Entaglement and hidden variables

[Demystifier]

Maybe I am missing something, but isn't equation (42) just the Schmidt decomposition?

http://en.wikipedia.org/wiki/Schmidt_decomposition

Yes, Eq. (42) is the Schmidt decomposition.

 

[Demystifier]

Since in real measurements the eigenfunctions of the electron momentum are not eigenfunctions of the interaction hamiltonian, the value of c_q(t=0) is different from c_q(t), which means that the two probabilities are different.
In other words, the de-broglie bohm theory does not reproduce the standard quantum mechanics results (as it is said also in eq. 39). Since experiments confirm standard quantum mechanics, this is again sufficent to prove that both "quantum theory of measurement" and hidden variable theories are wrong.

This, of course, is wrong, but if you are convinced that it is right, don't waste your time with us. Publish your new important result in a respectable journal, which will make you famous; you will destroy Bohmian interpretation, many-world interpretation, and theory of decoherence at once.

 

[andreab1987]

Yes, Eq. (42) is the Schmidt decomposition.

It evident that you do not understand what you are writing, since in the appendix it is written explicitly that the eigenvectors are eigenfunctions of the operator A; hence eq. 42 cannot be the Schmidt decomposition.

 

[andreab1987]

I gave you some arguments, but you ignored them.

Let me also note that Eq. (42) plays an important role in decoherence theory, which, by the way, is an observational fact. If you don't know about decoherence, see e.g.
http://xxx.lanl.gov/pdf/quant-ph/0312059 [Rev.Mod.Phys.76:1267-1305,2004]
which is published in a VERY respectable journal, and has MANY citations. Pay particular attention to Eq. (2.1).

You have given no arguments before and you are given no arguments now. You are only quoting some sources. It is well known that there are many people who do not accept standard quantum theory, in spite of its extraordinary experiemental evidences; these peoplea try to develope alternative theories, but this does not prove that their theories are correct.
Actually, from what I have seen, they are certainly wrong.

 

[andreab1987]

But in (42) v does not need to be the wave function of the electron before the measurement. Instead, v_i are BASIS states in which the wave function of the electron before the measurement can be expanded.

Nonsense

 

[andreab1987]

This, of course, is wrong, but if you are convinced that it is right, don't waste your time with us. Publish your new important result in a respectable journal, which will make you famous; you will destroy Bohmian interpretation, many-world interpretation, and theory of decoherence at once.

You seem not to understand that pratically all physicists use only standard quantum mechanics and consider hidden variables, bohemian mechanics, dechoerence, many worlds only as phylosofical speculations, and not as scientific theories.
I came here with sincere interest to understand if there was something good in hidden variables theories, but I have understood that such theories are totally inconsistent and based on serious mistakes.

 

[Demystifier]

You seem not to understand that pratically all physicists use only standard quantum mechanics and consider hidden variables, bohemian mechanics, dechoerence, many worlds only as phylosofical speculations, and not as scientific theories.

Decoherence is an experimental fact!

 

[comote]

No it isn't.

In fact Schmidt theorem says that:

For any vector v in the tensor product , there exist orthonormal sets u and v ...

this means that the two orthonormal sets are in general different for any different vector v; in other words they depend on the choice of v, but there are no fixed orthonormal sets which can be used for every vector v.

In the case of eq. 42 the orthonormal sets is chosen with eigenvectors of the operator A corresponding to the quantity to be measured; so this orhonormal sets does not depend on the vector v, i.e. the wave function of the electron.

I see what you are saying, but I don't see where he is doing that. One could do the decomposition for each time and then normalize it at each time. I am not a fan of Bohmian Mechanics but I do want to understand where exactly it differs from standard QM.

 

[andreab1987]

I see what you are saying, but I don't see where he is doing that. One could do the decomposition for each time and then normalize it at each time. I am not a fan of Bohmian Mechanics but I do want to understand where exactly it differs from standard QM.

