On Tuesday, May 13, 2025 at 4:54:55 AM UTC-6 Alan Grayson wrote:
On Monday, May 12, 2025 at 4:15:52 PM UTC-6 Brent Meeker wrote: On 5/12/2025 1:58 PM, Alan Grayson wrote: On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote: On 5/9/2025 7:08 PM, Alan Grayson wrote: *I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG* Yes. Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved. Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant. Brent *If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement, * Right, given that it's an ideal measurement. Most measurements don't leave the system in the eigenstate that is the measurement result. An ideal measurement is one that leaves the system in the state that the measurement yielded. *and the observable to be measured remains in that state on subsequent measurements? * Only if they're ideal measurements of that same variable or of other variables that commute with it. *If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread? * You don't need a change in the wf to "create the quantum spread". Having prepared in an eigenstate of A just measure some other variable B that doesn't commute with A. In general A will be a superposition of other variables, say A=xC+yD; that's just a change of coordinates. But the system is not in an eigenstate of C or D. Brent *Sorry, I really don't get it. Not at all! If we want to prepare a particle with some momentum p, why would we measure it with some non-commuting operator, and why would this, if done repeatedly, result in a spread of momentum? And what has this to do with a unitary operator which advances time? TY, AG * *Is the spread in momentum caused by an imprecision in preparing a particle in some particular momentum? Generally speaking, how is that done? TY, AG * -- 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/46017fb9-168c-4b0d-9824-8ca388f0da33n%40googlegroups.com.
