Re: Quantum theory of measurement
From: Ben Goertzel The paradox is as follows: One does the EPR thing of creating two particles with opposite spin. [...] More or less, it is the experiment by Birgit Dopfer (pdf on this page, unfortunately just in German) http://www.quantum.univie.ac.at/publications/thesis/
Re: Quantum theory of measurement
Hi Hal, After glancing over that paper it seems to me that Badagnani does not distinguish between classical and quantum forms of information. I strongly suspect that that is the reason why he thinks his idea would work. Onward! Stephen - Original Message - From: Hal Finney [EMAIL PROTECTED] To: everything-list@eskimo.com Sent: Wednesday, October 12, 2005 9:59 PM Subject: RE: Quantum theory of measurement Now that you are experts on this, try your hand on this FTL signalling device, http://arxiv.org/abs/quant-ph?0204108. The author, Daniel Badagnani, is apparently a genuine physicist, http://cabtep5.cnea.gov.ar/particulas/daniel/pag-db.html. Hal Finney
Re: Quantum theory of measurement
Well, as you can see here: http://cabtep5.cnea.gov.ar/particulas/daniel/curri/curreng.html He isn't very experienced yet. I know of some experienced professors of have made worse mistakes :) So, what goes wrong? Well, you don't get an interference pattern at one end even if you don't detect the photon at the other end. To see this, just write down the two particle state and add the phase shifts. If the detectors on the other sides are off, then the two contributions corresponding to the photon being detected at some position z consist of two orthogonal terms; one term correpsonds to the other photon in pipe 1 and the other for that photon in pipe 2 Suppose that you add a plate to detect the photon on that side as well. Then the probability that the photon at one end is detected at position z1 and the other is detected at z2 does contain an interference term of the form: Cos[delta1(z1) + delta2(z2)] If you don't detect where photon 2 is absorbed you have to integrate over z2 and the interference term vanishes. To see an interference term you must keep z2 fixed. This means that you must consider only those photons for which the entangled partners were detected at at some fixed z2. But this means that this value must be communicated by the observer there. - Original Message - From: Hal Finney [EMAIL PROTECTED] To: everything-list@eskimo.com Sent: Thursday, October 13, 2005 03:59 AM Subject: RE: Quantum theory of measurement Now that you are experts on this, try your hand on this FTL signalling device, http://arxiv.org/abs/quant-ph?0204108. The author, Daniel Badagnani, is apparently a genuine physicist, http://cabtep5.cnea.gov.ar/particulas/daniel/pag-db.html. Hal Finney
RE: Quantum theory of measurement
Hi, Oops, I gave the wrong link I said Specifically, I'll refer to the quantum eraser thought experiment summarized at http://grad.physics.sunysb.edu/~amarch/ but I meant http://www.dhushara.com/book/quantcos/qnonloc/eraser.htm Anyway, the essential idea of the two experiments is the same. -- Ben
Re: Quantum theory of measurement
Ben Goertzel writes about: http://grad.physics.sunysb.edu/~amarch/ The questions I have regard the replacement of the Coincidence Counter (from here on: CC) in the above experiment with a more complicated apparatus. What if we replace the CC with one of the following: 1) a carefully sealed, exquisitely well insulated box with a printer inside it. The printer is hooked up so that it prints, on paper, an exact record of everything that comes into the CC. Then, to erase the printed record, the whole box is melted, or annihilated using nuclear explosives, or whatever. The CC is not what is erased. Rather, the so-called erasure happens to the photons while they are flying through the apparatus. Nothing in the experiment proposes erasing data in the CC. So I don't really see what you are getting at. What will the outcome be in these experiments? It won't make any difference, because the CC is not used in the way you imagine. It doesn't have to produce a record and it doesn't have to erase any records. Let me tell you what really happens in the experiment above. It is actually not so mystical as people try to make it sound. We start off with the s photon going through a 2 slit experiment and getting interference. That is standard. Now we put two different polarization rotations in front of the two slits and interference goes away. The web page author professes amazement, but it is not really that surprising. After all, interference between two photons would typically be affected by messing with their polarizations. It is not all that surprising that putting polarizers into the paths could mess up the interference. But now comes the impressive part. He puts a polarizer in front of the other photon, the p photon, and suddenly the interference comes back! Surely something amazing and non-local has happened now, right? Not really. This new polarizer will eliminate some of the p photons. They won't get through. The result is that we will throw out some of the measurements of s photons, because if the p photon got eaten by its polarizer, the CC doesn't trigger as there is no coincidence. (This is the real reason for the CC in this experiment.) So now we are discarding some of the s photon measurements, and keeping some. It turns out that the ones we keep do show an interference pattern. If we had added back in the ones we discarded, it would blur out the interference fringes and there would be no pattern. The point is, there is no change to the s photon when we put the polarizer over by p. Its results do not visibly change from non-interference to interference, as the web page might imply. (If that did happen, we'd have the basis for a faster than light communicator.) No, all that is happening is that we are choosing to throw out half the data, and the half we keep does show interference. The only point of the CC, then, is to tell us which half of the data to throw out of the s photon measurements. Destroying the CC and all of the other crazy things you suggest have nothing to do with the experiment. The CC is not what is erased and it does not create a permanent record. It is only there to tell us whether a p photon got through its polarizer or not, so that we know whether to throw away the s photon measurement. Hal Finney