Please read the line just after eq. 40. The author says that psi_a are a complete normalized eigenfuntions of the operator A.
You must understand that in order to compare the probability distribution with the one of standard quantum mechanics, he must use the same basis of eigenvectors of the operator corresponding to quantity to be measured, forr example the electron momentum. If you change the basis, then your c_n does not represent any longer the probability for the electron to have a given value of momentum.
The point is that in eq. 39 the author says correctly that the bohemian distribution and the standard quantum distribution for momentum are different; then he tries to deny this, using eq. 40 and eq. 42 (see the statement on the sixth line after eq 42).

Bohemian mechanics is different from the standard because there are some new equations for the particles coordinates, in addictions to the standard quantum equations (see for example eq. 38)

 

[comote]

OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?

 

[yoda jedi]

You seem not to understand that pratically all physicists use only standard quantum mechanics and consider hidden variables, bohemian mechanics,
dechoerence,
many worlds only as phylosofical speculations, and not as scientific theories.
I came here with sincere interest to understand if there was something good in hidden variables theories, but I have understood that such theories are totally inconsistent and based on serious mistakes.

GOD !







-------------
Interesting...
bohemian mechanics

...lol... bohemian mechanics :rofl:

 

[andreab1987]

OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?

Correct

 

[Demystifier]

OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?

This is certainly an error, but whose error? Certainly not of the author of Eqs. (40) and (42), because there is no claim in the paper that they are the same.

 

[Demystifier]

Interesting...
bohemian mechanics

...lol... bohemian mechanics :rofl:

Have you seen the lirics after the Abstract of
http://xxx.lanl.gov/abs/physics/0702069 [Am.J.Phys.76:143-146,2008] ?

 

[Demystifier]

I am not a fan of Bohmian Mechanics but I do want to understand where exactly it differs from standard QM.

Fair enough! ALL equations of standard QM are also equations of BM. In particular, whatever andreab may say, (42) IS an equation of standard QM. (In fact, decoherence can be thought of as an indirect EXPERIMENTAL proof that (42) is correct.)

The difference is that BM denies the collapse of the wave function (which is postulated, but not described by any EQUATION in standard QM). Instead, to explain why collapse SEEMS to occur, BM postulates one ADDITIONAL EQUATION; the equation for particle trajectories. In terms of equations in the discussed Appendix, Eqs. (40)-(45) are common to both standard QM and BM, while the new equation of BM is Eq. (38).

So in essence, BM is an alternative formulation of QM in which the collapse postulate (which does not have a form of an equation) is replaced by another postulate (which does have a form of an equation). Due to the correctness of (42), the two formulations have equal measurable predictions.

 

[andreab1987]

This is certainly an error, but whose error? Certainly not of the author of Eqs. (40) and (42), because there is no claim in the paper that they are the same.

Actually the author claims they are the same; in fact read the first line after eq. 41
"According to standard QM, the probability of finding the state to have the value a of the observable Aˆ is equal to ..."

and then read the 5th line after eq. 42

"The probability for this to happen is, according to (42), ...

In fact, during a mathematical proof, you cannot change the meaning of the same symbols or functions.
By the way, if psi_a were not the same in eq. 40 and 42, the autor's proof would makes no sense at all.

 

[Demystifier]

Actually the author claims they are the same; in fact read the first line after eq. 41
"According to standard QM, the probability of finding the state to have the value a of the observable Aˆ is equal to ..."

and then read the 5th line after eq. 42

"The probability for this to happen is, according to (42), ...

In fact, during a mathematical proof, you cannot change the meaning of the same symbols or functions.
By the way, if psi_a were not the same in eq. 40 and 42, the autor's proof would makes no sense at all.

Jesus, I cannot believe that I have to explain this, but ...
the author uses the fact that c_a(t) and psi_a(x) are the same functions in both (40) and (42). Yet, he does not say that psi(x,t) in (40) is equal to Psi(x,y,t) in (42). Note the dependence on y in (42) absent in (40)!

 

[andreab1987]

Jesus, I cannot believe that I have to explain this, but ...
the author uses the fact that c_a(t) and psi_a(x) are the same functions in both (40) and (42). Yet, he does not say that psi(x,t) in (40) is equal to Psi(x,y,t) in (42). Note the dependence on y in (42) absent in (40)!

