How do particles become entangled?

[Eye_in_the_Sky]

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]

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]

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]

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]

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]

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.

 

[Louis Cypher]

Qm Entanglement



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

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 doesn't 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

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

 

[Louis Cypher]

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]

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]

would apreciate a reply before end of week

Any ideas guys?
see previous post

 

[Kea]

entanglement

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 no one understands it, but
this is a pretty good start.

 

[TheDonk]

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]

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]

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]

reply

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 doesn't 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

 

[reilly]

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]

tx

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 can't 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 couldn't 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 couldn't 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?

 

[alexepascual]

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]

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]

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 separate 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?

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]

thanks guys, yeah let's 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 don't people at least try it, if they are theyre doing small scale experiments, thinking small, let's 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 can't 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]

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]

(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]

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]

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

 

[alexepascual]

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 a_k|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]

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]

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]

Experimental facts

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's_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 realized: 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]

to Richard, back in post #60

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]

to Alex

... 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 a_k|ψ>|φ>_k?

In the above, the vectors |ψ_k> and |φ_k> 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> → Σ_k a_k|ψ_k>|φ_k> ,

where each a_k ≠ 0, and there are at least two distinct values for k (and, of course, the |ψ_k> (|φ_k>) 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> → Σ_k a_k|ψ_k>|φ_k> ,

I was working in the Schrödinger picture. The left-hand-side is at some initial time t_o, 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 a_i(t)|ψ_i>

and

|φ(t)> = ∑_j b_j(t)|φ_j>

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

Thus,

|ζ(t)> = |ψ(t)>|φ(t)>

= ∑_ij a_i(t)b_j(t) |ψ_i>|φ_j> .

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

|ψ_n><ψ_n| (x) 1 ,

and therefore the required probability is given by

<ζ(t)| |ψ_n><ψ_n| (x) 1 |ζ(t)>

= <ψ(t)|ψ_n><ψ_n|ψ(t)> ∙ <φ(t)|φ(t)>

= |a_n(t)|² .

... Does this clear up the confusion?

 

[Eye_in_the_Sky]

to DrChinese

Back in post #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]

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]

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]

"Two particlees 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