Schrödinger local and deterministic?

[DrChinese]

i know a model with spin 1/2 particles (but not MWI)
http://arxiv.org/PS_cache/quant-ph/pdf/9505/9505025v1.pdf

....A parametrized model, "Q", for the state vector evolution of spin-1/2 particles during measurement is developed...is local, deterministic, nonlinear and time asymmetric...that Q is not constrained by Bell’s inequality, locality and determinism notwithstanding.

Another meaningless reference. These are a dime a dozen. As Pio2001 says, this is actually anti-realistic and rests on semantic interpretation.

As with all candidate models: show me the realistic dataset! If you don't have one, please go back to the drawing board.

 

[Pio2001]

what is a realistic object ?

Among other things, it is an object that obeys physics' laws. For example, its position can't change faster than the speed of light. Or its properties can't be affcted by what is done outside its past light-cone.

 

[Frame Dragger]

Right ! irrelevant refutations or endorsing or spurious justifications...
or things like "is an old study, outdated" (Special Relativity dates back to 1905 ! so ?)

or the guys say solemnly "SHEER VOLUME OF EVIDENCE !", "is NOT generally accepted", "In probably 500+ papers in the past year alone" (ah ! then truth is matter of the number of papers, more papers, more truth), "ALL (and I mean 100%) of the experiments !", "I don't see the issue as being relevant"

hell ! they are cheer leaders or parrots ?

Wow, I take it someone pissed in your oatmeal this morning? Really yodajedi:

1.) WHO is "They"?
2.) What theory is it that you're pushing exactly?
3.) What part of the forum guidelines is hard to grasp?

As DrChinese has said, you're free to say what you want, but not everywhere you want to. If people such as Galileo can get their theories out in the face of politics and the (then) church, a patent clerk can produce E=MC^2, a quiet brit can produce (a LOT) of QM (Dirac), and a man trapped in his own body can share theories... you can manage.

Maybe the issue is the theory, and not the vehicle, or the "Them" those in the tinfoil deflector beanie crowd seem to blame for all of their ills.

As for what a realistic object is, most times that people use the term "Realism", they're speaking in terms of what Pio is saying, or more generally the Realism of EPR.

 

[Gerenuk]

Hey, anyone of you experts mind answering my last question #53? ;)

 

[Frame Dragger]

Hey, anyone of you experts mind answering my last question #53? ;)

Wouldn't it just be the usual time-dependant equation? a la http://www.britannica.com/EBchecked...ics/77513/Time-dependent-Schrodinger-equation

?

 

[Gerenuk]

I'm sure about the full form of the wavefunction which is connected to that singlet spin state.

So far it has been useful textbook quotes and references only, but no self-made thinking. Maybe someone can help me figure out the full wavefunction. Something like a plane wave position part and a time-dependent part which is connected to some energy? I mean really a full solution at least for the initial state
\Psi=?

It's important to make own thoughts. I once asked theoretical QM physics experts, what is the equivalent solution to singlets/triplet, but for three spins. They didn't have a clue and were even perplexed when I finally showed them the solution (no antisymmetric state and also you need an additional quantum number to distinguish degenerate states). So it seems its important to stroll around beyond textbook stuff.

PS: still trying to unpack the information in Pio's post. Maybe I miss some knowledge :)

 

[Frame Dragger]

I'm sure about the full form of the wavefunction which is connected to that singlet spin state.

So far it has been useful textbook quotes and references only, but no self-made thinking. Maybe someone can help me figure out the full wavefunction. Something like a plane wave position part and a time-dependent part which is connected to some energy? I mean really a full solution at least for the initial state
\Psi=?

It's important to make own thoughts. I once asked theoretical QM physics experts, what is the equivalent solution to singlets/triplet, but for three spins. They didn't have a clue and were even perplexed when I finally showed them the solution (no antisymmetric state and also you need an additional quantum number to distinguish degenerate states). So it seems its important to stroll around beyond textbook stuff.

PS: still trying to unpack the information in Pio's post. Maybe I miss some knowledge :)

Hmmm... that's beyond me frankly, but I wish you luck! Like you I'm still wading through Pio's posts (which are great btw, thanks Pio!). The amount of time I spend with reference books is truly sad.

