On Friday, November 16, 2018 at 4:39:42 PM UTC, scerir wrote: > > > Il 16 novembre 2018 alle 15.38 [email protected] <javascript:> ha > scritto: > > > > 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 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 [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.
