} basis, we can write

|+>|-> = (1/√2) [ |1,0> + |0,0> ] → (1/√2) [ eiα|1,0> + eiβ|0,0> ] ,

|->|+> = (1/√2) [ |1,0> - |0,0> ] → (1/√2) [ eiα|1,0> - eiβ|0,0> ] .

So, for these states, it is incorrect to say that the total spin (before or after) is zero; rather, these states are each a superposition of a J=1 state with the J=0 state.
----------------------------

... Okay, fine. So, we can use the example above, or we can consider your alternative suggestion:

Let's say that we have these two two-state systems, and that we ignore the physical nature of the states. These systems are not entangled. Then we come up with some matrix U, expressed in the same basis as the states, which transforms the compound system into an entangled system. I wonder if this is possible. I understand that it migh be hard to come up with a physical process that will achieve this, but I was wondering if at least in the abstract it is possible (using some unitary matrix).Ah ... this will make it easier. First, consider the following:

A unitary transformation is equivalent to the linear extension of a mapping of one orthonormal basis to another.

So, suppose we have two orthonormal bases {|uk>} and {|vk>}. We then define a mapping by

|uk> → |vk> , k = 1,2,... ,

and extend this mapping to the whole Hilbert space by linearity. This mapping is then a unitary transformation. (The converse is obviously also true.)

We are now equipped with all we need to build our own "unitary-entangling" operator.

Here, I'll build one that I like. Let system A have a basis {|0>, |1>} and let system B have a basis {|↑>, |↓>}. Suppose that when system A is in the |0> state, the interaction invokes no change ... but, when system A is in the |1> state, the interaction invokes a "flip-flop" in B. That is:

|0>|↑> → |0>|↑>

|0>|↓> → |0>|↓> ... i.e. no change ,

and

|1>|↑> → |1>|↓>

|1>|↓> → |1>|↑> ... i.e. "flip-flop" in B .

"But where is the entanglement?" you ask. Here is one spot (in the linear extension):

(a|0> + b|1>)|↓> → a|0>|↓> + b|1>|↑> .

------

Now, it seems to me you may want to ask:

What is a necessary and sufficient condition for a unitary operator to be able to induce entanglement?

Well ... a trivial response is:

There do not exist unitary operators V and W (on HA and HB , respectively) such that U = V (x) W.

Quite generally, whenever the Hamiltonian of the joint system involves terms which indicate interaction between the two subsystems, then we will not have U = V (x) W, and therefore, entanglement will occur for (at least) some initial states.
----------------------------------
----------------------------------

With respect to your example, I understand how that works physically, and I thank you for trying to find an example that is experimentally feasible. I think I understand your description. But the part where the wave packet splits is a little too complex for me. What I mean is that even I can understand that this is exactly what would happen in a Stern-Gerlach apparatus, working out the math in detail might be a little over my head. I am also affraid that we might need to consider the coupling between the field and the electron, and I don't know anything about quantum field theory.The "splitting" of the wave packet should not be too mind boggling. I will just sketch out the basic idea.

Fist of all, I should have picked a neutral particle (with a magnetic moment), rather than a charged one, like the electron. That way, there will be no Lorentz force on the particle to consider.

Okay. So let's use a hydrogen atom in the ground state (treating it as an electrically neutral particle with magnetic moment equal that of the electron (the magnetic moment of the proton can be neglected, and in the ground state L=0)).

Given that ... then, making now a simplification to 1 spatial dimension, the Hamiltonian can be written as

H = -(hbar2/2m)(∂/∂z)2 + (ehbar/2m)σzB(z) ,

where σz is the 2x2 matrix

1 0
0 -1 .

It does not take much effort to see that the time-dependent Schrödinger equation will (quite trivially) split into two decoupled equations, one for |z+> and one for |z->, the only difference between them being a "+" or a
"-" in front of the "interaction" part of the Hamiltonian. ... I hope this clears up some of the 'fog'.

... Also, there is no need here to quantize the magnetic field (QFT is not required in this approximation).

gptejms

Feb3-05, 05:12 AM

vanesch

Feb3-05, 06:12 AM

>Hans wrote :
>There is definitely faster then light correlation in the teleporting theories.

>The twisted argument is that it doesn't count because it can not
>be checked immediately if an event is correlated to another distant
>one. The check would have to go with the speed of light.

>With some real bad luck I could get killed instantaneously by some
>astronomical catastrophe some 100 lightyears away via entanglement.
>It would however take 100 years to check that my dead was correlated
>to the distant catastrophe.

>So, am I not dead then as long as the correlation is not checked?
>Doesn't it count?

>Regards, Hans

The above post by Hans went unanswered.What do you guys say to this--I mean how do you answer it?

Ok, there are 2 aspects. The first one is that you cannot "die by entanglement". You die locally, by something that came over to you at less than lightspeed, say a meteor. The probabilities of what happens locally are of course independent of whatever an entangled system far away might undergo or do. So the chance of Hans dying by the meteor is locally fixed.
However, indeed, in order to find out that this meteor was entangled with something else far away, we will indeed have to wait at least 100 years before we could tell about the correlation (and it wouldn't make a lot of sense since there has been only 1 event :-)

Again and again and again: entanglement does NOT imply "action at a distance". I've seen some stupid proposal of having entangled ions in a spacecraft, which could then be excited by exciting their partners on earth, but that is totally wrong of course. Whatever happens to the ions in the spacecraft has probability distributions which are independent of what you do to their partners on earth. You could at best do measurements on them, and find out later that they were correlated.

cheers,
Patrick.

alexepascual

Feb3-05, 11:19 AM

Eye_in_the_sky:
I am thinking about your post.
I'll let you know as soon as I have digested it.