Edit: This might be useful... https://www.physicsforums.com/showthread.php?t=147650

 

[ManyNames]

My posts where deleted for some reason.

Does this happen a lot round here?

 

[Pio2001]

The forum went offline for hours yesterday. It seems that they lost some posts. Mine are gone too.

 

[Frame Dragger]

Actually, I believe staff cleaned up the thread before the server had downtime.

 

[Gerenuk]

I know that you already know a lot about this, and I don't want to write you the stuff that you already know. Thus, you would help me if you could specify what EXACTLY you want me to write down mathematically.

Well, define some notion of non-locality and show how the wave-function evolution is non-local. So I think something similar to Pio's post. Just just still haven't followed all the indices and sometime I'm not sure what the next line means.
So show that anything at x changes, if I modify something at x' for a space-like interval. It should contain x and x', because without it, there could be uncontrolled simplifications.

 

[Gerenuk]

The non locality is mathematically represented by the fact that the amplitude of Alice's state vectors depends on \beta, and the amplitude of Bob's state vectors depends on \alpha, in a non-separable way, while \alpha was chosen in a space-time region spatially separated from Bob, as he is represented here, and \beta in a space-time region spatially separated from Alice, as she is represented here.

Are you saying that the wave-vector depends on the measuring device of the other person? I guess the actual probabilities won't depend this way, right? (otherwise it would be very easy to check for non-locality experimentally)

And is there a representation which doesn't have the phase freedom? Would it work with density matrices?

 

[Pio2001]

Actually, I believe staff cleaned up the thread before the server had downtime.

Yes, I saw about 4 messages cleaned up, with Yoda's smiley and DrChinese asking if he was making fun of him.
But then, some posts were lost during the technical operation, as reported here : https://www.physicsforums.com/showthread.php?t=389386

In these posts, Gerenuk asked if it was possible to write a complete wave function, with position and time coordinates.
I answered that the wavefunction depended a lot on the actual experiment, for example the speed of the particles.
Then I proposed a model where the particles position was very well know at all times, so that we could represent their wave function as eigenstates of the position : let's imagine that the particles travel nearly at the speed of light along the Ox axis between t0 and t1.
At t1, they reach the measurment device, and turn one direction or the other according to their spin.
They then go out of the device, still at the same speed, until at t2, they reach a screen where they are destroyed and leave a visible mark.
Between t1 and t2, you can represent their position with a superposition of two eigenstates of the position, entangled with the representation of their spin in the alpha or beta basis.
Experiment of recombination at the output of Stern-Gerlach devices have shown that the wave functions stay coherent long enough.
After t2, they are destroyed, so you just have to remove the kets that represent them in the final expression I gave, and keep only the kets representing the devices and the observers.

However, I don't think that this will bring something relevant to the debate. The spin wave function is enough to predict experimental results with accuracy, and it goes against local hidden variables.

Just just still haven't followed all the indices and sometime I'm not sure what the next line means.

Don't hesitate to ask. I sometimes went fast in my explanations.

Are you saying that the wave-vector depends on the measuring device of the other person?

What wave vector ? I took the omniscient point of view, and the initial state is such that I can't define a wave vector for one side only, that contains all the necessary information to predict the experimental results.

The global wave vector depends on both measuring devices after time t2.
Bewteen t1 and t2, it is also the case, but we would have to add the particles position in the equation to show why.
Before t1, the wave vector doesn't depend on the devices.

Are you saying that the wave-vector depends on the measuring device of the other person? I guess the actual probabilities won't depend this way, right? (otherwise it would be very easy to check for non-locality experimentally)

What probabilities ? The probabilities to get +\hbar/2 or -\hbar/2 on side i (i=1 or 2) are always equal to 50 %. You can't check for non locality this way.

However, the probability to get a given pair of result among the four possible pairs (++, +-, -+ or --) depends on both alpha (Alice's angle) and beta (Bob's angle), but you can't check it locally. You have to meet the other observer in order to know the result.

The mystery is that the results are already written down into macroscopic objects before the observers meet.

And is there a representation which doesn't have the phase freedom? Would it work with density matrices?

What do you mean ? If I change the relative phases in my wave function, my result in no more the same.

Would it work with density matrices?

