tag:blogger.com,1999:blog-3832136017893749497.post897514073756654078..comments2017-11-16T23:05:04.929-05:00Comments on Elliptic Composability: Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.comBlogger30125tag:blogger.com,1999:blog-3832136017893749497.post-77265627049514327672017-08-06T22:29:36.710-04:002017-08-06T22:29:36.710-04:00In my example the coefficients were in the ratio 1...In my example the coefficients were in the ratio 1:2, which is your S-G example with the numbers in the reverse order. (The overall factor of 1/sqrt(5) was ignored by construction.) You are suggesting I generate new coefficients, by cubing the old coefficients, which would put them in the ratio of 1:8. (The overall factor is now 5^-1.5, which, again, we can ignore by construction.) The relative number of orthonormal worlds is now 1:64 ;therefore the probabilities come out as 1/65 and 64/65, respectively.<br /><br />My system works for any rational coefficients. Irrational coefficients would require some additional Peano construction.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-43894340881387237782017-08-06T21:07:37.620-04:002017-08-06T21:07:37.620-04:00Can you prove this? I don't think so. What if ...Can you prove this? I don't think so. What if I consider the coefficients raised to an arbitrary power "r" to compute the probabilities (pick r=3 for definiteness sake). If the total probability is not one, I renormalize. In the 50-50 case I get the usual answer,but not in the 80-20 case.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-74800370902514866482017-08-06T00:49:32.794-04:002017-08-06T00:49:32.794-04:00Because the other constructions do not yield a dec...Because the other constructions do not yield a decomposition into a sum of orthonormal components and an *overall* multiplying factor. You have no overall symmetry to appeal to in deriving the Born rule.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-9753016945839848632017-08-06T00:04:46.788-04:002017-08-06T00:04:46.788-04:00"Anything you get with other constructions is..."Anything you get with other constructions is not relevant."<br /><br />Why not?Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-39962843017234755202017-08-05T09:13:04.715-04:002017-08-05T09:13:04.715-04:00You can imagine any imaginary degenerate splitting...You can imagine any imaginary degenerate splitting you like, but only the splitting as outlined is *relevant* to the Born rule. Anything you get with other constructions is not relevant.<br /><br />I fail to see the circularity involved. No physical assumptions have been made. michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-27759021522731728662017-08-05T08:49:55.980-04:002017-08-05T08:49:55.980-04:00Then I can "imagine" that each state is ...Then I can "imagine" that each state is split to a different degree and I recover something different than Born rule. <br /><br />The point when you "*imagine* each state is split to the required degrees" is where you introduce circularity in your proof.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-5665215253794244642017-08-05T07:48:21.367-04:002017-08-05T07:48:21.367-04:00Actually the problem arises even when there is a d...Actually the problem arises even when there is a degeneracy, since there is no physical reason for the magnitude or power of the degeneracy to match the required degree of subdivision for the demonstration to work. <br /><br />The proof works because the degeneracy is only fictitious or imaginary. We *imagine* each state is split to the required degrees, and then we recover reality by taking the limit as the splitting of the eigenvalues become infinitesimal.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-25751378418791257412017-08-04T09:30:50.242-04:002017-08-04T09:30:50.242-04:00Michael, I was reading your quora proof, and it do...Michael, I was reading your quora proof, and it does not hold water. What happens when there is no degeneracy? Simple counterexample: a S-G device measuring electron spin is rotated at say 20 degrees from vertical, and the incoming beam of electrons has all spins up on the vertical axis. The outcomes are not 50-50 and there is no degeneracy.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-56802364198865309852017-08-04T09:20:58.039-04:002017-08-04T09:20:58.039-04:00Hello, I am back. I will resume my regular posts S...Hello, I am back. I will resume my regular posts Sunday night. I will now read the comments and reply to some of them.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-5936870283954351062017-08-04T01:26:38.723-04:002017-08-04T01:26:38.723-04:00A loophole is a tiny omitted possibility that the ...A loophole is a tiny omitted possibility that the conclusion could be different despite an otherwise meaningful argument.<br /><br />Your paper can't have loopholes because it is an uninterrupted stream of nonsense.Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-57358841827670416622017-08-02T04:45:03.273-04:002017-08-02T04:45:03.273-04:00I do not agree with Dewitt that his (& Graham&...I do not agree with Dewitt that his (& Graham's) proof was incomplete. That aside, what was Caves and Schacks one valid criticism of Hartle, and does it apply to DeWitt? michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-74070490795428416492017-08-02T04:37:44.238-04:002017-08-02T04:37:44.238-04:00DeWitt's argumentation is not complete, and he...DeWitt's argumentation is not complete, and he conceded that Hartle's is the more complete version. But that version was heavily criticized by Caves and Schack. Their argumentation is flawed on all counts but one, but that was enough to kill it.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-57598370151950834002017-08-02T04:30:37.502-04:002017-08-02T04:30:37.502-04:00DeWitt's demonstration is the most complete; b...DeWitt's demonstration is the most complete; being frequestist it explains how probabilities arise. It makes no assumptions beyond the axioms of Hilbert space. The objections seem vague; uncomfortable with the infinite limit being taken in the frequentist derivation - but the frequentist approach only becomes rigorous in the infinite limit. <br />My little proof is only intended to be heuristic, not mathematically watertight.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-81490502746605020762017-08-02T03:12:50.376-04:002017-08-02T03:12:50.376-04:00I note your dogmatism and that you haven't fou...I note your dogmatism and that you haven't found any loop holes in my article. I will not respond further if your level of argumentation don't improve.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-85737755177163163642017-08-02T01:10:48.524-04:002017-08-02T01:10:48.524-04:00I am not ridiculing and I don't need to be rid...I am not ridiculing and I don't need to be ridiculing because it's already more ridiculous than almost all other ridiculous things.<br /><br />There is only one meaning of "probability" which has the well-known frequentist and Bayesian aspects - probability always has them. Quantum mechanics has shown that the predictions must fundamentally be in terms of probabilities. That's it.<br /><br />People had problems with quantum mechanics but it was always their psychological or religious problem, not an imperfection of quantum mechanics.Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-43646223093315219272017-07-31T01:59:24.711-04:002017-07-31T01:59:24.711-04:00Everett assumed that the probabilities should only...Everett assumed that the probabilities should only depend on the amplitudes. He continued with showing that the under that assumption it will only be the magnitude that matters and then that it will be equal to the magnitude squared. He failed also to show how probabilities come about in the many-worlds interpretation. Instead, he also assumed the existence of probabilities. <br /><br />It is necessary to show how probabilities arise, else we haven't shown that the many-worlds interpretation describe the measurement process. As I understand Florin's post, the point is that the way probabilities actually appear in MWI contradicts observation and this disproves MWI. Everett's work is not convincing enough to counter such statements.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-61433467868334979042017-07-30T14:58:09.836-04:002017-07-30T14:58:09.836-04:00Everett gave a convincing argument as to why the p...Everett gave a convincing argument as to why the probabilities should be a function of the modulus of the ampltidues alone.<br />My piece is for people who are looking to explain why the probabilities are proportional to the square of modulus of the amplitudes, rather than just the modulus. Nothing more.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-15921563224765154342017-07-30T11:47:47.095-04:002017-07-30T11:47:47.095-04:00"If the only difference between the states is..."If the only difference between the states is the amplititude, then what else can the probabilities depend upon?" - It is up to you to be able to argue if you want to convince anybody who doubts the validity of the many-worlds interpretation.<br /><br />You are right that Weinberg's argumentation is not very strong, but that is no reason for your argumentation to be at the same level. <br /><br />There are other better reasons to not buy Hartle's argumentation. In my article I bring up the argument that his derivation is for a coherent state which is quite different from the situation where probabilities arise. See also the article by Caves and Schack.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-61368242414899403952017-07-30T04:16:22.789-04:002017-07-30T04:16:22.789-04:00If the only difference between the states is the a...If the only difference between the states is the amplititude, then what else can the probabilities depend upon? <br />My argument is a heuristic, symmetry based argument. I am not interested in mathemetical quibbling, which also seems to be the objections to DeWitt's demonstration - as far as I can tell from your article. Weinberg merely says DeWitt's argument is "a stretch", which is not very helpful.michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-18934846391674020752017-07-30T03:40:03.927-04:002017-07-30T03:40:03.927-04:00It is not that easy to prove the Born rule. Your f...It is not that easy to prove the Born rule. Your first suggestion presumes that the probability only depends on the amplitudes. What's your argument for that?