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August 14, 2019 at 1:35 am #5627editorKeymaster
Thanks, Ruth!
August 12, 2019 at 7:59 am #5622editorKeymasterThanks for your useful comments, Jerry!
August 10, 2019 at 2:10 am #5616editorKeymasterThanks, Ruth. So, the collapse is not the superposition psi1 + psi2 randomly becomes psi1 or psi2 for the forwardpropagating state or the the backwardpropagating state?
August 8, 2019 at 1:56 am #5593editorKeymasterThanks, Ruth! I will need more time to understand your RTI. As far as I see, your RTI does not assume a real collapse of the real wave fucntions. Maybe in RTI the wave function is not ontic?
August 7, 2019 at 5:15 am #5586editorKeymasterOk, Ruth. I see. Then RTI is not really the usual collapse theory such as the GRW model for onevector. So, in RTI, there is no real dynamical collapse of the wave function. Then my analysis of collapse theories does not apply to RTI.
Now I think my analysis of unitary quantum theories will apply to RTI. See my paper section about the retrocausal theories. I guess Ken worried about this.
August 6, 2019 at 1:26 pm #5581editorKeymasterSo, you also agree that “in the second scenario, in which Bob’s measurement precedes (in the lab frame) all of Alice’s measurements”, “Since Alice cannot know Bob’s outcome before that event enters her back light cone, all she can predict with certainty at points on her worldline before making those measurements is that, whatever they are, they will be opposite to Bob’s.”
But this is just my analysis.
“In the second scenario Gao argued that the Born rules yields probability 1 for the sequence of outcomes of Alices repeated zspin measurements in which every outcome is the opposite of Bobs single outcome of a measurement at point 3, and probability 0 to every other possible sequence of outcomes of Alices zspin measurements.”
August 6, 2019 at 4:13 am #5579editorKeymasterThanks for your clarification, Richard!
But I cannot agree with you. I think the Born Rule must be applied at a time, while this would not necessarily introduce a preferred frame in relativity as some people think, although my new argument says the opposite.
The reason is that if the Born Rule is not applied at a time for a joint measurement of Alice and Bob on the EPR pair, then the theory will be unable to explain the perfect anticorrelation between the results of Alice’s and Bob’s spacelike measurements. At least in one frame, immediately after Bob’s measurement, the Born rules must yield probability 1 for the sequence of outcomes of Alices repeated zspin measurements in which every outcome is the opposite of Bobs outcome, whether for Alice or other observers in the frame. Otherwise the predictions will contradict experiments. This is my point in the last post.
August 6, 2019 at 3:43 am #5578editorKeymasterYou said “no way for Alice to check that when she detected, say particle #25, that Bob did not get that particle, because in order to do that, the two would have to exchange classical information faster than light.”
Yes, you are correct. But this is not necessary to verify the anticorrelation. They just need to record the results, and then compare them later. But a collapse theory must be able to explain the anticorrelation between the results of the two spacelike separated measurements.
As I argued in my last post, this explanation requires that the collapse should happen faster than light at least in one frame.
August 5, 2019 at 8:13 am #5574editorKeymasterThanks, Ruth! I will need to undertand your idea more deeply.
Here is what I thought. Suppose ALice measures the particle in box 1, and then Bob meqasures the particle in box 2. These two measurements are spacelike and almost simutaneous. Then, when Alice have dectected the particle in box 1, the superpostion should collapse also in the remote box 2, otherwiese the Born rule will be violated; it will predict that Bob has a nonzero probability to detect the particle in box 2.
I still cannot see how a collapse theory can explain this process without assuming the collapse happens faster than light in a frame.
August 5, 2019 at 2:32 am #5572editorKeymasterHi Richard,
Thanks for your comments! In your paper, you said:
“In the second scenario (i.e. Bob precedes Alice) Gao argued that the Born rules yields probability 1 for the sequence of outcomes of Alices repeated zspin measurements in which every outcome is the opposite of Bobs single outcome of a measurement at point 3, and probability 0 to every other possible sequence of outcomes of Alices zspin measurements. This is incorrect.”