With one density matrix, expressed in the (++, +-, -+, --) basis, it should work, but I think it would be more difficult to see the non-locality and anti-realism of the process.

You can also write the wave vectors I gave into matrices. It stays realist, and cleans up the heavy equations, but the non-locality gets difficult to see all the same.

 

[ManyNames]

Dr Chinese, i hope you can remember our discussion, so i shall continue despite the apparent technical difficulty.

As far as i am aware of the general physical side of the HUP, it remains true that statistical averages all contain an inherent uncertainty relative to our perspective of the experiment. There is no reason, like you suggested, to disallow the notion that what seems to be a level of uncertainty may not be indeterministic externally to our measurements. In effect, determinism in the Uncertainty Principle might seem contradictory, but it truly does lye down to how we perform our measurements and how much information we ascertain from it.

There is actually a lot of evidence to support it - models of physics which cannot escape some kind of predeterminism within the structure of spacetime, and the wave function itself.

Trivially, for instance, as i have noted before in the past, evolution of states (and perhaps even down to the evolution of the observable measurements of the HUP, systems evolve accordingly as:

\hat{H}|\psi>=i \hbar \partial_{t}|\psi>

In the mathematics of quantum mechanics, the Hamiltonian operator is self-adjoint so it's diagonalisable and all its eigenvalues are real. There is always atleast one family of orthogonal states |\phi_n> that span the state space:

\hat{H}|\psi_n>=E_n|\phi_n>

and the state |\phi_n> evolves as:

|\phi_n(t)>=e^{-i \omega_{n}t|\phi_n>}

These are called time-dependant evolutions of the schrodinger equation. But taking diffeomorphism invariants as solutions to General Relativity to account for energy as a whole in the universe, we find no energy changing - no relativistic clocks of motion - time essentially ceases to exist.

The Wheeler de-Witt eq. could make trivial statements such as nothing ever changes - or things exist but do not change on a cosmological scale. For something to be so defined must be something deterministic. Again, it's being about us; our role inside of the universe where apparently the ''illusion'' of consciousness can bring about a sense of detachment - one where we concurrently believe to be within our state of control. But if there is a lack of knowledge in quantum systems, then there is no rule suggesting that the entire wave function cannot be driven by pilot waves.

 

[Pio2001]

oh, and in the lost messages, Gerenuk was saying that he was trying to get through my calculus. I answered that if something was not clear, don't hesitate to ask. Sometimes I went fast with my explanations.

 

[Frame Dragger]

Dr Chinese, i hope you can remember our discussion, so i shall continue despite the apparent technical difficulty.

As far as i am aware of the general physical side of the HUP, it remains true that statistical averages all contain an inherent uncertainty relative to our perspective of the experiment. There is no reason, like you suggested, to disallow the notion that what seems to be a level of uncertainty may not be indeterministic externally to our measurements. In effect, determinism in the Uncertainty Principle might seem contradictory, but it truly does lye down to how we perform our measurements and how much information we ascertain from it.
...
...
These are called time-dependant evolutions of the schrodinger equation. But taking diffeomorphism invariants as solutions to General Relativity to account for energy as a whole in the universe, we find no energy changing - no relativistic clocks of motion - time essentially ceases to exist.

The Wheeler de-Witt eq. could make trivial statements such as nothing ever changes - or things exist but do not change on a cosmological scale. For something to be so defined must be something deterministic. Again, it's being about us; our role inside of the universe where apparently the ''illusion'' of consciousness can bring about a sense of detachment - one where we concurrently believe to be within our state of control. But if there is a lack of knowledge in quantum systems, then there is no rule suggesting that the entire wave function cannot be driven by pilot waves.

Ok, so you're a Bohmian, kind of... Hmm. Frankly, you seem to be missing the point that the HUP is fairly literally reflected in measurements of the CMB. The HUP does NOT merely place a limit on the amount of information available to US, but the nature and behaviour of nature.

 

[ManyNames]

Best part was, was that i was infractioned in my post saying it was a generally (academic resolution) over the ZPF. This is not the truth, and if a consensus was performed the ZPF would still be found in greater favour.

 

[ManyNames]

Ok, so you're a Bohmian, kind of... Hmm. Frankly, you seem to be missing the point that the HUP is fairly literally reflected in measurements of the CMB. The HUP does NOT merely place a limit on the amount of information available to US, but the nature and behaviour of nature.

