Meaning of the word 'instantaneous'

[Paul Colby]

No matter how you word or think about( not especially you), the problem is well established now decades later, otherwise we would not be talking about it

Yes, I understand all possible issues have been considered, discussed and understood better than I am ever likely to. This is more about me coming to grips with the physics and understand what people are saying, hopefully without pissing them off. Wiki pages are helpful but, like certain religious text, written by people with opinions that must be teased out and separated from the actual facts.

 

[Paul Colby]

And I would not agree that Alice and Bob's observables commute. For example: Alice measuring at 0 degrees and Bob at 45 degrees does not give the same result as Alice measuring at 45 degrees and Bob at 0 degrees. Any more than Alice measuring at 0 degrees and then at 45 degrees gives the same result as Alice measuring at 45 degrees and then at 0 degrees.

You must be operating with a different understanding of what constitutes an observable. My understanding is Bob's observable is $S_{1,\alpha}$ a component of spin for his particle (1) while Alice's is $S_{2,\beta}$ for particle (2). These occupy two separate and independent Hilbert spaces, $H_1$ and $H_2$. Therefore, $[S_{1,\alpha},S_{2,\beta}]=0.$ This remains true when entangled when the Hilbert space is the direct product, $H=H_1\otimes H_2$.

 

[DrChinese]

Wiki pages are helpful but, like certain religious text, written by people with opinions that must be teased out and separated from the actual facts.

Without defending Wikipedia: the usual criticism of the Wikipedia physics pages is not one of bias, but that the matters are presented at an insufficient depth level. I certainly don't think your "religious" comment applies to the section quoted above. This is pretty standard wording for describing entanglement in lay terms.

 

[DrChinese]

My understanding is Bob's observable is $S_{1,\alpha}$ a component of spin for his particle (1) while Alice's is $S_{2,\beta}$ for particle (2). These occupy two separate and independent Hilbert spaces, $H_1$ and $H_2$. Therefore, $[S_{1,\alpha},S_{2,\beta}]=0.$ This remains true when entangled when the Hilbert space is the direct product, $H=H_1\otimes H_2$.

Sorry, not true while they are entangled. There is just one system! If there were a product state (a combination of 2 separable particles) we would not be having this discussion.

 

[Paul Colby]

Sorry, not true while they are entangled. There is just one system! If there were a product state (a combination of 2 separable particles) we would not be having this discussion.

Hard to have a discussion if the rules aren't understood. Every quantum book I've read says the state space for a two particle system is $H_1\otimes H_2$ from which my statement follows. Hard for me to conclude you have a point since my working of the problem would contain this "flaw" from the outset.

 

[DrChinese]

Hard to have a discussion if the rules aren't understood. Every quantum book I've read says the state space for a two particle system is $H_1\otimes H_2$ from which my statement follows. Hard for me to conclude you have a point since my working of the problem would contain this "flaw" from the outset.

Your statement is true for 2 normal particles. But not true for entangled particles, at least in the space where they are entangled.

I can't stress enough that your point of view was the starting point for EPR, 1935. They looked at entangled particle pairs as belonging to different spaces as an assumption. Unfortunately, they (and everyone else for the next 30 years) missed out on the implications of such an assumption. This was ultimately noted by Bell. Today, we know entangled systems are not separable and they are not represented in that manner.

 

[Paul Colby]

I can't stress enough that your point of view was the starting point for EPR, 1935. They looked at entangled particle pairs as belonging to different spaces as an assumption. Unfortunately, they (and everyone else for the next 30 years) missed out on the implications of such an assumption. This was ultimately noted by Bell. Today, we know entangled systems are not separable and they are not represented in that manner.

Okay, so how do you compute $[S_{1,\alpha},S_{2,\beta}]$?

So, I might add this is an operator expression which holds independent of a give system state.

 

[DrChinese]

Okay, so how do you compute $[S_{1,\alpha},S_{2,\beta}]$?

It's actually nicely discussed in section 4.1 of the reference ftr provided (did you miss this part when you were dismissing the link?):

https://en.wikipedia.org/wiki/Quantum_entanglement#Pure_states

Again: IF you assume that the combined entangled system is in a (separable) Product state, THEN you can prove that the combined entangled system is in a... (separable) Product state. Not much there. Of course you cannot use this assumption AND get agreement with the predictions of QM for entangled systems. As Bell showed us.

 

[Paul Colby]

It's actually nicely discussed in section 4.1 of the reference ftr provided (did you miss this part when you were dismissing the link?):

Sorry, I thought you understood my question. The $S_{1,\alpha}$ and ##S_{2,\beta} operators commute independent of section 4.1 which correctly points out the entangled states are generally not separable into a product. Both these facts are known to me. Because they commute these operators may be simultaneously diagonalized. This yields a perfectly valid view of the entangle system. When expressed in terms of eigenstates of Alice and Bob's observables, one obtains an expression for the entangled $S=0$ state where the coefficients or probability amplitudes are angle dependent in accordance with Bell and the physics. So, each measurement Bob makes determines a single particle state whose coefficients for Alice are dependent on Bob's angle setting.

