On Mon, Jul 12, 2021 at 3:44 AM smitra <[email protected]> wrote: > On 11-07-2021 02:46, Bruce Kellett wrote: > > On Sun, Jul 11, 2021 at 10:21 AM smitra <[email protected]> wrote: > >>> > >> This is only true in practice, not in principle because the escaping > >> > >> photons be captured and detected in principle. > > > > Not true. You can;t ever catch up with the escaping photons to > capture them. >
> They can be reflected back using mirrors. You can't decide to put mirrors in place after the photons have been emitted. Of course, you could always decide in advance to perform all experiments in reflecting boxes, but that is not what is done in practice. So the general rule is that escaping photons are not recoverable. > One can therefore at least in > principle do experiments that rule out collapse theories for large > systems that are large enough to include observers. And once collapse > theories are ruled out in a few experiments it follows that in a > superposition all the sectors where observers find different results > objectively exist, regardless of whether or not in those particular > situations one could have done the same sorts of experiments that ruled > out collapse theories. > > > > > >> It's also irrelevant, > >> because the photons that escape continue to exist, the information > >> contained in the photons continues to exist, even if it were true that > >> they could not be recovered. Then if we were to conduct an > >> interference experiment with the balls then we wouldn't see an > interference > >> pattern. But if we write down where each balls lands on the screen then > this > >> information together with information that could be obtained by > >> performing certain measurements on the escaping photons emitted by > >> each ball would still yield an interference pattern. > > > > This is where you go seriously wrong. Simply recording where the > > photons land does not quantum erase the information they carried. Once > > the photons carry off the which way information, the interference > > pattern is restored only if the information carried by the photons is > > quantum erased. Simply running the photons into a screen (or the > > wall), even if you record where they land, is not quantum erasure. > > See, for example, the paper arXiv:1206.6578 on quantum erasure. In > > this paper they say "the presence of path information anywhere in the > > universe is sufficient to prohibit any possibility of interference. In > > other words, the atoms' path states alone are not in a coherent > > superposition due to the atom-photon entanglement." This transfers > > directly to the buckyball experiment under discussion. Running the > > photons into a screen, or the wall, does not destroy the ball-photon > > entanglement. > > > > Recording where the photons land on the screen is enough, this is very > easy to see. Let's consider an interference experiment where the > particle gets entangled with another particle that carries away the > which way information. If we work in the position representation then we > have a wavefunction: > > psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)] (1) > > where x is the position of particle 1 just before it hits the screen and > y the position of the other particle at that time, and psi_1(x,y) is the > wavefunction when only slit 1 is open while psi_2(x,y) the wavefunction > with only slit 2 open. We also assume that particle 2 carries the which > way information, this means that psi_1(x,y) and psi_2(x,y) for x kept > fixed and considered as a function of y are eigenfunctions with > different eigenvalues of an observable that corresponds to extracting > the which way information from the second particle. This then implies > that psi_1(x,y) and psi_2(x,y) are orthogonal for every x: > > Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2) > This is the same mistake that you have routinely made in all of these discussions. You calculate for particles through each slit independently. But if that is the case, then no interference can ever be seen. The possibility of interference relies on coherence between the amplitudes at the two slits. In your calculations, you have never allowed for this coherence -- the amplitudes at the two slits are not independent. Bruce So, if we compute the probability of observing particle 1 at some > position x, while we don't measure the position of particle 2, we find > by taking the modulus squared of (1) and integrating over y: > > Integral |psi(x,y)|^2 d^3y = 1/2 Integral |psi_1(x,y)|^2 d^3y + 1/2 > Integral |psi_2(x,y)|^2 d^3y > > + Re Integral psi_1(x,y)*psi_2(x,y) d^3y > > And the last term vanishes due to (2), so we don't detect interference. > But now suppose that we perform a simultaneous measurement of the > positions of both particles. The probability of detecting particle 1 at > position x and particle 2 at position y is: > > |psi(x,y)|^2 = 1/2 |psi_1(x,y)|^2 y + 1/2 |psi_2(x,y)|^2 + Re > psi_1(x,y)*psi_2(x,y) > > So, there now clearly an interference term. If we keep y fixed the > probability distribution of x will show interference., This requires one > to count the dots on the screen for particle 1 for the cases where > particle 2 ends up within some narrow range for some fixed y. If we > don't keep track of y and just add up all the dots at the positions on > the screen for particle 1, then that amounts to integrating the > interference term over all y, which will make it vanish die to > orthogonality. > > Saibal > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLSNjMuh4n-ct01AVhEdY3-09JB2gjjom3bVwyvn4rKLow%40mail.gmail.com.
