On Mon, Jul 28, 2025 at 3:50 PM Brent Meeker <[email protected]> wrote:
*>>> So if I illuminate a piece of iron with some photons that knock out >>> electrons via the photoelectric effect, then on passing them thru an SG >>> magnet and detecting their distribution, you say they must necessarily be >>> 50/50 UP/DN?* >> >> *>>Not necessarily. Electrons in magnetic iron are not randomly oriented, >> so the electrons knocked out of the iron may have some preferential spin >> orientation. Also because angular momentum is conserved, circularly >> polarized light can preferentially eject electrons of one spin orientation >> over the other. But both the magnetic orientation of iron atoms and the >> amount of polarization of the light are known quantities, at least >> theoretically, * > > *> What's that supposed to mean, besides "I need it to be right."?* > *Its meaning is self-evident. And I don't need to be right and often I'm wrong, but in this case I am right. And I think you're mad because I didn't fall for your trap. * >*You've just defined "unmeasured" to mean 50/50 distribution.* *That's the word I used to describe a single electron if I know nothing about its history. If it's not a single electron but there is a beam of them and I send many electrons through a Stern–Gerlach magnet then I may conclude that whatever the origin of the beam was, it's producing electrons in some distribution other than 50-50.* *> What's your definition of "measurement"?* > *Rather than a definition of measurement I will give you something much better, an example. Producing an electron with a known spin state, as a piece of magnetized iron does when, thanks to the photoelectric effect, it's exposed to circularly polarized light; or when an electron passes through a Stern–Gerlach magnet.* *> A bit of iron found in the ground is unlikely to have encountered an SG.* *True, but it's not unlikely that something equivalent could have occurred. * *John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis>* sqo -- 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 visit https://groups.google.com/d/msgid/everything-list/CAJPayv1%3D_%3D7Q5Quj3Gftmit5XuXu8VMqSWVHsfCXudSuSidE_Q%40mail.gmail.com.
