On Tue, Apr 26, 2016 at 9:16 PM, Bruce Kellett <[email protected]> wrote:
> On 27/04/2016 1:13 am, Jesse Mazer wrote: > > On Tue, Apr 26, 2016 at 6:45 AM, Bruce Kellett <[email protected]> > wrote: > >> >> You think that "the state of the other particle" refers to the quantum >> state that would be assigned to B given only knowledge of the state of A >> (as well as knowledge of how they were entangled originally). Actually, >> that is the interpretation I gave the words, except I teased out what that >> actually meant. From the entangled state, given A's state (result, say >> |+>), you would assign a state |-> to B. But this is wrong for spacelike >> separations -- the state B actually measures is exactly the same as the >> state A measured: |psi> = (|+>|-> - |->|+>)/sqrt(2). >> > > You use the full state if you just want to generate the total > probabilities for various possible *joint* outcomes. But if you want a > conditional probability of various outcomes *just for B* given knowledge of > what measurement A got, this can be done in QM, in the Schroedinger picture > you could project |psi> onto on eigenstate that corresponds to whatever > definite outcome was measured on A, resulting in a different state vector > for the combined system |psi'> which may lead to different probabilities of > getting various results for B, but which does not assume any knowledge of > what measurement was actually performed on B. I assume something similar is > possible in the Heisenberg picture which Rubin is using, so I was > speculating that he meant something like this when he talked about a label > on one particle which says something about the state of the other particle. > > > I am well aware of this, and I also thought that was probably what Rubin > had in mind. The problem is that this simply sneaks non-locality in the > back door -- neither Rubin nor you appear to realize this. This is often > the problem I find with these attempts to give a local account of EPR -- > non-locality is built in unobtrusively! > > That is why I said that, in any strictly local account, if A gets |+>, B > still measures the original |psi>. The measurement by A cannot *locally* > affect the state that B measures (or vice versa). > OK, let's say experimenter A measures particle 1, and experimenter B measures particle 2. Any given copy of particle 1 has a "label" that says something about the state of 2--we can imagine that the copy of particle 1 carries a little clipboard on which is written down both its own quantum state, and a quantum state it assigns to particle 2. When that copy of 1 is measured, it not only adjusts its own state (to an eigenstate of the measurement operator), it also adjusts the state it has written down for 2. You seem to be assuming, in effect, that when a copy of 1 adjusts what it has written down for the state of 2 on its own clipboard, this must mean that copies of 2 also instantaneously adjust what they have written down about *their* own state. However, in a copying-with-matching scheme, there's no reason this need be the case! The state that particle 1 assigns to particle 2 on its clipboard may just be for the purposes of later matching--deciding which copy of 2 to "partner up with" once it can meet them (or get some type of causal influence from them). The fraction of copies of 2 that show a given result when B measure can still be totally independent of what the various copies of 1 have written down on their clipboards about the state *they* assign to 2. For example, say we are using a particular setup where if particle 1 is measured along a spatial vector V (say, one parallel to to the x-axis and pointing in the +x direction) and gives a result +, that means if particle 2 is measured at a 120-degree angle from V, it will have a 75% chance of giving the result + and a 25% chance of giving the result -. So if a given copy of particle 1 is indeed measured along V and does give a result +, it can adjust the state it assigns to particle 2 on its clipboard accordingly, assigning 2 a state (or reduced density matrix) which has an amplitude-squared of 0.75 for + at an orientation of 120 degrees from V. It can pass on this clipboard information (Rubin's 'label') to copies of other systems it interacts with, like the experimenter, who carry their own clipboards/labels. Then if that copy of the experimenter later interacts with particle 2 (or with some other particle or system that conveys information about particle 2), the state assigned to 2 on the experimenter's clipboard is used to decide which copy of particle 2 it should be matched with. In this case, this could ensure that if it gets matched to a copy of particle 2 that was indeed measured at an angle of 120 degrees from V, there is a 75% chance it'll be matched to a copy of particle 2 that gave the result + (i.e. a copy of particle 2 that has an amplitude-squared of 1 for + at an orientation of 120 degrees from V), and a 25% chance it'll be matched to a copy of particle 2 that gave the result - (i.e. a copy of particle 2 that has an amplitude-squared of 0 for + at an orientation of 120 degrees from V). Is there anything in Rubin's words that clearly rules out the possibility that the "label" one particle carries about a second particle's state is only for the purpose of matching in this way, and that the label has no effect whatsoever on the actual fraction of copies of the second particle that showed a given measurement result? Or perhaps this interpretation just didn't occur to you? > It isn't obviously wrong in my interpretation above, and I think it's > wrongheaded to imagine you can be confident about the interpretation of any > verbal statement by a physicist if you don't have a detailed grasp on the > mathematics of the model the physicist is talking about--if you don't you > may miss possible interpretations, like the ones above that you don't seem > to have considered. > > > I think your arrogant patronizing here is a bit over the top! I have > perfectly well understood the mathematics of Rubin's paper-- that is why I > was confident that the paragraphs I quoted accurately summarized his > detailed