Entaglement and hidden variables

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Bohm's theory calculates entanglement by considering the global wave function, which includes both the particles and the measuring device, unlike standard quantum theory that relies on wave function collapse. The discussion highlights that while standard quantum mechanics does not require the global function for predictions, Bohm's approach can yield the same measurable outcomes through approximate calculations of this global function. A theorem asserts that both Bohm's and standard quantum theories provide identical predictions, though this claim requires convincing proof. The conversation also touches on the complexities of defining photon paths and wave functions, emphasizing the challenges in applying hidden variable theories to photons. Overall, the discussion reveals ongoing debates about the compatibility and implications of Bohm's theory in relation to standard quantum mechanics.
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  • #32
Demystifier said:
I am sure you are an expert in your field (whatever it is), but it does not mean that you know everything about QM.

I have explained the reason why eq.42 is wrong and you have raised no counter arguments; you have only said that it is correct because someone else think it is correct, and this is not a valid argument.

I give you another argument proving that the so called "theory of quantum measurement" of the appendix quoted before, is wrong .

In standard quantum mechanics the probability of measuring a specific value q of momentum is given by the square modulus of the coefficient c_q in the decomosition of the electron wave function. This value is calculated for the wave function of the electron BEFORE the measurement (say at t=0), i.e. for the wave funtion of the electron only, without any interaction with the detector.
On the contrary, in the hidden variable theory, this probability is calculated after the measument has occurred, since the needle is explicitly considered; so the value of c_q is calculated at an instant t successive AFTER the measuremet.
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.
 
  • #33
comote said:
Maybe I am missing something, but isn't equation (42) just the Schmidt decomposition?

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

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.
 
  • #34
andreab1987 said:
I have explained the reason why eq.42 is wrong and you have raised no counter arguments; you have only said that it is correct because someone else think it is correct, and this is not a valid argument.
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).
 
  • #35
andreab1987 said:
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.
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.
 
  • #37
andreab1987 said:
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.
 
  • #38
Demystifier said:
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.
 
  • #39
Demystifier said:
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.
 
  • #40
Demystifier said:
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
 
  • #41
Demystifier said:
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.
 
  • #42
andreab1987 said:
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!
 
  • #43
andreab1987 said:
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.
 
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  • #44
comote said:
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)
 
  • #45
OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?
 
  • #46
andreab1987 said:
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 :smile:
 
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  • #47
comote said:
OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?

Correct
 
  • #48
comote said:
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.
 
  • #50
comote said:
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.
 
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  • #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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