 

[DrChinese]

Sorry, I thought you understood my question. The $S_{1,\alpha}$ and ##S_{2,\beta} operators commute independent of section 4.1 which correctly points out the entangled states are generally not separable into a product. Both these facts are known to me. Because they commute these operators may be simultaneously diagonalized. This yields a perfectly valid view of the entangle system. When expressed in terms of eigenstates of Alice and Bob's observables, one obtains an expression for the entangled $S=0$ state where the coefficients or probability amplitudes are angle dependent in accordance with Bell and the physics. So, each measurement Bob makes determines a single particle state whose coefficients for Alice are dependent on Bob's angle setting.

You are mixing (often conflicting) subjects within sentences. So no, nothing you are saying makes sense. Your viewpoint simply does not follow standard physics: Entangled systems are NOT created in local separable states as you somehow imagine. Entangled states are in fact not separable, and product statistics (which do commute) do not accurately describe entangled systems. The bottom line is that you want to have your cake and eat it too - QM without quantum nonlocality.

You are welcome to your personal opinion, but this conversation has gone 'round in circles far too long. I will simply say that for anything you state contrary to standard QM (such as I have presented), please provide peer-reviewed or other suitable references to support such. Those are the forum rules in this situation. Readers may otherwise get the idea that there is scientific controversy in this realm, when there is not.

 

[Paul Colby]

You are mixing (often conflicting) subjects within sentences. So no, nothing you are saying makes sense.

Fair enough. I'm just trying to understand. Thank you for the help.

 

[Paul Colby]

Entangled states are in fact not separable,

I guess I fail to see where I ever said they are?

When expressed in terms of eigenstates of Alice and Bob's observables, one obtains an expression for the entangled $S=0$ state where the coefficients or probability amplitudes are angle dependent in accordance with Bell and the physics. So, each measurement Bob makes determines a single particle state whose coefficients for Alice are dependent on Bob's angle setting.

So, what exactly is wrong with my statement?

Ah, "in accordance with Bell" could be construed as incorrect.

 

[zonde]

I guess I fail to see where I ever said they are [separable]?

Here:

You must be operating with a different understanding of what constitutes an observable. My understanding is Bob's observable is $S_{1,\alpha}$ a component of spin for his particle (1) while Alice's is $S_{2,\beta}$ for particle (2). These occupy two separate and independent Hilbert spaces, $H_1$ and $H_2$. Therefore, $[S_{1,\alpha},S_{2,\beta}]=0.$ This remains true when entangled when the Hilbert space is the direct product, $H=H_1\otimes H_2$.

 

[zonde]

Still, I believe there is no instantaneous interaction involved in the way I would define one. Alice and Bob's observables commute and may be simultaneously diagonalized.

I don't think it's very smart thing to do - redefine terms just to look still right.

 

[Paul Colby]

Here:

Bob's observable is $\hat{n}_1\cdot S_1$ where $S_1$ is a 3-vector with the Pauli spin matrices as components. The unit vector, $\hat{n}_1$ is a unit vector along a direction chosen by Bob and is the alignment of his SG. This matrix operates on a 2 dimensional Hilbert space $H_1$. Likewise a for Alice. Her observable is $\hat{n}_2\cdot S_2$ which is an operator which lives an independent 2 dimensional Hilbert space, $H_2$. Alice and Bob's observables do in fact commute and may be simultaneously diagonalized. However, this does not imply in anyway that the $S=0$ is separable as DrChines insists[1].

These facts may be checked in virtually every book on QM. That the $S=0$ doesn't factor is trivial to show. Let Bob's SG be along the z-axis to save me some writing. The $S=0$ state is then,

$\frac{1}{\sqrt{2}}(\vert 1\rangle_{zB}\otimes\vert -1\rangle_{zA} - \vert -1\rangle_{zB}\otimes\vert 1\rangle_{zA})$
​

where, the observant reader will instantly accuse me of using Bob's z-axis to expand Alice's vectors. Because I'm free to use Alice's coordinates for her particle I may expand them as such,

$\vert 1\rangle_{zA} = a\vert 1\rangle_{\alpha A} + b\vert -1\rangle_{\alpha A}$ [2]

$\vert -1\rangle_{zA} = c\vert 1\rangle_{\alpha A} + d\vert -1\rangle_{\alpha A}$
​

where $a, b, c$ and $d$ are complex coefficients which very much depend on the choice of $\hat{n}_1$ and $\hat{n}_2$ just as DrChines has pointed out numerous times. Substitution of these expressions into the one for the $S=0$ above yields (a very much not separable) expression for the $S=0$ state in terms of both Bob's and Alice's particle eigenvectors. Before anyone replies asserting that I am claiming this somehow removes all EPR mysteries consider the fact that I do not claim such nor do I ever feel I have in the past. I have always assumed a level of mathematical sophistication which is perhaps unwarranted.