arguments and results. > If you understand the mathematics, why not point to the actual equations that you think are incorrect rather than an ambiguous verbal summary? I didn't think I was being patronizing, since I said several times that I myself didn't understand the detailed mathematics of his paper myself (not having much familiarity with the Heisenberg picture of quantum mechanics which the paper is using). > Also, do you plan to respond to the rest of my comment? In particular, do > you think you can come up with any simple numerical examples that show a > local-copies-with-matching model can't correctly reproduce some observed > statistics at a given location if we assume that location has been > "shielded" from any physical influences from Alice or Bob (and assuming > 'matching' between copies of Alice and copies of Bob can only be done in > regions that have received measurable physical signals from them), as you > seemed to claim earlier? > > > No, I have no intention of replying to any of this because it is all > beside the point. If you can't follow the simple physical and conceptual > arguments that I make, numerical examples are not going to help much. > Why not try it and see how I respond, instead of immediately assuming that because I find your English-language arguments ambiguous or unconvincing, that implies I would find it impossible to follow a numerical example? (talk about patronizing!) I did get my undergraduate degree in physics, I'm pretty sure I could follow a numerical example involving the statistics of entangled particles, and I think I would find it much *easier* to follow that a purely verbal argument because there'd be less room for ambiguity in interpreting what you mean. > If Alice and Bob are truly local, and fully independent, then no matching > scheme can ever have the necessary information to reproduce the quantum > probabilities -- where do you find the cos^2(theta/2) basis for the > probabilities if they are truly independent? > The numerical example I gave earlier, showing how copies-with-later-matching could reproduce a certain set of statistics that violate a Bell inequality, was actually based on an experiment with entangled particles where the probabilities are derived from the cos^2(theta/2) relation you mention. In a message I posted on April 19, I said: 'For example, one Bell-inequality-violating quantum experiment would involve Alice and Bob each choosing from one of three detector angles, with the result that when they choose the same angle they are guaranteed to get opposite results with probability 1, whereas when they measure different angles they only have a 1/4 probability of getting opposite results (the corresponding Bell inequality says that in any local realist theory, if they get opposite results with probability 1 when they use the same setting, the probability of getting opposite results on different settings must be greater than or equal to 1/3--if anyone's interested, I did a little derivation of this in a post at http://physics.stackexchange.com/a/140883/59406 ). So in the simulation, let's say we have 360 copies of Alice and Bob each, and 120 copies of Alice used each of the three settings 1,2,3, likewise with Bob. Of the 120 copies of Alice who used setting 1, 60 got the result + and 60 got the result -. If we just look at the 60 copies of Alice who used setting 1 and got result +, then when the collection of messages from copies of Bob arrives at the computer simulating the copies of Alice, it will assign 20 of these copies of Alice to get the message "Bob used setting 1 and got result -", 15 of them to get the message "Bob used setting 2 and got result +", 5 of them to get the message "Bob used setting 2 and got result -", 15 of them to get the message "Bob used setting 3 and got the result +", and 5 of them to get the message "Bob used setting 3 and got the result -". So indeed, we find that the Alice-copies who learn that Bob used the same detector setting as her will always learn that Bob got the opposite result with probability 1, whereas the Alice-copies who learn that Bob used a different detector setting will only have a 1/4 chance of hearing that Bob got the opposite result from their own.' This particular result can be seen in an experiment where the stern-gerlach devices that measure polarization can be oriented at three possible angles at 120-degree intervals--one device could be oriented in the direction of some vector V, the second orientated along a vector 120 degrees from V, and the third along a vector 240 degrees from V (all in the same plane). For a pair of spin-entangled electrons, if both electrons are measured at the same orientation, they have a probability of cos^2(0) = 1 of giving opposite results (if one gives result +, the other gives result - with probability 1). On the other hand if they are measured at different orientations, the angle between the two detectors must be ±120 or ±240, so they have a probability of cos^2(±60) = cos^2(±120) = 0.25 of giving opposite results. And these are exactly the statistics that would be seen by a randomly-selected copy of Alice or a randomly-selected copy of Bob using the rules above. These rules first locally determine the number of copies of Alice and number of copies of Bob that see each result for their own measurement, without any knowledge of the orientation or result for the other experimenter (in this case, the rule can be very simple--however many copies of Alice use a given detector angle, exactly half of those copies get + and half get -, and likewise for the Bob-copies). Only later do the rules do any matching to determine which copy of Alice gets matched to which copy of Bob, and they do it in a way that gives the statistics above. So, the final probability that a randomly-selected matched pair will have a given pair of results will match the probabilities you get if you use the cos^2(theta/2) rule to predict the real-world probabilities for this particular experiment (and as I said, the resulting probabilities are ones which violate a Bell inequality). Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