But this has nothing to do with my previous statement.
My point is that the psi_a are the same functions both in (40) and in (42), and these psi_a are defined as the eigenfunctions of the operator A.

In order to apply the schmidt decomposition for Psi(x,y,t) you must choose a different basis both for the x-space and the y-space, and therefore you cannot write eq. 42 using the psi_a of equation 40. Every student in Mathematics or Physics of the first year can explain this to you.

 

[comote]

Jesus, I cannot believe that I have to explain this, but ...
the author uses the fact that c_a(t) and psi_a(x) are the same functions in both (40) and (42). Yet, he does not say that psi(x,t) in (40) is equal to Psi(x,y,t) in (42). Note the dependence on y in (42) absent in (40)!

If you insist that the $\psi_a(x)$ in equations (40) and (42) are the same then you can not say that the $c_a(t)\chi_a(y)$ are orthogonal, likewise if you insist that $\chi_a(y)$ are orthonormal then you can't say that the $\psi_a(x)$ in (40) and (42) are the same.

 

[Demystifier]

If you insist that the $\psi_a(x)$ in equations (40) and (42) are the same

Yes, I insist on that ...

then you can not say that the $c_a(t)\chi_a(y)$ are orthogonal

For a general interaction between two subsystems, you are right that $\chi_a(y)$ do not need to be mutually orthogonal. But is it possible that, for some SPECIAL interaction, $\chi_a(y)$ turn out to BE orthogonal? I hope you can agree that it is possible.

If you can agree with that, then the point is that a MEASUREMENT of A, by definition, is nothing but such a special interaction. In general, for interactions which do not lead to (42) with orthogonal $\chi_a(y)$ (which is the case with almost all interactions), we cannot say that such interactions measure A. But they are not of our interest, because the Appendix does not talk about general interactions, but about special interactions that do correspond to the measurement of A.

If, on the other hand, you cannot agree that the above is possible even for a single special interaction, then can you explain why?

 

[comote]

When you do the Schmidt decomposition in eqn (42) you can't just apriori pick what vectors you get in one of the spaces, ie: you can't just pick the $$\psi_a(x)$$ beforehand they are prescribed by taking the eigenvectors of the partial trace of your product state. A good reference for this is p236 of "Geometry of Quantum States".

They go through it here
http://cua.mit.edu/8.422_S05/NOTES-schmidt-decomposition-and-epr-from-nielsen-and-chuang-p109.pdf

But this hides some of the mechanics of the result under linear algebra.

It seems to me, that you are still picking the $$\psi_a(x)$$ beforehand. Yes it would work, for one specific set of eigenvectors on a given product state, but not any measurement. It would depend also on your hidden variables.

In order to make QM work we want to be able to make predictions for any observable in the state.

 

[yoda jedi]

Have you seen the lirics after the Abstract of
http://xxx.lanl.gov/abs/physics/0702069 [Am.J.Phys.76:143-146,2008] ?

very INTERESTING.

Is this the real life
Is this just fantasy
Caught in a landslide
No escape from reality
Freddie Mercury, “Boh(e)mian Rhapsody”

to me too, this one:

...Or is it just because of the <inertia> of pragmatic physicists
[STRIKE]who do not want to waste much time on (for them) irrelevant interpretational issues[/STRIKE], so that it is the simplest for them to (uncritically) accept the interpretation to which they were exposed first?...


or better yet "ignorance", literally (a true ignorance).
simplest = naive

exe: you can eat and don't know too much about nutrition... lol



pd: the Copenhagen and Many World Interpretations are "there is nothing more to say"
cos we don't know.

 

[yoda jedi]

I am not a fan of Bohmian Mechanics but I do want to understand where exactly it differs from standard QM.

Bohmian Mechanics - Non Linear.
SQM - Linear.

 

[Demystifier]

Bohmian Mechanics - Non Linear.
SQM - Linear.

SQM - wf Collapse - Non Linear.