No because, according to many here, Bell's in-equa. presents a non-local universe - quantum entanglement, desolving the idea that it cannot be local*. However, the universe AND I STRESS THIS PART, has a secondary energy state cosmologically where low energy states must correspond or even proportional to local events within the universe.

*Just because a vacuum can give non-local events, yet equally present a local event, must be an indication that both are somehow complimentary, by the Copenhagen Intereprtation; meaning that the cosmological model has non specfic favour for either local or non-local, but perhaps both inexorably simultaneously. Non-locality would pertain to no geometry:

''In a standard course of geometry, you learn pythagora's theorem. It relates the lengths of the side of a right triangle, where if we take the sqaure of two sides of the triangle and their sum, they will equal the same as the remaining length. The three dimensional nature of the universe obides bythis simple and conscise rule. Now, since Phythagora's theorem, it has been applied into the geometry which is most commonly associated with Minkowskian Spacetime. It is a four dimensionsional vector space, with one imaginary quantity which is time. Time then becomes under the description as the imaginary space dimension. This imaginary dimension, which is not meaning it is ethereal in anyway without proof, takes off this spacetime triangle making the four-dimensional manifold of spacetime what it is, and what unifying them means. The math which described this was a new geometry:

s^2=-(c\Delta t)^2+(\Delta x)^2+(\Delta y)^2+(\Delta z)^2

This equation is a Cartesian Coordinate of spacetime. In a Minkowskian Row Vector Notation in a bilinear form can be given as: V=(0,0,0,1). The Row Value of the Matrix is given as:

\eta=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}

This makes a smooth manifold consistent of time and space as single entities. This is why time cannot disappear from some small square unit of space for it is a universal invariant. The presence of space according to the new physics generalized this to mean that it also included the appearance of time. Such an example was the big bang itself, it was not just the beginning of primal space, but also of primal time. ''

Thus, non-locality is when super-high potentials can exist for a form of quantization (aka. quantum graphity), where non-locality seems to only exist in high valus.

 

[Gerenuk]

In these posts, Gerenuk asked if it was possible to write a complete wave function, with position and time coordinates.
I answered that the wavefunction depended a lot on the actual experiment, for example the speed of the particles.

Hmm, that's not very satisfactory :) I'm fine with all simplifications you can think of, but still prefer an answer rather than "it depends" ;) OK, here is my first own guess
\Psi=\exp(i(kx_1-k^2t/2m))\exp(i(-kx_2-k^2t/2m))(|+->-|-+>)

Don't hesitate to ask. I sometimes went fast in my explanations.

They are fine I think. It was just a bit frustrating to me to dig through the conventions at first. Maybe some more concise and consistent notation would be easier. But I think I get the answer now :)

What wave vector ? I took the omniscient point of view, and the initial state is such that I can't define a wave vector for one side only, that contains all the necessary information to predict the experimental results.

I actually meant to full wavevector, but now I see that the concept of "depends on other observer" is ill-defined if you cannot separate the wavefunction.

What probabilities ? The probabilities to get +\hbar/2 or -\hbar/2 on side i (i=1 or 2) are always equal to 50 %. You can't check for non locality this way.

I was imagining that the wavefunction is nicely local, but the procedure of finding <\Psi|\Psi> causes all the trouble.

The mystery is that the results are already written down into macroscopic objects before the observers meet.

and the results cannot be written individually, but the writing depends on what will be in the future, right?

What do you mean ? If I change the relative phases in my wave function, my result in no more the same.

Yes, and I thought it would be more natural to find a representation where there is no arbitrary phase possible, but a state is unique.

 

[Frame Dragger]

@Gerenuk: Your second to last question really hits one of the bullseyes of "quantum weirdness"... or the appearance of it. You might enjoy some reading of Delayed Choice Quantum Eraser experiments. While it seems to be the case that in a seemingly Realistic (of EPR realism) world we experience macrocosmically time has an arrow, it really is not clear that such is the case in the (quantum) microcosm. I don't believe there is an answer right now, although I like DrChinese's view of DCQE, which is that it isn't as weird as it seems to be. The "Delay" in the measurement may be the only real delay, but then, some people argue for atemporal collapse, Retarded and Advancing waves in the Transactional Interpretation, etc. The issue seems to be that the behaviour in a lab is as yet inexplicable without observational data... and that would seem to be difficult.