I cannot agree. I think amost all people will agree that this is a correction application of the Born rule (at least in one frame).
If this is not correct, then QM will be unable to explain the perfect anticorrelation between the results of Alice’s and Bob’s spacelike measurements.
Best,
Shan
August 5, 2019 at 2:20 am #5571editorKeymasterI think the issue is not really related to the detials. Consider a particle is in a superposition of two wellseparated boxes. When one measures the particle in one box, then the whole superposition collapses to the state in one of the two boxes. Now if the collapses of the superposition in the two boxes are not spacelike seperated in any frame, then the theory cannot explain the violation of Bell’s inequalities in this case; the results of two spacelike measurements on the particle in the two boxes will not be correlated.
August 4, 2019 at 2:17 am #5558editorKeymasterHi Ruth,
Thanks for your interesting comments! In my paper, I said:
“In collapse theories, the collapse of the wave function is simultaneous in different regions of space in a preferred Lorentz frame.”
I think this is correct. Otherwise, if the collapses of the wave function in different regions of space are not spacelike seperated in any frame, then the theory cannot explain the violation of Bell’s inequalities.
Can RTI explain the violation of Bell’s inequalities without spacelike seperated collapses?
Best,
Shan
November 8, 2018 at 11:40 am #5269editorKeymasterHi Richard,
In your post, you said: “To correctly calculate the correlation between his outcome and the outcome of Alice’s actual measurement, Bob must assign a state at a time (in his frame) before the superobserver’s intervention.” Could you explain why? Thanks!
Shan
PS. Gijs Leegwater has just posted a paper in arXiv (https://arxiv.org/abs/1811.02442), in which he discussed a similar thought experiment as your third argument, but he reached the same conclusion as mine. He may later participate in our discussions.
November 6, 2018 at 1:11 am #5257editorKeymasterThanks, Ruth. Your pointed out a potential issue. My argument does not reply on Alice’s result being erased, but relies on the state of the particles being recovered before Bob’s measurement. Shan
November 5, 2018 at 2:16 am #5251editorKeymasterHi Richard,
Thanks for your further clarification!
I think you did not understand my objection concerning the nonSR case. In the nonSR case, since the time order of spacelike separated events is invariant in different frames, you cannot derive the relation $E(b,c)=cos(bc)$ in Bob’s frame using your derivation in the SR case, since just like in Alice’s frame, when Bob obtains his result, Carol’s result has been erased. I think you should at least admit this point.
However, one should be able to derive the correlation function $E(b,c)$ in the nonSR case. After all, these correlation functions in nonrelativistic QM should not depend on SR. This suggests that your derivation of the relation $E(b,c)=cos(bc)$ in the SR case is problematic.
Concerning my derivation of $E(b,c)$, I think it does not depend on any assumption about the measurement process such as the collapse of the wave function. We just analyze the joint probability distribution of the results (a,b,c,d) in a large number of trials. The distribution already shows that when $c=+1$, the probability of $d=+1$ is $sin^2[(cd)/2]$, etc, and then we can use these probability relations to derive $E(b,c)$ as I have done in my last post.
I have just written a draft paper which tries to solve these issues from a new angle. Your comments are very welcome!
Unitary quantum theory is incompatible with special relativity
Shan
November 1, 2018 at 9:00 am #5245editorKeymasterHi Richard,
Many thanks for your further clarification! Here is my responses:
> 3. I don’t understand how you have derived your alternative value for E(b,c) in the special case c=b, a=d. Can you explain?
In Alice’s frame, we have the quantum links $c —> d —> a —>b$. We can then derive $E(b, c)$ from the correlation functions $E(c, d) = −cos(c − d)$, $E(d, a) = −cos(d − a)$, and $E(a, b) = −cos(a − b)$, which have been derived correctly in your paper.
