tag:blogger.com,1999:blog-3832136017893749497.post9117325471450070395..comments2023-09-29T08:49:30.765-04:00Comments on Elliptic Composability: Florin Moldoveanuhttp://www.blogger.com/profile/01087655914212705768noreply@blogger.comBlogger36125tag:blogger.com,1999:blog-3832136017893749497.post-38931796045093048122016-01-28T16:51:52.613-05:002016-01-28T16:51:52.613-05:00Hi Al,
Indeed I don't think we can make any m...Hi Al,<br /><br />Indeed I don't think we can make any more progress and we can agree to disagree. I bet we will meet in person one day and I am looking forward to it.<br /><br />All Best,<br />FlorinFlorin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-63944009487953992412016-01-26T22:07:34.101-05:002016-01-26T22:07:34.101-05:00Hi Florin,
Sorry for the extensive delay. I think...Hi Florin,<br /><br />Sorry for the extensive delay. I think we've reached an impasse now, since I simply don't understand why you continue to insist that proportions of numbers worlds and probabilities have to go together. The only reason I can imagine for thinking this is the assumption that we should treat our world as randomly picked from all the worlds there are. But quantum mechanics is not like a lottery machine, so there's no reason to treat all the individual worlds that exist as equally probable.<br /><br />You talk about falsifiability, but I don't see how the assumption that probabilities is given by number of worlds is any more or less falsifiable than the assumption that probability is given by branch weights. It's not like we can directly measure numbers of worlds either.<br /><br />Thanks for the interesting discussion, hope to get to talk more in person someday.<br /><br />all best,<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-84810670160766160512016-01-17T22:12:43.800-05:002016-01-17T22:12:43.800-05:00Not at all, please be my guest.Not at all, please be my guest.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-84049627140320214342016-01-17T22:11:01.038-05:002016-01-17T22:11:01.038-05:00Hi Al,
“I don't see any reason to accept your...Hi Al,<br /><br />“I don't see any reason to accept your claim that 'This association should be 1-to-1.' What is the argument for this?”<br /><br />I explained why you need this: to have a well-defined meaning of probabilities, otherwise the concept is vacuous just like unicorns or tooth ferries. Physics is an experimental science and you need to attach experimental consequences to concepts. No 1-to-1 association means no experimental distinction between weights and no way to tell apart a fair from a biased coin. Simply stating that you watch the outcomes to determine a fair from an unfair coin is not good enough. This has to follow from theory and it does not because there is no 1-to-1 association. If it does not follow from theory, then fundamentally MWI is not different than: “In the beginning God created the heavens and the earth.” Is MWI faith based, or is it a falsifiable scientific theory?<br /><br />“Similarly, I don't see why the example you give at the end is absurd. There's just no reason to think that branch numbers have anything to do with probabilities, since they're not physically fundamental. ”<br /><br />It is absurd because you get odds impossible to be realized in our universe.<br /><br />“As Saunders and Wallace have argued, and as Gleason's theorem shows in the Everettian context, there's something physically special about the branch weights which distinguishes them from 'toothferiness' and makes them appropriate candidates to be identified with probabilities. ”<br /><br />I never had the pleasure of meeting Saunders, but I listen to many discussions Wallace took part in it and I never comprehended his arguments. If they will humor me and repeat their arguments here in this blog, I will give them my full attention and I will reply to them. But since you are a MWI expert, I presume you can repeat their arguments here. <br /><br />Last year I was having lunch with a leading MWI expert (I will not name him because I do not have his permission). Invariably I have asked him about this probability business. The answer was something like: it must be true because it passed peer review and got published. One may say he had a bad day and did not want to elaborate, but he could have said: not in the mood to talk about this right now, but we can exchange emails about this in the future. My gut feeling was however that the emperor is naked and the appeal to authority was only designed to cover it up and continue the naked precession with the beautiful “emperor’s new clothes”. However, I keep an open mind and I am ready to accept MWI claims if proper arguments are given.<br /><br />Best,<br />Florin<br />Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-76458314515531298542016-01-17T19:25:33.563-05:002016-01-17T19:25:33.563-05:00This is interesting. Would you have any objection ...This is interesting. Would you have any objection to me sharing this with my friends on Google plus?Andrew Wellshttps://www.blogger.com/profile/11256719572679458530noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-73332366063887226602016-01-17T00:27:09.659-05:002016-01-17T00:27:09.659-05:00Hi Florin,