As for the wavefunction being local, wouldn't that make it something OTHER than \Psi or the concept of a continuous probablistic function? If not, I'd love it if someone explained that to me, becuase clearly I'm missing something (more).

 

[Pio2001]

here is my first own guess
\Psi=\exp(i(kx_1-k^2t/2m))\exp(i(-kx_2-k^2t/2m))(|+->-|-+>)

You're missing the gaussian shape that allow the particles to be localized. According to your expression, the mean amplitude is the same everywhere in the universe.

http://en.wikipedia.org/wiki/User:Brews_ohare/Brews_ohare/Wavepacket

and the results cannot be written individually, but the writing depends on what will be in the future, right?

Something like that. But the writing is also essentially random.

 

[Gerenuk]

You're missing the gaussian shape that allow the particles to be localized. According to your expression, the mean amplitude is the same everywhere in the universe.

Feel free to correct it ;-) I'm also not sure if the spatial part and the spin should be separated. But basically I was asking for a straight answer with an equation similar to this.

 

[Pio2001]

I'm not sure what you are trying to do with such a wave function. After position, we'll have also to add energy, electric charge, and maybe other things (leptonic number ? Strangeness ?...).
We 'll have to introduce at least diagonal trajectories at the output of the detectors, if not curved ones, all put into equations. That's going to be a lot of work.

What's the goal ? Predict something else ?

 

[Gerenuk]

I really appreciate your earlier explanation and I learned something new from it!

But it's incomprehensible to me why you are always dodging my last question.
You either say "I think your question is uninteresting" (no! my question is interesting to me) or you say "Your question is not what you want to know. I know better what you really want to know and for that the answer is too difficult" (no! I know what my question is and it should have spin, position and time only; no need for something else). See it this way: if it were an exam question you would get no marks.

If you think Gaussian is something nice to add then OK. Now I can figure it out myself, I think, if I find out a 3D Gaussian. Thanks for the suggestion.

Not sure if you were aware of it, but when someone is only "questioning the question" this is usually because he doesn't want the uncomfortable feeling of not knowing the answer.

 

[Frame Dragger]

I'll freely admit that I'm out of my depth now, but it strikes me that Pio outlines a good reason not to make this equation: it would be a lot of work, and for what? You may be right Gerenuk, and he's dodging you, but maybe if you tell him why you want that work done he'd be more motivated?

 

[Gerenuk]

I don't know if spin variables and position should be mixed. To me it's not clear whether I shouldn't use:
\phi_1(x_1)\phi_2(x_2)|+->+\phi_2(x_1)\phi_1(x_2)|-+>
Also I expected that it would be easy to write down the equation?! I mean if it's only Gaussian shape that's missing, then I thought experts could easily quote it.

And not all questions must be from textbooks. I find it interesting and it could be a new-mini-thread. Isn't one allowed to ask questions without proofing that they will lead to the TOE?

Earlier I wasn't addressing Pio alone. I just wasn't very pleased that self-announced experts hijacked the thread to throw textbook passages at each other, but didn't proof their knowledge but giving solution to something that is the same level, but just not written in their textbooks.

 

[Pio2001]

You either say "I think your question is uninteresting" (no! my question is interesting to me) or you say "Your question is not what you want to know. I know better what you really want to know and for that the answer is too difficult"

I certainly won't say that I know better than you. I'm just trying to help the discussion advance, proposing solutions in order to work fast.

(no! I know what my question is and it should have spin, position and time only; no need for something else). See it this way: if it were an exam question you would get no marks.

I'm not trying to get some marks. I'm trying to stay on-topic.
The topic is about non-locality and determinism. You asked where non-local processes were showing up in the wave function formalism. I gave an answer.
You want to go further and write a wave function about spin and position. OK, I have nothing against it.

I mean if it's only Gaussian shape that's missing, then I thought experts could easily quote it

Sure. But I'm no expert. I studied physics for four years, then I became programmer. I've never written or see written a spin-position wave function in my life. I'm searching together with you. I suggested that a gaussian part was missing after having read the wikipedia article about wave packets. I had merely any idea about this before you asked the question.
That's why I may seem a bit reluctant : finding your wave function would be at least as much work for me as it is for you.

