On Sunday, November 18, 2018 at 12:19:20 PM UTC, Bruno Marchal wrote: > > > On 16 Nov 2018, at 15:38, [email protected] <javascript:> wrote: > > > > On Friday, November 16, 2018 at 10:14:32 AM UTC, scerir wrote: >> >> >> Il 16 novembre 2018 alle 10.19 [email protected] ha scritto: >> >> >> >> On Thursday, November 15, 2018 at 2:14:48 PM UTC, scerir wrote: >> >> >> Il 15 novembre 2018 alle 14.29 [email protected] ha scritto: >> >> >> >> On Thursday, November 15, 2018 at 8:04:53 AM UTC, scerir wrote: >> >> Imagine a spin-1/2 particle described by the state psi = sqrt(1/2) >> [(s+)_z + (s-)_z] . >> >> If the x-component of spin is measured by passing the spin-1/2 particle >> through a Stern-Gerlach with its field oriented along the x-axis, the >> particle will ALWAYS emerge 'up'. >> >> >> *Why? Won't the measured value be along the x axis in both directions, >> in effect Up or Dn? AG* >> >> "Hence we must conclude that the system described by the |+>x state is >> not the >> same as a mixture of atoms in the |+> and !-> states. This means that >> each atom in the >> beam is in a state that itself is a combination of the |+> and |-> >> states. A superposition >> state is often called a coherent superposition since the relative phase >> of the two terms is >> important." >> >> .see pages 18-19 here *https://tinyurl.com/ybm56whu >> <https://tinyurl.com/ybm56whu>* >> >> >> *Try answering in your own words. When the SG device is oriented along >> the x axis, now effectively the z-axix IIUC, and we're dealing with >> superpositions, the outcomes will be 50-50 plus and minus. Therefore, >> unless I am making some error, what you stated above is incorrect. AG * >> >> sqrt(1/2) [(s+)_z +(s-)_z] is a superposition, but since sqrt(1/2) >> [(s+)_z +(s-)_z] = (s+)_x the particle will always emerge 'up' >> > > I'll probably get back to on the foregoing. In the meantime, consider > this; I claim one can never MEASURE Up + Dn or Up - Dn with a SG apparatus > regardless of how many other instruments one uses to create a composite > measuring apparatus (Bruno's claim IIUC). The reason is simple. We know > that the spin operator > > > Which one? >
*Good question. AG* There are spin operator for each direction in space. The superposition of > up and down is a precise pure state, with precise eigenvalues, when > measuring state in the complementary directions. > *As I wrote earlier, based on scerir's superpositions on different axes, and simulation, I now think that Up + Dn and Up - Dn can be measured along the x axis but not along the z axis (which I was focused on). You were probably correct about x axis measurements, but perhaps were not clear enough. You were not explicit that measurements along the x axis is a different SG experiment from along z axis. I thought you meant do them in succession, not as separate experiments. Also introducing an infinity of universes seems extraneous and confusing for a solution to this problem. AG * > > has exactly two eigenstates, each with probability of .5. We can write > them down. We also know that every quantum measurement gives up an > eigenvalue of some eigenstate. Therefore, if there existed an Up + Dn or Up > - Dn eigenstate, it would have to have probability ZERO since the Up and Dn > eigenstates have probabilities which sum to unity. Do you agree or not, and > if not, why? TIA, AG > > > You add the probabilities, but you need to add the amplitudes of > probabilities instead, and then take their square. You simply dismiss the > quantum formalism, it seems to me. > *I did not; an incorrect inference on your part. I* * never mentioned Born's rule (it wasn't necessary), from which one cannot infer I am criticizing QM itself. AG * The states constituted a vector space: the sum (superposition) of > orthogonal states are pure state, after a change of base, and I did give > you the corresponding operator. You are not criticising an interpretation > of QM, but QM itself. > > Bruno > > > > > >> >> >> >> In fact (s+)_z = sqrt(1/2) [(s+)_x + (s-)_x] >> >> and (s-)_z = sqrt(1/2) [(s+)_x - (s-)_x] >> >> (where _z, _x, are the z-component and the x-component of spin) >> >> so that psi = sqrt(1/2)[(s+)_z +(s-)_z] = (s+)_x. (pure state, not >> mixture state).. >> >> AGrayson2000 asked "If a system is in a superposition of states, whatever >> value measured, will be repeated if the same system is repeatedly >> measured. But what happens if the system is in a mixed state?" >> >> Does Everett's "relative state interpretation" show how to interpret a >> real superposition (like the above, in which the particle will always >> emerge 'up') and how to interpret a mixture (in which the particle will >> emerge 50% 'up' or 50% 'down')? >> >> >> -- >> 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 post to this group, send email to [email protected]. >> Visit this group at https://groups.google.com/group/everything-list. >> For more options, visit 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 [email protected]. >> To post to this group, send email to [email protected]. >> Visit this group at https://groups.google.com/group/everything-list. >> For more options, visit 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 [email protected] <javascript:>. > To post to this group, send email to [email protected] > <javascript:>. > Visit this group at https://groups.google.com/group/everything-list. > For more options, visit 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 [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