I don't see any reason to accept y...Hi Florin,<br /><br />I don't see any reason to accept your claim that 'This association should be 1-to-1.' What is the argument for this?<br /><br />Similarly, I don't see why the example you give at the end is absurd. There's just no reason to think that branch numbers have anything to do with probabilities, since they're not physically fundamental.<br /><br />As Saunders and Wallace have argued, and as Gleason's theorem shows in the Everettian context, there's something physically special about the branch weights which distinguishes them from 'toothferiness' and makes them appropriate candidates to be identified with probabilities.<br /><br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-7179523640177739492016-01-14T16:41:53.106-05:002016-01-14T16:41:53.106-05:00Hi Al,
The branch weight is meta-information abou...Hi Al,<br /><br />The branch weight is meta-information about the branch: suppose I have two branches and I attach a number to the branch 1 say 0.8 and to branch 2: 0.2. But I can also attach some other meta-information to the branches like color, or a bank account number, or toothferryness. To make them meaningful there should be a way to associate those meta-information to experimental outcomes because this is what an experimentalist actually measures. Toothferiness is an absurd notion but I am making a point. In the case of MWI if the associated weight number is .8 the way to associate the weight with the experimental outcomes should be that if I flip a coin 1 million times, I will get 800 thousands head and 200 thousands tail plus or minus one thousand. <br /><br />This association should be 1-to-1. The problem in MWI is that the prescription is not 1-to-1. No matter which weight I choose, the branches are always the same given by the binomial coefficients. <br /><br />To take this to the extreme, suppose the branch weight for heads is 0.999999999… and the one for tails is 10 to the power of -100, both numbers adding up to 1. If I flip this coin one million times I will get for sure all heads (the odds against it are 1 to 10 to the power 94. For comparison the mass of the visible universe measured in electron masses is less than 10 to the power of 90, four orders of magnitude smaller). But in MWI only 1 out of 2 to the power 1 million (2 to the power of one thousand is 10 to the power of 300) cases I will get all heads. This is absurd. <br /><br />Best,<br />Florin<br /><br />Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-19722785676015461242016-01-12T21:42:37.588-05:002016-01-12T21:42:37.588-05:00Thanks, Florin. I didn't mean anything special...Thanks, Florin. I didn't mean anything special by 'gets a branch weight' - I just meant that the branch has a particular property. So I don't think that it's circular, but perhaps there are other problems with the claim.<br /><br />In fact I think I understand your worry a bit better now. Your concern is that it makes no sense to say that a particular universe is characterized by a parameter like the branch weight / squared-amplitude / co-efficient (or whatever you want to call it). I personally see no problem with this. The branch weight is just a property that the universe has. Quantum mechanics itself describes that property and tells us how to calculate it. Perhaps your concern is that it can't be directly observed - I agree with that. But quantum mechanics still tells us how to estimate it through measurement of relative frequencies as I described above. <br /><br />You say 'information is lost in MWI' - I don't see why this should be. The information about the coefficients isn't encoded in the number of branches, no. But the information doesn't disappear either. Why should it? It's encoded in the coefficient of each branch. There is no mechanism in QM where these coefficients get deleted, or set to zero, or whatever, on a measurement. A process like that is added by some interpretations, but not by Everettians.<br /><br />all best,<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-59982875332082553002016-01-12T18:08:32.632-05:002016-01-12T18:08:32.632-05:00Now I got the definition, but the circularity crit...Now I got the definition, but the circularity criticism is not on the definition but on the action: "a branch gets a..." How does a branch get its weight? This is what I contend it is circular. Do you agree with this statement? <br /><br />I flip a biased coin which lands with probability p heads, and (1-p) tails. In MWI there is a universe where the coin lands heads, and a universe where the coin lands tails. The probability p information is lost in MWI as I always have two branches. And it is lost every single time I repeat the coin flipping. It does not matter if I flip the coin once or a million times. To preserve the p information you have to add it by hand in MWI in a circular way. Your statement is basically this: our universe (meaning our branch) gets the p weight because p is associated with our universe.