1
$\begingroup$

I had always thought similarly and then came across a paper here that argues this.

The abstract is as follows:

Many authors state that quantum nonlocality could not involve any controllable superluminal transmission of momentum-energy, signals, or information. We claim that most or all no-signalling proofs to date are question-begging, in that they depend upon assumptions about the locality of the measurement process that needed to be established in the first place. We analyse no-signalling arguments by Bohm and Hiley, and Shimony, which illustrate the problem in an especially striking way.

The gist of the paper is that FTL communication is attempted to be disproved by assuming the non existence of FTL communication…which is the very thing that is attempted to be disproved.

Note that it seems that for the purposes of this paper, it includes the notion of any non local causal influences occurring between any particles, and not just whether if say, Alice, wanted to send information to Bob after measurement.

Are there any discussions on this?

Improve this question
$\endgroup$
6
  • $\begingroup$ The critic of the paper indeed applies. See also P. Mittelstaedt. Can EPR-correlations be used for the transmission of sup erluminal signals? (1998). The book by C. Beck on local quantum measurements is also very good. The long story short is that in (relativistic) QFT, for example, you build in the micro-causality condition to get relativistic consistency and no-signaling. In non-relativistic QM, you use the tensor product structure of subsystems (which however is an axiom itself) + the condition that both parts are non-interacting. $\endgroup$
    – Tobias Fünke
    Commented Apr 13 at 11:26
  • $\begingroup$ See also the papers by A. Peres (and Peres and Terno) on quantum mechanics and relativity and classical intervention (I don't remember the exact titles, but you should find something). Note also that the formalism of QM, by itself, is completely agnostic about relativistics. As said, in relativistic QFT, you put additional assumptions, like what are symmetries of your theory, and put extra conditions on observables, to be compatible with SR. Anyway, what is your question, actually? Are there any discussions on this? --is not a suitable question for this site, I think. $\endgroup$
    – Tobias Fünke
    Commented Apr 13 at 11:29
  • $\begingroup$ Does this ultimately imply that the theorem assumes no violation of SR to conclude there is no violation of SR? Or does it assume no FTL communication to conclude there is none? Are these the same for this context? $\endgroup$
    – inquisitive
    Commented Apr 13 at 11:29
  • $\begingroup$ It ultimately means that within a certain framework of QM/QFT, you can show that you cannot communicate using just entanglement, roughly speaking (no-communication/signaling theorem). The way you build your theory, however, includes already conditions which, more or less trivially, leads to the conclusion. Then, yes, one might ask if these assumptions are equivalent, or stronger, or weaker, than assuming something like the no-communication theorem to begin with; see e.g. the book of Beck and the paper by Mittelstaedt.---Again, I would suggest to rephrase the question. It seems off-topic. $\endgroup$
    – Tobias Fünke
    Commented Apr 13 at 11:32
  • $\begingroup$ How do you recommend to rephrase the question? You can edit it if that helps $\endgroup$
    – inquisitive
    Commented Apr 13 at 11:35

1 Answer 1

Reset to default
0
$\begingroup$

Let's look at a quote from the paper you linked:

Does their proof amount to anything more than an illustration of the fact that an operator that doesn’t operate on a wave-function doesn’t change the wave-function? (Kennedy argues that virtually all no-signalling arguments within nonrelativistic quantum mechanics boil down to this unexceptionable claim, at least mathematically. [24]) That would not seem to be especially illuminating.

Here is a more charitable reading: even though proofs of this sort cannot show that there is no direct causal interaction between left and right particles, they do show that there is no inconsistency in the formalism of quantum mechanics, such that we would get evidence of a superluminal causal interaction if we assume there is none. In other words, one cannot beat the house merely by some sort of statistical trickery.

It was, no doubt, a salutary exercise to have shown this, but the use of such a calculation in support of a general no-signalling claim is completely question-begging. This is because it is very hard to see how any sort of signal from A to B would not require the disturbance of B by A, albeit in some fashion “that is perhaps not yet known”.

If you want to know whether QM is local or not then you should use the equations of motion of QM not some other theory. Since we have local quantum field theories that are used successfully to explain a very wide range of predictions it would make little sense to go looking for a non-local theory.

In addition the literature has included an entirely local explanation of Bell correlations in terms of quantum theory without collapse for more than 20 years:

https://arxiv.org/abs/quant-ph/9906007

So what problem is solved by going looking for a theory that violates no signalling?

Improve this answer
$\endgroup$
7
  • $\begingroup$ So, in your opinion, the violation of Bells's inequality isn't a proof of the non-locality of standard QM? It only shows that there is no local hidden variables theory that could explain its results? $\endgroup$
    – freecharly
    Commented May 4 at 16:59
  • $\begingroup$ No. Bell's inequality shows that if the results of measurements are described by stochastic variables, numbers picked at random, then to reproduce the predictions of quantum theory those variables must change non-locally. This means that hidden variables theories are non-local and so are collapse theories. This doesn't show that QM is non-local because in QM the results of measurements are described by observables represented by Hermitian operators, not stochastic variables. $\endgroup$
    – alanf
    Commented May 4 at 20:07
  • $\begingroup$ Thank you! Also for pointing out the interesting paper! By "collapse theories" you mean theories that expand QM to include the collapse process? With "local" you mean that (causal) effects can only propagate in space from one point to the next with the speed maximum of light? $\endgroup$
    – freecharly
    Commented May 5 at 2:26
  • $\begingroup$ @freecharly Yes, collapse theories modify quantum theory to include collapse, e.g. spontaneous collapse theories arxiv.org/abs/2310.14969 "With "local" you mean that (causal) effects can only propagate in space from one point to the next with the speed maximum of light?" Yes. $\endgroup$
    – alanf
    Commented May 5 at 14:50
  • $\begingroup$ To understand you correctly, by collapse theories you do not mean the "normal" wave function collapse upon measurement, which is part of standard QM (Kopenhagen), but theories beyond standard QM, which describe the process of wave function collapse mathematically. Or do you mean all theories which include a collapse as opposed to non-collapse interpretations like the MWI? $\endgroup$
    – freecharly
    Commented May 5 at 18:00