Re: [kevintr...@hotmail.com: Jacques Mallah]
On Sun, Feb 08, 2009 at 09:34:30PM -0500, Jesse Mazer wrote: > > > > > > Date: Mon, 9 Feb 2009 13:02:31 +1100 > > From: li...@hpcoders.com.au > > To: everything-l...@googlegroups.com > > Subject: Re: [kevintr...@hotmail.com: Jacques Mallah] > > > > > All I have ever said was that effective probability given by the > > squared norm of the projected eigenvector does not follow from Born's > > rule. It can't follow, because Born's rule says nothing about what the > > normalisation of the state vector after observation should be. It is a > > conditional probability only. > I still don't understand the connection you're making. When people say the > effective probability is equal to the amplitude squared, it doesn't require > you to assume anything about the state vector *after* observation (in > particular you don't have to assume an objective collapse), it's just the > square of the norm of the vector you get when you project the system's > (normalized) state vector at the instant *before* observation onto an > eigenvector. > Jesse Sure. What you've just said is just an interpretation-free description of the mathematics. Whether it is a helpful description is another matter. But Jacques Mallah is making a metaphysical claim when saying it is equal to the squared amplitude of the branch. Which is a completely different beast. He must have some model in mind which tells us how the "amplitude" of the branches relates to the "amplitude" of the original state. Cheers -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: [kevintr...@hotmail.com: Jacques Mallah]
> Date: Mon, 9 Feb 2009 13:02:31 +1100 > From: li...@hpcoders.com.au > To: everything-l...@googlegroups.com > Subject: Re: [kevintr...@hotmail.com: Jacques Mallah] > > All I have ever said was that effective probability given by the > squared norm of the projected eigenvector does not follow from Born's > rule. It can't follow, because Born's rule says nothing about what the > normalisation of the state vector after observation should be. It is a > conditional probability only. I still don't understand the connection you're making. When people say the effective probability is equal to the amplitude squared, it doesn't require you to assume anything about the state vector *after* observation (in particular you don't have to assume an objective collapse), it's just the square of the norm of the vector you get when you project the system's (normalized) state vector at the instant *before* observation onto an eigenvector. Jesse --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
On Sun, Feb 08, 2009 at 08:33:52PM -0500, Jesse Mazer wrote: > > > > > > Date: Mon, 9 Feb 2009 11:47:02 +1100 > > From: li...@hpcoders.com.au > > To: everything-l...@googlegroups.com > > Subject: Re: [kevintr...@hotmail.com: Jacques Mallah] > > > > > > Jesse, you need to fix up your email client to follow the usual > > quoting conventions, wrap lines etc. > I'm using hotmail, any idea what I'd need to do to make it work correctly? Looking at how the posts show up on google groups, it looks like my previous posts didn't have this problem, I'm not sure what went wrong with the last one. This time it looks better, although you might want to enable word wrap on 80 columns for the text only mode. I've never used Hotmail, so can't comment on specifics. I would hope that it doesn't trash email conventions built up over a number of years, but given that its Microsoft owned, and they did precisely that with the abomination called Outlook, one can't be too optimistic. > > > > > By working *effectively*, I think you mean the subjective experience (or > > 1st person experience) should be as though \psi is transformed to > > \psi_i, where \psi_i is an eigenvector of the relevant observable. > > > > Of course no such thing happens in the MWI, ie the 3rd person > > description. \psi remains unaltered in this case, and is undergoing > > regular unitary evolution. > Yes, my understanding is that decoherence might be of some use in explaining > the appearance of collapse even if we assume that an experimenter measuring a > quantum system merely becomes entangled with it, with the wavefunction of the > combined experimenter/quantum system continuing to evolve in a unitary way. > There's a discussion of this on Greg Egan's page at > http://gregegan.customer.netspace.net.au/SCHILD/Decoherence/Decoherence.html I think the einselection issue should be orthogonal to issues of observer measure, but suspect that Mallah may be using it. Einselection does have its problems, though. > > > > > In any case, there is no requirement for to be given > > by the product of the Born rule probability with . > I don't understand, why is this implied by what Jacques or I said? My comment > was that the "Born rule probability" is equal to , and since is the > amplitude, it's accurate to say the probability is the amplitude-squared > (although after editing the wikipedia article to say this, I was told that > this only works for operators where the eigenstates are 1D eigenvectors, and > that there can be cases where the eigenstate is an 'eigenspace' with more > than one dimension in which case the Born rule probability can't be written > this way). I don't see how this implies that = (Born rule probability)*, > i.e. = , which is what you