<br /><br />Best,<br />Florin<br />Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-75030384657053373742016-01-12T06:28:54.042-05:002016-01-12T06:28:54.042-05:00Hi - that was just a part of a definition of the t...Hi - that was just a part of a definition of the term 'branch weight', which you'd asked for! So, I don't think it's fair to call it circular - unless all definitions are circular.<br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-27372885226521989222016-01-11T23:50:21.530-05:002016-01-11T23:50:21.530-05:00Hi Al,
Thank you for your reply. You state: “a br...Hi Al,<br /><br />Thank you for your reply. You state: “a branch gets a high branch weight by a high coefficient being associated with that branch” <br /><br />Now I understand your point, but this statement is circular in my opinion. You stated something above about circularity. <br /><br />Do you agree this particular statement is circular? <br /><br />Best,<br />Florin Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-2476690435548806002016-01-11T01:56:37.590-05:002016-01-11T01:56:37.590-05:00Hi Florin,
Branch weight is just an Everettian te...Hi Florin,<br /><br />Branch weight is just an Everettian term for the coefficient, or squared-amplitude, associated with the branch. So a branch gets a high branch weight by a high coefficient being associated with that branch, and a set of branches gets high total weight by the sum of the coefficients of each branch in the set being high.<br /><br />So, my claim is that as the biased coin is tossed many times, the sum of the coefficients associated with the set of branches in which the relative frequency of Heads approaches 0.8 itself approaches 1. This is what Everett's original argument shows. And it's no more or less convincing than the equivalent argument in a non-Everettian interpretation: as the biased coin is tossed many times, the probability of the relative frequency of heads approaching 0.8 itself approaches 1.<br /><br />As I say, all this is a basic part of the modern Everettian approach due to Saunders, Wallace, etc. They also try to go beyond this using the decision-theoretic arguments, but I maintain that those additional arguments are optional and not required for probability to make sense. We don't require any such additional arguments in the non-Everettian context.<br /><br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-55804309773786662782016-01-10T23:49:46.103-05:002016-01-10T23:49:46.103-05:00Hi Al,
Sorry for the delay in response, I was ver...Hi Al,<br /><br />Sorry for the delay in response, I was very busy working on a paper. <br /><br />You state: <br />“You can't define probabilities in terms of frequencies given Everettian QM. But I say that you can't define them in terms of frequencies at all, no matter what your interpretation. No matter how times you toss your biased coin, there's always a chance that they'll all come up heads. Sure, this chance gets very small as you toss it many times, but the fact remains: probabilities don't align with relative frequencies automatically, but *only with very high probability*. Since the connection between probabilities and frequencies itself presupposes probability, it can't be used to define probability. That's true in all interpretations.”<br /><br />This is why I emphasized the operational point of view. Vaidman asks about self-location, I ask a similar question: can you detect the biased coin in a MWI universe? Suppose I work in a factory producing coins and my job is that of quality assurance: eliminate the biased coins. How do I do it in a MWI universe? <br /><br />Your answer is: <br />“You can do the same as in the non-Everettian case. Toss it many times, observe the frequencies. […] In the Everettian case, branches in which frequencies approach branch weights get high branch weight.”<br /><br />I believe we are at an impasse. I do not understand your statement: “branches in which frequencies approach branch weights get high branch weight”. How does a branch get a branch weight? What does this mean? What is a branch weight?<br /><br />Best,<br />Florin <br />Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-75892723160988058452016-01-09T23:58:35.091-05:002016-01-09T23:58:35.091-05:00Hi Florin,