RE: Quantum theory of measurement
Hal, What will the outcome be in these experiments? It won't make any difference, because the CC is not used in the way you imagine. It doesn't have to produce a record and it doesn't have to erase any records. OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC that records things, but still the records could be kept somewhere, and one can ask what would happen if the records were kept somewhere else (e.g. in a macroscopic medium). No? The point is, there is no change to the s photon when we put the polarizer over by p. Its results do not visibly change from non-interference to interference, as the web page might imply. (If that did happen, we'd have the basis for a faster than light communicator.) No, all that is happening is that we are choosing to throw out half the data, and the half we keep does show interference. Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... Ben
RE: Quantum theory of measurement
What if instead of throwing out the information you shoot it into a black hole? Then presumably the information is really gone so the result should be as if the information were quantum erased?? Unless there are white holes of course!! ;-) Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... Ben
RE: Quantum theory of measurement
Ben Goertzel writes: Hal, It won't make any difference, because the CC is not used in the way you imagine. It doesn't have to produce a record and it doesn't have to erase any records. OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC that records things, but still the records could be kept somewhere, and one can ask what would happen if the records were kept somewhere else (e.g. in a macroscopic medium). No? I don't think this makes sense, at least I can't understand it. The point is, there is no change to the s photon when we put the polarizer over by p. Its results do not visibly change from non-interference to interference, as the web page might imply. (If that did happen, we'd have the basis for a faster than light communicator.) No, all that is happening is that we are choosing to throw out half the data, and the half we keep does show interference. Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... You have to try to stop thinking of this in mystical terms. IMO people present a rather prosaic phenomenon in a misleading and confusing way, and this is giving you an incorrect idea. Nothing is un-happening. No records are destroyed after they were gathered. Forget that anybody told you this was a quantum eraser and think about what really happens. When all the polarizers are in place, half of the p photons get eaten and half get through. This gives us a way to split up the s measurements into two halves. It turns out that each half independently shows interference, but that the two interference patterns are the opposite of each other. When you combine the two halves back together, the peaks of one half fill in the valleys of the other, and the data set as a whole shows no interference. Look at it concretely as it might happen in the lab. We record a bunch of s measurements and also record whether we get a coincidence with a p photon getting through, in the CC. Maybe we write a little check mark next to the s measurements where there was a p photon coincidence. We go through afterwards to analyze the data. If we just plot all the s measurements we see a smooth curve, no interference. Now we go through and cross off the ones where there was no p coincidence. We cross off s measurement number 1, then numbers 3 and 4, then 5, 7, 10 through 12, and so on. When we plot the remaining measurements, now we see an interference pattern. In other words, the coincidence with the p photon identifies a subset of the s measurements which shows interference. The total collection of s measurements still shows no interference. There is no real erasing going on. Whoever coined the term quantum eraser was a master of public relations, but unfortunately he confused millions of lay people into getting the wrong idea about the physics. Hal Finney
Re: Quantum theory of measurement