seem to be accusing Jacques of claiming above. > (By the way, this would probably be easier to read if a different symbol than > \psi was used for eigenstates, as I did in my previous post) All I have ever said was that effective probability given by the squared norm of the projected eigenvector does not follow from Born's rule. It can't follow, because Born's rule says nothing about what the normalisation of the state vector after observation should be. It is a conditional probability only. Now Jacques must have a model for what happens to the state vector after observation - I made an interpretation that the normalisation might be given by chaining Born rule probabilities with the squared norm of the original state just to try an advance understandinf, but regardless of whether that is correct, Jacques needs to come clean about what model he is using. What does "total squared amplitude of a branch" actually mean? > Jesse > -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: [kevintr...@hotmail.com: Jacques Mallah]
> From: laserma...@hotmail.com> To: everything-l...@googlegroups.com> Subject: RE: [kevintr...@hotmail.com: Jacques Mallah]> Date: Sun, 8 Feb 2009 20:33:52 -0500>> I don't understand, why is this implied by what Jacques or I said? My comment was that the "Born rule probability" is equal to , and since is the amplitude, it's accurate to say the probability is the amplitude-squared (although after editing the wikipedia article to say this, I was told that this only works for operators where the eigenstates are 1D eigenvectors, and that there can be cases where the eigenstate is an 'eigenspace' with more than one dimension in which case the Born rule probability can't be written this way). I don't see how this implies that = (Born rule probability)*, i.e. = , which is what you seem to be accusing Jacques of claiming above. (By the way, this would probably be easier to read if a different symbol than \psi was used for eigenstates, as I did in my previous post)Arrrgh, now it looks like it snipped out my bra-ket notation for no discernable reason. Trying again:My comment was that the "Born rule probability" is equal to <\psi|\psi_i><\psi_i|\psi>, and since <\psi_i|\psi> is the amplitude, it's accurate to say the probability is the amplitude-squared (although after editing the wikipedia article to say this, I was told that this only works for operators where the eigenstates are 1D eigenvectors, and that there can be cases where the eigenstate is an 'eigenspace' with more than one dimension in which case the Born rule probability can't be written this way). I don't see how this implies that <\psi_i|\psi_i> = (Born rule probability)*<\psi|\psi>, i.e. <\psi_i|\psi_i> = <\psi|\psi_i><\psi_i|\psi><\psi|\psi>, which is what you seem to be accusing Jacques of claiming above. (By the way, this would probably be easier to read if a different symbol than \psi was used for eigenstates, as I did in my previous post) --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: [kevintr...@hotmail.com: Jacques Mallah]
> Date: Mon, 9 Feb 2009 11:47:02 +1100 > From: li...@hpcoders.com.au > To: everything-l...@googlegroups.com > Subject: Re: [kevintr...@hotmail.com: Jacques Mallah] > > > Jesse, you need to fix up your email client to follow the usual > quoting conventions, wrap lines etc. I'm using hotmail, any idea what I'd need to do to make it work correctly? Looking at how the posts show up on google groups, it looks like my previous posts didn't have this problem, I'm not sure what went wrong with the last one. > > By working *effectively*, I think you mean the subjective experience (or > 1st person experience) should be as though \psi is transformed to > \psi_i, where \psi_i is an eigenvector of the relevant observable. > > Of course no such thing happens in the MWI, ie the 3rd person > description. \psi remains unaltered in this case, and is undergoing > regular unitary evolution. Yes, my understanding is that decoherence might be of some use in explaining the appearance of collapse even if we assume that an experimenter measuring a quantum system merely becomes entangled with it, with the wavefunction of the combined experimenter/quantum system continuing to evolve in a unitary way. There's a discussion of this on Greg Egan's page at http://gregegan.customer.netspace.net.au/SCHILD/Decoherence/Decoherence.html > > In any case, there is no requirement for to be given > by the product of the Born rule probability with . I don't understand, why is this implied by what Jacques or I said? My comment was that the "Born rule probability" is equal to , and since is the amplitude, it's accurate to say the probability is the amplitude-squared (although after editing the wikipedia article to say this, I was told that this only works for operators where the eigenstates are 1D eigenvectors, and that there can be cases where the eigenstate is an 'eigenspace' with more than one dimension in which case the Born rule probability can't be written this way). I don't see how this implies that = (Born rule probability)*, i.e. = , which is what you seem to be accusing Jacques of claiming above. (By the way, this would probably be easier to read if a different symbol than \psi was used for eigenstates, as I did in my previous post) Jesse --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