Thanks for the reply. Given your claim...Hi Florin,<br /><br />Thanks for the reply. Given your claims about operational meaning, I think I now see where you're coming from with this critique. It may be that operationalism/frequentism about probability is what we disagree about, rather than anything specifically quantum.<br /><br /><br />"Let’s consider this: let’s have an unfair (quantum) coin (corresponding to a vertical spin 1/2 measurement) which lands 80% heads and 20% tails. In a non-MWI description (Copenhagen, etc), when I flip the coin I get either heads or tails, and if I repeat it I get on average 80 heads and 20 tails out on 100 tries (up to the expected statistical fluctuations; if you want to reduce them repeat say 1 million times). Now in an Everettian description, at each flip the worlds splits and in each split I get both the heads and tails outcomes. How can I define probabilities in an Everettian description for my unfair coin?"<br /><br />You can't define probabilities in terms of frequencies given Everettian QM. But I say that you can't define them in terms of frequencies at all, no matter what your interpretation. No matter how times you toss your biased coin, there's always a chance that they'll all come up heads. Sure, this chance gets very small as you toss it many times, but the fact remains: probabilities don't align with relative frequencies automatically, but *only with very high probability*. Since the connection between probabilities and frequencies itself presupposes probability, it can't be used to define probability. That's true in all interpretations.<br /><br /><br />"*In other words, in an Everettian universe outcomes by themselves cannot tell a fair from an unfair coin.* And so the concept of probability in an Everettian universe is meaningless."<br /><br />Again, I agree, but I think that probabilities cannot and should not be reduced to 'outcomes by themselves'. We need to bring in the branch weights also.<br /><br /><br />"How can you tell apart a fair from an unfair (quantum) coin in an Everettian universe?"<br /><br />You can do the same as in the non-Everettian case. Toss it many times, observe the frequencies. In the non-Everettian case, worlds in which frequencies approach probabilities get high probability. In the Everettian case, branches in which frequencies approach branch weights get high branch weight. Both approaches are just as circular as one another.<br /><br />By the way, I think Wallace would agree with everything I say above. Look at Saunders, 'What is Probability?', in the 'Many Worlds?' edited collection, for a more detailed presentation of this sort of position.<br /><br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-77759998008432693182016-01-06T22:48:05.978-05:002016-01-06T22:48:05.978-05:00PPS: My argument above is not an argument against ...PPS: My argument above is not an argument against MWI, and I believe Wallace would reply in the same way. It is only an argument against the contention that Born rule need not be derived in MWI. Once the need for deriving Born rule in MWI is established, we need to look at the validity of such a proof. I contend that if you give me such a proof I will either<br />(a) find a mathematical mistake in it<br />or<br />(b) find a place where you assume Born rule and hence your derivation is circular<br /><br />If I cannot do either (a) or (b) I will admit defeat. <br /><br />Since I managed to derive QM from physical principles, I am confident in my QM knowledge to put this challenge to the MWI community. I can also bet you cannot defeat Bell correlations using LHV, or square the circle, but I don't think there are sane takes on those bets.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-27684659220919516832016-01-06T22:10:43.450-05:002016-01-06T22:10:43.450-05:00Hi Al,
Thank you for your reply. Let me respond:
...Hi Al,<br /><br />Thank you for your reply. Let me respond:<br /><br />“The assumption of the existence of a probability function is just as strong an assumption in e.g. Copenhagen as it is in Everett.”<br /><br />I think this is the main contention point. I do not know what probabilities mean in MWI. <br /><br />Let’s consider this: let’s have an unfair (quantum) coin (corresponding to a vertical spin 1/2 measurement) which lands 80% heads and 20% tails. In a non-MWI description (Copenhagen, etc), when I flip the coin I get either heads or tails, and if I repeat it I get on average 80 heads and 20 tails out on 100 tries (up to the expected statistical fluctuations; if you want to reduce them repeat say 1 million times). Now in an Everettian description, at each flip the worlds splits and in each split I get both the heads and tails outcomes. How can I define probabilities in an Everettian description for my unfair coin? <br /><br />To me the probabilities in this case should let me know if the coin is fair or not. <br /><br />If I do the same thing with a fair coin, I get the very same outcomes and word splits as in the unfair case. *In other words, in an Everettian universe outcomes by themselves cannot tell a fair from an unfair coin.* And so the concept of probability in an Everettian universe is meaningless. <br /><br />All other points are secondary to this point and I will not reply to them to preserve clarity. How can you tell apart a fair from an unfair (quantum) coin in an Everettian universe? <br /><br />Best,<br />Florin<br /><br />PS: please note my operational meaning attached to probabilities. I can say probabilities in MWI just as I can say flux capacitor, but without a concrete way to test its experimental consequences, the concept is vacuous, just meaningless marks on paper. Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-7299020682689766882016-01-05T22:41:33.336-05:002016-01-05T22:41:33.336-05:00Hi Florin,
