On Friday, November 16, 2018 at 4:39:42 PM UTC, scerir wrote: > > > Il 16 novembre 2018 alle 15.38 agrays...@gmail.com <javascript:> ha > scritto: > > > > On Friday, November 16, 2018 at 10:14:32 AM UTC, scerir wrote: > > > Il 16 novembre 2018 alle 10.19 agrays...@gmail.com ha scritto: > > > > On Thursday, November 15, 2018 at 2:14:48 PM UTC, scerir wrote: > > > Il 15 novembre 2018 alle 14.29 agrays...@gmail.com 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 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 > > I think the question should rather be how to prepare a superposition state > like sqrt(1/2) [(s+)_z +(s-)_z] . But when you have this specific state, > and when you orient the SG along "x", you always get "up". >
*I'm still not sure I understand your comment. I will think about it some more. But back to my original question; Is there any circumstance where the result could be an eigenvalue of Up + Dn or Up - Dn? Alternately, can Up + Dn or Up - Dn ever be an eigenstate of the spin vector? TIA, AG* > > > > > 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 everything-li...@googlegroups.com. > To post to this group, send email to everyth...@googlegroups.com. > 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 everything-li...@googlegroups.com. > To post to this group, send email to everyth...@googlegroups.com. > 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 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. > 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 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.
