Entaglement and hidden variables

  • #51
Demystifier said:
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.
 
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  • #52
andreab1987 said:
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)!
 
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  • #53
Demystifier said:
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.
 
  • #54
Demystifier said:
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.
 
  • #55
comote said:
If you insist that the $\psi_a(x)$ in equations (40) and (42) are the same
Yes, I insist on that ...
comote said:
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?
 
  • #56
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.
 
  • #57
Demystifier said:
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.
 
  • #58
comote said:
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.
 
  • #59
yoda jedi said:
Bohmian Mechanics - Non Linear.
SQM - Linear.
SQM - wf Collapse - Non Linear.
 
  • #60
comote said:
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?
 
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  • #61
Demystifier said:
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.
 
  • #62
Demystifier said:
SQM - wf Collapse - Non Linear.

SQM - wf Collapse.
which one ? specific model please...
 
  • #63
RUTA said:
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?
 
  • #64
RUTA said:
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.
 
  • #65
yoda jedi said:
SQM - wf Collapse.
which one ? specific model please...
SQM - wf collapse, but no model
GRW, Penrose (or some other) specific model - no SQM
 
  • #66
Demystifier said:
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 ?
 
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  • #67
yoda jedi said:
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."
 
  • #68
Demystifier said:
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.
 
  • #69
Demystifier said:
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
 
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