The EPR experiment was the subject I chose for the personal research in my 4th year, and I've been interested in it since then. That's why I'm still capable of writing entangled wave functions and understand Bell's theorem.
I was incredibly lucky to get my hand on a paper written by John Bell where he summarizes the CHSH inequality while I was messing around in a restricted area of the university library, where 4th year students are not supposed to go. (Bell's original paper was archived and available only on request by 5th years students and above). That helped me a lot understanding the local hidden variables problem.

I don't know if spin variables and position should be mixed. To me it's not clear whether I shouldn't use:
\phi_1(x_1)\phi_2(x_2)|+->+\phi_2(x_1)\phi_1(x_2)|-+>

Let's do it step by step.
First, are we shure that we can multiply a spin ket with a position ket ? Measurments can give us a position, and / or a spin. So one correct representation at least would be using a state space whose basis would include eigenstates of the position (an R3-dimention vectorial space), in addition with spin values (two more dimentions). A ket from this space would look like
|x,\, y, \,z, \,spin\rangle
I'm not knowledgeable enough about maths to state if this can be dealt as
|x,\, y, \,z\rangle \, |spin\rangle
And what kind of product operation should be between (tensorial product ?).
So let's see if someone else can help us with this.

Then, let's simplify the problem as much as we can. I suggest not to write the whole equation of the movement, but to keep it as f(t), f being a function from R to R3, that associates a position (x, y, z) to a time (t).
We know that the two particles move in opposite directions, towards the devices, then that their trajectory is deviated according to their spins.
Let's define f_i(t) the function giving the position of particle i between t0 and t1, when it goes from the source to the device, then g_i(+, \phi, t) and g_i(-, \phi, t) the trajectories of the particles between t1 and t2 according respectively to the projection of their spin along the \phi axis (+ or -), the angle \phi between Oz and the detector, and the time t.
The position of Alice's particle will be then given by

|f_1(t)\rangle between t0 and t1, then by

|g_1(+, \alpha, t)\rangle - |g_1(-, \alpha, t)\rangle between t1 and t2 (because its spin is |+\rangle - |-\rangle for any angle).

Then by nothing after t2, since the particle is destroyed when it hits the screen.

I'm not sure if a phase must be present in front of the position kets.

 

[Pio2001]

Here. That's not a good drawing, and I certainly have messed up the signums and the orientations of the devices, but the attached picture should clarify the position functions.

I don't think that it will bring something new because between t0 and t1, the f functions are completely independant from the spin, and between t1 and t2, the g functions are trivially entagled with spins.

Thus, if we want to write the position wave function between t1 and t2, it will be exactly the same as the one with spins, but with

|g_1(+, \alpha, t)\rangle_1 instead of |+\alpha\rangle_1

|g_1(-, \alpha, t)\rangle_1 instead of |-\alpha\rangle_1

|g_2(+, \beta, t)\rangle_2 instead of |+\beta\rangle_2

|g_2(-, \beta, t)\rangle_2 instead of |-\beta\rangle_2

Which is nothing else than the fact that you don't actually measure the particle's spin, but the position of the particle at the output of the spin detector, translated into mathematical terms.

The complete position wave function between t1 and t2 is thus

<br /> \frac{1}{\sqrt{2}} <br /> (\,<br /> <br /> sin{\frac{\beta-\alpha}{2}} <br /> ({|g_1(+, \alpha, t)\rangle_1} <br /> \otimes <br /> {|g_2(+, \beta, t)\rangle_2} ) <br />

<br /> +<br /> \,<br /> cos{\frac{\beta-\alpha}{2}}<br /> ({g_1(+, \alpha, t)\rangle_1} <br /> \otimes <br /> {|g_2(-, \beta, t)\rangle_2} )<br />

<br /> - <br /> \,<br /> cos{\frac{\beta-\alpha}{2}}<br /> ({|g_1(-, \alpha, t)\rangle_1} <br /> \otimes <br /> {|g_2(+, \beta, t)\rangle_2} )<br />

<br /> +<br /> \,<br /> sin{\frac{\beta-\alpha}{2}}<br /> ({|g_1(-, \alpha, t)\rangle_1} <br /> \otimes <br /> {|g_2(-, \beta, t)\rangle_2})<br /> <br /> )<br />

I still don't know exactly how to write the spin together with it (maybe just adding the spins kets beside the position kets), but you can see that the mathematical form is exactly the same.