Consider the simple case where $c=b$ and $a=d$. When the result of Carol is $c=+1$, the conditional probability of $d=+1$ is $sin^2[(cd)/2]$, the conditional probability of $a=1$ when $d=+1$ is 1, and the conditional probability of $b=+1$ when $a=1$ is $cos^2[(ab)/2]$. When the result of Carol is $c=+1$, the conditional probability of $d=1$ is $cos^2[(cd)/2]$, the conditional probability of $a=+1$ when $d=1$ is 1, and the conditional probability of $b=+1$ when $a=+1$ is $sin^2[(ab)/2]$. Then, when the result of Carol is $c=+1$, the total probability of $b=+1$ is $2sin^2[(ab)/2]cos^2[(ab)/2]$.
Similarly, when the result of Carol is $c=+1$, the total probability of $b=1$ is $12sin^2[(ab)/2]cos^2[(ab)/2]$, which is not equal to $1$ in general. This already shows that the relation $E(b, c) = −cos(b − c)$ does not hold in Alice’s frame or in the rest frame where the particles and labs of the experimenters in the Gedankenexperiment are at relative rest.
On the other hand, when the result of Carol is $c=1$, the total probability of $b=1$ is $2sin^2[(ab)/2]cos^2[(ab)/2]$, and the total probability of $b=+1$ is $12sin^2[(ab)/2]cos^2[(ab)/2]$. Then we can calculate the correlation function $E(b,c)$, which turns out to be $E(b,c)=4sin^2[(ab)/2]cos^2[(ab)/2]1$. It can be readily checked that the inequality (32) is not violated when using this correlation function $E(b,c)$.
> 2. The boosts (and rotations) must be Lorentz transformations, not Galilean boosts here. This is important because the whole discussion must be set in Minkowski spacetime, not Galilean spacetime: if it were not, then there could be no switching of time order of spacelike separated events between frames. Such Lorentz transformations are standard in discussions like that of my reference (17).
Yes, I agree. My point is that one should be able to derive the correlation function $E(b,c)$ by using the Galileo transformations. After all, these correlation functions in *nonrelativistic* QM should not depend on SR (Special Relativity). But, in this nonSR case, your derivation in Bob’s frame (by the switching of time order of spacelike separated events) will not go through.
> 1. It is important to notice that a quantum state assignment on a fixed spacelike hyperplane (like the hyperplane t*^3) may itself be made with respect to different inertial frames (say, Alice’s and Bob’s). Quantum states on different spacelike hyperplanes (like t^3 and t*^3) are not related by a boost transformation. So a derivation of E(b,c)= cos(bc) in Alice’s frame would still proceed from an assignment of a quantum state on t*^3 even though that is not a simultaneity slice in Alice’s frame. To derive the relation E(b,c)= cos(bc) in Alice’s frame you would first have to transform both the state on t*^3 as well as the angles b,c from Bob’s frame (in which the derivation is easiest) to Alice’s frame. This cannot affect the result of the calculation, since E(b,c) is invariant under changes of frame. But the calculation would be unnecessarily complicated.
Yes, I agree. But I think you ignored the important influence of each measurement process on the entangled wave function (I just realized this). The influence is nonlocal, as clearly manifested in collapse theories or Bohm’s theory. If there were no such influences, then the relation $E(b,c)= cos(bc)$ would indeed hold also in Alice’s frame as you have derived. Then it would be not surprised that the Bell inequality (31) is violated by QM. This is consistent with the Bell theorem.
On the other hand, when considering the underlying nonlocal processes, the relation $E(b,c)= cos(bc)$ will not hold in Bob’s frame, as well as in Alice’s frame, and we will again have the above relation $E(b,c)=4sin^2[(ab)/2]cos^2[(ab)/2]1$ in Alice’s frame. As noted above, this does not lead to the violation of the inequality (32).
Shan
October 30, 2018 at 4:02 am #5234editorKeymasterHi Richard,
I think Alice’s reasoning is right. My worry is still that the inequality (32) should be defined (and can also be calculated) in one frame such as Alice’s frame. But in this frame it seems that QM does not require the relation E(b, c) = −cos(b − c), since the result of Carol has been erased by Alice and replaced by Alice’s result before Bob’s measurement, and thus E(b, c) = −cos(b − c) cannot be derived by using QM in Alice’s frame.