Jesse, you need to fix up your email client to follow the usual quoting conventions, wrap lines etc. Below is how your text appears in mine: On Sun, Feb 08, 2009 at 06:46:04AM -0500, Jesse Mazer wrote: > > Russell Standish wrote:> > According to Wikipedia, Born's rule is that the > probability of an> observed result \lambda_i is given by <\psi|P_i|\psi>, > where P_i is the> projection onto the eigenspace corresponding to \lambda_i > of the> observable. > > This formula is only correct if \psi is normalised. > More correctly,> the above formula should be divided by <\psi|\psi>.> > This > probability can be interpreted as a conditional probability - the> > probability of observing outcome \lambda_i for some observation A,> _given_ a > pre-measurment state \psi.> > What is important here is that it says nothing > about what the state> vector is after the measurement occurs. There is a (von > Neumann)> projection postulate, which says that after measurement, the > system> will be found in the state P_i|\psi>, but as I said before, this is> > independent of the Born rule, and also it does not state what the> > "amplitude" (ie magnitude) of the state is. The v-N PP is also distinctly > not> a feature of the MWI (it is basically the Copenhagen collapse).The > projection postulate needs to work *effectively* in the MWI though, in order > for it to make the same predictions about actual experimental results as the > Copenhagen interpretation. If an observer makes one measurement and then > makes a later measurement of the same system, they can collapse the system's > quantum state onto an eigenstate at the moment of the first measurement and > evolve that new state forward, and this will give correct predictions about > the probabilities of different outcomes when the second measurement is made. > I think part of the difficulty with connecting the MWI to actual observed > probabilities is explaining *why* this rule should work in an effective > sense.> > > I think the quote I was responding to was the following:> > "In > an ordinary quantum mechanical situation (without deaths), and> assuming the > Born Rule holds, the effective probability is proportional> to the total > squared amplitude of a branch."> > If you compare it with the description of > the Born rule above (which> computes a conditional probability), there is no > sense in which one> can say that "the effective probability is proportional > to the total> squared amplitude of a branch" follows directly from the Born> > rule. Jacques is assuming something else entirely - perhaps> einselection?The > notion that probability is proportional to squared amplitude is equivalent to > the wikipedia definition you posted above. The projection operator P_i onto a > given eigenstate |\lambda_i> is really just |\lambda_i><\lambda_i| (see the > text immediately above 'Postulate 5' near the bottom of the page at > http://xbeams.chem.yale.edu/~batista/vvv/node2.html ), which means that when > this operator acts on a given state vector |\psi>, it gives > (|\lambda_i><\lambda_i|)|\psi> = |\lambda_i>(<\lambda_i|\psi>), and > (<\lambda_i|\psi>) is just a scalar c_i, so this becomes c_i * |\lambda_i>. > This c_i is referred to as the "amplitude" that the original state |\psi> > assigns to the eigenstate |\lambda_i> (by something called the 'expansion > postulate' it's possible to write |psi> as a weighted sum of all the > eigenstates of a measurement operator, and the weights on each eigenstate are > just the amplitudes for each eigenstate). So, if the probability is > <\psi|P_i|\psi> (which I think assumes that |\psi> has been normalized), then > since P_i = |\lambda_i><\lambda_i|, this is the same as saying the > probability is <\psi|\lambda_i><\lambda_i|\psi>. If <\lambda_i|\psi> is the > amplitude c_i, then <\psi|\lambda_i> is just the complex conjugate (c_i)*, so > the probability is just the amplitude times its own complex conjugate, often > referred to as the "amplitude-squared".Jesse By working *effectively*, I think you mean the subjective experience (or 1st person experience) should be as though \psi is transformed to \psi_i, where \psi_i is an eigenvector of the relevant observable. Of course no such thing happens in the MWI, ie the 3rd person description. \psi remains unaltered in this case, and is undergoing regular unitary evolution. In any case, there is no requirement for <\psi_i|\psi_i> to be given by the product of the Born rule probability with <\psi|
RE: [kevintr...@hotmail.com: Jacques Mallah]
Russell Standish wrote:> > According to Wikipedia, Born's rule is that the probability of an> observed result \lambda_i is given by <\psi|P_i|\psi>, where P_i is the> projection onto the eigenspace corresponding to \lambda_i of the> observable. > > This formula is only correct if \psi is normalised. More correctly,> the above formula should be divided by <\psi|\psi>.> > This probability can be interpreted as a conditional probability - the> probability of observing outcome \lambda_i for some observation A,> _given_ a pre-measurment state \psi.