Alice and Bob are no more measuring Bell's inequality from the spins of their particles, but from the way their particles got out of their respective spin detectors... which is exactly the same thing.
Of course, they write exactly the same thing on their notebooks (the words "up" or "down"), and Bell's inequality is violated exactly the same way because the amplitudes of each pair of terms are the same as before, leading to the same prediction.

 

[eaglelake]

I believe you are correct - the spin state and position state is written as a juxtaposition.

The point is that the wave function depends 1) on state preparation as well as 2) the representation, which is determined by the observable(s) to be measured. In fact, it need not be a "wave function" at all. It can be as simple as <br /> \psi = {1 \over {\sqrt {b - a} }}<br />, a &lt; x &lt; b

when we know it is equally probable that the particle will be found at any location x in the given interval.

The comments made at the beginning of http://arxiv.org/pdf/quant-ph/0703126 are interesting.

 

[Pio2001]

Thanks for the link. That's a good example of what Gerenuk would like to do.

In my simplified version above, I completely discarded the wave nature of my particles. An acceptable simplification, since in this experiment, their wave-like behaviour does not show up.

 

[yoda jedi]

Hmm, that question seems really interesting to me:
What is the full (position and time dependent) wave-function of that singlet state, where the particles fly apart?

...in the presently known formulation of quantum theory (SQM): the unsatisfactory presence of an external classical time in the formulation......The entangled state acts as the initial state, which is to be evolved further by the nonlinear equation. The non-linearity breaks the superposition, and only one out of the two outcomes is realized; as a result the electron goes either through the upper slit, or the lower one, but not both.
But we have still not answered a crucial question : what decides which slit the electron will go through?! The answer lies in the nature of the nonlinear terms in the Schrodinger equation. These terms contain the phase of the state. It can be shown that depending on the value the phase takes at the onset of measurement, one out of the two superposed states grows exponentially with time, while the other one damps exponentially. Now, since repeated measurements are made at random times, the phase is effectively a random variable. Thus the outcome is random. It can be shown that if an appropriate probability distribution is associated with the random phase, the outcome of the
quantum measurement obeys the Born probability rule. That is, the probability of any particular outcome being realized is proportional to the square of the amplitude for the electron to be in the corresponding state. This analysis is easily generalized to the situation when the system is in a superposition of more than two states - depending on the value of the initial phase, one out of the many states grows exponentially, while all others are damped......
...The upshot is that, as a result of the non-linearity, quantum mechanics (NLQM) is a deterministic random theory. The outcome of a measurement is determined by the value of the phase, and the randomness of the phase leads to different outcomes, in consistency with the Born rule. Probabilities are dispensed with, once and for all, and furthermore, one no longer needs to invoke the untestable ‘many words interpretation’ to ‘hide’ the superpositions which the process of decoherence inherently preserves.....

 

[Pio2001]

Except that such a hidden-variables model necessarily violates special relativity.

 

[yoda jedi]

Except that such a hidden-variables model necessarily violates special relativity.

wrong, unifies quantum theory and relavity, is nonlocal and so on.
you have to read and research more... very ignorant of your part...

 

[DrChinese]

very ignorant of your part...

That was unnecessarily rude.

I realize that you believe your argument to be strong. But actually they are just some disjointed statements. If you want to make that argument (hidden variables are required to be nonlocal), you MUST address Bell. That was what Pio2001 was doing. Without that, how do you expect to be taken seriously?

 

[Frame Dragger]

That was unnecessarily rude.

I realize that you believe your argument to be strong. But actually they are just some disjointed statements. If you want to make that argument (hidden variables are required to be nonlocal), you MUST address Bell. That was what Pio2001 was doing. Without that, how do you expect to be taken seriously?

It seems his strategy is broken sentences and acting like a prick. Not the best way to prove your beliefs, but then, who cares?

By the way, if you're going to insult Pio for no damned reason, at least get it right, "very ignorant ON your part"... not of. You don't need to actually type like Yoda talked...