In Alice’s frame, we have the quantum links c —> d —> a —>b, but no the required quantum link b —> c or c —> b for directly deriving the relation E(b, c) = −cos(b − c). For example, suppose the four directions have the relations: c=b and a=d but c is not equal to a. Then when the result c=+1, your relation E(b, c) = −cos(b − c) requires that the result b must be 1. But QM only restricts the correlations E(c, d), E(a, d) and E(a, b) in Alice’s frame, while these correlations do not require that b must be 1. In fact, when c=b and a=d, it can be derived that E(b,c)=4sin^2[(ab)/2]cos^2[(ab)/2]1, and there is no violation of the inequality (32).
In your paper, you derived the relation E(b, c) = −cos(b − c) in another frame (i.e. Bob’s frame) by using the Lorentz transformations (where exchanging the time orders of events seems to be a trick). But one should be able to derive this relation in Alice’s frame (if the relation is true), and moreover, one should be able to derive this relation by using the Galileo transformations (in this case your derivation in Bob’s frame will not go through). After all, these correlation functions in QM should not depend on SR.
I think this is a potential issue, and maybe if this issue is solved, then there will be no contradiction.
Shan
PS. I would like to know how you can derive the relation E(b, c) = −cos(b − c) in Alice’s frame. Could you tell me more details? Thanks a lot!
October 29, 2018 at 4:53 am #5229editorKeymasterPS. Moreover, when Bob and Carol are in different frames, it seems that the relation E(b, c) = −cos(b − c) cannot be consistently defined in QM, since the measured states for them, which extends to two spacelikeseparated regions, are not the same.
October 28, 2018 at 12:10 pm #5228editorKeymasterHi Richard,
Thanks a lot for your further explanation! I now understand your third argument more clearly. I think the potential issue is that you used results from two different frames in the same inequality (32). Concretely speaking, I think in Alice’s frame corr(b, c) is not equal to E(b, c) = −cos(b − c). It seems that QM does not require this relation in Alice’s frame (we need a more careful analysis here). In other words, the predictions of QM should be complete in each inertial frame. And we should be able to derive corr(b, c) in Alice’s frame if QM permits. This can be seen from the nonSR cases. If we assume the Galileo transformations as in QM, then we cannot derive corr(b, c) in Bob’s frame as you did in your paper. I would like to know what you think about this. Thanks!
Shan
October 27, 2018 at 8:37 am #5223editorKeymasterHi Richard,
Thanks for your kind explanation!
I understand that the inequality (31) is derived from the hidden variable assumption in Fine’s paper. But the statistical correlations between pairs of *actual experimental outcomes* does not necessarily satisfy the inequality. It is the assumption that these outcomes directly reflect the values of the *measured observables* (i.e. the hidden variable assumption) that leads to the inequality, which is violated by QM and the experimental observations.
Moreover, I think the thought experiment discussed in your third argument is essentially the same as the original Bell experiment. In the thought experiment, the EPR pair of spin1/2 particles is measured two times in the same state. It seems to me that this is equivalent to measuring two EPR pair of spin1/2 particles in the same state.
I understand that you tried to use the extended Wigner’s friend thought experiment to extend the BellKG theorem for hidden variables to prove the nonexistence of the experimental outcomes of different observables. I also had the same idea after FR’s paper appeared. But I think it may not go through. For actual experimental outcomes of observables are different from the values of these observables, and these outcomes are only different values of the same pointer observable.
Shan
October 26, 2018 at 3:28 am #5215editorKeymasterHi Richard,
I think there are two concerns about your third argument. The first is that in order to derive the Bell inequality (31), one needs a locality assumption for the factorization, e.g. of the term corr(c,d), which is the same as that in the original Bell inequality. The second concern is that in your thought experiment, it seems that the two terms corr(b, c) and corr(a, d) in (31) do not exist. The reason is that the measurements of Alice and Bob undo the measurements of Carol and Dan, and thus when Alice and Bob have obtained the results a and b, the other two results c and d already disappeared. In other words, c and d do not exist in any device or the memories of Carol and Dan after the undo procedure; they are erased by the undo procedure.