Thanks for the reply. Some responses b...Hi Florin,<br /><br />Thanks for the reply. Some responses below:<br /><br /><br />"You state: “To compare: no other interpretation can derive the Born rule either! They all just postulate it as part of the theory.” This is simply false. Gleason’s theorem derives Born rule given its assumption: the existence of a probability function respecting some natural requirements. No such thing exists in MWI..."<br /><br />I don't understand why you say this. Gleason's theorem is just as applicable to MWI as it is to other interpretations. The assumption of the existence of a probability function is just as strong an assumption in e.g. Copenhagen as it is in Everett.<br /><br /><br />"You also state: “There's no reason why Everettians should not do the same.” This is easier said than done. If you postulate Born rule you have to explain probabilities in MWI and this is where you run into trouble. Show me how you do it. Take the S-G example from the post and show what it means for the probabilities to change when I rotate the device."<br /><br />For Everettians just like for others, for the probabilities to change is just for the squared-amplitude co-efficients to change - no more, no less. I don't see the difference, or the special problem for Everettians.<br /><br /><br />"“Everettians should simply say that branch counting sometimes gives the right answer by coincidence even though it is not the correct rule to use.” I agree with this statement, but this does not reject my claim."<br /><br />I had misunderstood your argument, then. It seemed to me that you were arguing that since branch counting sometimes agrees with the Born rule, Everettians could not consistently maintain that branch counting was incorrect and the Born rule was correct. I pointed out that this would be a bad argument, using the clock analogy. But if your point is just that all derivations of the Born rule are either incorrect or circular, then my objection does not apply.<br /><br />As it happens I'm agnostic about whether Everettians can derive the Born rule. I'm not sure whether the existing derivations work, and I'm not sure whether other derivations might arise in the future. I thought you were arguing that no derivation could possibly work, and that was the argument I was objecting to.<br /><br /><br /><br />"I run the experiment N times and I get N+1 worlds with distinct total numbers of up and down. I rotate the device and repeat the same experiment N times: I get the same N+1 worlds with distinct total numbers of up and down. *No change*." <br /><br />No change in the number of worlds, no, but Everettians don't have to agree that this means there are no changes at all. They can say that there are changes in the coefficients so there are changes in the probabilities. <br /><br /><br />"The value of the coefficients simply do not matter."<br /><br />What is the argument for this? All the Everettians I know would deny it. It seems to me that you're presupposing that the only thing that could possibly be relevant to probabilities for Everettians is the branch numbers. And I don't think there's any reason to make that presupposition.<br /><br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-76041679679215814512016-01-03T11:04:12.486-05:002016-01-03T11:04:12.486-05:00Hi Kashyap,
Here it is (from Ingemat Bengtsson an...Hi Kashyap,<br /><br />Here it is (from Ingemat Bengtsson and Karol Zyczkowski book: Geometry of Quantum States, page 143):<br /><br />Assumption 1: normalization: The elements \(|e_i \rangle\) of every orthonormal basis are assigned probabilities such that:<br />\(\sum_{i=1}^{N} p_i |e_i\rangle \langle e_i| = 1\)<br /><br />Assumption 2: non-contextuality: Every vector is an element of many orthonormal bases. The probability of its ray is independent of how the remaining vectors of the basis are chosen.<br /><br />Gleason's theorem: under the conditions stated, and provided the dimension of N of the Hilbert space obeys N>2, there exists a density matrix \(\rho \) such that \(p_i = Tr (\rho |e_i \rangle \langle e_i|\))Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-65300560487151504162016-01-03T10:43:28.326-05:002016-01-03T10:43:28.326-05:00Hi Florin,
Can you summarize Gleason's assumpt...Hi Florin,<br />Can you summarize Gleason's assumptions for lazy people like me (!)<br />who do not want to read his papers? Are they reasonable and acceptable right away?kashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-76800080807816286892016-01-03T09:42:07.613-05:002016-01-03T09:42:07.613-05:00Hello Al.