> > What is important here is that it says nothing about what the state> vector is after the measurement occurs. There is a (von Neumann)> projection postulate, which says that after measurement, the system> will be found in the state P_i|\psi>, but as I said before, this is> independent of the Born rule, and also it does not state what the> "amplitude" (ie magnitude) of the state is. The v-N PP is also distinctly not> a feature of the MWI (it is basically the Copenhagen collapse).The projection postulate needs to work *effectively* in the MWI though, in order for it to make the same predictions about actual experimental results as the Copenhagen interpretation. If an observer makes one measurement and then makes a later measurement of the same system, they can collapse the system's quantum state onto an eigenstate at the moment of the first measurement and evolve that new state forward, and this will give correct predictions about the probabilities of different outcomes when the second measurement is made. I think part of the difficulty with connecting the MWI to actual observed probabilities is explaining *why* this rule should work in an effective sense.> > > I think the quote I was responding to was the following:> > "In an ordinary quantum mechanical situation (without deaths), and> assuming the Born Rule holds, the effective probability is proportional> to the total squared amplitude of a branch."> > If you compare it with the description of the Born rule above (which> computes a conditional probability), there is no sense in which one> can say that "the effective probability is proportional to the total> squared amplitude of a branch" follows directly from the Born> rule. Jacques is assuming something else entirely - perhaps> einselection?The notion that probability is proportional to squared amplitude is equivalent to the wikipedia definition you posted above. The projection operator P_i onto a given eigenstate |\lambda_i> is really just |\lambda_i><\lambda_i| (see the text immediately above 'Postulate 5' near the bottom of the page at http://xbeams.chem.yale.edu/~batista/vvv/node2.html ), which means that when this operator acts on a given state vector |\psi>, it gives (|\lambda_i><\lambda_i|)|\psi> = |\lambda_i>(<\lambda_i|\psi>), and (<\lambda_i|\psi>) is just a scalar c_i, so this becomes c_i * |\lambda_i>. This c_i is referred to as the "amplitude" that the original state |\psi> assigns to the eigenstate |\lambda_i> (by something called the 'expansion postulate' it's possible to write |psi> as a weighted sum of all the eigenstates of a measurement operator, and the weights on each eigenstate are just the amplitudes for each eigenstate). So, if the probability is <\psi|P_i|\psi> (which I think assumes that |\psi> has been normalized), then since P_i = |\lambda_i><\lambda_i|, this is the same as saying the probability is <\psi|\lambda_i><\lambda_i|\psi>. If <\lambda_i|\psi> is the amplitude c_i, then <\psi|\lambda_i> is just the complex conjugate (c_i)*, so the probability is just the amplitude times its own complex conjugate, often referred to as the "amplitude-squared".Jesse --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
On Fri, Feb 06, 2009 at 08:59:44AM -0500, Jesse Mazer wrote: > > Ah, never mind, rereading your post I think I see where I misunderstood > you--you weren't saying "nothing in QM says anything about" the amplitude of > an eigenvector that you square to get the probability of measuring that > eigenvector's eigenvalue, you were saying "nothing in QM says anything about" > how the length of the state vector immediately after the measurement > "collapses" the system's quantum state is related to the length of the > eigenvector it collapses onto (since the probabilities given by squaring the > amplitudes of the eignevectors always get normalized I think it doesn't > matter, the 'direction' of the state vector is all that's important). > > Still, I don't quite see where Mallah makes the mistake about the Born rule > you accuse him of making, what specific quote are you referring to? > > Jesse > According to Wikipedia, Born's rule is that the probability of an observed result \lambda_i is given by <\psi|P_i|\psi>, where P_i is the projection onto the eigenspace corresponding to \lambda_i of the observable. This formula is only correct if \psi is normalised. More correctly, the above formula should be divided by <\psi|\psi>. This probability can be interpreted as a conditional probability - the probability of observing outcome \lambda_i for some observation A, _given_ a pre-measurment state \psi. What is important here is that it says nothing about what the state vector is after the measurement occurs. There is a (von Neumann) projection postulate, which says that after measurement, the system will be found in the state P_i|\psi>, but as I said before, this is independent of the Born rule, and also it does not state what the "amplitude" (ie magnitude) of the state is. The v-N PP is also distinctly not a feature of the MWI (it is basically the Copenhagen collapse). I think the quote I was responding to was the following: "In an ordinary quantum mechanical situation (without deaths), and assuming the Born Rule holds, the effective probability is proportional to the total squared amplitude of a branch." If you compare it with the description of the Born rule above (which computes a conditional probability), there is no sense in which one can say that "the effective probability is proportional to the total squared amplitude of a branch" follows directly from the Born rule. Jacques is assuming something else entirely - perhaps einselection? It may be true that if the Born rule is false, then the effective probability is not proportional to the norm squared (yes I was having a little dig there, amplitude is a somewhat ambiguous term in this context, but one could interpret it as meaning norm (or L2-norm, to be even more precise), but without seeing Jacques's starting assumptions, and the logic he uses to derive his statement, it is really hard to know if that is the case. Cheers -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