[1] Weather he's insisting $[S_1,S_2]=0$ implies factorability of the $S=0$ state or insisting that I have insisted such is unclear. Both positions are in fact wrong.
[2] My notation here is very confusing. The basis on the right are eigenstates of Alice's SG while those on the left are eigenstates assuming she had aligned hers with Bob's which is along the z-axis. Okay, I've attempted to repair it. I've added a subscript $z$ to denote z-axis eigenvectors and an $\alpha$ to denote eigenvectors for Alice's SG direction which is arbitrary.

 

[Paul Colby]

I don't think it's very smart thing to do - redefine terms just to look still right.

I don't believe I am. Interaction usually implies the presents of a term in the Hamiltonian for the interacting systems. By it's very definition the EPR example has no such term.

 

[zonde]

Bob's observable is $\hat{n}_1\cdot S_1$ where $S_1$ is a 3-vector with the Pauli spin matrices as components. The unit vector, $\hat{n}_1$ is a unit vector along a direction chosen by Bob and is the alignment of his SG. This matrix operates on a 2 dimensional Hilbert space $H_1$. Likewise a for Alice. Her observable is $\hat{n}_2\cdot S_2$ which is an operator which lives an independent 2 dimensional Hilbert space, $H_2$. Alice and Bob's observables do in fact commute and may be simultaneously diagonalized. However, this does not imply in anyway that the $S=0$ is separable as DrChines insists[1].

The point is that $H_1$ and $H_2$ are not independent. If you have two states in two independent Hilbert spaces then it is not entangled state. The two states have to be antisymmetrized and you can't have that in two independent Hilbert spaces.

 

[Paul Colby]

The point is that H1H1H_1 and H2H2H_2 are not independent. If you have two states in two independent Hilbert spaces then it is not entangled state. The two states have to be antisymmetrized and you can't have that in two independent Hilbert spaces.

There is more than one common and accepted usage of the word independent in both math and physics. The meaning here was clear as two separate degrees of freedom. I was clearly not referring to entanglement. The symbols $H_1$ and $H_2$ clearly refer to sets of all possible 1 particle states and can't in anyway be confused with being dependent on one another.

 

[zonde]

There is more than one common and accepted usage of the word independent in both math and physics. The meaning here was clear as two separate degrees of freedom. I was clearly not referring to entanglement. The symbols $H_1$ and $H_2$ clearly refer to sets of all possible 1 particle states and can't in anyway be confused with being dependent on one another.

Direct product $H_1\otimes H_2$ clearly refers to all possible 2 particle states:

This remains true when entangled when the Hilbert space is the direct product, $H=H_1\otimes H_2$.

And btw I am not saying that measurements of entangled particles do not commute (as it would directly lead to FTL communication).

 

[Paul Colby]

Direct product $H_1\otimes H_2$ clearly refers to all possible 2 particle states:

Good to see you're catching on.

The point is that $H_1$ and $H_2$ are not independent. If you have two states in two independent Hilbert spaces then it is not entangled state. The two states have to be antisymmetrized and you can't have that in two independent Hilbert spaces.

So this would be an example of incorrect statement.

 

[nortonian]

Yes, I know perfectly well that yours appears to be a good explanation - at least for the so-called perfect correlations. Those are the cases in which Alice and Bob measure at the same angle. Please note that the outcomes are essentially redundant in that case - and it is a special case. This special case certainly suggests strongly that there is no question about "about actions occurring faster than light". And in fact, this is essentially the premise of EPR (1935).

I have followed your explanation of Bell's theorem which gave me my first understanding in physical terms. Although the math is still a bit foggy your arguments seem quite clear in the two posts you mention. My confusion, or perhaps misunderstanding is at a more fundamental level. Wave functions were first used to explain elements of atomic structure and specifically they describe the (statistical) interface between an observer and photon detection in the Schroedinger equations. Photon pair production (an emission rather than detection process) used in Bell-type experiments are done in free space; that is, in the absence of an observer. What precedent allows theoreticians to say that isolated events may be assigned a wave function with only the knowledge of their existence but without actually detecting them? This to me seems a leap of faith. Just because we know something does not mean we can assign objective meaning to it.

 

[gnnmartin]

"Instantaneous" is used to mean that two events appear correlated, neither has a purely local causal history, and we are unable to define and demonstrate which happened first . Paul's example (post 10) is not instantaneous by this definition, because the causal history of each gyroscope is local to that gyroscope.

It is not obvious how to demonstrate that the causal history of an event is not local, but John Bell came up with a proof that in some circumstances the causal history of entangled photons cannot be local. (see ON THE EINSTEIN PODOLSKY ROSEN PARADOX*)

 

[gnnmartin]

I apologise for posting that (post 112) without realising that there was already a long history of replies, so I may well just be duplicating what has already been said. I saw the first page of this thread, and did not realize it was the first page of many!

 

[nortonian]

It is not obvious how to demonstrate that the causal history of an event is not local, but John Bell came up with a proof that in some circumstances the causal history of entangled photons cannot be local. (see ON THE EINSTEIN PODOLSKY ROSEN PARADOX*)

No need to apologize. You provided me with a link to Bells original paper which I had not seen. this leads into the subject matter I was asking about which is way off topic from instantaneous. Because the subject matter has changed I am starting a new thread called Is entanglement based on first principles? Perhaps you can help me sort some questions out that I have