 

[Demystifier]

It seems to me, that you are still picking the $$\psi_a(x)$$ beforehand. Yes it would work, for one specific set of eigenvectors on a given product state, but not any measurement.

Fine, it works for one specific choice of the measured observable A only.

But then for another choice of the observable B (B not equal to A), I choose ANOTHER basis $$\varphi_b(x)$$, so instead of (40) now I can write
$$\psi(x,t) = \sum_b d_b(t) \varphi_b(x)$$
To measure B (rather than A) I have to apply a different interaction, so now (42) will no longer be true. Instead, with that different interaction, instead of (42) I will have
$$\Phi(x,z,t) = \sum_b d_b(t) \varphi_b(x) \xi_b(z)$$
This is different from (42). Yet, it has the same FORM as (42).

The physical point is that there is no measurement without interaction, and each kind of measurement requires a different kind of interaction. Consequently, each kind of measurement will lead to a different wave function. Yet, as long as each of these measuremts is "ideal", the wave function after the interaction always takes the FORM (42).

And all this does not depend at all on hidden variables.

Does it help?

 

[RUTA]

And all this does not depend at all on hidden variables.

But it does depend on superluminal information exchange, even in its Lorentz invariant manifestations. At least that's what they told me here at the Hiley Symposium this week.

 

[yoda jedi]

SQM - wf Collapse - Non Linear.

SQM - wf Collapse.
which one ? specific model please...

 

[zenith8]

At least that's what they told me here at the Hiley Symposium this week.

You mean Bohm's chief collaborator Hiley - I presume..
Anything interesting to report? New results? Gossip?

 

[Demystifier]

But it does depend on superluminal information exchange, even in its Lorentz invariant manifestations. At least that's what they told me here at the Hiley Symposium this week.

No, what I said IN THE POST ABOVE, does not depend on superluminal information exchange.

 

[Demystifier]

SQM - wf Collapse.
which one ? specific model please...

SQM - wf collapse, but no model
GRW, Penrose (or some other) specific model - no SQM

 

[yoda jedi]

SQM - wf collapse, but no model
GRW, Penrose (or some other) specific model - no SQM

i know, GRW (a CSL version, other versions: Adler, Pearle, Bassi, Diosi, Tumulka and others), Penrose, are objective collapse theories.
...And Singh, Elze, Svetlichny, Zloshchastiev, Hansson, Nattermann, Khrennikov and others are non linear but no standard -linear-.
but you said Standard Quantum Mechanics.

you mean SQM without collapse ?

 

[Demystifier]

you mean SQM without collapse ?

By SQM, I meant SQM with collapse, but without mathematical description of collapse in terms of a precise model. Instead, collapse is introduced as a vague postulate. Something like: "When a measurement is performed, the wave function collapses to an eigenstate of the measured observable."

 

[yoda jedi]

By SQM, I meant SQM with collapse, but without mathematical description of collapse in terms of a precise model. Instead, collapse is introduced as a vague postulate. Something like: "When a measurement is performed, the wave function collapses to an eigenstate of the measured observable."

and no model existent.

 

[Demystifier]

Fine, it works for one specific choice of the measured observable A only.

But then for another choice of the observable B (B not equal to A), I choose ANOTHER basis $$\varphi_b(x)$$, so instead of (40) now I can write
$$\psi(x,t) = \sum_b d_b(t) \varphi_b(x)$$
To measure B (rather than A) I have to apply a different interaction, so now (42) will no longer be true. Instead, with that different interaction, instead of (42) I will have
$$\Phi(x,z,t) = \sum_b d_b(t) \varphi_b(x) \xi_b(z)$$
This is different from (42). Yet, it has the same FORM as (42).

The physical point is that there is no measurement without interaction, and each kind of measurement requires a different kind of interaction. Consequently, each kind of measurement will lead to a different wave function. Yet, as long as each of these measuremts is "ideal", the wave function after the interaction always takes the FORM (42).

And all this does not depend at all on hidden variables.

The measurement I am talking here about is known also under the name non-demolition measurement:
http://en.wikipedia.org/wiki/Quantum_nondemolition_measurement