With respect to the last two posts:
I would argue that it is possible to die by entanglement. You could device a kind of Schrodinger's cat experiment with two boxes that contain atoms with entangled spins and a person in each box. You take the boxes far appart. Then, within each of the boxes there is a mechanism that kills the person depending on the result of a spin measurement. The death of one person would be correlated to the death of the other person. Well, actually you don't need to put the person in a box.
Now, this does not mean that you can do anything at one end to provoke the death of the person at the other end.
With respect to the "faster than light" issue, yes the events could be space-like related, but as there is nothing going from one end to the other (at the time between events) , it is missleading to talk about "faster than light".
I don't agree that the fact that you may have to wait 100 years to check the correlation really matters. you can confirm the correlation 100 years later and thus show that an experiment that was done 100 years ago proved or disproved whaterver is was designed to prove.
The fact that it is only one event is not relevant. You could design your experiment with 1000 events. You could send 1000 people in a spaceship, each of which carries an atom whose spin is entangled to that of an atom on earth. Their death depends on their finding the spin to be up or down. Of course there might be practical difficulties that might prevent an experiment like this being done. The main problem might be finding the money. Sending 1000 people to a possible death might not be a problem from a political point of view. If that were a problem you can always do it with people from a country you don't like, which automatically makes them less human. (as they would be in outer space, their death would not be under the jurisdiction of any eartly court). Well, enough about this, let's leave politics to the politicians and get back to the science. Yes, I started it. Sorry for the digression.

RandallB

Feb3-05, 11:50 AM

entanglement does NOT imply "action at a distance". Patrick.
Maybe you can help me be sure I understand entangement correctly.Lets say we generate two balls A & B that are entangled going left and right.
We know we have a ball coming to us at the left we may even be fairly sure that it is Ball A, but there is some probability that it is ball B. The QM superposition point is that it isn't ether ball A or ball B it can be either one until a commitment is required by some interaction or test. Hit it with a BAT and see it. Now if both A and B are completely identical as many are we cannot tell if we actually got ball B when we expected A.
But if there is a characteristic that is different, A is RED and B is GREEN then our testing will show if we hit the unexpected GREEN ball. AND we know anyone in the right area is going to "or already has" seen a RED Ball. But they a free to catch it or hit it as they local choose.

Now if it is true that most likely the RED Ball is the most likely result in the left area; nothing in QM allows us to be tipped to what that most likely one is. Thus it could also be true that the chance is the GREEN Ball instead can be very high say .49 at first and reduce to a smaller chance as the pair separate. Even down to such a low chance that entanglement could be considered to have been lost. (i.e. You can't stick them in separate jars for later use somehow) . Since it cannot be shown if RED really was most likely to start with, it should be very hard to prove or disprove the "decay" of entanglement.
The question - - is the idea of entanglement decay viable within the math of QM?

Personally I'd think it should be expected.
Randall B

alexepascual

Feb3-05, 12:15 PM

I don't know much about the possibility of a gradual decline in the degree of entanglement. I'll let other members answer to that.
But I would like to tell you that in my opinion, it would be more advantageous for you to think about superpositions not in terms of a superposition of different "balls" but a superposition of different properties in the same ball.
It migh not be a good idea either to give "balls" as an example as this suggests that you are dealing with macroscopic objects, which have never been found in a state of superposition.
I also think that while your question about a "decay" in entanglement might be a valid one, you might not be able to understand this issue without studying QM in depth, including all the math and jargon.

DrChinese

Feb3-05, 03:03 PM

The question - - is the idea of entanglement decay viable within the math of QM?

Personally I'd think it should be expected.
Randall B

There is no decay of entanglement due to distance in the ideal case, but... In the real world, a lot of things can cause the entanglement to end. A measurement does, and a lot of things can act as if they are measurements.

Weihs et al (http://arxiv.org/PS_cache/quant-ph/pdf/9810/9810080.pdf) performed testing over a distance of 1 kilometer and violated Bell's Inequality. I think Tittel et al did it across 10 kilometers and saw high correlations, but less than Weihs.

vanesch

Feb4-05, 03:38 AM

Eye_in_the_sky:
I am thinking about your post.
I'll let you know as soon as I have digested it.

With respect to the last two posts:
I would argue that it is possible to die by entanglement. You could device a kind of Schrodinger's cat experiment with two boxes that contain atoms with entangled spins and a person in each box. You take the boxes far appart. Then, within each of the boxes there is a mechanism that kills the person depending on the result of a spin measurement. The death of one person would be correlated to the death of the other person.

Yes, the deaths would be correlated. But not "caused by". If the local person only looked locally at his spin, he knows that he has 50% chance of dying. And locally, whether this local atom is entangled or not with a faraway atom doesn't change anything for the probabilities of dying.
So, yes, entanglement can make it such that people die in correlated ways. But they do not die more or less because there is entanglement. We just observe that there are correlations in their deaths, when we know the result of both.
I had understood (maybe wrongly) that "to die by entanglement" would mean that you somehow die because the particle was entangled, and if it weren't, you wouldn't die. Maybe this was not what the original question was about, in which case my answer was next to the point.

In fact, in an MWI-like interpretation, everything is about entangled with everything else, and this is the basis of decoherence, which makes us avoid seeing superpositions of classical states of macroscopic subsystems, except in rare circumstances

cheers,
patrick.