> On 31 Jul 2018, at 06:21, agrayson2...@gmail.com wrote: > > > > On Tuesday, July 31, 2018 at 1:34:58 AM UTC, Brent wrote: > > > On 7/30/2018 4:40 PM, agrays...@gmail.com <javascript:> wrote: >> >> >> On Monday, July 30, 2018 at 7:50:47 PM UTC, Brent wrote: >> >> >> On 7/30/2018 8:02 AM, Bruno Marchal wrote: >>>> and claims the system being measured is physically in all eigenstates >>>> simultaneously before measurement. >>> >>> >>> Nobody claims that this is true. But most of us would I think agree that >>> this is what happens if you describe the couple “observer particle” by QM, >>> i.e by the quantum wave. It is a consequence of elementary quantum >>> mechanics (unless of course you add the unintelligible collapse of the >>> wave, which for me just means that QM is false). >> >> This talk of "being in eigenstates" is confused. An eigenstate is relative >> to some operator. The system can be in an eigenstate of an operator. Ideal >> measurements are projection operators that leave the system in an eigenstate >> of that operator. But ideal measurements are rare in QM. All the >> measurements you're discussing in Young's slit examples are destructive >> measurements. You can consider, as a mathematical convenience, using a >> complete set of commuting operators to define a set of eigenstates that will >> provide a basis...but remember that it's just mathematics, a certain choice >> of basis. The system is always in just one state and the mathematics says >> there is some operator for which that is the eigenstate. But in general we >> don't know what that operator is and we have no way of physically >> implementing it. >> >> Brent >> >> I can only speak for myself, but when I write that a system in a >> superposition of states is in all component states simultaneously, I am >> assuming the existence of an operator with eigenstates that form a complete >> set and basis, that the wf is written as a sum using this basis, and that >> this representation corresponds to the state of the system before >> measurement. > > In general you need a set of operators to have the eigenstates form a > complete basis...but OK. > >> I am also assuming that the interpretation of a quantum superposition is >> that before measurement, the system is in all eigenstates simultaneously, >> one of which represents the system after measurement. I do allow for >> situations where we write a superposition as a sum of eigenstates even if we >> don't know what the operator is, such as the Up + Dn state of a spin >> particle. In the case of the cat, using the hypothesis of superposition I >> argue against, we have two eigenstates, which if "occupied" by the system >> simultaneously, implies the cat is alive and dead simultaneously. AG > > Yes, you can write down the math for that. But to realize that physically > would require that the cat be perfectly isolated and not even radiate IR > photons (c.f. C60 Bucky ball experiment). So it is in fact impossible to > realize (which is why Schroedinger considered if absurd). > > CMIIAW, but as I have argued, in decoherence theory it is assumed the cat is > initially isolated and decoheres in a fraction of a nano second.
But decoherence is only entanglement with the environment. The cat decoder relatively to the observer, but that is exactly what the SWE describes. The superposed state of the cat just get very quickly contagious to the environment and then the observer, which reports decoherence or collapse according to its philosophy. Decoherence is the best explanation why we don’t feel the split/differentiation. > So, IMO, the problem with the interpretation of superposition remains. Certainly, if you postulate a unique well defined physical reality. > It doesn't go away because the decoherence time is exceedingly short. And for > this reason I still conclude that Schroedinger correctly pointed out the > fallacy in the standard interpretation of superposition; namely, that the > system represented by a superposition, is in all components states > simultaneously. AG Without that simultaneity there would be no interference, which we observe in all case. Give me a two state system u and d, u will be equal to u’ + d”, and d will be equal to u’ - d’. A qubit is only a bit that we can rotate in the hilbert space. A qu-register is only a register of qubit, and we can put it in the state (u’+d')(u’+d')(u’+d') ... (u’+d’), making the full register of length n containing the superposition of all sequence of bit, when seen or considered in the u and d base. We can test if the result of 2^n computations, get the same results or not, for example, by NOT observing each bit in the u/d base, but instead making such result interfering in another base, and then measuring them. If someone can explain me Shor quantum algorithm for factoring number, even a small number as 15 (this has been done experimentally) without a physical (in a Bohm non local potential or in the MW) rendering of the 2^16 computations, be my guest. Bruno > > Brent > >> -- >> 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 everything-li...@googlegroups.com <javascript:>. >> To post to this group, send email to everyth...@googlegroups.com >> <javascript:>. >> Visit this group at https://groups.google.com/group/everything-list >> <https://groups.google.com/group/everything-list>. >> For more options, visit https://groups.google.com/d/optout >> <https://groups.google.com/d/optout>. > > > -- > 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 everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to everything-list@googlegroups.com > <mailto:everything-list@googlegroups.com>. > Visit this group at https://groups.google.com/group/everything-list > <https://groups.google.com/group/everything-list>. > For more options, visit https://groups.google.com/d/optout > <https://groups.google.com/d/optout>. -- 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