2009/2/7 Bruno Marchal > > > Le 06-févr.-09, à 12:06, Quentin Anciaux a écrit : > > > Hi, > > > > 2009/2/6 russell standish > >> He also mentions Tegmark's amoeba croaks argument, which is not > >> actually an argument against QI, but rather a discussion of what QI > >> might actually mean. Contrary to what some people might think, QI > >> doesn't predict one would necessarily experience being vastly older > >> than the rest of the population. It just predicts that we should all > >> experience a "good innings", and that what happens after that is > >> rather unpredictable - it may be lapsing into senesence, it may be > >> followed > >> by rebirth into a different consciousness, it may be a form of > >> afterlife, or of uploading Singulatarian style. > > > > Well if you are "rebirth" in another consciousness for me it means > > you're dead, so rebirth without memory is equal to real death. > > > But what if you rebirth with the same consciousness? What if there is > only one consciousness, or one person? > What does it mean to be rebirth in the "same" consciousness (without memories) ? I understand this as meaning there is a property "me" which is somehow transferred and totally independant of memories... I don't think it makes sense. "me" is a feeling and is attached to the memories of being me... without that there is no "me". > > > > > > But if comp is true (hence RSSA) and no cul-de-sac hold, > > By definition: in the observable or probability "hypostases". They are > defined by Bp & Dp, given that Gödel's incompleteness makes Bp not > implying Dp "on earth". But BP itself entails there are cul-de-sac > everywhere (cf the realist multiverse, which I have called also the > Papaoiannou multiverse in preceding discussions). That is why we put Dp > (and DDp, and DDDp, etc.). OK it is a bit technical, and we will > probably come back on this. > > > > then there always exists a continuation of you with your memory... if > > you loose your memory I don't see how it can count as a continuation ? > > neither causaly nor does it shows similarity. > > > Are you sure about this? > I don't like to much thought experiment involving amnesia, and I have > eliminated them in my publications and/or theses. Yet it is hard to > avoid them in the immortality discussion. Actually I have change my > mind often on this issue: the "truth" is hard to believe here. So let > me propose a thought experiment involving amnesia, and I will just ask > you a question. Suppose you are read and annihilate in Brussels, and > then reconstituted in 1000 versions. 999 of them are partially amnesic > (some memories have been blocked or suppressed), and one version keeps > its memory. What is your expectation to live the experience of an > amnesic? > Is there a notion of statistically normal immortality? > > Bruno > Well if the 999 have partial memory of me... then they are partial continuation of me and I could agree to a certain degree that they could depend on the "me" right now (provided they remember this me right now). Secondly If I have to take your though experiment and that I would be warned in advance of what you'll be doing, the copy who is plainly me (with all of my memories) just have to do perfect copy of itself 1000 times after being awaken to outnumber the umperfect copies... But anyhow, I don't think it makes sense. But my answer would be anyway (even without making the 1000 copies) to be the one with full memory... Or I should say the one with all the memories knows everything I know then it is me. I can accept that degree exists in the definition of me... but if there always exists a continuation of me with full memories it's all I need to ensure I will be. The only thing I'm sure in this setting is that I'm in hell forever and I'm unable to escape that. Regards, Quentin -- All those moments will be lost in time, like tears in rain. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
Le 06-févr.-09, à 12:06, Quentin Anciaux a écrit : > Hi, > > 2009/2/6 russell standish >> He also mentions Tegmark's amoeba croaks argument, which is not >> actually an argument against QI, but rather a discussion of what QI >> might actually mean. Contrary to what some people might think, QI >> doesn't predict one would necessarily experience being vastly older >> than the rest of the population. It just predicts that we should all >> experience a "good innings", and that what happens after that is >> rather unpredictable - it may be lapsing into senesence, it may be >> followed >> by rebirth into a different consciousness, it may be a form of >> afterlife, or of uploading Singulatarian style. > > Well if you are "rebirth" in another consciousness for me it means > you're dead, so rebirth without memory is equal to real death. But what if you rebirth with the same consciousness? What if there is