<br /><br />DeWitt's approach, which in a more stringent form was given by Hartree in 1968 Am J Phys vol 8 page 704, is heavily criticized in the literature, see S. Weinberg, Quantum Mechanics, and my article http://arxiv.org/abs/1603.01625 .<br />Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-86294588591829699132017-07-29T04:01:47.164-04:002017-07-29T04:01:47.164-04:00To derive probabilities in MWI you just compare th...To derive probabilities in MWI you just compare the relative number of orthonormmal worlds for each outcome.<br /><br />If you don't like that approach use DeWitt's 1970 derivation, which shows that the norm of the sum of the worlds in which the Born rule is violated vanishes.<br /><br />See my Quora answer.<br /><br />https://www.quora.com/How-does-the-Many-Worlds-Interpretation-of-Quantum-Mechanics-explain-the-emergence-of-probabilities-according-to-the-Born-rule/answer/Michael-Price-29michael.price917https://www.blogger.com/profile/06228097084087632117noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-35865915906043689682017-07-25T04:47:04.231-04:002017-07-25T04:47:04.231-04:00Since the early years of quantum mechanics, the me...Since the early years of quantum mechanics, the measurement process that give a stochastic outcome has been problematic to understand in view of the unitary development in all other processes. Any good physicist should be open to that the probability concept that appear in quantum physics is fundamentally different from the classical probability concept. That is, it works in many ways as a classical probability, but fundamentally quantum probabilities is something different. <br /><br />You need to do better then ridiculing that possibility. If you don't find any actual loopholes in my argumentation, then you should concede.<br /><br />Your response is LIKE telling a string theoretician that electrons appear to be a fundamental particle and it is therefore wrong to describe it as a state of string. Something cannot simultaneously be a fundamental particle and not a fundamental particle.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-4594510085990680202017-07-25T01:40:02.130-04:002017-07-25T01:40:02.130-04:00Something's being the probability and somethin...Something's being the probability and something's appearing to be the probability is exactly the same thing from a physics viewpoint because physics studies things that can be observed, i.e. how they ultimately appear.<br /><br />It's pure fog to suggest that one can stand on a third way in between where something simultaneously is a probability and isn't a probability.Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-40711739851971247032017-07-20T06:30:28.707-04:002017-07-20T06:30:28.707-04:00What is it about Born rule that physicists actuall...What is it about Born rule that physicists actually know? We have seen that in measurements the relative frequency of a particular value is given by the (or rather very close to) the amplitude absolute squared. That is what my theory says we should see. Thus my theory agrees with observation and is so far not falsified by experiments.<br /><br />I have proven that the amplitude absolute square appear to be a probability for physicists that believe in a single world outcome. <br /><br />My theory also supports the use of probability theory to analyze quantum physics experimental data. The theory is so far neutral with respect to the choice between Bayesian and frequentist probability theory. That is, for all practical purposes the amplitude absolute square can be considered to be a probability.<br /><br />In the context of this debate, your statement "the rho and psi-squared *are* probabilities" is problematic in two ways. First, in science we can only describe nature. We can do it better and better, but we can never claim that we know how nature really is. Second, probability is a very difficult concept and philosophers haven't yet agreed upon how to understand it.Per Arvehttps://www.blogger.com/profile/00320681832870710523noreply@blogger.comtag:blogger.com,1999:blog-3832136017893749497.post-56815252963551902392017-07-20T02:02:17.495-04:002017-07-20T02:02:17.495-04:00If the density matrix or the wave function squared...If the density matrix or the wave function squared doesn't determine the probabilities in your theory, and instead, it is something analogous to a diluted distribution of a continuous charge, then your theory is just immediately falsified by the experiments because the experiments prove that the rho and psi-squared *are* probabilities.<br /><br />You are just a total hack pretending to be a physicist.Luboš Motlhttps://www.blogger.com/profile/17487263983247488359noreply@blogger.com