You have a very interesting take on the...Hello Al.<br /><br />You have a very interesting take on the issues, and I’ll try to reply. Let me start by quoting Wallace from “Elegance and Enigma”, in the “Big Issues” chapter on page 55:<br /><br />“As it happens, I *do* think the measurement problem is solvable within ordinary quantum mechanics: I think the Everett (“many worlds”) interpretation solves it in a fully satisfactory way, and while I think there are some philosophical puzzles thrown up by that solution—mostly concerned with probability and with emergence—that would benefit more thought, I wouldn’t call them *pressing*.”<br /><br />Now I contrast this sentence with Kelvin’s two clouds speech from 1900:<br /><br />“The beauty and clearness of the dynamical theory, which asserts heat and light to be modes of motion, is at present obscured by two clouds.”<br /><br />Clouds or not, pressing or not, the issues of probability and Born rule emergence sinks the whole MWI approach and I am putting the spotlight on the core issues of MWI.<br /><br />You state: “To compare: no other interpretation can derive the Born rule either! They all just postulate it as part of the theory.” This is simply false. Gleason’s theorem derives Born rule given its assumption: the existence of a probability function respecting some natural requirements. No such thing exists in MWI and that is why Born rule is not derivable in MWI. And this is why all known valid derivations in MWI are circular.<br /><br />You also state: “There's no reason why Everettians should not do the same.” This is easier said than done. If you postulate Born rule you have to explain probabilities in MWI and this is where you run into trouble. Show me how you do it. Take the S-G example from the post and show what it means for the probabilities to change when I rotate the device.<br /><br />Now onto your other points. <br /><br />First “your claim about mathematical inconsistency doesn't hold up.” Perhaps, but I see no real rebuttal of my claim. <br /><br />“But that doesn't mean an Everettian who embraces the Born rule but rejects branch counting is being inconsistent”. This is a wishful ex cathedra statement without proof. As far as I can tell, the only proof offered was: <br /><br />“Everettians should simply say that branch counting sometimes gives the right answer by coincidence even though it is not the correct rule to use.” I agree with this statement, but this does not reject my claim. <br /><br />Here is my claim again: “when not circular, Born rule derivations in MWI are mathematically incorrect”. All the evidence so far is consistent with my claim: I know only invalid or valid circular derivations of Born rule in MWI. I think on this aspect there should be no debate. The real test of my claim is based on future attempts to derive the Born rule. My claim is falsifiable: show me a valid non-circular derivation of Born rule in MWI and I stand corrected. I already rebutted above: “There's no reason why Everettians should not do the same.” and therefore the only way to disprove/reject my claim is to produce a valid non-circular derivation of Born rule in MWI.<br /><br />“Everettians can simply take it as a primitive claim of the theory that the Born rule specifies probabilities (or their functional equivalents).”<br /><br />Again, easier said than done. Take the S-G example from the post and show what it means for the probabilities to change when I rotate the device.<br /><br />“Saying that probabilities are given by the squared-amplitudes is a theoretical identification, like saying that water=H20. Such identifications don't have to be derived but are judged on their theoretical consequences.” <br /><br />This is fine with me. What are the consequences in my example from the main text? I run the experiment N times and I get N+1 worlds with distinct total numbers of up and down. I rotate the device and repeat the same experiment N times: I get the same N+1 worlds with distinct total numbers of up and down. *No change*. The value of the coefficients simply do not matter.<br /><br />Best,<br />Florin<br />Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-51593135283423243342016-01-02T20:52:41.683-05:002016-01-02T20:52:41.683-05:00Hello Florin,