only one consciousness, or one person? > > But if comp is true (hence RSSA) and no cul-de-sac hold, By definition: in the observable or probability "hypostases". They are defined by Bp & Dp, given that Gödel's incompleteness makes Bp not implying Dp "on earth". But BP itself entails there are cul-de-sac everywhere (cf the realist multiverse, which I have called also the Papaoiannou multiverse in preceding discussions). That is why we put Dp (and DDp, and DDDp, etc.). OK it is a bit technical, and we will probably come back on this. > then there always exists a continuation of you with your memory... if > you loose your memory I don't see how it can count as a continuation ? > neither causaly nor does it shows similarity. Are you sure about this? I don't like to much thought experiment involving amnesia, and I have eliminated them in my publications and/or theses. Yet it is hard to avoid them in the immortality discussion. Actually I have change my mind often on this issue: the "truth" is hard to believe here. So let me propose a thought experiment involving amnesia, and I will just ask you a question. Suppose you are read and annihilate in Brussels, and then reconstituted in 1000 versions. 999 of them are partially amnesic (some memories have been blocked or suppressed), and one version keeps its memory. What is your expectation to live the experience of an amnesic? Is there a notion of statistically normal immortality? Bruno > > Regards, > Quentin > >> >> So sorry Jacques - you need to do better. I'm sure you can! >> >> Cheers >> -- >> >> >> -- >> -- >> A/Prof Russell Standish Phone 0425 253119 (mobile) >> Mathematics >> UNSW SYDNEY 2052 hpco...@hpcoders.com.au >> Australia http://www.hpcoders.com.au >> >> -- >> -- >> >> > > > > -- > All those moments will be lost in time, like tears in rain. > > > > http://iridia.ulb.ac.be/~marchal/ --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: [kevintr...@hotmail.com: Jacques Mallah]
Ah, never mind, rereading your post I think I see where I misunderstood you--you weren't saying "nothing in QM says anything about" the amplitude of an eigenvector that you square to get the probability of measuring that eigenvector's eigenvalue, you were saying "nothing in QM says anything about" how the length of the state vector immediately after the measurement "collapses" the system's quantum state is related to the length of the eigenvector it collapses onto (since the probabilities given by squaring the amplitudes of the eignevectors always get normalized I think it doesn't matter, the 'direction' of the state vector is all that's important). Still, I don't quite see where Mallah makes the mistake about the Born rule you accuse him of making, what specific quote are you referring to? Jesse From: laserma...@hotmail.com To: everything-l...@googlegroups.com Subject: RE: [kevintr...@hotmail.com: Jacques Mallah] Date: Fri, 6 Feb 2009 08:16:20 -0500 > His discussion of the Born rule is incorrect. The probability given by > the Born rule is not the square of the state vector, but rather the square > modulus of the inner product of some eigenvector with the original > state, appropriately normalised to make it a probability. After > observation, the state vector describing the new will be proportional > to the eigenvector corresponding the measured eigenvalue, but nothing > in QM says anything about its amplitude. I may be misunderstanding you, but I don't think it's correct to say that "nothing in QM says anything about its amplitude"--in QM every state vector can be expressed as a weighted *sum* of the eigenvectors for any measurement operator (vaguely similar to Fourier analysis), and the Born rule says the probability the system will be measured in a given eigenstate should be given by the square of the amplitude assigned to that eigenvector in the sum which corresponds to the state vector at the moment before the measurement (technically the probability the amplitude multiplied by its own complex conjugate rather than the amplitude squared, so if the amplitude is x + iy you multiply by x - iy to get a probability of x^2 + y^2, but it's common to just say 'amplitude squared' as shorthand). Jesse --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: [kevintr...@hotmail.com: Jacques Mallah]
> His discussion of the Born rule is incorrect. The probability given by > the Born rule is not the square of the state vector, but rather the square > modulus of the inner product of some eigenvector with the original > state, appropriately normalised to make it a probability. After > observation, the state vector describing the new will be proportional > to the eigenvector corresponding the measured eigenvalue, but nothing > in QM says anything about its amplitude. I may be misunderstanding you, but I don't think it's correct to say that "nothing in QM says anything about its amplitude"--in QM every state vector can be expressed as a weighted *sum* of the eigenvectors for any measurement operator (vaguely similar to Fourier analysis), and the Born rule says the probability the system will be measured in a given eigenstate should be given by the square of the amplitude assigned to that eigenvector in the sum which corresponds to the state vector at the moment before the measurement (technically the probability the amplitude multiplied by its own complex conjugate rather than the amplitude squared, so if the amplitude is x + iy you multiply by x - iy to get a probability of x^2 + y^2, but it's common to just say 'amplitude squared' as shorthand). Jesse --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: [kevintr...@hotmail.com: Jacques Mallah]