Hal gives the correct explanation of what's going on. In general, all you have to do to analyze the problem is to consider all contributions to a particular state and add up the amplitudes. The absolute value squared of the amplitude gives the probability, which may or may not contain an interference term. A simple minded formal description can be given as follows: If you pass a photon through two slits then close to the screen its state would be of the form: Integral over z of [1 + Exp(i delta(z))] |s,z Here z denotes the postion on the screen, s is just a label for the photon and Exp(i delta(z)) is the phase shift between the two paths which gives rise to the interference term. Delta(z) will be zero exactly inbetween the two slits and will be nonzero elsewhere. The probability of having the photon at z is obtained (up to normalization) by taking the absolute value squared of the prefactor of |s,z, which is 2 + 2 cos(delta(z)). Let's do the same for the two entagled photons. The entangled photon pair can be denoted as: |p_x,s_y + |p_y,s_x here x and y denote the polarization states. If you pass s through the two slits then the state becomes: [1 + Exp(i delta(z))] |p_x,s_y,z + [1 + Exp(i delta(z))] ||p_y,s_x,z This has to be integrated over z, but let's focus only at the contribution at some fixed position z. The prefactors of both state vectors |p_x,s_y,z and |p_y,s_x,z are of the same form as in the single photon case and thus you get an interference term Cos(delta(z) as above. If you put the quarter wave plate in then instead of [1 + Exp(i delta(z))] |p_x,s_y,z you get: |p_x,s_r,z + Exp(i delta(z)) |p_x,s_l,z And the complete state vector becomes: |p_x,s_r,z + Exp(i delta(z)) |p_x,s_l,z+ |p_y,s_l,z + Exp(i delta(z)) |p_y,s_r,z All the four state vectors are orthogonal. The probability that you detect the photon s at z is just the sum of the absolute value squared of the four terms, which is constant and doesn't contain an interference term. Now let's pass photon p through the polarizer (45 degrees w.r.t. x). This amounts to measuring the polarization state of photon p in the basis |p_x + p_y and |p_x - p_y. If you don't discard one of these two states and keep them both then, as Hal rightly points out, nothing changes. If you substitute: |p_x = |a + |b; |p_y = |a - |b in the state vector above you get: |a,s_r,z + Exp(i delta(z)) |a,s_l,z+ |a,s_l,z + Exp(i delta(z)) |a,s_r,z+ |b,s_r,z + Exp(i delta(z)) |b,s_l,z+ - |b,s_l,z - Exp(i delta(z)) |b,s_r,z = (collecting the prefactors of like state vectors) = [1 + Exp(i delta(z))]|a,s_r,z + [1 + Exp(i delta(z))]|a,s_l,z + [1 - Exp(i delta(z))]|b,s_r,z - [1 - Exp(i delta(z))] |b,s_l,z The absolute value squared of the terms with the p photons in the |a state contains the term 2 Cos (delta(z)), but for the b terms this is - 2 Cos(delta(z). If you don't observe the polarization of the p photon, the interference terms would thus cancel. This is obvious since all we have done is to write down the same state in a different basis. But if you do observe the polarization of the p photon, then can measure the probability of detecting s at position z and p in polarization state |a then you have to add up the absolute value squared of |a,s_r,z and |a,s_l,z. Then you do get the Cos(delta(z) interference term. - Original Message - From: Hal Finney [EMAIL PROTECTED] To: [EMAIL PROTECTED]; everything-list@eskimo.com Sent: Wednesday, October 12, 2005 08:16 PM Subject: Re: Quantum theory of measurement Ben Goertzel writes about: http://grad.physics.sunysb.edu/~amarch/ The questions I have regard the replacement of the Coincidence Counter (from here on: CC) in the above experiment with a more complicated apparatus. What if we replace the CC with one of the following: 1) a carefully sealed, exquisitely well insulated box with a printer inside it. The printer is hooked up so that it prints, on paper, an exact record of everything that comes into the CC. Then, to erase the printed record, the whole box is melted, or annihilated using nuclear explosives, or whatever. The CC is not what is erased. Rather, the so-called erasure happens to the photons while they are flying through the apparatus. Nothing in the experiment proposes erasing data in the CC. So I don't really see what you are getting at. What will the outcome be in these experiments? It won't make any difference, because the CC is not used in the way you imagine. It doesn't have to produce a record and it doesn't have to erase any records. Let me tell you what really happens in the experiment above. It is actually not so mystical as people try to make it sound. We start off with the s photon going through a 2 slit experiment and getting interference. That is standard. Now we put two different polarization rotations in front of the two slits and interference goes away. The web page author professes
RE: Quantum theory of measurement
Hal Finney wrote: Ben Goertzel writes: Hal, It won't make any difference, because the CC is not used in the way you imagine. It doesn't have to produce a record and it doesn't have to erase any records. OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC that records things, but still the records could be kept somewhere, and one can ask what would happen if the records were kept somewhere else (e.g. in a macroscopic medium). No? I don't think this makes sense, at least I can't understand it. I think Ben's question here does make sense. See below... The point is, there is no change to the s photon when we put the polarizer over by p. Its results do not visibly change from non-interference to interference, as the web page might imply. (If that did happen, we'd have the basis for a faster than light communicator.) No, all that is happening is that we are choosing to throw out half the data, and the half we keep does show interference. Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... You have to try to stop thinking of this in mystical terms. IMO people present a rather prosaic phenomenon in a misleading and confusing way, and this is giving you an incorrect idea. Nothing is un-happening. No records are destroyed after they were gathered. Although it may be true that no records are destroyed after they're gathered, what is true is that an *opportunity* to find out retroactively which path the signal photon took is eliminated when you choose to combine the paths of the idler photons instead of measuring them. For reference, look at the diagram of the setup in fig. 1 of this paper: http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment when you don't measure which slit the particle came through. Note that if you removed the beam-splitters BSA and BSB you could guarantee that the idler would be detected at D3 or D4 and thus that the path of the signal photon would be known; likewise, if you replaced the beam-splitters BSA and BSB with mirrors, then you could guarantee that the idler would be detected at D1 or D2 and thus that the path of the signal photon would be unknown. By making the distances large enough you could even choose whether to make sure the idlers go to D3D4 or to go to D1D2 *after* you have already observed the position that the signal photon was detected, so in this sense you have the choice whether or not to retroactively erase your opportunity to know which atom the signal photon came from, after the signal photon's position has already been detected. This confused me for a while since it seemed like this would imply your later choice determines whether or not you observe interference in the signal photons earlier, until I got into a discussion about it online and someone showed me the trick. In the same paper, look at the graphs in Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D1, and Fig. 4 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D2 (the two cases where the idler's 'which-path' information is lost). They do both show interference, but if you line the graphs up you see that the peaks of one interference pattern line
Re: Quantum theory of measurement
From: Ben Goertzel [...] but still the records could be kept somewhere, and one can ask what would happen if the records were kept somewhere else [...] Not sure, but the quote below - about the information 'in principle' - might be helpful. The superposition of amplitudes is only valid if there is no way to know, even in principle, which path the particle took. It is important to realize that this does not imply that an observer actually takes note of what happens. It is sufficient to destroy the interference pattern, if the path information is accessible in principle from the experiment or even if it is dispersed in the environment and beyond any technical possibility to be recovered, but in principle 'still out there'. - Anton Zeilinger, (Rev.Mod.Phys.,1999, p.S-288)
RE: Quantum theory of measurement
Thanks very much Jesse! You answered the question I *would have* asked had I rememberd my quantum physics better ;-) I think your answer is related to a paradox a friend mentioned to me. The paradox is as follows: One does the EPR thing of creating two particles with opposite spin. Send one far away to Alpha Centauri and send the other through a Stern-Gerlach magnet and let the spin up and spin down outputs interfere to form a double slit. If the far away particle is measured up vs. down, our local particle must definitely go through the up hole or the down hole and we get no interference pattern. If he measures the far away particle sideways we get a superposition of states and we get interference. Thus by rotating his measurement he should be able to communicate to us faster than the speed of light. We should see our pattern blinking between interference and not. What's wrong with that argument? Along the lines of your solution to my other, related puzzle, I'll try to analogize a solution to this puzzle. I guess the idea must be: there is no change to what a particular particle does when you observe its faraway coupled pair in a certain way. Its individual results do not visibly change from non-interference to interference. (If that did happen, you'd have the basis for a faster than light communicator, as you say.) Instead, when you observe some of the particles sideways and some vertically, you must be creating correlational information that exists only statistically as a correlation between that's happening in Alpha Centauri and what's happening locally. So, maybe there is some weird cancellation here, like in the case you described in your email. Perhaps, if one restricts attention to the cases where the faraway particle is measured right then interference is seen, and if one restricts attention to the cases where the faraway particle is measured left then interference is seen; but if one looks across all cases where the faraway particle is measured sideways, then the peaks and troughs of the different cases might cancel out and you'd get no interference? -- Ben I think Ben's question here does make sense. See below... The point is, there is no change to the s photon when we put the polarizer over by p. Its results do not visibly change from non-interference to interference, as the web page might imply. (If that did happen, we'd have the basis for a faster than light communicator.) No, all that is happening is that we are choosing to throw out half the data, and the half we keep does show interference. Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... You have to try to stop thinking of this in mystical terms. IMO people present a rather prosaic phenomenon in a misleading and confusing way, and this is giving you an incorrect idea. Nothing is un-happening. No records are destroyed after they were gathered. Although it may be true that no records are destroyed after they're gathered, what is true is that an *opportunity* to find out retroactively which path the signal photon took is eliminated when you choose to combine the paths of the idler photons instead of measuring them. For reference, look at the diagram of the setup in fig. 1 of this paper: http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment
RE: Quantum theory of measurement
Ben Goertzel wrote: Thanks very much Jesse! You answered the question I *would have* asked had I rememberd my quantum physics better ;-) I think your answer is related to a paradox a friend mentioned to me. The paradox is as follows: One does the EPR thing of creating two particles with opposite spin. Send one far away to Alpha Centauri and send the other through a Stern-Gerlach magnet and let the spin up and spin down outputs interfere to form a double slit. If the far away particle is measured up vs. down, our local particle must definitely go through the up hole or the down hole and we get no interference pattern. If he measures the far away particle sideways we get a superposition of states and we get interference. Thus by rotating his measurement he should be able to communicate to us faster than the speed of light. We should see our pattern blinking between interference and not. What's wrong with that argument? Along the lines of your solution to my other, related puzzle, I'll try to analogize a solution to this puzzle. I guess the idea must be: there is no change to what a particular particle does when you observe its faraway coupled pair in a certain way. Its individual results do not visibly change from non-interference to interference. (If that did happen, you'd have the basis for a faster than light communicator, as you say.) Instead, when you observe some of the particles sideways and some vertically, you must be creating correlational information that exists only statistically as a correlation between that's happening in Alpha Centauri and what's happening locally. So, maybe there is some weird cancellation here, like in the case you described in your email. Perhaps, if one restricts attention to the cases where the faraway particle is measured right then interference is seen, and if one restricts attention to the cases where the faraway particle is measured left then interference is seen; but if one looks across all cases where the faraway particle is measured sideways, then the peaks and troughs of the different cases might cancel out and you'd get no interference? -- Ben Yeah, I'd agree with your guess about what would happen here. Certainly you won't see interference vs. non-interference on the screen depending on what measurement was made on the other particle, since it's been proven that quantum effects cannot be used to transmit information faster than light; so, I think you would see no interference in the total pattern on the screen no matter what, but if you looked at specific subsets of cases once you found out the result of the measurement on the second particle, you'd probably be able to see interference patterns in those subsets where the measurement of the second particle did not allow you to determine which slit the first one went through. Jesse
RE: Quantum theory of measurement
Ben Goertzel wrote: My own understanding is that whether Fred, a pigeon or a printer is involved in the experiment should be basically irrelevant. That is, I don't think registration in consciousness (whatever that means) is the important thing, but rather registration in the sense of statistical correlation with some macroscopic system effectively obeying classical probability theory. However, I realize that not everyone agrees with me on this; my reading of Penrose, for instance, is that he would predict a different outcome for 2 versus 1, because he believes that Fred's brain (via unspecified quantum gravity related effects) does something special to collapse the wave function, which the printer does not. By the way, I also wanted to point out that this actually isn't Penrose's position, although given his unusual views about the relationship between consciousness and QM it's a pretty common misunderstanding. It's been a while since I read his book, but my memory is that Penrose advocates a gravitationally-induced collapse view where the wavefunction of an entangled system spontaneously collapses once it exceeds a certain threshold in mass (due to interactions with the outside world which result in more and more particles becoming entangled with the original system), which he speculates is likely to be around the Planck mass, about 5*10^-8 kg (close to the mass of a flea). He also speculates that the theory of quantum gravity which deals with this may be noncomputable, and that the human brain may somehow take advantage of such noncomputable effects to solve problems that could not be solved by a computational system. The consciousness collapses the wavefunction view has been advocated by a few serious physicists like Eugene Wigner, and of course by lots of new-agey types. Jesse
Re: Quantum theory of measurement
Hi Ben, Could it be that black holes do not destroy information but merely randomize it, ala decoherence? For a hint see: http://www.arxiv.org/abs/quant-ph/0410172 Onward! Stephen - Original Message - From: Ben Goertzel [EMAIL PROTECTED] To: everything-list@eskimo.com Sent: Wednesday, October 12, 2005 2:52 PM Subject: RE: Quantum theory of measurement What if instead of throwing out the information you shoot it into a black hole? Then presumably the information is really gone so the result should be as if the information were quantum erased?? Unless there are white holes of course!! ;-) Yes but we are choosing which half to throw out in a very peculiar way -- i.e. we are throwing it out by un-happening it after it happened, by destroying some records that were only gathered after the events recorded in the data already happened... Ben