As a proponent of the many-world ap...Hello Florin,<br /><br />As a proponent of the many-world approach I have a couple of responses to your arguments here.<br /><br />Firstly, a specific point: your claim about mathematical inconsistency doesn't hold up. It's true that in cases of equal coefficients, the Born rule probabilities coincide with the probabilities that would be derived from branch counting given an appropriate decomposition. But that doesn't mean an Everettian who embraces the Born rule but rejects branch counting is being inconsistent. Even a stopped clock is right twice a day - but we shouldn't use it to tell the time. In the same way, Everettians should simply say that branch counting sometimes gives the right answer by coincidence even though it is not the correct rule to use. <br /><br />Now to the more general point. You challenge Everettians to produce a derivation of the Born rule. While some (Deutch, Wallace, Carroll, Sebens) do try to do this, I agree with Saunders, Papineau and others that in fact no such derivation is necessary for the many-worlds approach to be adequate. Everettians can simply take it as a primitive claim of the theory that the Born rule specifies probabilities (or their functional equivalents). Saying that probabilities are given by the squared-amplitudes is a theoretical identification, like saying that water=H20. Such identifications don't have to be derived but are judged on their theoretical consequences. To compare: no other interpretation can derive the Born rule either! They all just postulate it as part of the theory. There's no reason why Everettians should not do the same.<br /><br />all best<br />AlAlastair Wilsonhttp://alastairwilson.orgnoreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-1734521398182907992015-12-23T10:44:36.661-05:002015-12-23T10:44:36.661-05:00Hi Kashyap,
First I don't quite know what to ...Hi Kashyap,<br /><br />First I don't quite know what to call Carroll's proposal, perhaps MWI^2 (MWI squared) because he seems to have something new: a MWI split within a MWI branch. Adrian Kent's criticism from the link in Lubos' reply is rather flimsy and if I were a MWI believer I could very easily dismiss it: the jump from eq 2 to 3 is simply distributivity regardless of how I interpret it. <br /><br />There is ambiguity in Caroll's proposal: does a world branch split internally or not? There is language to suggest that the branch split happens, but there is also language that suggests that the usual (Everett) MWI picture is considered.<br /><br />Criticizing Carroll's proposal requires the ambiguity to be settled first. Until that time I do not consider it a fully formed proposal. It is possible that Carroll already removed this ambiguity in recent papers I am not aware of yet. After the ambiguity is resolved, my arguments from the post above applies. Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-55003552887867405502015-12-23T09:34:23.921-05:002015-12-23T09:34:23.921-05:00Thanks Lubos. We are old buddies on your blog!Thanks Lubos. We are old buddies on your blog!kashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-63025996714012815532015-12-23T05:34:59.134-05:002015-12-23T05:34:59.134-05:00Dear Kashyap, you haven't asked me but there&#...Dear Kashyap, you haven't asked me but there's some literature on that stuff. The Carroll-Sebens "derivation" has one followup that discusses the content in detail, <a href="http://arxiv.org/abs/1408.1944" rel="nofollow">one by Adrian Kent</a>, which explains that the paper by Carroll-Sebens is a collection of internally inconsistent ill-defined and vague assertions taken out of thin air. There is absolutely no valuable content in those papers.Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-27555452562242685262015-12-22T17:54:32.846-05:002015-12-22T17:54:32.846-05:00Hi Florin,
What is your opinion about Sean Carroll...Hi Florin,<br />What is your opinion about Sean Carroll and collaborators' derivation of Born rule? Does that also assume circular arguments?kashyap vasavadahttps://www.blogger.com/profile/10732897306667764590noreply@blogger.com