Hi, 2009/2/6 russell standish > > - Forwarded message from Kevin Tryon - > > I see that one of the earlier participants on the Everything list has now > taken it upon himself to educate the masses because the "cat is out of the > bag" and QI has become a familiar topic to many. > > > http://arxiv.org/ftp/arxiv/papers/0902/0902.0187.pdf > > > Does he say anything in this article that he hasn't said on the Everything > list in his struggles against QI? > > > > - End forwarded message - > > I have now read the whole of Jacques Mallah's "paper", and to put it > mildly, it is disappointing. I would have expected more from him. It > is neither the "definitive debunking" hoped for by the author, nor is > it persuasive in the rhetorical sense. What little technical detail he > provides obscures, rather than illuminates the issue. > > So what is the paper? I mentioned the interesting comment on how we > should expect to find ourselves a Boltzmann brain shortly after the > big bang, but there was no follow up to this. I have no idea how he > came up with that notion. > > His discussion of the Born rule is incorrect. The probability given by > the Born rule is not the square of the state vector, but rather the square > modulus of the inner product of some eigenvector with the original > state, appropriately normalised to make it a probability. After > observation, the state vector describing the new will be proportional > to the eigenvector corresponding the measured eigenvalue, but nothing > in QM says anything about its amplitude. Indeed it is conventional to > normalise the resulting state vector, as a computational convenience - > but this is an entirely different proposition to Mallah's. > > What I think he is trying to discuss, somewhat clumsily, in the > section on measure, is the ASSA notion of a unique well-defined > measure for all observer moments. This has been discussed in this > list extensively, and also summarised in my book. But it would sure > confuse anyone not familiar with the notion. > I've read too and all his argument is an argument against ASSA not QI nor RSSA. > > He goes on to mention rather briefly in passing his doomsday style > argument against QI, but not in detail. Which is just as well, as that > argument predicts that we should be neonatal infants! > > He also mentions Tegmark's amoeba croaks argument, which is not > actually an argument against QI, but rather a discussion of what QI > might actually mean. Contrary to what some people might think, QI > doesn't predict one would necessarily experience being vastly older > than the rest of the population. It just predicts that we should all > experience a "good innings", and that what happens after that is > rather unpredictable - it may be lapsing into senesence, it may be followed > by rebirth into a different consciousness, it may be a form of > afterlife, or of uploading Singulatarian style. > Well if you are "rebirth" in another consciousness for me it means you're dead, so rebirth without memory is equal to real death. But if comp is true (hence RSSA) and no cul-de-sac hold, then there always exists a continuation of you with your memory... if you loose your memory I don't see how it can count as a continuation ? neither causaly nor does it shows similarity. Regards, Quentin > So sorry Jacques - you need to do better. I'm sure you can! > > Cheers > -- > > > > A/Prof Russell Standish Phone 0425 253119 (mobile) > Mathematics > UNSW SYDNEY 2052 hpco...@hpcoders.com.au > Australiahttp://www.hpcoders.com.au > > > > > > -- All those moments will be lost in time, like tears in rain. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
[KevinTryon: Jacques Mallah]
- Forwarded message from Kevin Tryon - I see that one of the earlier participants on the Everything list has now taken it upon himself to educate the masses because the "cat is out of the bag" and QI has become a familiar topic to many. http://arxiv.org/ftp/arxiv/papers/0902/0902.0187.pdf Does he say anything in this article that he hasn't said on the Everything list in his struggles against QI? - End forwarded message - I have now read the whole of Jacques Mallah's "paper", and to put it mildly, it is disappointing. I would have expected more from him. It is neither the "definitive debunking" hoped for by the author, nor is it persuasive in the rhetorical sense. What little technical detail he provides obscures, rather than illuminates the issue. So what is the paper? I mentioned the interesting comment on how we should expect to find ourselves a Boltzmann brain shortly after the big bang, but there was no follow up to this. I have no idea how he came up with that notion. His discussion of the Born rule is incorrect. The probability given by the Born rule is not the square of the state vector, but rather the square modulus of the inner product of some eigenvector with the original state, appropriately normalised to make it a probability. After observation, the state vector describing the new will be proportional to the eigenvector