Maybe my analysis is not right. I would like to know what you think about these two concerns. Thanks!
Shan
October 25, 2018 at 1:22 am #5206editorKeymasterA good paper, Dustin. Your analysis is consistent with the analyses of others, including those of Tony and Richard, which are listed in the “References related” post. Shan
October 25, 2018 at 12:58 am #5205editorKeymasterYes, Ruth. You may start a new topic to address the measurement problem.
October 25, 2018 at 12:58 am #5204editorKeymasterThanks for your comments, Nikolay.
May 12, 2018 at 12:17 am #4805editorKeymasterThanks, Ruth! This is an interesting topic.
May 1, 2018 at 8:12 am #4732editorKeymasterI think one of these pressing problems is to determine what physical state is eligible to represent a measurement result. And a deep analysis of the psychophysical connection is still needed to solve this problem.
December 31, 2014 at 12:55 am #1740editorKeymasterHi Richard,
Thanks again for your further clarification!
Best,
ShanDecember 20, 2014 at 2:44 am #1656editorKeymasterMy proof of Bell’s theorem can be summarized in one sentence: since local evolution can change nosignaling to nonlocal signaling, the nosignaling also contains nonlocality, the same nonlocality in nonlocal signaling.
This new proof of the existence of nonlocality seems too simple to be true. Critical comments are very welcome!
December 16, 2014 at 3:14 am #1611editorKeymasterThanks to Harvey and Chris for their stimulating paper, and to Travis and Howard, whose comments have kicked off the John Bell Workshop 2014. I hope everyone will be inspired by these discussions, either via email notifications or by online text chat.
October 31, 2014 at 3:37 am #1112editorKeymasterHi Bob,
Thanks for your further comments!
I agree with you. It is just the meanings of words that differ between us. The “reality of the wavefunction” is usually understood as “the wavefunction applies to a single system” (e.g. in the PBR theorem).
As you have seen from my previous posts, my interpretation in terms of RDM of particles indeed says that “the wavefunction tells us something about statistical properties of the motion of a single system”. But this does not mean they belong to the epistemological side of things. the psiepistemic view usually denotes that the wavefunction does not apply to a single system.
Best,
ShanOctober 31, 2014 at 3:29 am #1111editorKeymasterFor example, in his scheme, one get many inaccurate expectation values, from which the right one can only be constructed (which is very like the standard tomography based on projective measurements). But in a PM, one directly get the right expectation value with arbitrary accuracy.
October 31, 2014 at 3:26 am #1110editorKeymasterHi Max,
Thanks for your comments! I think you will agree that Matt’s scheme is not a PM. Right?
Shan
October 29, 2014 at 5:27 am #984editorKeymasterThanks, Max. See you then!
October 27, 2014 at 2:07 am #877editorKeymasterOK, we may discuss Valia’s papers later on this forum. Thank you all for your participance!
October 27, 2014 at 2:03 am #875editorKeymasterSorry, my internet contact for Skype is broken. I am trying to restore the system.
October 27, 2014 at 1:55 am #870editorKeymasterI should improve this in the future. I think Google Hangout is better. Thanks Matt!
October 27, 2014 at 1:44 am #867editorKeymasterValia is online now. If anybody has questions about her papers, we can discuss here.
October 27, 2014 at 1:10 am #866editorKeymasterWe may discuss Valia’s papers here after her talk on Skype, which is on live at URL http://www.ijqf.org/archives/841. You may listen to her talk and see her ppt slides (see above attachments) on your computer at the same time.
October 26, 2014 at 3:03 am #801editorKeymasterThank you all for your participance in this online workshop! I hope everyone will be inspired by these intense and stimulating discussions. I will try my best to provide technical support and make the forum provide you better experience.
Best,
Shan 
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