corresponding the measured eigenvalue, but nothing in QM says anything about its amplitude. Indeed it is conventional to normalise the resulting state vector, as a computational convenience - but this is an entirely different proposition to Mallah's. What I think he is trying to discuss, somewhat clumsily, in the section on measure, is the ASSA notion of a unique well-defined measure for all observer moments. This has been discussed in this list extensively, and also summarised in my book. But it would sure confuse anyone not familiar with the notion. He goes on to mention rather briefly in passing his doomsday style argument against QI, but not in detail. Which is just as well, as that argument predicts that we should be neonatal infants! He also mentions Tegmark's amoeba croaks argument, which is not actually an argument against QI, but rather a discussion of what QI might actually mean. Contrary to what some people might think, QI doesn't predict one would necessarily experience being vastly older than the rest of the population. It just predicts that we should all experience a "good innings", and that what happens after that is rather unpredictable - it may be lapsing into senesence, it may be followed by rebirth into a different consciousness, it may be a form of afterlife, or of uploading Singulatarian style. So sorry Jacques - you need to do better. I'm sure you can! Cheers -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
[kevintr...@hotmail.com: Jacques Mallah]
- Forwarded message from Kevin Tryon - I see that one of the earlier participants on the Everything list has now taken it upon himself to educate the masses because the "cat is out of the bag" and QI has become a familiar topic to many. http://arxiv.org/ftp/arxiv/papers/0902/0902.0187.pdf Does he say anything in this article that he hasn't said on the Everything list in his struggles against QI? - End forwarded message - I have now read the whole of Jacques Mallah's "paper", and to put it mildly, it is disappointing. I would have expected more from him. It is neither the "definitive debunking" hoped for by the author, nor is it persuasive in the rhetorical sense. What little technical detail he provides obscures, rather than illuminates the issue. So what is the paper? I mentioned the interesting comment on how we should expect to find ourselves a Boltzmann brain shortly after the big bang, but there was no follow up to this. I have no idea how he came up with that notion. His discussion of the Born rule is incorrect. The probability given by the Born rule is not the square of the state vector, but rather the square modulus of the inner product of some eigenvector with the original state, appropriately normalised to make it a probability. After observation, the state vector describing the new will be proportional to the eigenvector corresponding the measured eigenvalue, but nothing in QM says anything about its amplitude. Indeed it is conventional to normalise the resulting state vector, as a computational convenience - but this is an entirely different proposition to Mallah's. What I think he is trying to discuss, somewhat clumsily, in the section on measure, is the ASSA notion of a unique well-defined measure for all observer moments. This has been discussed in this list extensively, and also summarised in my book. But it would sure confuse anyone not familiar with the notion. He goes on to mention rather briefly in passing his doomsday style argument against QI, but not in detail. Which is just as well, as that argument predicts that we should be neonatal infants! He also mentions Tegmark's amoeba croaks argument, which is not actually an argument against QI, but rather a discussion of what QI might actually mean. Contrary to what some people might think, QI doesn't predict one would necessarily experience being vastly older than the rest of the population. It just predicts that we should all experience a "good innings", and that what happens after that is rather unpredictable - it may be lapsing into senesence, it may be followed by rebirth into a different consciousness, it may be a form of afterlife, or of uploading Singulatarian style. So sorry Jacques - you need to do better. I'm sure you can! Cheers -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Jacques Mallah
- Forwarded message from Kevin Tryon - I see that one of the earlier participants on the Everything list has now taken it upon himself to educate the masses because the "cat is out of the bag" and QI has become a familiar topic to many. http://arxiv.org/ftp/arxiv/papers/0902/0902.0187.pdf Does he say anything in this article that he hasn't said on the Everything list in his struggles against QI? - End forwarded message - Thanks Kevin - I wasn't previously aware of this paper. I had a brief glance through it, and while much of it is on the everything list, the argument of page 4 that we should expect to be a Boltzmann brain born near the big bang is new to me. I'll need a bit of time to read it to comment properly. It obviously has to be read in terms of the ASSA, which Jacques doesn't make explicit. I'm not sure how effective his debunking is as a result. Cheers -- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpco...@hpcoders.com.au Australiahttp://www.hpcoders.com.au --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
